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

Optimal. Leaf size=267 \[ -\frac {\sqrt {b} (a+b)^{3/2} (3 a+8 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^5 f}+\frac {(9 a+8 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac {b \left (19 a^2+52 a b+32 b^2\right ) \tan (e+f x)}{16 a^4 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\left (33 a^2+82 a b+48 b^2\right ) \sin (e+f x) \cos (e+f x)}{48 a^3 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac {x \left (5 a^3+60 a^2 b+120 a b^2+64 b^3\right )}{16 a^5}+\frac {\sin ^3(e+f x) \cos ^3(e+f x)}{6 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

1/16*(5*a^3+60*a^2*b+120*a*b^2+64*b^3)*x/a^5-1/2*(a+b)^(3/2)*(3*a+8*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*
b^(1/2)/a^5/f-1/48*(33*a^2+82*a*b+48*b^2)*cos(f*x+e)*sin(f*x+e)/a^3/f/(a+b+b*tan(f*x+e)^2)+1/24*(9*a+8*b)*cos(
f*x+e)^3*sin(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)+1/6*cos(f*x+e)^3*sin(f*x+e)^3/a/f/(a+b+b*tan(f*x+e)^2)-1/16*b*(
19*a^2+52*a*b+32*b^2)*tan(f*x+e)/a^4/f/(a+b+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.43, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {b \left (19 a^2+52 a b+32 b^2\right ) \tan (e+f x)}{16 a^4 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\left (33 a^2+82 a b+48 b^2\right ) \sin (e+f x) \cos (e+f x)}{48 a^3 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac {x \left (60 a^2 b+5 a^3+120 a b^2+64 b^3\right )}{16 a^5}-\frac {\sqrt {b} (a+b)^{3/2} (3 a+8 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^5 f}+\frac {(9 a+8 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\sin ^3(e+f x) \cos ^3(e+f x)}{6 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^3 + 60*a^2*b + 120*a*b^2 + 64*b^3)*x)/(16*a^5) - (Sqrt[b]*(a + b)^(3/2)*(3*a + 8*b)*ArcTan[(Sqrt[b]*Tan[
e + f*x])/Sqrt[a + b]])/(2*a^5*f) - ((33*a^2 + 82*a*b + 48*b^2)*Cos[e + f*x]*Sin[e + f*x])/(48*a^3*f*(a + b +
b*Tan[e + f*x]^2)) + ((9*a + 8*b)*Cos[e + f*x]^3*Sin[e + f*x])/(24*a^2*f*(a + b + b*Tan[e + f*x]^2)) + (Cos[e
+ f*x]^3*Sin[e + f*x]^3)/(6*a*f*(a + b + b*Tan[e + f*x]^2)) - (b*(19*a^2 + 52*a*b + 32*b^2)*Tan[e + f*x])/(16*
a^4*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 )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^4 \left (a+b+b x^2\right )^2} \, 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 )}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 (a+b)+(b-6 (a+b)) x^2\right )}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{6 a f}\\ &=\frac {(9 a+8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {(a+b) (9 a+8 b)+\left (-24 a^2-65 a b-40 b^2\right ) x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{24 a^2 f}\\ &=-\frac {\left (33 a^2+82 a b+48 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 )}+\frac {(9 a+8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {3 (a+b) \left (5 a^2+22 a b+16 b^2\right )-3 b \left (33 a^2+82 a b+48 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 )}{48 a^3 f}\\ &=-\frac {\left (33 a^2+82 a b+48 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 )}+\frac {(9 a+8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {b \left (19 a^2+52 a b+32 b^2\right ) \tan (e+f x)}{16 a^4 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {6 (a+b)^2 \left (5 a^2+36 a b+32 b^2\right )-6 b (a+b) \left (19 a^2+52 a b+32 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{96 a^4 (a+b) f}\\ &=-\frac {\left (33 a^2+82 a b+48 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 )}+\frac {(9 a+8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {b \left (19 a^2+52 a b+32 b^2\right ) \tan (e+f x)}{16 a^4 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\left (b (a+b)^2 (3 a+8 b)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^5 f}+\frac {\left (5 a^3+60 a^2 b+120 a b^2+64 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 a^5 f}\\ &=\frac {\left (5 a^3+60 a^2 b+120 a b^2+64 b^3\right ) x}{16 a^5}-\frac {\sqrt {b} (a+b)^{3/2} (3 a+8 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^5 f}-\frac {\left (33 a^2+82 a b+48 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 )}+\frac {(9 a+8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {b \left (19 a^2+52 a b+32 b^2\right ) \tan (e+f x)}{16 a^4 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [C]  time = 23.51, size = 2738, normalized size = 10.25 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/512*((a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(16*x + ((-a^3 + 6*a^2*b + 24*a*b^2 + 16*b^3)*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*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + ((a^2 + 8*a*b + 8*b
^2)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[e] - Sin[e])*(Cos[e]
+ Sin[e]))))/(a^2*(a + b*Sec[e + f*x]^2)^2) + (3*(a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(-64*(a + 2*
b)*x + ((a^4 - 16*a^3*b - 144*a^2*b^2 - 256*a*b^3 - 128*b^4)*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
*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (16*a*Cos[2*f*x]*Sin[2*e])/f + (16*a*Cos[2*e]*Sin[2*f*x])/f
- ((a^3 + 18*a^2*b + 48*a*b^2 + 32*b^3)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(
e + f*x)])*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(4096*a^3*(a + b*Sec[e + f*x]^2)^2) + (3*(a + 2*b + a*Cos[2*
e + 2*f*x])^2*Sec[e + f*x]^4*(((a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2) - (a*Sqrt[b
]*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)]))))/(2048*b^(3/2)*f*(a + b*Sec[e + f*x]^2)^2) - ((a
+ 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(-((a*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2))
+ (Sqrt[b]*(a + 2*b)*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)]))))/(2048*b^(3/2)*f*(a + b*Sec[
e + f*x]^2)^2) + ((a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(-(((a^5 - 30*a^4*b - 480*a^3*b^2 - 1600*a^2
*b^3 - 1920*a*b^4 - 768*b^5)*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]*Sqrt[b*(Cos[e] - I*Si
n[e])^4])) + (Sec[2*e]*(32*b*(5*a^4 + 39*a^3*b + 106*a^2*b^2 + 120*a*b^3 + 48*b^4)*f*x*Cos[2*e] + 16*a*b*(5*a^
3 + 29*a^2*b + 48*a*b^2 + 24*b^3)*f*x*Cos[2*f*x] + 80*a^4*b*f*x*Cos[4*e + 2*f*x] + 464*a^3*b^2*f*x*Cos[4*e + 2
*f*x] + 768*a^2*b^3*f*x*Cos[4*e + 2*f*x] + 384*a*b^4*f*x*Cos[4*e + 2*f*x] + a^5*Sin[2*e] + 34*a^4*b*Sin[2*e] +
224*a^3*b^2*Sin[2*e] + 576*a^2*b^3*Sin[2*e] + 640*a*b^4*Sin[2*e] + 256*b^5*Sin[2*e] - a^5*Sin[2*f*x] - 62*a^4
*b*Sin[2*f*x] - 318*a^3*b^2*Sin[2*f*x] - 512*a^2*b^3*Sin[2*f*x] - 256*a*b^4*Sin[2*f*x] - 12*a^4*b*Sin[2*(e + 2
*f*x)] - 36*a^3*b^2*Sin[2*(e + 2*f*x)] - 24*a^2*b^3*Sin[2*(e + 2*f*x)] - 30*a^4*b*Sin[4*e + 2*f*x] - 158*a^3*b
^2*Sin[4*e + 2*f*x] - 256*a^2*b^3*Sin[4*e + 2*f*x] - 128*a*b^4*Sin[4*e + 2*f*x] - 12*a^4*b*Sin[6*e + 4*f*x] -
36*a^3*b^2*Sin[6*e + 4*f*x] - 24*a^2*b^3*Sin[6*e + 4*f*x] + 2*a^4*b*Sin[4*e + 6*f*x] + 2*a^3*b^2*Sin[4*e + 6*f
*x] + 2*a^4*b*Sin[8*e + 6*f*x] + 2*a^3*b^2*Sin[8*e + 6*f*x]))/(a + 2*b + a*Cos[2*(e + f*x)])))/(2048*a^4*b*(a
+ b)*f*(a + b*Sec[e + f*x]^2)^2) + ((a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(-(((a^6 - 48*a^5*b - 1200
*a^4*b^2 - 6400*a^3*b^3 - 13440*a^2*b^4 - 12288*a*b^5 - 4096*b^6)*((ArcTan[Sec[f*x]*(Cos[2*e]/(2*Sqrt[a + b]*S
qrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2*e])/(Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]))*(-(a*Sin[f*x
]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Cos[2*e])/(8*a^5*b*Sqrt[a + b]*f*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I
/8)*ArcTan[Sec[f*x]*(Cos[2*e]/(2*Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2*e])/(Sqrt[a + b]*
Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]))*(-(a*Sin[f*x]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Sin[2*e])/(a^5*b*Sqrt[a +
b]*f*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]])))/(a + b)) - (Sec[2*e]*(-960*a^5*b*f*x*Cos[2*e] - 10944*a^4*b^2*f*x*Cos
[2*e] - 44544*a^3*b^3*f*x*Cos[2*e] - 83712*a^2*b^4*f*x*Cos[2*e] - 73728*a*b^5*f*x*Cos[2*e] - 24576*b^6*f*x*Cos
[2*e] - 480*a^5*b*f*x*Cos[2*f*x] - 4512*a^4*b^2*f*x*Cos[2*f*x] - 13248*a^3*b^3*f*x*Cos[2*f*x] - 15360*a^2*b^4*
f*x*Cos[2*f*x] - 6144*a*b^5*f*x*Cos[2*f*x] - 480*a^5*b*f*x*Cos[4*e + 2*f*x] - 4512*a^4*b^2*f*x*Cos[4*e + 2*f*x
] - 13248*a^3*b^3*f*x*Cos[4*e + 2*f*x] - 15360*a^2*b^4*f*x*Cos[4*e + 2*f*x] - 6144*a*b^5*f*x*Cos[4*e + 2*f*x]
- 3*a^6*Sin[2*e] - 156*a^5*b*Sin[2*e] - 1500*a^4*b^2*Sin[2*e] - 5760*a^3*b^3*Sin[2*e] - 10560*a^2*b^4*Sin[2*e]
- 9216*a*b^5*Sin[2*e] - 3072*b^6*Sin[2*e] + 3*a^6*Sin[2*f*x] + 366*a^5*b*Sin[2*f*x] + 3000*a^4*b^2*Sin[2*f*x]
+ 8400*a^3*b^3*Sin[2*f*x] + 9600*a^2*b^4*Sin[2*f*x] + 3840*a*b^5*Sin[2*f*x] + 216*a^5*b*Sin[4*e + 2*f*x] + 18
00*a^4*b^2*Sin[4*e + 2*f*x] + 5040*a^3*b^3*Sin[4*e + 2*f*x] + 5760*a^2*b^4*Sin[4*e + 2*f*x] + 2304*a*b^5*Sin[4
*e + 2*f*x] + 76*a^5*b*Sin[2*e + 4*f*x] + 460*a^4*b^2*Sin[2*e + 4*f*x] + 768*a^3*b^3*Sin[2*e + 4*f*x] + 384*a^
2*b^4*Sin[2*e + 4*f*x] + 76*a^5*b*Sin[6*e + 4*f*x] + 460*a^4*b^2*Sin[6*e + 4*f*x] + 768*a^3*b^3*Sin[6*e + 4*f*
x] + 384*a^2*b^4*Sin[6*e + 4*f*x] - 16*a^5*b*Sin[4*e + 6*f*x] - 48*a^4*b^2*Sin[4*e + 6*f*x] - 32*a^3*b^3*Sin[4
*e + 6*f*x] - 16*a^5*b*Sin[8*e + 6*f*x] - 48*a^4*b^2*Sin[8*e + 6*f*x] - 32*a^3*b^3*Sin[8*e + 6*f*x] + 4*a^5*b*
Sin[6*e + 8*f*x] + 4*a^4*b^2*Sin[6*e + 8*f*x] + 4*a^5*b*Sin[10*e + 8*f*x] + 4*a^4*b^2*Sin[10*e + 8*f*x]))/(24*
a^5*b*(a + b)*f*(a + 2*b + a*Cos[2*e + 2*f*x]))))/(512*(a + b*Sec[e + f*x]^2)^2)

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fricas [A]  time = 0.68, size = 674, normalized size = 2.52 \[ \left [\frac {3 \, {\left (5 \, a^{4} + 60 \, a^{3} b + 120 \, a^{2} b^{2} + 64 \, a b^{3}\right )} f x \cos \left (f x + e\right )^{2} + 3 \, {\left (5 \, a^{3} b + 60 \, a^{2} b^{2} + 120 \, a b^{3} + 64 \, b^{4}\right )} f x + 6 \, {\left (3 \, a^{2} b + 11 \, a b^{2} + 8 \, b^{3} + {\left (3 \, a^{3} + 11 \, a^{2} b + 8 \, a b^{2}\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 ) - {\left (8 \, a^{4} \cos \left (f x + e\right )^{7} - 2 \, {\left (13 \, a^{4} + 8 \, a^{3} b\right )} \cos \left (f x + e\right )^{5} + {\left (33 \, a^{4} + 82 \, a^{3} b + 48 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (19 \, a^{3} b + 52 \, a^{2} b^{2} + 32 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, {\left (a^{6} f \cos \left (f x + e\right )^{2} + a^{5} b f\right )}}, \frac {3 \, {\left (5 \, a^{4} + 60 \, a^{3} b + 120 \, a^{2} b^{2} + 64 \, a b^{3}\right )} f x \cos \left (f x + e\right )^{2} + 3 \, {\left (5 \, a^{3} b + 60 \, a^{2} b^{2} + 120 \, a b^{3} + 64 \, b^{4}\right )} f x + 12 \, {\left (3 \, a^{2} b + 11 \, a b^{2} + 8 \, b^{3} + {\left (3 \, a^{3} + 11 \, a^{2} b + 8 \, a b^{2}\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^{4} \cos \left (f x + e\right )^{7} - 2 \, {\left (13 \, a^{4} + 8 \, a^{3} b\right )} \cos \left (f x + e\right )^{5} + {\left (33 \, a^{4} + 82 \, a^{3} b + 48 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (19 \, a^{3} b + 52 \, a^{2} b^{2} + 32 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, {\left (a^{6} f \cos \left (f x + e\right )^{2} + a^{5} b 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)^2,x, algorithm="fricas")

[Out]

[1/48*(3*(5*a^4 + 60*a^3*b + 120*a^2*b^2 + 64*a*b^3)*f*x*cos(f*x + e)^2 + 3*(5*a^3*b + 60*a^2*b^2 + 120*a*b^3
+ 64*b^4)*f*x + 6*(3*a^2*b + 11*a*b^2 + 8*b^3 + (3*a^3 + 11*a^2*b + 8*a*b^2)*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)) - (8*a^
4*cos(f*x + e)^7 - 2*(13*a^4 + 8*a^3*b)*cos(f*x + e)^5 + (33*a^4 + 82*a^3*b + 48*a^2*b^2)*cos(f*x + e)^3 + 3*(
19*a^3*b + 52*a^2*b^2 + 32*a*b^3)*cos(f*x + e))*sin(f*x + e))/(a^6*f*cos(f*x + e)^2 + a^5*b*f), 1/48*(3*(5*a^4
+ 60*a^3*b + 120*a^2*b^2 + 64*a*b^3)*f*x*cos(f*x + e)^2 + 3*(5*a^3*b + 60*a^2*b^2 + 120*a*b^3 + 64*b^4)*f*x +
12*(3*a^2*b + 11*a*b^2 + 8*b^3 + (3*a^3 + 11*a^2*b + 8*a*b^2)*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^4*cos(f*x + e)^7 - 2*(13*a^4 +
8*a^3*b)*cos(f*x + e)^5 + (33*a^4 + 82*a^3*b + 48*a^2*b^2)*cos(f*x + e)^3 + 3*(19*a^3*b + 52*a^2*b^2 + 32*a*b^
3)*cos(f*x + e))*sin(f*x + e))/(a^6*f*cos(f*x + e)^2 + a^5*b*f)]

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giac [A]  time = 0.38, size = 311, normalized size = 1.16 \[ \frac {\frac {3 \, {\left (5 \, a^{3} + 60 \, a^{2} b + 120 \, a b^{2} + 64 \, b^{3}\right )} {\left (f x + e\right )}}{a^{5}} - \frac {24 \, {\left (3 \, a^{3} b + 14 \, a^{2} b^{2} + 19 \, a b^{3} + 8 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{5}} - \frac {24 \, {\left (a^{2} b \tan \left (f x + e\right ) + 2 \, a b^{2} \tan \left (f x + e\right ) + b^{3} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a^{4}} - \frac {33 \, a^{2} \tan \left (f x + e\right )^{5} + 108 \, a b \tan \left (f x + e\right )^{5} + 72 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 192 \, a b \tan \left (f x + e\right )^{3} + 144 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) + 84 \, a b \tan \left (f x + e\right ) + 72 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{4}}}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(5*a^3 + 60*a^2*b + 120*a*b^2 + 64*b^3)*(f*x + e)/a^5 - 24*(3*a^3*b + 14*a^2*b^2 + 19*a*b^3 + 8*b^4)*(
pi*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^5) - 24*(a^2*
b*tan(f*x + e) + 2*a*b^2*tan(f*x + e) + b^3*tan(f*x + e))/((b*tan(f*x + e)^2 + a + b)*a^4) - (33*a^2*tan(f*x +
e)^5 + 108*a*b*tan(f*x + e)^5 + 72*b^2*tan(f*x + e)^5 + 40*a^2*tan(f*x + e)^3 + 192*a*b*tan(f*x + e)^3 + 144*
b^2*tan(f*x + e)^3 + 15*a^2*tan(f*x + e) + 84*a*b*tan(f*x + e) + 72*b^2*tan(f*x + e))/((tan(f*x + e)^2 + 1)^3*
a^4))/f

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maple [B]  time = 0.95, size = 555, normalized size = 2.08 \[ -\frac {b \tan \left (f x +e \right )}{2 a^{2} f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {b^{2} \tan \left (f x +e \right )}{f \,a^{3} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {b^{3} \tan \left (f x +e \right )}{2 f \,a^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f \,a^{2} \sqrt {\left (a +b \right ) b}}-\frac {7 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{3} \sqrt {\left (a +b \right ) b}}-\frac {19 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f \,a^{4} \sqrt {\left (a +b \right ) b}}-\frac {4 b^{4} \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 {9 \left (\tan ^{5}\left (f x +e \right )\right ) b}{4 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {3 \left (\tan ^{5}\left (f x +e \right )\right ) b^{2}}{2 f \,a^{4} \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^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {4 \left (\tan ^{3}\left (f x +e \right )\right ) b}{f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {3 \left (\tan ^{3}\left (f x +e \right )\right ) b^{2}}{f \,a^{4} \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^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {5 \tan \left (f x +e \right )}{16 f \,a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {7 \tan \left (f x +e \right ) b}{4 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {3 \tan \left (f x +e \right ) b^{2}}{2 f \,a^{4} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}+\frac {15 \arctan \left (\tan \left (f x +e \right )\right ) b}{4 f \,a^{3}}+\frac {15 \arctan \left (\tan \left (f x +e \right )\right ) b^{2}}{2 f \,a^{4}}+\frac {4 \arctan \left (\tan \left (f x +e \right )\right ) b^{3}}{f \,a^{5}}+\frac {5 \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*tan(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)-1/f*b^2/a^3*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)-1/2/f*b^3/a^4*tan(f*x
+e)/(a+b+b*tan(f*x+e)^2)-3/2/f*b/a^2/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-7/f*b^2/a^3/((a+b)*b
)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-19/2/f*b^3/a^4/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2
))-4/f*b^4/a^5/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-9/4/f/a^3/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5*
b-3/2/f/a^4/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5*b^2-11/16/f/a^2/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5-4/f/a^3/(tan(f*x+e
)^2+1)^3*tan(f*x+e)^3*b-3/f/a^4/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3*b^2-5/6/f/a^2/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3-
5/16/f/a^2/(tan(f*x+e)^2+1)^3*tan(f*x+e)-7/4/f/a^3/(tan(f*x+e)^2+1)^3*tan(f*x+e)*b-3/2/f/a^4/(tan(f*x+e)^2+1)^
3*tan(f*x+e)*b^2+15/4/f/a^3*arctan(tan(f*x+e))*b+15/2/f/a^4*arctan(tan(f*x+e))*b^2+4/f/a^5*arctan(tan(f*x+e))*
b^3+5/16/f/a^2*arctan(tan(f*x+e))

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maxima [A]  time = 0.46, size = 302, normalized size = 1.13 \[ -\frac {\frac {3 \, {\left (19 \, a^{2} b + 52 \, a b^{2} + 32 \, b^{3}\right )} \tan \left (f x + e\right )^{7} + {\left (33 \, a^{3} + 253 \, a^{2} b + 516 \, a b^{2} + 288 \, b^{3}\right )} \tan \left (f x + e\right )^{5} + {\left (40 \, a^{3} + 319 \, a^{2} b + 564 \, a b^{2} + 288 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{3} + 41 \, a^{2} b + 68 \, a b^{2} + 32 \, b^{3}\right )} \tan \left (f x + e\right )}{a^{4} b \tan \left (f x + e\right )^{8} + {\left (a^{5} + 4 \, a^{4} b\right )} \tan \left (f x + e\right )^{6} + a^{5} + a^{4} b + 3 \, {\left (a^{5} + 2 \, a^{4} b\right )} \tan \left (f x + e\right )^{4} + {\left (3 \, a^{5} + 4 \, a^{4} b\right )} \tan \left (f x + e\right )^{2}} - \frac {3 \, {\left (5 \, a^{3} + 60 \, a^{2} b + 120 \, a b^{2} + 64 \, b^{3}\right )} {\left (f x + e\right )}}{a^{5}} + \frac {24 \, {\left (3 \, a^{3} b + 14 \, a^{2} b^{2} + 19 \, a b^{3} + 8 \, b^{4}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{5}}}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*((3*(19*a^2*b + 52*a*b^2 + 32*b^3)*tan(f*x + e)^7 + (33*a^3 + 253*a^2*b + 516*a*b^2 + 288*b^3)*tan(f*x +
e)^5 + (40*a^3 + 319*a^2*b + 564*a*b^2 + 288*b^3)*tan(f*x + e)^3 + 3*(5*a^3 + 41*a^2*b + 68*a*b^2 + 32*b^3)*t
an(f*x + e))/(a^4*b*tan(f*x + e)^8 + (a^5 + 4*a^4*b)*tan(f*x + e)^6 + a^5 + a^4*b + 3*(a^5 + 2*a^4*b)*tan(f*x
+ e)^4 + (3*a^5 + 4*a^4*b)*tan(f*x + e)^2) - 3*(5*a^3 + 60*a^2*b + 120*a*b^2 + 64*b^3)*(f*x + e)/a^5 + 24*(3*a
^3*b + 14*a^2*b^2 + 19*a*b^3 + 8*b^4)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*a^5))/f

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mupad [B]  time = 6.66, size = 1461, normalized size = 5.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((75*b^3*tan(e + f*x)*(- 3*a*b^3 - a^3*b - b^4 - 3*a^2*b^2)^(1/2))/(256*((211*a*b^4)/128 + (811*b^5)/256
+ (75*a^2*b^3)/256 + (41*b^6)/(16*a) + (3*b^7)/(4*a^2))) + (17*b^4*tan(e + f*x)*(- 3*a*b^3 - a^3*b - b^4 - 3*
a^2*b^2)^(1/2))/(16*((811*a*b^5)/256 + (41*b^6)/16 + (211*a^2*b^4)/128 + (75*a^3*b^3)/256 + (3*b^7)/(4*a))) +
(3*b^5*tan(e + f*x)*(- 3*a*b^3 - a^3*b - b^4 - 3*a^2*b^2)^(1/2))/(4*((41*a*b^6)/16 + (3*b^7)/4 + (811*a^2*b^5)
/256 + (211*a^3*b^4)/128 + (75*a^4*b^3)/256)))*(-b*(a + b)^3)^(1/2)*(3*a + 8*b))/(2*a^5*f) - (atan(((((((8*a^1
0*b^5 + 17*a^11*b^4 + (41*a^12*b^3)/4 + (5*a^13*b^2)/4)/a^12 - (tan(e + f*x)*(2048*a^10*b^3 + 1024*a^11*b^2)*(
a*b^2*120i + a^2*b*60i + a^3*5i + b^3*64i))/(4096*a^13))*(a*b^2*120i + a^2*b*60i + a^3*5i + b^3*64i))/(32*a^5)
- (tan(e + f*x)*(34816*a*b^8 + 8192*b^9 + 59520*a^2*b^7 + 52160*a^3*b^6 + 24640*a^4*b^5 + 5976*a^5*b^4 + 601*
a^6*b^3))/(128*a^8))*(a*b^2*120i + a^2*b*60i + a^3*5i + b^3*64i)*1i)/(32*a^5) - (((((8*a^10*b^5 + 17*a^11*b^4
+ (41*a^12*b^3)/4 + (5*a^13*b^2)/4)/a^12 + (tan(e + f*x)*(2048*a^10*b^3 + 1024*a^11*b^2)*(a*b^2*120i + a^2*b*6
0i + a^3*5i + b^3*64i))/(4096*a^13))*(a*b^2*120i + a^2*b*60i + a^3*5i + b^3*64i))/(32*a^5) + (tan(e + f*x)*(34
816*a*b^8 + 8192*b^9 + 59520*a^2*b^7 + 52160*a^3*b^6 + 24640*a^4*b^5 + 5976*a^5*b^4 + 601*a^6*b^3))/(128*a^8))
*(a*b^2*120i + a^2*b*60i + a^3*5i + b^3*64i)*1i)/(32*a^5))/((376*a*b^10 + 64*b^11 + 937*a^2*b^9 + (10285*a^3*b
^8)/8 + (33701*a^4*b^7)/32 + (8333*a^5*b^6)/16 + (38085*a^6*b^5)/256 + (2765*a^7*b^4)/128 + (285*a^8*b^3)/256)
/a^12 + (((((8*a^10*b^5 + 17*a^11*b^4 + (41*a^12*b^3)/4 + (5*a^13*b^2)/4)/a^12 - (tan(e + f*x)*(2048*a^10*b^3
+ 1024*a^11*b^2)*(a*b^2*120i + a^2*b*60i + a^3*5i + b^3*64i))/(4096*a^13))*(a*b^2*120i + a^2*b*60i + a^3*5i +
b^3*64i))/(32*a^5) - (tan(e + f*x)*(34816*a*b^8 + 8192*b^9 + 59520*a^2*b^7 + 52160*a^3*b^6 + 24640*a^4*b^5 + 5
976*a^5*b^4 + 601*a^6*b^3))/(128*a^8))*(a*b^2*120i + a^2*b*60i + a^3*5i + b^3*64i))/(32*a^5) + (((((8*a^10*b^5
+ 17*a^11*b^4 + (41*a^12*b^3)/4 + (5*a^13*b^2)/4)/a^12 + (tan(e + f*x)*(2048*a^10*b^3 + 1024*a^11*b^2)*(a*b^2
*120i + a^2*b*60i + a^3*5i + b^3*64i))/(4096*a^13))*(a*b^2*120i + a^2*b*60i + a^3*5i + b^3*64i))/(32*a^5) + (t
an(e + f*x)*(34816*a*b^8 + 8192*b^9 + 59520*a^2*b^7 + 52160*a^3*b^6 + 24640*a^4*b^5 + 5976*a^5*b^4 + 601*a^6*b
^3))/(128*a^8))*(a*b^2*120i + a^2*b*60i + a^3*5i + b^3*64i))/(32*a^5)))*(a*b^2*120i + a^2*b*60i + a^3*5i + b^3
*64i)*1i)/(16*a^5*f) - ((tan(e + f*x)*(68*a*b^2 + 41*a^2*b + 5*a^3 + 32*b^3))/(16*a^4) + (tan(e + f*x)^5*(516*
a*b^2 + 253*a^2*b + 33*a^3 + 288*b^3))/(48*a^4) + (tan(e + f*x)^3*(564*a*b^2 + 319*a^2*b + 40*a^3 + 288*b^3))/
(48*a^4) + (b*tan(e + f*x)^7*(52*a*b + 19*a^2 + 32*b^2))/(16*a^4))/(f*(a + b + tan(e + f*x)^2*(3*a + 4*b) + ta
n(e + f*x)^4*(3*a + 6*b) + b*tan(e + f*x)^8 + tan(e + f*x)^6*(a + 4*b)))

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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)**2,x)

[Out]

Timed out

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