3.60 \(\int \frac {\sin ^6(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=314 \[ -\frac {5 \sqrt {b} \sqrt {a+b} (a+4 b) (3 a+4 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^6 f}+\frac {(9 a+10 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac {5 x (a+2 b) \left (a^2+16 a b+16 b^2\right )}{16 a^6}-\frac {5 b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{16 a^5 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac {5 b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{48 a^4 f \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\left (33 a^2+110 a b+80 b^2\right ) \sin (e+f x) \cos (e+f x)}{48 a^3 f \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac {\sin ^3(e+f x) \cos ^3(e+f x)}{6 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

5/16*(a+2*b)*(a^2+16*a*b+16*b^2)*x/a^6-5/8*(a+4*b)*(3*a+4*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^(1/2)*(a
+b)^(1/2)/a^6/f-1/48*(33*a^2+110*a*b+80*b^2)*cos(f*x+e)*sin(f*x+e)/a^3/f/(a+b+b*tan(f*x+e)^2)^2+1/24*(9*a+10*b
)*cos(f*x+e)^3*sin(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^2+1/6*cos(f*x+e)^3*sin(f*x+e)^3/a/f/(a+b+b*tan(f*x+e)^2)^
2-5/48*b*(9*a^2+32*a*b+24*b^2)*tan(f*x+e)/a^4/f/(a+b+b*tan(f*x+e)^2)^2-5/16*b*(5*a^2+20*a*b+16*b^2)*tan(f*x+e)
/a^5/f/(a+b+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.50, antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4132, 470, 578, 527, 522, 203, 205} \[ -\frac {5 b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{16 a^5 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac {5 b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{48 a^4 f \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\left (33 a^2+110 a b+80 b^2\right ) \sin (e+f x) \cos (e+f x)}{48 a^3 f \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac {5 x (a+2 b) \left (a^2+16 a b+16 b^2\right )}{16 a^6}-\frac {5 \sqrt {b} \sqrt {a+b} (a+4 b) (3 a+4 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^6 f}+\frac {(9 a+10 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac {\sin ^3(e+f x) \cos ^3(e+f x)}{6 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

(5*(a + 2*b)*(a^2 + 16*a*b + 16*b^2)*x)/(16*a^6) - (5*Sqrt[b]*Sqrt[a + b]*(a + 4*b)*(3*a + 4*b)*ArcTan[(Sqrt[b
]*Tan[e + f*x])/Sqrt[a + b]])/(8*a^6*f) - ((33*a^2 + 110*a*b + 80*b^2)*Cos[e + f*x]*Sin[e + f*x])/(48*a^3*f*(a
+ b + b*Tan[e + f*x]^2)^2) + ((9*a + 10*b)*Cos[e + f*x]^3*Sin[e + f*x])/(24*a^2*f*(a + b + b*Tan[e + f*x]^2)^
2) + (Cos[e + f*x]^3*Sin[e + f*x]^3)/(6*a*f*(a + b + b*Tan[e + f*x]^2)^2) - (5*b*(9*a^2 + 32*a*b + 24*b^2)*Tan
[e + f*x])/(48*a^4*f*(a + b + b*Tan[e + f*x]^2)^2) - (5*b*(5*a^2 + 20*a*b + 16*b^2)*Tan[e + f*x])/(16*a^5*f*(a
+ b + b*Tan[e + f*x]^2))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps

\begin {align*} \int \frac {\sin ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^4 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 (a+b)+(-6 a-7 b) x^2\right )}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 a f}\\ &=\frac {(9 a+10 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {\operatorname {Subst}\left (\int \frac {(a+b) (9 a+10 b)+\left (-24 a^2-91 a b-70 b^2\right ) x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{24 a^2 f}\\ &=-\frac {\left (33 a^2+110 a b+80 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(9 a+10 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {5 (a+b) \left (3 a^2+18 a b+16 b^2\right )-5 b \left (33 a^2+110 a b+80 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{48 a^3 f}\\ &=-\frac {\left (33 a^2+110 a b+80 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(9 a+10 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {5 b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{48 a^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {60 (a+b)^2 \left (a^2+8 a b+8 b^2\right )-60 b (a+b) \left (9 a^2+32 a b+24 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{192 a^4 (a+b) f}\\ &=-\frac {\left (33 a^2+110 a b+80 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(9 a+10 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {5 b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{48 a^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {5 b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{16 a^5 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {120 (a+b)^3 \left (a^2+12 a b+16 b^2\right )-120 b (a+b)^2 \left (5 a^2+20 a b+16 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{384 a^5 (a+b)^2 f}\\ &=-\frac {\left (33 a^2+110 a b+80 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(9 a+10 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {5 b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{48 a^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {5 b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{16 a^5 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {(5 b (a+b) (a+4 b) (3 a+4 b)) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^6 f}+\frac {\left (5 (a+2 b) \left (a^2+16 a b+16 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 a^6 f}\\ &=\frac {5 (a+2 b) \left (a^2+16 a b+16 b^2\right ) x}{16 a^6}-\frac {5 \sqrt {b} \sqrt {a+b} (a+4 b) (3 a+4 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^6 f}-\frac {\left (33 a^2+110 a b+80 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(9 a+10 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {5 b \left (9 a^2+32 a b+24 b^2\right ) \tan (e+f x)}{48 a^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {5 b \left (5 a^2+20 a b+16 b^2\right ) \tan (e+f x)}{16 a^5 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [C]  time = 18.87, size = 1639, normalized size = 5.22 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

(5*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(((3*a^2 + 8*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqr
t[a + b]])/(a + b)^(5/2) - (a*Sqrt[b]*(3*a^2 + 16*a*b + 16*b^2 + 3*a*(a + 2*b)*Cos[2*(e + f*x)])*Sin[2*(e + f*
x)])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2)))/(65536*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^3) - (15*(a + 2*b
+ a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((-6*a^2*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]
) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sq
rt[b*(Cos[e] - I*Sin[e])^4]) + (a*Sec[2*e]*((-9*a^4 - 16*a^3*b + 48*a^2*b^2 + 128*a*b^3 + 64*b^4)*Sin[2*f*x] +
a*(-3*a^3 + 2*a^2*b + 24*a*b^2 + 16*b^3)*Sin[2*(e + 2*f*x)] + (3*a^4 - 64*a^2*b^2 - 128*a*b^3 - 64*b^4)*Sin[4
*e + 2*f*x]) + (9*a^5 + 18*a^4*b - 64*a^3*b^2 - 256*a^2*b^3 - 320*a*b^4 - 128*b^5)*Tan[2*e])/(a^2*(a + 2*b + a
*Cos[2*(e + f*x)])^2)))/(262144*b^2*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^3) + (3*(a + 2*b + a*Cos[2*e + 2*f*x])^
3*Sec[e + f*x]^6*(-1536*(a + 2*b)*x - (3*(a^6 - 8*a^5*b + 120*a^4*b^2 + 1280*a^3*b^3 + 3200*a^2*b^4 + 3072*a*b
^5 + 1024*b^6)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a
+ b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(b^2*(a + b)^(5/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^
4]) + (4*(a^4 + 32*a^3*b + 160*a^2*b^2 + 256*a*b^3 + 128*b^4)*Sec[2*e]*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b
*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])^2) + (256*a*Sin[2*(e + f*x)])/f + (a*(-3*a^5 + 26*a^4*b + 736*a^3*b^
2 + 2624*a^2*b^3 + 3200*a*b^4 + 1280*b^5)*Sec[2*e]*Sin[2*f*x] + (3*a^6 - 24*a^5*b - 920*a^4*b^2 - 4864*a^3*b^3
- 10112*a^2*b^4 - 9216*a*b^5 - 3072*b^6)*Tan[2*e])/(b^2*(a + b)^2*f*(a + 2*b + a*Cos[2*(e + f*x)]))))/(65536*
a^4*(a + b*Sec[e + f*x]^2)^3) - ((a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(-6144*(7*a^3 + 54*a^2*b + 12
0*a*b^2 + 80*b^3)*x - (3*(3*a^8 - 64*a^7*b + 2240*a^6*b^2 + 53760*a^5*b^3 + 313600*a^4*b^4 + 802816*a^3*b^5 +
1032192*a^2*b^6 + 655360*a*b^7 + 163840*b^8)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) +
a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(b^2*(a + b)^(5/2)
*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (12*(a^6 + 72*a^5*b + 840*a^4*b^2 + 3584*a^3*b^3 + 6912*a^2*b^4 + 6144*a*b
^5 + 2048*b^6)*Sec[2*e]*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])^2) +
(1152*a*(7*a^2 + 32*a*b + 32*b^2)*((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)]))/f + (1152*a*(7*a^2 + 32*a*b + 32
*b^2)*(I*Cos[2*(e + f*x)] + Sin[2*(e + f*x)]))/f + (192*a^2*(a + 2*b)*((-6*I)*Cos[4*(e + f*x)] - 6*Sin[4*(e +
f*x)]))/f + ((1152*I)*a^2*(a + 2*b)*(Cos[4*(e + f*x)] + I*Sin[4*(e + f*x)]))/f + (256*a^3*Sin[6*(e + f*x)])/f
+ (3*(3*a*(-a^7 + 22*a^6*b + 1352*a^5*b^2 + 11312*a^4*b^3 + 37120*a^3*b^4 + 57856*a^2*b^5 + 43008*a*b^6 + 1228
8*b^7)*Sec[2*e]*Sin[2*f*x] + (3*a^8 - 64*a^7*b - 4480*a^6*b^2 - 45696*a^5*b^3 - 196928*a^4*b^4 - 438272*a^3*b^
5 - 528384*a^2*b^6 - 327680*a*b^7 - 81920*b^8)*Tan[2*e]))/(b^2*(a + b)^2*f*(a + 2*b + a*Cos[2*(e + f*x)]))))/(
393216*a^6*(a + b*Sec[e + f*x]^2)^3)

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fricas [A]  time = 0.63, size = 930, normalized size = 2.96 \[ \left [\frac {30 \, {\left (a^{5} + 18 \, a^{4} b + 48 \, a^{3} b^{2} + 32 \, a^{2} b^{3}\right )} f x \cos \left (f x + e\right )^{4} + 60 \, {\left (a^{4} b + 18 \, a^{3} b^{2} + 48 \, a^{2} b^{3} + 32 \, a b^{4}\right )} f x \cos \left (f x + e\right )^{2} + 30 \, {\left (a^{3} b^{2} + 18 \, a^{2} b^{3} + 48 \, a b^{4} + 32 \, b^{5}\right )} f x + 15 \, {\left ({\left (3 \, a^{4} + 16 \, a^{3} b + 16 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} + 16 \, a b^{3} + 16 \, b^{4} + 2 \, {\left (3 \, a^{3} b + 16 \, a^{2} b^{2} + 16 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 2 \, {\left (8 \, a^{5} \cos \left (f x + e\right )^{9} - 2 \, {\left (13 \, a^{5} + 10 \, a^{4} b\right )} \cos \left (f x + e\right )^{7} + {\left (33 \, a^{5} + 110 \, a^{4} b + 80 \, a^{3} b^{2}\right )} \cos \left (f x + e\right )^{5} + 20 \, {\left (6 \, a^{4} b + 23 \, a^{3} b^{2} + 18 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (5 \, a^{3} b^{2} + 20 \, a^{2} b^{3} + 16 \, a b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{96 \, {\left (a^{8} f \cos \left (f x + e\right )^{4} + 2 \, a^{7} b f \cos \left (f x + e\right )^{2} + a^{6} b^{2} f\right )}}, \frac {15 \, {\left (a^{5} + 18 \, a^{4} b + 48 \, a^{3} b^{2} + 32 \, a^{2} b^{3}\right )} f x \cos \left (f x + e\right )^{4} + 30 \, {\left (a^{4} b + 18 \, a^{3} b^{2} + 48 \, a^{2} b^{3} + 32 \, a b^{4}\right )} f x \cos \left (f x + e\right )^{2} + 15 \, {\left (a^{3} b^{2} + 18 \, a^{2} b^{3} + 48 \, a b^{4} + 32 \, b^{5}\right )} f x + 15 \, {\left ({\left (3 \, a^{4} + 16 \, a^{3} b + 16 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} + 16 \, a b^{3} + 16 \, b^{4} + 2 \, {\left (3 \, a^{3} b + 16 \, a^{2} b^{2} + 16 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - {\left (8 \, a^{5} \cos \left (f x + e\right )^{9} - 2 \, {\left (13 \, a^{5} + 10 \, a^{4} b\right )} \cos \left (f x + e\right )^{7} + {\left (33 \, a^{5} + 110 \, a^{4} b + 80 \, a^{3} b^{2}\right )} \cos \left (f x + e\right )^{5} + 20 \, {\left (6 \, a^{4} b + 23 \, a^{3} b^{2} + 18 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (5 \, a^{3} b^{2} + 20 \, a^{2} b^{3} + 16 \, a b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, {\left (a^{8} f \cos \left (f x + e\right )^{4} + 2 \, a^{7} b f \cos \left (f x + e\right )^{2} + a^{6} b^{2} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/96*(30*(a^5 + 18*a^4*b + 48*a^3*b^2 + 32*a^2*b^3)*f*x*cos(f*x + e)^4 + 60*(a^4*b + 18*a^3*b^2 + 48*a^2*b^3
+ 32*a*b^4)*f*x*cos(f*x + e)^2 + 30*(a^3*b^2 + 18*a^2*b^3 + 48*a*b^4 + 32*b^5)*f*x + 15*((3*a^4 + 16*a^3*b + 1
6*a^2*b^2)*cos(f*x + e)^4 + 3*a^2*b^2 + 16*a*b^3 + 16*b^4 + 2*(3*a^3*b + 16*a^2*b^2 + 16*a*b^3)*cos(f*x + e)^2
)*sqrt(-a*b - b^2)*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a + 2*b)
*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b - b^2)*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e
)^2 + b^2)) - 2*(8*a^5*cos(f*x + e)^9 - 2*(13*a^5 + 10*a^4*b)*cos(f*x + e)^7 + (33*a^5 + 110*a^4*b + 80*a^3*b^
2)*cos(f*x + e)^5 + 20*(6*a^4*b + 23*a^3*b^2 + 18*a^2*b^3)*cos(f*x + e)^3 + 15*(5*a^3*b^2 + 20*a^2*b^3 + 16*a*
b^4)*cos(f*x + e))*sin(f*x + e))/(a^8*f*cos(f*x + e)^4 + 2*a^7*b*f*cos(f*x + e)^2 + a^6*b^2*f), 1/48*(15*(a^5
+ 18*a^4*b + 48*a^3*b^2 + 32*a^2*b^3)*f*x*cos(f*x + e)^4 + 30*(a^4*b + 18*a^3*b^2 + 48*a^2*b^3 + 32*a*b^4)*f*x
*cos(f*x + e)^2 + 15*(a^3*b^2 + 18*a^2*b^3 + 48*a*b^4 + 32*b^5)*f*x + 15*((3*a^4 + 16*a^3*b + 16*a^2*b^2)*cos(
f*x + e)^4 + 3*a^2*b^2 + 16*a*b^3 + 16*b^4 + 2*(3*a^3*b + 16*a^2*b^2 + 16*a*b^3)*cos(f*x + e)^2)*sqrt(a*b + b^
2)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f*x + e))) - (8*a^5*cos(f*x + e
)^9 - 2*(13*a^5 + 10*a^4*b)*cos(f*x + e)^7 + (33*a^5 + 110*a^4*b + 80*a^3*b^2)*cos(f*x + e)^5 + 20*(6*a^4*b +
23*a^3*b^2 + 18*a^2*b^3)*cos(f*x + e)^3 + 15*(5*a^3*b^2 + 20*a^2*b^3 + 16*a*b^4)*cos(f*x + e))*sin(f*x + e))/(
a^8*f*cos(f*x + e)^4 + 2*a^7*b*f*cos(f*x + e)^2 + a^6*b^2*f)]

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giac [A]  time = 0.61, size = 372, normalized size = 1.18 \[ \frac {\frac {15 \, {\left (a^{3} + 18 \, a^{2} b + 48 \, a b^{2} + 32 \, b^{3}\right )} {\left (f x + e\right )}}{a^{6}} - \frac {30 \, {\left (3 \, a^{3} b + 19 \, a^{2} b^{2} + 32 \, a b^{3} + 16 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{6}} - \frac {6 \, {\left (7 \, a^{2} b^{2} \tan \left (f x + e\right )^{3} + 23 \, a b^{3} \tan \left (f x + e\right )^{3} + 16 \, b^{4} \tan \left (f x + e\right )^{3} + 9 \, a^{3} b \tan \left (f x + e\right ) + 34 \, a^{2} b^{2} \tan \left (f x + e\right ) + 41 \, a b^{3} \tan \left (f x + e\right ) + 16 \, b^{4} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2} a^{5}} - \frac {33 \, a^{2} \tan \left (f x + e\right )^{5} + 162 \, a b \tan \left (f x + e\right )^{5} + 144 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 288 \, a b \tan \left (f x + e\right )^{3} + 288 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) + 126 \, a b \tan \left (f x + e\right ) + 144 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{5}}}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

1/48*(15*(a^3 + 18*a^2*b + 48*a*b^2 + 32*b^3)*(f*x + e)/a^6 - 30*(3*a^3*b + 19*a^2*b^2 + 32*a*b^3 + 16*b^4)*(p
i*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/(sqrt(a*b + b^2)*a^6) - 6*(7*a^2*
b^2*tan(f*x + e)^3 + 23*a*b^3*tan(f*x + e)^3 + 16*b^4*tan(f*x + e)^3 + 9*a^3*b*tan(f*x + e) + 34*a^2*b^2*tan(f
*x + e) + 41*a*b^3*tan(f*x + e) + 16*b^4*tan(f*x + e))/((b*tan(f*x + e)^2 + a + b)^2*a^5) - (33*a^2*tan(f*x +
e)^5 + 162*a*b*tan(f*x + e)^5 + 144*b^2*tan(f*x + e)^5 + 40*a^2*tan(f*x + e)^3 + 288*a*b*tan(f*x + e)^3 + 288*
b^2*tan(f*x + e)^3 + 15*a^2*tan(f*x + e) + 126*a*b*tan(f*x + e) + 144*b^2*tan(f*x + e))/((tan(f*x + e)^2 + 1)^
3*a^5))/f

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maple [B]  time = 1.28, size = 689, normalized size = 2.19 \[ -\frac {7 b^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \,a^{3} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {23 b^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \,a^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {9 b \tan \left (f x +e \right )}{8 a^{2} f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {17 b^{2} \tan \left (f x +e \right )}{4 f \,a^{3} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {41 b^{3} \tan \left (f x +e \right )}{8 f \,a^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {15 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{8 f \,a^{3} \sqrt {\left (a +b \right ) b}}-\frac {95 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{8 f \,a^{4} \sqrt {\left (a +b \right ) b}}-\frac {20 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{5} \sqrt {\left (a +b \right ) b}}-\frac {2 b^{4} \left (\tan ^{3}\left (f x +e \right )\right )}{f \,a^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {2 b^{4} \tan \left (f x +e \right )}{f \,a^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {10 b^{4} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{6} \sqrt {\left (a +b \right ) b}}-\frac {27 \left (\tan ^{5}\left (f x +e \right )\right ) b}{8 f \,a^{4} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {3 \left (\tan ^{5}\left (f x +e \right )\right ) b^{2}}{f \,a^{5} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {11 \left (\tan ^{5}\left (f x +e \right )\right )}{16 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {6 \left (\tan ^{3}\left (f x +e \right )\right ) b}{f \,a^{4} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {6 \left (\tan ^{3}\left (f x +e \right )\right ) b^{2}}{f \,a^{5} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {5 \left (\tan ^{3}\left (f x +e \right )\right )}{6 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {5 \tan \left (f x +e \right )}{16 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {21 \tan \left (f x +e \right ) b}{8 f \,a^{4} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {3 \tan \left (f x +e \right ) b^{2}}{f \,a^{5} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}+\frac {45 \arctan \left (\tan \left (f x +e \right )\right ) b}{8 f \,a^{4}}+\frac {15 \arctan \left (\tan \left (f x +e \right )\right ) b^{2}}{f \,a^{5}}+\frac {10 \arctan \left (\tan \left (f x +e \right )\right ) b^{3}}{f \,a^{6}}+\frac {5 \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x)

[Out]

-7/8/f*b^2/a^3/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-23/8/f*b^3/a^4/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-9/8*b*ta
n(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^2-17/4/f*b^2/a^3/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)-41/8/f*b^3/a^4/(a+b+b*t
an(f*x+e)^2)^2*tan(f*x+e)-15/8/f*b/a^3/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-95/8/f*b^2/a^4/((a
+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-20/f*b^3/a^5/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(
1/2))-2/f*b^4/a^5/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-2/f*b^4/a^5/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)-10/f*b^4/a
^6/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-27/8/f/a^4/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5*b-3/f/a^5/(
tan(f*x+e)^2+1)^3*tan(f*x+e)^5*b^2-11/16/f/a^3/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5-6/f/a^4/(tan(f*x+e)^2+1)^3*tan(
f*x+e)^3*b-6/f/a^5/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3*b^2-5/6/f/a^3/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3-5/16/f/a^3/(t
an(f*x+e)^2+1)^3*tan(f*x+e)-21/8/f/a^4/(tan(f*x+e)^2+1)^3*tan(f*x+e)*b-3/f/a^5/(tan(f*x+e)^2+1)^3*tan(f*x+e)*b
^2+45/8/f/a^4*arctan(tan(f*x+e))*b+15/f/a^5*arctan(tan(f*x+e))*b^2+10/f/a^6*arctan(tan(f*x+e))*b^3+5/16/f/a^3*
arctan(tan(f*x+e))

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maxima [A]  time = 0.45, size = 418, normalized size = 1.33 \[ -\frac {\frac {15 \, {\left (5 \, a^{2} b^{2} + 20 \, a b^{3} + 16 \, b^{4}\right )} \tan \left (f x + e\right )^{9} + 40 \, {\left (3 \, a^{3} b + 19 \, a^{2} b^{2} + 39 \, a b^{3} + 24 \, b^{4}\right )} \tan \left (f x + e\right )^{7} + {\left (33 \, a^{4} + 470 \, a^{3} b + 1910 \, a^{2} b^{2} + 2880 \, a b^{3} + 1440 \, b^{4}\right )} \tan \left (f x + e\right )^{5} + 40 \, {\left (a^{4} + 14 \, a^{3} b + 46 \, a^{2} b^{2} + 57 \, a b^{3} + 24 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + 15 \, {\left (a^{4} + 14 \, a^{3} b + 41 \, a^{2} b^{2} + 44 \, a b^{3} + 16 \, b^{4}\right )} \tan \left (f x + e\right )}{a^{5} b^{2} \tan \left (f x + e\right )^{10} + {\left (2 \, a^{6} b + 5 \, a^{5} b^{2}\right )} \tan \left (f x + e\right )^{8} + a^{7} + 2 \, a^{6} b + a^{5} b^{2} + {\left (a^{7} + 8 \, a^{6} b + 10 \, a^{5} b^{2}\right )} \tan \left (f x + e\right )^{6} + {\left (3 \, a^{7} + 12 \, a^{6} b + 10 \, a^{5} b^{2}\right )} \tan \left (f x + e\right )^{4} + {\left (3 \, a^{7} + 8 \, a^{6} b + 5 \, a^{5} b^{2}\right )} \tan \left (f x + e\right )^{2}} - \frac {15 \, {\left (a^{3} + 18 \, a^{2} b + 48 \, a b^{2} + 32 \, b^{3}\right )} {\left (f x + e\right )}}{a^{6}} + \frac {30 \, {\left (3 \, a^{3} b + 19 \, a^{2} b^{2} + 32 \, a b^{3} + 16 \, b^{4}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{6}}}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

-1/48*((15*(5*a^2*b^2 + 20*a*b^3 + 16*b^4)*tan(f*x + e)^9 + 40*(3*a^3*b + 19*a^2*b^2 + 39*a*b^3 + 24*b^4)*tan(
f*x + e)^7 + (33*a^4 + 470*a^3*b + 1910*a^2*b^2 + 2880*a*b^3 + 1440*b^4)*tan(f*x + e)^5 + 40*(a^4 + 14*a^3*b +
46*a^2*b^2 + 57*a*b^3 + 24*b^4)*tan(f*x + e)^3 + 15*(a^4 + 14*a^3*b + 41*a^2*b^2 + 44*a*b^3 + 16*b^4)*tan(f*x
+ e))/(a^5*b^2*tan(f*x + e)^10 + (2*a^6*b + 5*a^5*b^2)*tan(f*x + e)^8 + a^7 + 2*a^6*b + a^5*b^2 + (a^7 + 8*a^
6*b + 10*a^5*b^2)*tan(f*x + e)^6 + (3*a^7 + 12*a^6*b + 10*a^5*b^2)*tan(f*x + e)^4 + (3*a^7 + 8*a^6*b + 5*a^5*b
^2)*tan(f*x + e)^2) - 15*(a^3 + 18*a^2*b + 48*a*b^2 + 32*b^3)*(f*x + e)/a^6 + 30*(3*a^3*b + 19*a^2*b^2 + 32*a*
b^3 + 16*b^4)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*a^6))/f

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mupad [B]  time = 7.72, size = 2117, normalized size = 6.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^6/(a + b/cos(e + f*x)^2)^3,x)

[Out]

(5*atan(((5*((tan(e + f*x)*(179200*a*b^8 + 51200*b^9 + 249600*a^2*b^7 + 176000*a^3*b^6 + 65800*a^4*b^5 + 12300
*a^5*b^4 + 925*a^6*b^3))/(128*a^10) - (((20*a^12*b^5 + 35*a^13*b^4 + (65*a^14*b^3)/4 + (5*a^15*b^2)/4)/a^15 -
(tan(e + f*x)*(2048*a^12*b^3 + 1024*a^13*b^2)*(a + 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(4096*a^16))*(a + 2*b)*(16
*a*b + a^2 + 16*b^2)*5i)/(32*a^6))*(a + 2*b)*(16*a*b + a^2 + 16*b^2))/(32*a^6) + (5*((tan(e + f*x)*(179200*a*b
^8 + 51200*b^9 + 249600*a^2*b^7 + 176000*a^3*b^6 + 65800*a^4*b^5 + 12300*a^5*b^4 + 925*a^6*b^3))/(128*a^10) +
(((20*a^12*b^5 + 35*a^13*b^4 + (65*a^14*b^3)/4 + (5*a^15*b^2)/4)/a^15 + (tan(e + f*x)*(2048*a^12*b^3 + 1024*a^
13*b^2)*(a + 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(4096*a^16))*(a + 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(32*a^6))*(a
+ 2*b)*(16*a*b + a^2 + 16*b^2))/(32*a^6))/((4750*a*b^10 + 1000*b^11 + (18875*a^2*b^9)/2 + (40625*a^3*b^8)/4 +
(204875*a^4*b^7)/32 + (305125*a^5*b^6)/128 + (256125*a^6*b^5)/512 + (53125*a^7*b^4)/1024 + (1875*a^8*b^3)/1024
)/a^15 - (((tan(e + f*x)*(179200*a*b^8 + 51200*b^9 + 249600*a^2*b^7 + 176000*a^3*b^6 + 65800*a^4*b^5 + 12300*a
^5*b^4 + 925*a^6*b^3))/(128*a^10) - (((20*a^12*b^5 + 35*a^13*b^4 + (65*a^14*b^3)/4 + (5*a^15*b^2)/4)/a^15 - (t
an(e + f*x)*(2048*a^12*b^3 + 1024*a^13*b^2)*(a + 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(4096*a^16))*(a + 2*b)*(16*a
*b + a^2 + 16*b^2)*5i)/(32*a^6))*(a + 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(32*a^6) + (((tan(e + f*x)*(179200*a*b^
8 + 51200*b^9 + 249600*a^2*b^7 + 176000*a^3*b^6 + 65800*a^4*b^5 + 12300*a^5*b^4 + 925*a^6*b^3))/(128*a^10) + (
((20*a^12*b^5 + 35*a^13*b^4 + (65*a^14*b^3)/4 + (5*a^15*b^2)/4)/a^15 + (tan(e + f*x)*(2048*a^12*b^3 + 1024*a^1
3*b^2)*(a + 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(4096*a^16))*(a + 2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(32*a^6))*(a +
2*b)*(16*a*b + a^2 + 16*b^2)*5i)/(32*a^6)))*(a + 2*b)*(16*a*b + a^2 + 16*b^2))/(16*a^6*f) - ((5*tan(e + f*x)^
3*(57*a*b^3 + 14*a^3*b + a^4 + 24*b^4 + 46*a^2*b^2))/(6*a^5) + (5*tan(e + f*x)^7*(39*a*b^3 + 3*a^3*b + 24*b^4
+ 19*a^2*b^2))/(6*a^5) + (tan(e + f*x)^5*(2880*a*b^3 + 470*a^3*b + 33*a^4 + 1440*b^4 + 1910*a^2*b^2))/(48*a^5)
+ (5*tan(e + f*x)*(44*a*b^3 + 14*a^3*b + a^4 + 16*b^4 + 41*a^2*b^2))/(16*a^5) + (5*b*tan(e + f*x)^9*(20*a*b^2
+ 5*a^2*b + 16*b^3))/(16*a^5))/(f*(2*a*b + tan(e + f*x)^6*(8*a*b + a^2 + 10*b^2) + a^2 + b^2 + tan(e + f*x)^8
*(2*a*b + 5*b^2) + b^2*tan(e + f*x)^10 + tan(e + f*x)^2*(8*a*b + 3*a^2 + 5*b^2) + tan(e + f*x)^4*(12*a*b + 3*a
^2 + 10*b^2))) + (atan((((a + 4*b)*(3*a + 4*b)*(-b*(a + b))^(1/2)*((tan(e + f*x)*(179200*a*b^8 + 51200*b^9 + 2
49600*a^2*b^7 + 176000*a^3*b^6 + 65800*a^4*b^5 + 12300*a^5*b^4 + 925*a^6*b^3))/(128*a^10) - (5*(a + 4*b)*((20*
a^12*b^5 + 35*a^13*b^4 + (65*a^14*b^3)/4 + (5*a^15*b^2)/4)/a^15 - (5*tan(e + f*x)*(2048*a^12*b^3 + 1024*a^13*b
^2)*(a + 4*b)*(3*a + 4*b)*(-b*(a + b))^(1/2))/(2048*a^16))*(3*a + 4*b)*(-b*(a + b))^(1/2))/(16*a^6))*5i)/(16*a
^6) + ((a + 4*b)*(3*a + 4*b)*(-b*(a + b))^(1/2)*((tan(e + f*x)*(179200*a*b^8 + 51200*b^9 + 249600*a^2*b^7 + 17
6000*a^3*b^6 + 65800*a^4*b^5 + 12300*a^5*b^4 + 925*a^6*b^3))/(128*a^10) + (5*(a + 4*b)*((20*a^12*b^5 + 35*a^13
*b^4 + (65*a^14*b^3)/4 + (5*a^15*b^2)/4)/a^15 + (5*tan(e + f*x)*(2048*a^12*b^3 + 1024*a^13*b^2)*(a + 4*b)*(3*a
+ 4*b)*(-b*(a + b))^(1/2))/(2048*a^16))*(3*a + 4*b)*(-b*(a + b))^(1/2))/(16*a^6))*5i)/(16*a^6))/((4750*a*b^10
+ 1000*b^11 + (18875*a^2*b^9)/2 + (40625*a^3*b^8)/4 + (204875*a^4*b^7)/32 + (305125*a^5*b^6)/128 + (256125*a^
6*b^5)/512 + (53125*a^7*b^4)/1024 + (1875*a^8*b^3)/1024)/a^15 - (5*(a + 4*b)*(3*a + 4*b)*(-b*(a + b))^(1/2)*((
tan(e + f*x)*(179200*a*b^8 + 51200*b^9 + 249600*a^2*b^7 + 176000*a^3*b^6 + 65800*a^4*b^5 + 12300*a^5*b^4 + 925
*a^6*b^3))/(128*a^10) - (5*(a + 4*b)*((20*a^12*b^5 + 35*a^13*b^4 + (65*a^14*b^3)/4 + (5*a^15*b^2)/4)/a^15 - (5
*tan(e + f*x)*(2048*a^12*b^3 + 1024*a^13*b^2)*(a + 4*b)*(3*a + 4*b)*(-b*(a + b))^(1/2))/(2048*a^16))*(3*a + 4*
b)*(-b*(a + b))^(1/2))/(16*a^6)))/(16*a^6) + (5*(a + 4*b)*(3*a + 4*b)*(-b*(a + b))^(1/2)*((tan(e + f*x)*(17920
0*a*b^8 + 51200*b^9 + 249600*a^2*b^7 + 176000*a^3*b^6 + 65800*a^4*b^5 + 12300*a^5*b^4 + 925*a^6*b^3))/(128*a^1
0) + (5*(a + 4*b)*((20*a^12*b^5 + 35*a^13*b^4 + (65*a^14*b^3)/4 + (5*a^15*b^2)/4)/a^15 + (5*tan(e + f*x)*(2048
*a^12*b^3 + 1024*a^13*b^2)*(a + 4*b)*(3*a + 4*b)*(-b*(a + b))^(1/2))/(2048*a^16))*(3*a + 4*b)*(-b*(a + b))^(1/
2))/(16*a^6)))/(16*a^6)))*(a + 4*b)*(3*a + 4*b)*(-b*(a + b))^(1/2)*5i)/(8*a^6*f)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**6/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

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