3.61 \(\int \frac {\sin ^4(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)
Optimal. Leaf size=238 \[ -\frac {3 b (a+2 b) \tan (e+f x)}{2 a^4 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac {b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {(5 a+8 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {3 \sqrt {b} \left (5 a^2+20 a b+16 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^5 f \sqrt {a+b}}+\frac {3 x \left (a^2+12 a b+16 b^2\right )}{8 a^5}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]
[Out]
3/8*(a^2+12*a*b+16*b^2)*x/a^5-3/8*(5*a^2+20*a*b+16*b^2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^(1/2)/a^5/f/(
a+b)^(1/2)-1/8*(5*a+8*b)*cos(f*x+e)*sin(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^2+1/4*cos(f*x+e)^3*sin(f*x+e)/a/f/(a
+b+b*tan(f*x+e)^2)^2-1/8*b*(7*a+12*b)*tan(f*x+e)/a^3/f/(a+b+b*tan(f*x+e)^2)^2-3/2*b*(a+2*b)*tan(f*x+e)/a^4/f/(
a+b+b*tan(f*x+e)^2)
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Rubi [A] time = 0.34, antiderivative size = 238, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used =
{4132, 470, 527, 522, 203, 205} \[ -\frac {3 \sqrt {b} \left (5 a^2+20 a b+16 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^5 f \sqrt {a+b}}+\frac {3 x \left (a^2+12 a b+16 b^2\right )}{8 a^5}-\frac {3 b (a+2 b) \tan (e+f x)}{2 a^4 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac {b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {(5 a+8 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]
Antiderivative was successfully verified.
[In]
Int[Sin[e + f*x]^4/(a + b*Sec[e + f*x]^2)^3,x]
[Out]
(3*(a^2 + 12*a*b + 16*b^2)*x)/(8*a^5) - (3*Sqrt[b]*(5*a^2 + 20*a*b + 16*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqr
t[a + b]])/(8*a^5*Sqrt[a + b]*f) - ((5*a + 8*b)*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*f*(a + b + b*Tan[e + f*x]^2)
^2) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*a*f*(a + b + b*Tan[e + f*x]^2)^2) - (b*(7*a + 12*b)*Tan[e + f*x])/(8*a^
3*f*(a + b + b*Tan[e + f*x]^2)^2) - (3*b*(a + 2*b)*Tan[e + f*x])/(2*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 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 ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {\operatorname {Subst}\left (\int \frac {a+b+(-4 a-7 b) x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=-\frac {(5 a+8 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {(a+b) (3 a+8 b)-5 b (5 a+8 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{8 a^2 f}\\ &=-\frac {(5 a+8 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {12 (a+b)^2 (a+4 b)-12 b (a+b) (7 a+12 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{32 a^3 (a+b) f}\\ &=-\frac {(5 a+8 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {3 b (a+2 b) \tan (e+f x)}{2 a^4 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {24 (a+b)^2 \left (a^2+8 a b+8 b^2\right )-96 b (a+b)^2 (a+2 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{64 a^4 (a+b)^2 f}\\ &=-\frac {(5 a+8 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {3 b (a+2 b) \tan (e+f x)}{2 a^4 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\left (3 \left (a^2+12 a b+16 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a^5 f}-\frac {\left (3 b \left (5 a^2+20 a b+16 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^5 f}\\ &=\frac {3 \left (a^2+12 a b+16 b^2\right ) x}{8 a^5}-\frac {3 \sqrt {b} \left (5 a^2+20 a b+16 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^5 \sqrt {a+b} f}-\frac {(5 a+8 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {3 b (a+2 b) \tan (e+f x)}{2 a^4 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}
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Mathematica [C] time = 25.01, size = 2469, normalized size = 10.37 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sin[e + f*x]^4/(a + b*Sec[e + f*x]^2)^3,x]
[Out]
(3*(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)))/(16384*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^3) + ((a + 2*b + a
*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((-3*a*(a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(5/2)
+ (Sqrt[b]*(3*a^3 + 14*a^2*b + 24*a*b^2 + 16*b^3 + a*(3*a^2 + 4*a*b + 4*b^2)*Cos[2*(e + f*x)])*Sin[2*(e + f*x
)])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2)))/(16384*b^(5/2)*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*((2*(3*a^5 - 10*a^4*b + 80*a^3*b^2 + 480*a^2*b^3 + 640*a*b^4 + 256*b^5)*A
rcTan[(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*Sin[e])^4]) + (Sec[2*e]*(256*b^2
*(a + b)^2*(3*a^2 + 8*a*b + 8*b^2)*f*x*Cos[2*e] + 512*a*b^2*(a + b)^2*(a + 2*b)*f*x*Cos[2*f*x] + 128*a^4*b^2*f
*x*Cos[2*(e + 2*f*x)] + 256*a^3*b^3*f*x*Cos[2*(e + 2*f*x)] + 128*a^2*b^4*f*x*Cos[2*(e + 2*f*x)] + 512*a^4*b^2*
f*x*Cos[4*e + 2*f*x] + 2048*a^3*b^3*f*x*Cos[4*e + 2*f*x] + 2560*a^2*b^4*f*x*Cos[4*e + 2*f*x] + 1024*a*b^5*f*x*
Cos[4*e + 2*f*x] + 128*a^4*b^2*f*x*Cos[6*e + 4*f*x] + 256*a^3*b^3*f*x*Cos[6*e + 4*f*x] + 128*a^2*b^4*f*x*Cos[6
*e + 4*f*x] - 9*a^6*Sin[2*e] + 12*a^5*b*Sin[2*e] + 684*a^4*b^2*Sin[2*e] + 2880*a^3*b^3*Sin[2*e] + 5280*a^2*b^4
*Sin[2*e] + 4608*a*b^5*Sin[2*e] + 1536*b^6*Sin[2*e] + 9*a^6*Sin[2*f*x] - 14*a^5*b*Sin[2*f*x] - 608*a^4*b^2*Sin
[2*f*x] - 2112*a^3*b^3*Sin[2*f*x] - 2560*a^2*b^4*Sin[2*f*x] - 1024*a*b^5*Sin[2*f*x] + 3*a^6*Sin[2*(e + 2*f*x)]
- 12*a^5*b*Sin[2*(e + 2*f*x)] - 204*a^4*b^2*Sin[2*(e + 2*f*x)] - 384*a^3*b^3*Sin[2*(e + 2*f*x)] - 192*a^2*b^4
*Sin[2*(e + 2*f*x)] - 3*a^6*Sin[4*e + 2*f*x] + 10*a^5*b*Sin[4*e + 2*f*x] + 304*a^4*b^2*Sin[4*e + 2*f*x] + 1056
*a^3*b^3*Sin[4*e + 2*f*x] + 1280*a^2*b^4*Sin[4*e + 2*f*x] + 512*a*b^5*Sin[4*e + 2*f*x]))/(a + 2*b + a*Cos[2*(e
+ f*x)])^2))/(65536*a^3*b^2*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^3) - ((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]*Sqrt[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)))/(8192*b
^2*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^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]*(Co
s[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 + 128
0*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 - 307
2*b^6)*Tan[2*e])/(b^2*(a + b)^2*f*(a + 2*b + a*Cos[2*(e + f*x)]))))/(16384*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*(768*(7*a^2 + 32*a*b + 32*b^2)*x + (3*(a^7 - 14*a^6*b + 336*a^5*
b^2 + 5600*a^4*b^3 + 22400*a^3*b^4 + 37632*a^2*b^5 + 28672*a*b^6 + 8192*b^7)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Si
n[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^5 + 50*a^4*b + 400*a^3*b^2 + 1120*a^
2*b^3 + 1280*a*b^4 + 512*b^5)*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) - ((768*I)*a*(a + 2*b)*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)]))/f + ((768*I)*a*(a + 2*b)*(Cos[2*(
e + f*x)] + I*Sin[2*(e + f*x)]))/f + (128*a^2*Sin[4*(e + f*x)])/f + (a*(3*a^6 - 44*a^5*b - 1900*a^4*b^2 - 1088
0*a^3*b^3 - 23360*a^2*b^4 - 21504*a*b^5 - 7168*b^6)*Sec[2*e]*Sin[2*f*x] + (-3*a^7 + 42*a^6*b + 2192*a^5*b^2 +
16480*a^4*b^3 + 51200*a^3*b^4 + 77824*a^2*b^5 + 57344*a*b^6 + 16384*b^7)*Tan[2*e])/(b^2*(a + b)^2*f*(a + 2*b +
a*Cos[2*(e + f*x)]))))/(32768*a^5*(a + b*Sec[e + f*x]^2)^3)
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fricas [A] time = 0.60, size = 803, normalized size = 3.37 \[ \left [\frac {12 \, {\left (a^{4} + 12 \, a^{3} b + 16 \, a^{2} b^{2}\right )} f x \cos \left (f x + e\right )^{4} + 24 \, {\left (a^{3} b + 12 \, a^{2} b^{2} + 16 \, a b^{3}\right )} f x \cos \left (f x + e\right )^{2} + 12 \, {\left (a^{2} b^{2} + 12 \, a b^{3} + 16 \, b^{4}\right )} f x + 3 \, {\left ({\left (5 \, a^{4} + 20 \, a^{3} b + 16 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 5 \, a^{2} b^{2} + 20 \, a b^{3} + 16 \, b^{4} + 2 \, {\left (5 \, a^{3} b + 20 \, a^{2} b^{2} + 16 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a + b}} \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} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \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 ) + 4 \, {\left (2 \, a^{4} \cos \left (f x + e\right )^{7} - {\left (5 \, a^{4} + 8 \, a^{3} b\right )} \cos \left (f x + e\right )^{5} - {\left (19 \, a^{3} b + 36 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} - 12 \, {\left (a^{2} b^{2} + 2 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left (a^{7} f \cos \left (f x + e\right )^{4} + 2 \, a^{6} b f \cos \left (f x + e\right )^{2} + a^{5} b^{2} f\right )}}, \frac {6 \, {\left (a^{4} + 12 \, a^{3} b + 16 \, a^{2} b^{2}\right )} f x \cos \left (f x + e\right )^{4} + 12 \, {\left (a^{3} b + 12 \, a^{2} b^{2} + 16 \, a b^{3}\right )} f x \cos \left (f x + e\right )^{2} + 6 \, {\left (a^{2} b^{2} + 12 \, a b^{3} + 16 \, b^{4}\right )} f x + 3 \, {\left ({\left (5 \, a^{4} + 20 \, a^{3} b + 16 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 5 \, a^{2} b^{2} + 20 \, a b^{3} + 16 \, b^{4} + 2 \, {\left (5 \, a^{3} b + 20 \, a^{2} b^{2} + 16 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 2 \, {\left (2 \, a^{4} \cos \left (f x + e\right )^{7} - {\left (5 \, a^{4} + 8 \, a^{3} b\right )} \cos \left (f x + e\right )^{5} - {\left (19 \, a^{3} b + 36 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} - 12 \, {\left (a^{2} b^{2} + 2 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left (a^{7} f \cos \left (f x + e\right )^{4} + 2 \, a^{6} b f \cos \left (f x + e\right )^{2} + a^{5} b^{2} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
[Out]
[1/32*(12*(a^4 + 12*a^3*b + 16*a^2*b^2)*f*x*cos(f*x + e)^4 + 24*(a^3*b + 12*a^2*b^2 + 16*a*b^3)*f*x*cos(f*x +
e)^2 + 12*(a^2*b^2 + 12*a*b^3 + 16*b^4)*f*x + 3*((5*a^4 + 20*a^3*b + 16*a^2*b^2)*cos(f*x + e)^4 + 5*a^2*b^2 +
20*a*b^3 + 16*b^4 + 2*(5*a^3*b + 20*a^2*b^2 + 16*a*b^3)*cos(f*x + e)^2)*sqrt(-b/(a + b))*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 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2
)*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + 4*(2
*a^4*cos(f*x + e)^7 - (5*a^4 + 8*a^3*b)*cos(f*x + e)^5 - (19*a^3*b + 36*a^2*b^2)*cos(f*x + e)^3 - 12*(a^2*b^2
+ 2*a*b^3)*cos(f*x + e))*sin(f*x + e))/(a^7*f*cos(f*x + e)^4 + 2*a^6*b*f*cos(f*x + e)^2 + a^5*b^2*f), 1/16*(6*
(a^4 + 12*a^3*b + 16*a^2*b^2)*f*x*cos(f*x + e)^4 + 12*(a^3*b + 12*a^2*b^2 + 16*a*b^3)*f*x*cos(f*x + e)^2 + 6*(
a^2*b^2 + 12*a*b^3 + 16*b^4)*f*x + 3*((5*a^4 + 20*a^3*b + 16*a^2*b^2)*cos(f*x + e)^4 + 5*a^2*b^2 + 20*a*b^3 +
16*b^4 + 2*(5*a^3*b + 20*a^2*b^2 + 16*a*b^3)*cos(f*x + e)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e
)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e))) + 2*(2*a^4*cos(f*x + e)^7 - (5*a^4 + 8*a^3*b)*cos(f*x
+ e)^5 - (19*a^3*b + 36*a^2*b^2)*cos(f*x + e)^3 - 12*(a^2*b^2 + 2*a*b^3)*cos(f*x + e))*sin(f*x + e))/(a^7*f*co
s(f*x + e)^4 + 2*a^6*b*f*cos(f*x + e)^2 + a^5*b^2*f)]
________________________________________________________________________________________
giac [A] time = 0.51, size = 325, normalized size = 1.37 \[ \frac {\frac {3 \, {\left (a^{2} + 12 \, a b + 16 \, b^{2}\right )} {\left (f x + e\right )}}{a^{5}} - \frac {3 \, {\left (5 \, a^{2} b + 20 \, a b^{2} + 16 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{5}} - \frac {12 \, a b^{2} \tan \left (f x + e\right )^{7} + 24 \, b^{3} \tan \left (f x + e\right )^{7} + 19 \, a^{2} b \tan \left (f x + e\right )^{5} + 72 \, a b^{2} \tan \left (f x + e\right )^{5} + 72 \, b^{3} \tan \left (f x + e\right )^{5} + 5 \, a^{3} \tan \left (f x + e\right )^{3} + 46 \, a^{2} b \tan \left (f x + e\right )^{3} + 108 \, a b^{2} \tan \left (f x + e\right )^{3} + 72 \, b^{3} \tan \left (f x + e\right )^{3} + 3 \, a^{3} \tan \left (f x + e\right ) + 27 \, a^{2} b \tan \left (f x + e\right ) + 48 \, a b^{2} \tan \left (f x + e\right ) + 24 \, b^{3} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{4} + a \tan \left (f x + e\right )^{2} + 2 \, b \tan \left (f x + e\right )^{2} + a + b\right )}^{2} a^{4}}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
[Out]
1/8*(3*(a^2 + 12*a*b + 16*b^2)*(f*x + e)/a^5 - 3*(5*a^2*b + 20*a*b^2 + 16*b^3)*(pi*floor((f*x + e)/pi + 1/2)*s
gn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/(sqrt(a*b + b^2)*a^5) - (12*a*b^2*tan(f*x + e)^7 + 24*b^3*tan(
f*x + e)^7 + 19*a^2*b*tan(f*x + e)^5 + 72*a*b^2*tan(f*x + e)^5 + 72*b^3*tan(f*x + e)^5 + 5*a^3*tan(f*x + e)^3
+ 46*a^2*b*tan(f*x + e)^3 + 108*a*b^2*tan(f*x + e)^3 + 72*b^3*tan(f*x + e)^3 + 3*a^3*tan(f*x + e) + 27*a^2*b*t
an(f*x + e) + 48*a*b^2*tan(f*x + e) + 24*b^3*tan(f*x + e))/((b*tan(f*x + e)^4 + a*tan(f*x + e)^2 + 2*b*tan(f*x
+ e)^2 + a + b)^2*a^4))/f
________________________________________________________________________________________
maple [A] time = 1.18, size = 423, normalized size = 1.78 \[ -\frac {7 b^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \,a^{3} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {3 b^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{2 f \,a^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {9 b \tan \left (f x +e \right )}{8 a^{2} f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {21 b^{2} \tan \left (f x +e \right )}{8 f \,a^{3} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {3 b^{3} \tan \left (f x +e \right )}{2 f \,a^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {15 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{8 f \,a^{3} \sqrt {\left (a +b \right ) b}}-\frac {15 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f \,a^{4} \sqrt {\left (a +b \right ) b}}-\frac {6 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{5} \sqrt {\left (a +b \right ) b}}-\frac {3 \left (\tan ^{3}\left (f x +e \right )\right ) b}{2 f \,a^{4} \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}-\frac {5 \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}-\frac {3 \tan \left (f x +e \right )}{8 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}-\frac {3 \tan \left (f x +e \right ) b}{2 f \,a^{4} \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}+\frac {9 \arctan \left (\tan \left (f x +e \right )\right ) b}{2 f \,a^{4}}+\frac {6 \arctan \left (\tan \left (f x +e \right )\right ) b^{2}}{f \,a^{5}}+\frac {3 \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(f*x+e)^4/(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)^3-3/2/f*b^3/a^4/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-9/8*b*tan
(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^2-21/8/f*b^2/a^3/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)-3/2/f*b^3/a^4/(a+b+b*tan
(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))-15/2/f*b^2/a^4/((a+b
)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-6/f*b^3/a^5/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2
))-3/2/f/a^4/(tan(f*x+e)^2+1)^2*tan(f*x+e)^3*b-5/8/f/a^3/(tan(f*x+e)^2+1)^2*tan(f*x+e)^3-3/8/f/a^3/(tan(f*x+e)
^2+1)^2*tan(f*x+e)-3/2/f/a^4/(tan(f*x+e)^2+1)^2*tan(f*x+e)*b+9/2/f/a^4*arctan(tan(f*x+e))*b+6/f/a^5*arctan(tan
(f*x+e))*b^2+3/8/f/a^3*arctan(tan(f*x+e))
________________________________________________________________________________________
maxima [A] time = 0.44, size = 299, normalized size = 1.26 \[ -\frac {\frac {12 \, {\left (a b^{2} + 2 \, b^{3}\right )} \tan \left (f x + e\right )^{7} + {\left (19 \, a^{2} b + 72 \, a b^{2} + 72 \, b^{3}\right )} \tan \left (f x + e\right )^{5} + {\left (5 \, a^{3} + 46 \, a^{2} b + 108 \, a b^{2} + 72 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{3} + 9 \, a^{2} b + 16 \, a b^{2} + 8 \, b^{3}\right )} \tan \left (f x + e\right )}{a^{4} b^{2} \tan \left (f x + e\right )^{8} + 2 \, {\left (a^{5} b + 2 \, a^{4} b^{2}\right )} \tan \left (f x + e\right )^{6} + a^{6} + 2 \, a^{5} b + a^{4} b^{2} + {\left (a^{6} + 6 \, a^{5} b + 6 \, a^{4} b^{2}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{6} + 3 \, a^{5} b + 2 \, a^{4} b^{2}\right )} \tan \left (f x + e\right )^{2}} - \frac {3 \, {\left (a^{2} + 12 \, a b + 16 \, b^{2}\right )} {\left (f x + e\right )}}{a^{5}} + \frac {3 \, {\left (5 \, a^{2} b + 20 \, a b^{2} + 16 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{5}}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
[Out]
-1/8*((12*(a*b^2 + 2*b^3)*tan(f*x + e)^7 + (19*a^2*b + 72*a*b^2 + 72*b^3)*tan(f*x + e)^5 + (5*a^3 + 46*a^2*b +
108*a*b^2 + 72*b^3)*tan(f*x + e)^3 + 3*(a^3 + 9*a^2*b + 16*a*b^2 + 8*b^3)*tan(f*x + e))/(a^4*b^2*tan(f*x + e)
^8 + 2*(a^5*b + 2*a^4*b^2)*tan(f*x + e)^6 + a^6 + 2*a^5*b + a^4*b^2 + (a^6 + 6*a^5*b + 6*a^4*b^2)*tan(f*x + e)
^4 + 2*(a^6 + 3*a^5*b + 2*a^4*b^2)*tan(f*x + e)^2) - 3*(a^2 + 12*a*b + 16*b^2)*(f*x + e)/a^5 + 3*(5*a^2*b + 20
*a*b^2 + 16*b^3)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*a^5))/f
________________________________________________________________________________________
mupad [B] time = 7.19, size = 1317, normalized size = 5.53 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(e + f*x)^4/(a + b/cos(e + f*x)^2)^3,x)
[Out]
(atan(((((tan(e + f*x)*(4608*a*b^6 + 2304*b^7 + 3312*a^2*b^5 + 1008*a^3*b^4 + 117*a^4*b^3))/(16*a^8) - (3*((12
*a^10*b^4 + 12*a^11*b^3 + (3*a^12*b^2)/2)/a^12 - (3*tan(e + f*x)*(256*a^10*b^3 + 128*a^11*b^2)*(-b*(a + b))^(1
/2)*(20*a*b + 5*a^2 + 16*b^2))/(256*a^8*(a^5*b + a^6)))*(-b*(a + b))^(1/2)*(20*a*b + 5*a^2 + 16*b^2))/(16*(a^5
*b + a^6)))*(-b*(a + b))^(1/2)*(20*a*b + 5*a^2 + 16*b^2)*3i)/(16*(a^5*b + a^6)) + (((tan(e + f*x)*(4608*a*b^6
+ 2304*b^7 + 3312*a^2*b^5 + 1008*a^3*b^4 + 117*a^4*b^3))/(16*a^8) + (3*((12*a^10*b^4 + 12*a^11*b^3 + (3*a^12*b
^2)/2)/a^12 + (3*tan(e + f*x)*(256*a^10*b^3 + 128*a^11*b^2)*(-b*(a + b))^(1/2)*(20*a*b + 5*a^2 + 16*b^2))/(256
*a^8*(a^5*b + a^6)))*(-b*(a + b))^(1/2)*(20*a*b + 5*a^2 + 16*b^2))/(16*(a^5*b + a^6)))*(-b*(a + b))^(1/2)*(20*
a*b + 5*a^2 + 16*b^2)*3i)/(16*(a^5*b + a^6)))/((540*a*b^7 + 216*b^8 + (999*a^2*b^6)/2 + (837*a^3*b^5)/4 + (121
5*a^4*b^4)/32 + (135*a^5*b^3)/64)/a^12 - (3*((tan(e + f*x)*(4608*a*b^6 + 2304*b^7 + 3312*a^2*b^5 + 1008*a^3*b^
4 + 117*a^4*b^3))/(16*a^8) - (3*((12*a^10*b^4 + 12*a^11*b^3 + (3*a^12*b^2)/2)/a^12 - (3*tan(e + f*x)*(256*a^10
*b^3 + 128*a^11*b^2)*(-b*(a + b))^(1/2)*(20*a*b + 5*a^2 + 16*b^2))/(256*a^8*(a^5*b + a^6)))*(-b*(a + b))^(1/2)
*(20*a*b + 5*a^2 + 16*b^2))/(16*(a^5*b + a^6)))*(-b*(a + b))^(1/2)*(20*a*b + 5*a^2 + 16*b^2))/(16*(a^5*b + a^6
)) + (3*((tan(e + f*x)*(4608*a*b^6 + 2304*b^7 + 3312*a^2*b^5 + 1008*a^3*b^4 + 117*a^4*b^3))/(16*a^8) + (3*((12
*a^10*b^4 + 12*a^11*b^3 + (3*a^12*b^2)/2)/a^12 + (3*tan(e + f*x)*(256*a^10*b^3 + 128*a^11*b^2)*(-b*(a + b))^(1
/2)*(20*a*b + 5*a^2 + 16*b^2))/(256*a^8*(a^5*b + a^6)))*(-b*(a + b))^(1/2)*(20*a*b + 5*a^2 + 16*b^2))/(16*(a^5
*b + a^6)))*(-b*(a + b))^(1/2)*(20*a*b + 5*a^2 + 16*b^2))/(16*(a^5*b + a^6))))*(-b*(a + b))^(1/2)*(20*a*b + 5*
a^2 + 16*b^2)*3i)/(8*f*(a^5*b + a^6)) - (atan((27*b^2*tan(e + f*x))/(256*((27*b^2)/256 + (81*b^3)/(64*a) + (27
*b^4)/(16*a^2))) + (81*b^3*tan(e + f*x))/(64*((27*a*b^2)/256 + (81*b^3)/64 + (27*b^4)/(16*a))) + (27*b^4*tan(e
+ f*x))/(16*((81*a*b^3)/64 + (27*b^4)/16 + (27*a^2*b^2)/256)))*(a*b*12i + a^2*1i + b^2*16i)*3i)/(8*a^5*f) - (
(tan(e + f*x)^5*(72*a*b^2 + 19*a^2*b + 72*b^3))/(8*a^4) + (tan(e + f*x)^3*(108*a*b^2 + 46*a^2*b + 5*a^3 + 72*b
^3))/(8*a^4) + (3*tan(e + f*x)*(16*a*b^2 + 9*a^2*b + a^3 + 8*b^3))/(8*a^4) + (3*b*tan(e + f*x)^7*(a*b + 2*b^2)
)/(2*a^4))/(f*(2*a*b + tan(e + f*x)^4*(6*a*b + a^2 + 6*b^2) + a^2 + b^2 + tan(e + f*x)^6*(2*a*b + 4*b^2) + b^2
*tan(e + f*x)^8 + tan(e + f*x)^2*(6*a*b + 2*a^2 + 4*b^2)))
________________________________________________________________________________________
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)**4/(a+b*sec(f*x+e)**2)**3,x)
[Out]
Timed out
________________________________________________________________________________________