3.3.93 \(\int \cos ^4(e+f x) (a+b \sin ^2(e+f x))^2 \, dx\) [293]

3.3.93.1 Optimal result
3.3.93.2 Mathematica [A] (verified)
3.3.93.3 Rubi [A] (verified)
3.3.93.4 Maple [A] (verified)
3.3.93.5 Fricas [A] (verification not implemented)
3.3.93.6 Sympy [B] (verification not implemented)
3.3.93.7 Maxima [A] (verification not implemented)
3.3.93.8 Giac [A] (verification not implemented)
3.3.93.9 Mupad [B] (verification not implemented)

3.3.93.1 Optimal result

Integrand size = 23, antiderivative size = 156 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {1}{128} \left (48 a^2+16 a b+3 b^2\right ) x+\frac {\left (48 a^2+16 a b+3 b^2\right ) \cos (e+f x) \sin (e+f x)}{128 f}+\frac {\left (48 a^2+16 a b+3 b^2\right ) \cos ^3(e+f x) \sin (e+f x)}{192 f}-\frac {b (10 a+3 b) \cos ^5(e+f x) \sin (e+f x)}{48 f}-\frac {b \cos ^7(e+f x) \sin (e+f x) \left (a+(a+b) \tan ^2(e+f x)\right )}{8 f} \]

output
1/128*(48*a^2+16*a*b+3*b^2)*x+1/128*(48*a^2+16*a*b+3*b^2)*cos(f*x+e)*sin(f 
*x+e)/f+1/192*(48*a^2+16*a*b+3*b^2)*cos(f*x+e)^3*sin(f*x+e)/f-1/48*b*(10*a 
+3*b)*cos(f*x+e)^5*sin(f*x+e)/f-1/8*b*cos(f*x+e)^7*sin(f*x+e)*(a+(a+b)*tan 
(f*x+e)^2)/f
 
3.3.93.2 Mathematica [A] (verified)

Time = 2.36 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.62 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {24 \left (48 a^2+16 a b+3 b^2\right ) (e+f x)+96 a (8 a+b) \sin (2 (e+f x))+24 \left (4 a^2-4 a b-b^2\right ) \sin (4 (e+f x))-32 a b \sin (6 (e+f x))+3 b^2 \sin (8 (e+f x))}{3072 f} \]

input
Integrate[Cos[e + f*x]^4*(a + b*Sin[e + f*x]^2)^2,x]
 
output
(24*(48*a^2 + 16*a*b + 3*b^2)*(e + f*x) + 96*a*(8*a + b)*Sin[2*(e + f*x)] 
+ 24*(4*a^2 - 4*a*b - b^2)*Sin[4*(e + f*x)] - 32*a*b*Sin[6*(e + f*x)] + 3* 
b^2*Sin[8*(e + f*x)])/(3072*f)
 
3.3.93.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3670, 315, 298, 215, 215, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (e+f x)^4 \left (a+b \sin (e+f x)^2\right )^2dx\)

\(\Big \downarrow \) 3670

\(\displaystyle \frac {\int \frac {\left ((a+b) \tan ^2(e+f x)+a\right )^2}{\left (\tan ^2(e+f x)+1\right )^5}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 315

\(\displaystyle \frac {\frac {1}{8} \int \frac {(a+b) (8 a+3 b) \tan ^2(e+f x)+a (8 a+b)}{\left (\tan ^2(e+f x)+1\right )^4}d\tan (e+f x)-\frac {b \tan (e+f x) \left ((a+b) \tan ^2(e+f x)+a\right )}{8 \left (\tan ^2(e+f x)+1\right )^4}}{f}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (48 a^2+16 a b+3 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)-\frac {b (10 a+3 b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}\right )-\frac {b \tan (e+f x) \left ((a+b) \tan ^2(e+f x)+a\right )}{8 \left (\tan ^2(e+f x)+1\right )^4}}{f}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (48 a^2+16 a b+3 b^2\right ) \left (\frac {3}{4} \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)+\frac {\tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {b (10 a+3 b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}\right )-\frac {b \tan (e+f x) \left ((a+b) \tan ^2(e+f x)+a\right )}{8 \left (\tan ^2(e+f x)+1\right )^4}}{f}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (48 a^2+16 a b+3 b^2\right ) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)+\frac {\tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {\tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {b (10 a+3 b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}\right )-\frac {b \tan (e+f x) \left ((a+b) \tan ^2(e+f x)+a\right )}{8 \left (\tan ^2(e+f x)+1\right )^4}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (48 a^2+16 a b+3 b^2\right ) \left (\frac {3}{4} \left (\frac {1}{2} \arctan (\tan (e+f x))+\frac {\tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {\tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {b (10 a+3 b) \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}\right )-\frac {b \tan (e+f x) \left ((a+b) \tan ^2(e+f x)+a\right )}{8 \left (\tan ^2(e+f x)+1\right )^4}}{f}\)

input
Int[Cos[e + f*x]^4*(a + b*Sin[e + f*x]^2)^2,x]
 
output
(-1/8*(b*Tan[e + f*x]*(a + (a + b)*Tan[e + f*x]^2))/(1 + Tan[e + f*x]^2)^4 
 + (-1/6*(b*(10*a + 3*b)*Tan[e + f*x])/(1 + Tan[e + f*x]^2)^3 + ((48*a^2 + 
 16*a*b + 3*b^2)*(Tan[e + f*x]/(4*(1 + Tan[e + f*x]^2)^2) + (3*(ArcTan[Tan 
[e + f*x]]/2 + Tan[e + f*x]/(2*(1 + Tan[e + f*x]^2))))/4))/6)/8)/f
 

3.3.93.3.1 Defintions of rubi rules used

rule 215
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 298
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 315
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3670
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Su 
bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e 
 + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
 
3.3.93.4 Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.62

method result size
parallelrisch \(\frac {\left (96 a^{2}-96 a b -24 b^{2}\right ) \sin \left (4 f x +4 e \right )+\left (768 a^{2}+96 a b \right ) \sin \left (2 f x +2 e \right )-32 a b \sin \left (6 f x +6 e \right )+3 b^{2} \sin \left (8 f x +8 e \right )+1152 f \left (a^{2}+\frac {1}{3} a b +\frac {1}{16} b^{2}\right ) x}{3072 f}\) \(96\)
risch \(\frac {3 a^{2} x}{8}+\frac {a b x}{8}+\frac {3 b^{2} x}{128}+\frac {b^{2} \sin \left (8 f x +8 e \right )}{1024 f}-\frac {a b \sin \left (6 f x +6 e \right )}{96 f}+\frac {\sin \left (4 f x +4 e \right ) a^{2}}{32 f}-\frac {\sin \left (4 f x +4 e \right ) a b}{32 f}-\frac {\sin \left (4 f x +4 e \right ) b^{2}}{128 f}+\frac {\sin \left (2 f x +2 e \right ) a^{2}}{4 f}+\frac {\sin \left (2 f x +2 e \right ) a b}{32 f}\) \(135\)
derivativedivides \(\frac {a^{2} \left (\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+2 a b \left (-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )}{6}+\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{24}+\frac {f x}{16}+\frac {e}{16}\right )+b^{2} \left (-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) \left (\sin ^{3}\left (f x +e \right )\right )}{8}-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )}{16}+\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{64}+\frac {3 f x}{128}+\frac {3 e}{128}\right )}{f}\) \(167\)
default \(\frac {a^{2} \left (\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+2 a b \left (-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )}{6}+\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{24}+\frac {f x}{16}+\frac {e}{16}\right )+b^{2} \left (-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) \left (\sin ^{3}\left (f x +e \right )\right )}{8}-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )}{16}+\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{64}+\frac {3 f x}{128}+\frac {3 e}{128}\right )}{f}\) \(167\)
norman \(\frac {\left (\frac {3}{8} a^{2}+\frac {1}{8} a b +\frac {3}{128} b^{2}\right ) x +\left (3 a^{2}+a b +\frac {3}{16} b^{2}\right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (3 a^{2}+a b +\frac {3}{16} b^{2}\right ) x \left (\tan ^{14}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (21 a^{2}+7 a b +\frac {21}{16} b^{2}\right ) x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (21 a^{2}+7 a b +\frac {21}{16} b^{2}\right ) x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {3}{8} a^{2}+\frac {1}{8} a b +\frac {3}{128} b^{2}\right ) x \left (\tan ^{16}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {21}{2} a^{2}+\frac {7}{2} a b +\frac {21}{32} b^{2}\right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {21}{2} a^{2}+\frac {7}{2} a b +\frac {21}{32} b^{2}\right ) x \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {105}{4} a^{2}+\frac {35}{4} a b +\frac {105}{64} b^{2}\right ) x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {\left (80 a^{2}-16 a b -3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{64 f}-\frac {\left (80 a^{2}-16 a b -3 b^{2}\right ) \left (\tan ^{15}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{64 f}+\frac {\left (432 a^{2}-496 a b -2013 b^{2}\right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{192 f}-\frac {\left (432 a^{2}-496 a b -2013 b^{2}\right ) \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{192 f}+\frac {\left (816 a^{2}+656 a b -69 b^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{192 f}-\frac {\left (816 a^{2}+656 a b -69 b^{2}\right ) \left (\tan ^{13}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{192 f}+\frac {\left (1008 a^{2}+208 a b +999 b^{2}\right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{192 f}-\frac {\left (1008 a^{2}+208 a b +999 b^{2}\right ) \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{192 f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{8}}\) \(503\)

input
int(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
 
output
1/3072*((96*a^2-96*a*b-24*b^2)*sin(4*f*x+4*e)+(768*a^2+96*a*b)*sin(2*f*x+2 
*e)-32*a*b*sin(6*f*x+6*e)+3*b^2*sin(8*f*x+8*e)+1152*f*(a^2+1/3*a*b+1/16*b^ 
2)*x)/f
 
3.3.93.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.73 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} f x + {\left (48 \, b^{2} \cos \left (f x + e\right )^{7} - 8 \, {\left (16 \, a b + 9 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{384 \, f} \]

input
integrate(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^2,x, algorithm="fricas")
 
output
1/384*(3*(48*a^2 + 16*a*b + 3*b^2)*f*x + (48*b^2*cos(f*x + e)^7 - 8*(16*a* 
b + 9*b^2)*cos(f*x + e)^5 + 2*(48*a^2 + 16*a*b + 3*b^2)*cos(f*x + e)^3 + 3 
*(48*a^2 + 16*a*b + 3*b^2)*cos(f*x + e))*sin(f*x + e))/f
 
3.3.93.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (146) = 292\).

Time = 0.73 (sec) , antiderivative size = 481, normalized size of antiderivative = 3.08 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {3 a^{2} x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {3 a^{2} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {5 a^{2} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac {a b x \sin ^{6}{\left (e + f x \right )}}{8} + \frac {3 a b x \sin ^{4}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{8} + \frac {3 a b x \sin ^{2}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{8} + \frac {a b x \cos ^{6}{\left (e + f x \right )}}{8} + \frac {a b \sin ^{5}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {a b \sin ^{3}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {a b \sin {\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{8 f} + \frac {3 b^{2} x \sin ^{8}{\left (e + f x \right )}}{128} + \frac {3 b^{2} x \sin ^{6}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{32} + \frac {9 b^{2} x \sin ^{4}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{64} + \frac {3 b^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{6}{\left (e + f x \right )}}{32} + \frac {3 b^{2} x \cos ^{8}{\left (e + f x \right )}}{128} + \frac {3 b^{2} \sin ^{7}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{128 f} + \frac {11 b^{2} \sin ^{5}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{128 f} - \frac {11 b^{2} \sin ^{3}{\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{128 f} - \frac {3 b^{2} \sin {\left (e + f x \right )} \cos ^{7}{\left (e + f x \right )}}{128 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin ^{2}{\left (e \right )}\right )^{2} \cos ^{4}{\left (e \right )} & \text {otherwise} \end {cases} \]

input
integrate(cos(f*x+e)**4*(a+b*sin(f*x+e)**2)**2,x)
 
output
Piecewise((3*a**2*x*sin(e + f*x)**4/8 + 3*a**2*x*sin(e + f*x)**2*cos(e + f 
*x)**2/4 + 3*a**2*x*cos(e + f*x)**4/8 + 3*a**2*sin(e + f*x)**3*cos(e + f*x 
)/(8*f) + 5*a**2*sin(e + f*x)*cos(e + f*x)**3/(8*f) + a*b*x*sin(e + f*x)** 
6/8 + 3*a*b*x*sin(e + f*x)**4*cos(e + f*x)**2/8 + 3*a*b*x*sin(e + f*x)**2* 
cos(e + f*x)**4/8 + a*b*x*cos(e + f*x)**6/8 + a*b*sin(e + f*x)**5*cos(e + 
f*x)/(8*f) + a*b*sin(e + f*x)**3*cos(e + f*x)**3/(3*f) - a*b*sin(e + f*x)* 
cos(e + f*x)**5/(8*f) + 3*b**2*x*sin(e + f*x)**8/128 + 3*b**2*x*sin(e + f* 
x)**6*cos(e + f*x)**2/32 + 9*b**2*x*sin(e + f*x)**4*cos(e + f*x)**4/64 + 3 
*b**2*x*sin(e + f*x)**2*cos(e + f*x)**6/32 + 3*b**2*x*cos(e + f*x)**8/128 
+ 3*b**2*sin(e + f*x)**7*cos(e + f*x)/(128*f) + 11*b**2*sin(e + f*x)**5*co 
s(e + f*x)**3/(128*f) - 11*b**2*sin(e + f*x)**3*cos(e + f*x)**5/(128*f) - 
3*b**2*sin(e + f*x)*cos(e + f*x)**7/(128*f), Ne(f, 0)), (x*(a + b*sin(e)** 
2)**2*cos(e)**4, True))
 
3.3.93.7 Maxima [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.08 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} {\left (f x + e\right )} + \frac {3 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{7} + 11 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + {\left (624 \, a^{2} + 80 \, a b - 33 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (80 \, a^{2} - 16 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{8} + 4 \, \tan \left (f x + e\right )^{6} + 6 \, \tan \left (f x + e\right )^{4} + 4 \, \tan \left (f x + e\right )^{2} + 1}}{384 \, f} \]

input
integrate(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^2,x, algorithm="maxima")
 
output
1/384*(3*(48*a^2 + 16*a*b + 3*b^2)*(f*x + e) + (3*(48*a^2 + 16*a*b + 3*b^2 
)*tan(f*x + e)^7 + 11*(48*a^2 + 16*a*b + 3*b^2)*tan(f*x + e)^5 + (624*a^2 
+ 80*a*b - 33*b^2)*tan(f*x + e)^3 + 3*(80*a^2 - 16*a*b - 3*b^2)*tan(f*x + 
e))/(tan(f*x + e)^8 + 4*tan(f*x + e)^6 + 6*tan(f*x + e)^4 + 4*tan(f*x + e) 
^2 + 1))/f
 
3.3.93.8 Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.67 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {1}{128} \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} x + \frac {b^{2} \sin \left (8 \, f x + 8 \, e\right )}{1024 \, f} - \frac {a b \sin \left (6 \, f x + 6 \, e\right )}{96 \, f} + \frac {{\left (4 \, a^{2} - 4 \, a b - b^{2}\right )} \sin \left (4 \, f x + 4 \, e\right )}{128 \, f} + \frac {{\left (8 \, a^{2} + a b\right )} \sin \left (2 \, f x + 2 \, e\right )}{32 \, f} \]

input
integrate(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^2,x, algorithm="giac")
 
output
1/128*(48*a^2 + 16*a*b + 3*b^2)*x + 1/1024*b^2*sin(8*f*x + 8*e)/f - 1/96*a 
*b*sin(6*f*x + 6*e)/f + 1/128*(4*a^2 - 4*a*b - b^2)*sin(4*f*x + 4*e)/f + 1 
/32*(8*a^2 + a*b)*sin(2*f*x + 2*e)/f
 
3.3.93.9 Mupad [B] (verification not implemented)

Time = 14.90 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=x\,\left (\frac {3\,a^2}{8}+\frac {a\,b}{8}+\frac {3\,b^2}{128}\right )+\frac {\left (\frac {3\,a^2}{8}+\frac {a\,b}{8}+\frac {3\,b^2}{128}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^7+\left (\frac {11\,a^2}{8}+\frac {11\,a\,b}{24}+\frac {11\,b^2}{128}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (\frac {13\,a^2}{8}+\frac {5\,a\,b}{24}-\frac {11\,b^2}{128}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {5\,a^2}{8}-\frac {a\,b}{8}-\frac {3\,b^2}{128}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^8+4\,{\mathrm {tan}\left (e+f\,x\right )}^6+6\,{\mathrm {tan}\left (e+f\,x\right )}^4+4\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )} \]

input
int(cos(e + f*x)^4*(a + b*sin(e + f*x)^2)^2,x)
 
output
x*((a*b)/8 + (3*a^2)/8 + (3*b^2)/128) + (tan(e + f*x)^7*((a*b)/8 + (3*a^2) 
/8 + (3*b^2)/128) - tan(e + f*x)*((a*b)/8 - (5*a^2)/8 + (3*b^2)/128) + tan 
(e + f*x)^3*((5*a*b)/24 + (13*a^2)/8 - (11*b^2)/128) + tan(e + f*x)^5*((11 
*a*b)/24 + (11*a^2)/8 + (11*b^2)/128))/(f*(4*tan(e + f*x)^2 + 6*tan(e + f* 
x)^4 + 4*tan(e + f*x)^6 + tan(e + f*x)^8 + 1))