3.1.77 \(\int (a+b \sin ^2(x))^4 \, dx\) [77]

3.1.77.1 Optimal result
3.1.77.2 Mathematica [A] (verified)
3.1.77.3 Rubi [A] (verified)
3.1.77.4 Maple [A] (verified)
3.1.77.5 Fricas [A] (verification not implemented)
3.1.77.6 Sympy [B] (verification not implemented)
3.1.77.7 Maxima [A] (verification not implemented)
3.1.77.8 Giac [A] (verification not implemented)
3.1.77.9 Mupad [B] (verification not implemented)

3.1.77.1 Optimal result

Integrand size = 10, antiderivative size = 140 \[ \int \left (a+b \sin ^2(x)\right )^4 \, dx=\frac {1}{128} \left (128 a^4+256 a^3 b+288 a^2 b^2+160 a b^3+35 b^4\right ) x-\frac {1}{384} b \left (608 a^3+808 a^2 b+480 a b^2+105 b^3\right ) \cos (x) \sin (x)-\frac {1}{192} b^2 \left (104 a^2+104 a b+35 b^2\right ) \cos (x) \sin ^3(x)-\frac {7}{48} b (2 a+b) \cos (x) \sin (x) \left (a+b \sin ^2(x)\right )^2-\frac {1}{8} b \cos (x) \sin (x) \left (a+b \sin ^2(x)\right )^3 \]

output
1/128*(128*a^4+256*a^3*b+288*a^2*b^2+160*a*b^3+35*b^4)*x-1/384*b*(608*a^3+ 
808*a^2*b+480*a*b^2+105*b^3)*cos(x)*sin(x)-1/192*b^2*(104*a^2+104*a*b+35*b 
^2)*cos(x)*sin(x)^3-7/48*b*(2*a+b)*cos(x)*sin(x)*(a+b*sin(x)^2)^2-1/8*b*co 
s(x)*sin(x)*(a+b*sin(x)^2)^3
 
3.1.77.2 Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.81 \[ \int \left (a+b \sin ^2(x)\right )^4 \, dx=\frac {24 \left (128 a^4+256 a^3 b+288 a^2 b^2+160 a b^3+35 b^4\right ) x-96 b (2 a+b) \left (16 a^2+16 a b+7 b^2\right ) \sin (2 x)+24 b^2 \left (24 a^2+24 a b+7 b^2\right ) \sin (4 x)-32 b^3 (2 a+b) \sin (6 x)+3 b^4 \sin (8 x)}{3072} \]

input
Integrate[(a + b*Sin[x]^2)^4,x]
 
output
(24*(128*a^4 + 256*a^3*b + 288*a^2*b^2 + 160*a*b^3 + 35*b^4)*x - 96*b*(2*a 
 + b)*(16*a^2 + 16*a*b + 7*b^2)*Sin[2*x] + 24*b^2*(24*a^2 + 24*a*b + 7*b^2 
)*Sin[4*x] - 32*b^3*(2*a + b)*Sin[6*x] + 3*b^4*Sin[8*x])/3072
 
3.1.77.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3659, 3042, 3649, 3042, 3648}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3659

\(\displaystyle \frac {1}{8} \int \left (b \sin ^2(x)+a\right )^2 \left (7 b (2 a+b) \sin ^2(x)+a (8 a+b)\right )dx-\frac {1}{8} b \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^3\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} \int \left (b \sin (x)^2+a\right )^2 \left (7 b (2 a+b) \sin (x)^2+a (8 a+b)\right )dx-\frac {1}{8} b \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^3\)

\(\Big \downarrow \) 3649

\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \int \left (b \sin ^2(x)+a\right ) \left (b \left (104 a^2+104 b a+35 b^2\right ) \sin ^2(x)+a \left (48 a^2+20 b a+7 b^2\right )\right )dx-\frac {7}{6} b (2 a+b) \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^2\right )-\frac {1}{8} b \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^3\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \int \left (b \sin (x)^2+a\right ) \left (b \left (104 a^2+104 b a+35 b^2\right ) \sin (x)^2+a \left (48 a^2+20 b a+7 b^2\right )\right )dx-\frac {7}{6} b (2 a+b) \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^2\right )-\frac {1}{8} b \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^3\)

\(\Big \downarrow \) 3648

\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (-\frac {1}{4} b^2 \left (104 a^2+104 a b+35 b^2\right ) \sin ^3(x) \cos (x)-\frac {1}{8} b \left (608 a^3+808 a^2 b+480 a b^2+105 b^3\right ) \sin (x) \cos (x)+\frac {3}{8} x \left (128 a^4+256 a^3 b+288 a^2 b^2+160 a b^3+35 b^4\right )\right )-\frac {7}{6} b (2 a+b) \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^2\right )-\frac {1}{8} b \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^3\)

input
Int[(a + b*Sin[x]^2)^4,x]
 
output
-1/8*(b*Cos[x]*Sin[x]*(a + b*Sin[x]^2)^3) + ((-7*b*(2*a + b)*Cos[x]*Sin[x] 
*(a + b*Sin[x]^2)^2)/6 + ((3*(128*a^4 + 256*a^3*b + 288*a^2*b^2 + 160*a*b^ 
3 + 35*b^4)*x)/8 - (b*(608*a^3 + 808*a^2*b + 480*a*b^2 + 105*b^3)*Cos[x]*S 
in[x])/8 - (b^2*(104*a^2 + 104*a*b + 35*b^2)*Cos[x]*Sin[x]^3)/4)/6)/8
 

3.1.77.3.1 Defintions of rubi rules used

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

rule 3648
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)*((A_.) + (B_.)*sin[(e_.) + (f_ 
.)*(x_)]^2), x_Symbol] :> Simp[(4*A*(2*a + b) + B*(4*a + 3*b))*(x/8), x] + 
(-Simp[b*B*Cos[e + f*x]*(Sin[e + f*x]^3/(4*f)), x] - Simp[(4*A*b + B*(4*a + 
 3*b))*Cos[e + f*x]*(Sin[e + f*x]/(8*f)), x]) /; FreeQ[{a, b, e, f, A, B}, 
x]
 

rule 3649
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]^2), x_Symbol] :> Simp[(-B)*Cos[e + f*x]*Sin[e + f*x]*((a + b* 
Sin[e + f*x]^2)^p/(2*f*(p + 1))), x] + Simp[1/(2*(p + 1))   Int[(a + b*Sin[ 
e + f*x]^2)^(p - 1)*Simp[a*B + 2*a*A*(p + 1) + (2*A*b*(p + 1) + B*(b + 2*a* 
p + 2*b*p))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && G 
tQ[p, 0]
 

rule 3659
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*C 
os[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p - 1)/(2*f*p)), x] + Sim 
p[1/(2*p)   Int[(a + b*Sin[e + f*x]^2)^(p - 2)*Simp[a*(b + 2*a*p) + b*(2*a 
+ b)*(2*p - 1)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[ 
a + b, 0] && GtQ[p, 1]
 
3.1.77.4 Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.79

method result size
default \(b^{4} \left (-\frac {\left (\sin ^{7}\left (x \right )+\frac {7 \left (\sin ^{5}\left (x \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (x \right )\right )}{24}+\frac {35 \sin \left (x \right )}{16}\right ) \cos \left (x \right )}{8}+\frac {35 x}{128}\right )+4 a \,b^{3} \left (-\frac {\left (\sin ^{5}\left (x \right )+\frac {5 \left (\sin ^{3}\left (x \right )\right )}{4}+\frac {15 \sin \left (x \right )}{8}\right ) \cos \left (x \right )}{6}+\frac {5 x}{16}\right )+6 a^{2} b^{2} \left (-\frac {\left (\sin ^{3}\left (x \right )+\frac {3 \sin \left (x \right )}{2}\right ) \cos \left (x \right )}{4}+\frac {3 x}{8}\right )+4 a^{3} b \left (-\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+a^{4} x\) \(110\)
parts \(b^{4} \left (-\frac {\left (\sin ^{7}\left (x \right )+\frac {7 \left (\sin ^{5}\left (x \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (x \right )\right )}{24}+\frac {35 \sin \left (x \right )}{16}\right ) \cos \left (x \right )}{8}+\frac {35 x}{128}\right )+4 a \,b^{3} \left (-\frac {\left (\sin ^{5}\left (x \right )+\frac {5 \left (\sin ^{3}\left (x \right )\right )}{4}+\frac {15 \sin \left (x \right )}{8}\right ) \cos \left (x \right )}{6}+\frac {5 x}{16}\right )+6 a^{2} b^{2} \left (-\frac {\left (\sin ^{3}\left (x \right )+\frac {3 \sin \left (x \right )}{2}\right ) \cos \left (x \right )}{4}+\frac {3 x}{8}\right )+4 a^{3} b \left (-\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}\right )+a^{4} x\) \(110\)
parallelrisch \(\frac {\left (-32 a^{3} b -48 a^{2} b^{2}-30 a \,b^{3}-7 b^{4}\right ) \sin \left (2 x \right )}{32}+\frac {\left (24 a^{2} b^{2}+24 a \,b^{3}+7 b^{4}\right ) \sin \left (4 x \right )}{128}+\frac {\left (-2 a \,b^{3}-b^{4}\right ) \sin \left (6 x \right )}{96}+\frac {b^{4} \sin \left (8 x \right )}{1024}+\left (a^{4}+2 a^{3} b +\frac {9}{4} a^{2} b^{2}+\frac {5}{4} a \,b^{3}+\frac {35}{128} b^{4}\right ) x\) \(118\)
risch \(a^{4} x +2 a^{3} b x +\frac {9 a^{2} b^{2} x}{4}+\frac {5 a \,b^{3} x}{4}+\frac {35 b^{4} x}{128}+\frac {b^{4} \sin \left (8 x \right )}{1024}-\frac {\sin \left (6 x \right ) a \,b^{3}}{48}-\frac {\sin \left (6 x \right ) b^{4}}{96}+\frac {3 \sin \left (4 x \right ) a^{2} b^{2}}{16}+\frac {3 \sin \left (4 x \right ) a \,b^{3}}{16}+\frac {7 \sin \left (4 x \right ) b^{4}}{128}-\sin \left (2 x \right ) a^{3} b -\frac {3 \sin \left (2 x \right ) a^{2} b^{2}}{2}-\frac {15 \sin \left (2 x \right ) a \,b^{3}}{16}-\frac {7 \sin \left (2 x \right ) b^{4}}{32}\) \(136\)
norman \(\frac {\left (-36 a^{3} b -\frac {153}{2} a^{2} b^{2}-\frac {383}{6} a \,b^{3}-\frac {2681}{192} b^{4}\right ) \left (\tan ^{5}\left (\frac {x}{2}\right )\right )+\left (-20 a^{3} b -\frac {93}{2} a^{2} b^{2}-\frac {283}{6} a \,b^{3}-\frac {5053}{192} b^{4}\right ) \left (\tan ^{7}\left (\frac {x}{2}\right )\right )+\left (-20 a^{3} b -\frac {69}{2} a^{2} b^{2}-\frac {115}{6} a \,b^{3}-\frac {805}{192} b^{4}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+\left (-4 a^{3} b -\frac {9}{2} a^{2} b^{2}-\frac {5}{2} a \,b^{3}-\frac {35}{64} b^{4}\right ) \tan \left (\frac {x}{2}\right )+\left (4 a^{3} b +\frac {9}{2} a^{2} b^{2}+\frac {5}{2} a \,b^{3}+\frac {35}{64} b^{4}\right ) \left (\tan ^{15}\left (\frac {x}{2}\right )\right )+\left (20 a^{3} b +\frac {69}{2} a^{2} b^{2}+\frac {115}{6} a \,b^{3}+\frac {805}{192} b^{4}\right ) \left (\tan ^{13}\left (\frac {x}{2}\right )\right )+\left (20 a^{3} b +\frac {93}{2} a^{2} b^{2}+\frac {283}{6} a \,b^{3}+\frac {5053}{192} b^{4}\right ) \left (\tan ^{9}\left (\frac {x}{2}\right )\right )+\left (36 a^{3} b +\frac {153}{2} a^{2} b^{2}+\frac {383}{6} a \,b^{3}+\frac {2681}{192} b^{4}\right ) \left (\tan ^{11}\left (\frac {x}{2}\right )\right )+\left (a^{4}+2 a^{3} b +\frac {9}{4} a^{2} b^{2}+\frac {5}{4} a \,b^{3}+\frac {35}{128} b^{4}\right ) x +\left (a^{4}+2 a^{3} b +\frac {9}{4} a^{2} b^{2}+\frac {5}{4} a \,b^{3}+\frac {35}{128} b^{4}\right ) x \left (\tan ^{16}\left (\frac {x}{2}\right )\right )+\left (8 a^{4}+16 a^{3} b +18 a^{2} b^{2}+10 a \,b^{3}+\frac {35}{16} b^{4}\right ) x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+\left (8 a^{4}+16 a^{3} b +18 a^{2} b^{2}+10 a \,b^{3}+\frac {35}{16} b^{4}\right ) x \left (\tan ^{14}\left (\frac {x}{2}\right )\right )+\left (28 a^{4}+56 a^{3} b +63 a^{2} b^{2}+35 a \,b^{3}+\frac {245}{32} b^{4}\right ) x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )+\left (28 a^{4}+56 a^{3} b +63 a^{2} b^{2}+35 a \,b^{3}+\frac {245}{32} b^{4}\right ) x \left (\tan ^{12}\left (\frac {x}{2}\right )\right )+\left (56 a^{4}+112 a^{3} b +126 a^{2} b^{2}+70 a \,b^{3}+\frac {245}{16} b^{4}\right ) x \left (\tan ^{6}\left (\frac {x}{2}\right )\right )+\left (56 a^{4}+112 a^{3} b +126 a^{2} b^{2}+70 a \,b^{3}+\frac {245}{16} b^{4}\right ) x \left (\tan ^{10}\left (\frac {x}{2}\right )\right )+\left (70 a^{4}+140 a^{3} b +\frac {315}{2} a^{2} b^{2}+\frac {175}{2} a \,b^{3}+\frac {1225}{64} b^{4}\right ) x \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{8}}\) \(616\)

input
int((a+b*sin(x)^2)^4,x,method=_RETURNVERBOSE)
 
output
b^4*(-1/8*(sin(x)^7+7/6*sin(x)^5+35/24*sin(x)^3+35/16*sin(x))*cos(x)+35/12 
8*x)+4*a*b^3*(-1/6*(sin(x)^5+5/4*sin(x)^3+15/8*sin(x))*cos(x)+5/16*x)+6*a^ 
2*b^2*(-1/4*(sin(x)^3+3/2*sin(x))*cos(x)+3/8*x)+4*a^3*b*(-1/2*cos(x)*sin(x 
)+1/2*x)+a^4*x
 
3.1.77.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.88 \[ \int \left (a+b \sin ^2(x)\right )^4 \, dx=\frac {1}{128} \, {\left (128 \, a^{4} + 256 \, a^{3} b + 288 \, a^{2} b^{2} + 160 \, a b^{3} + 35 \, b^{4}\right )} x + \frac {1}{384} \, {\left (48 \, b^{4} \cos \left (x\right )^{7} - 8 \, {\left (32 \, a b^{3} + 25 \, b^{4}\right )} \cos \left (x\right )^{5} + 2 \, {\left (288 \, a^{2} b^{2} + 416 \, a b^{3} + 163 \, b^{4}\right )} \cos \left (x\right )^{3} - 3 \, {\left (256 \, a^{3} b + 480 \, a^{2} b^{2} + 352 \, a b^{3} + 93 \, b^{4}\right )} \cos \left (x\right )\right )} \sin \left (x\right ) \]

input
integrate((a+b*sin(x)^2)^4,x, algorithm="fricas")
 
output
1/128*(128*a^4 + 256*a^3*b + 288*a^2*b^2 + 160*a*b^3 + 35*b^4)*x + 1/384*( 
48*b^4*cos(x)^7 - 8*(32*a*b^3 + 25*b^4)*cos(x)^5 + 2*(288*a^2*b^2 + 416*a* 
b^3 + 163*b^4)*cos(x)^3 - 3*(256*a^3*b + 480*a^2*b^2 + 352*a*b^3 + 93*b^4) 
*cos(x))*sin(x)
 
3.1.77.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (146) = 292\).

Time = 0.50 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.93 \[ \int \left (a+b \sin ^2(x)\right )^4 \, dx=a^{4} x + 2 a^{3} b x \sin ^{2}{\left (x \right )} + 2 a^{3} b x \cos ^{2}{\left (x \right )} - 2 a^{3} b \sin {\left (x \right )} \cos {\left (x \right )} + \frac {9 a^{2} b^{2} x \sin ^{4}{\left (x \right )}}{4} + \frac {9 a^{2} b^{2} x \sin ^{2}{\left (x \right )} \cos ^{2}{\left (x \right )}}{2} + \frac {9 a^{2} b^{2} x \cos ^{4}{\left (x \right )}}{4} - \frac {15 a^{2} b^{2} \sin ^{3}{\left (x \right )} \cos {\left (x \right )}}{4} - \frac {9 a^{2} b^{2} \sin {\left (x \right )} \cos ^{3}{\left (x \right )}}{4} + \frac {5 a b^{3} x \sin ^{6}{\left (x \right )}}{4} + \frac {15 a b^{3} x \sin ^{4}{\left (x \right )} \cos ^{2}{\left (x \right )}}{4} + \frac {15 a b^{3} x \sin ^{2}{\left (x \right )} \cos ^{4}{\left (x \right )}}{4} + \frac {5 a b^{3} x \cos ^{6}{\left (x \right )}}{4} - \frac {11 a b^{3} \sin ^{5}{\left (x \right )} \cos {\left (x \right )}}{4} - \frac {10 a b^{3} \sin ^{3}{\left (x \right )} \cos ^{3}{\left (x \right )}}{3} - \frac {5 a b^{3} \sin {\left (x \right )} \cos ^{5}{\left (x \right )}}{4} + \frac {35 b^{4} x \sin ^{8}{\left (x \right )}}{128} + \frac {35 b^{4} x \sin ^{6}{\left (x \right )} \cos ^{2}{\left (x \right )}}{32} + \frac {105 b^{4} x \sin ^{4}{\left (x \right )} \cos ^{4}{\left (x \right )}}{64} + \frac {35 b^{4} x \sin ^{2}{\left (x \right )} \cos ^{6}{\left (x \right )}}{32} + \frac {35 b^{4} x \cos ^{8}{\left (x \right )}}{128} - \frac {93 b^{4} \sin ^{7}{\left (x \right )} \cos {\left (x \right )}}{128} - \frac {511 b^{4} \sin ^{5}{\left (x \right )} \cos ^{3}{\left (x \right )}}{384} - \frac {385 b^{4} \sin ^{3}{\left (x \right )} \cos ^{5}{\left (x \right )}}{384} - \frac {35 b^{4} \sin {\left (x \right )} \cos ^{7}{\left (x \right )}}{128} \]

input
integrate((a+b*sin(x)**2)**4,x)
 
output
a**4*x + 2*a**3*b*x*sin(x)**2 + 2*a**3*b*x*cos(x)**2 - 2*a**3*b*sin(x)*cos 
(x) + 9*a**2*b**2*x*sin(x)**4/4 + 9*a**2*b**2*x*sin(x)**2*cos(x)**2/2 + 9* 
a**2*b**2*x*cos(x)**4/4 - 15*a**2*b**2*sin(x)**3*cos(x)/4 - 9*a**2*b**2*si 
n(x)*cos(x)**3/4 + 5*a*b**3*x*sin(x)**6/4 + 15*a*b**3*x*sin(x)**4*cos(x)** 
2/4 + 15*a*b**3*x*sin(x)**2*cos(x)**4/4 + 5*a*b**3*x*cos(x)**6/4 - 11*a*b* 
*3*sin(x)**5*cos(x)/4 - 10*a*b**3*sin(x)**3*cos(x)**3/3 - 5*a*b**3*sin(x)* 
cos(x)**5/4 + 35*b**4*x*sin(x)**8/128 + 35*b**4*x*sin(x)**6*cos(x)**2/32 + 
 105*b**4*x*sin(x)**4*cos(x)**4/64 + 35*b**4*x*sin(x)**2*cos(x)**6/32 + 35 
*b**4*x*cos(x)**8/128 - 93*b**4*sin(x)**7*cos(x)/128 - 511*b**4*sin(x)**5* 
cos(x)**3/384 - 385*b**4*sin(x)**3*cos(x)**5/384 - 35*b**4*sin(x)*cos(x)** 
7/128
 
3.1.77.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77 \[ \int \left (a+b \sin ^2(x)\right )^4 \, dx=\frac {1}{48} \, {\left (4 \, \sin \left (2 \, x\right )^{3} + 60 \, x + 9 \, \sin \left (4 \, x\right ) - 48 \, \sin \left (2 \, x\right )\right )} a b^{3} + \frac {1}{3072} \, {\left (128 \, \sin \left (2 \, x\right )^{3} + 840 \, x + 3 \, \sin \left (8 \, x\right ) + 168 \, \sin \left (4 \, x\right ) - 768 \, \sin \left (2 \, x\right )\right )} b^{4} + \frac {3}{16} \, a^{2} b^{2} {\left (12 \, x + \sin \left (4 \, x\right ) - 8 \, \sin \left (2 \, x\right )\right )} + a^{3} b {\left (2 \, x - \sin \left (2 \, x\right )\right )} + a^{4} x \]

input
integrate((a+b*sin(x)^2)^4,x, algorithm="maxima")
 
output
1/48*(4*sin(2*x)^3 + 60*x + 9*sin(4*x) - 48*sin(2*x))*a*b^3 + 1/3072*(128* 
sin(2*x)^3 + 840*x + 3*sin(8*x) + 168*sin(4*x) - 768*sin(2*x))*b^4 + 3/16* 
a^2*b^2*(12*x + sin(4*x) - 8*sin(2*x)) + a^3*b*(2*x - sin(2*x)) + a^4*x
 
3.1.77.8 Giac [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.84 \[ \int \left (a+b \sin ^2(x)\right )^4 \, dx=\frac {1}{1024} \, b^{4} \sin \left (8 \, x\right ) + \frac {1}{128} \, {\left (128 \, a^{4} + 256 \, a^{3} b + 288 \, a^{2} b^{2} + 160 \, a b^{3} + 35 \, b^{4}\right )} x - \frac {1}{96} \, {\left (2 \, a b^{3} + b^{4}\right )} \sin \left (6 \, x\right ) + \frac {1}{128} \, {\left (24 \, a^{2} b^{2} + 24 \, a b^{3} + 7 \, b^{4}\right )} \sin \left (4 \, x\right ) - \frac {1}{32} \, {\left (32 \, a^{3} b + 48 \, a^{2} b^{2} + 30 \, a b^{3} + 7 \, b^{4}\right )} \sin \left (2 \, x\right ) \]

input
integrate((a+b*sin(x)^2)^4,x, algorithm="giac")
 
output
1/1024*b^4*sin(8*x) + 1/128*(128*a^4 + 256*a^3*b + 288*a^2*b^2 + 160*a*b^3 
 + 35*b^4)*x - 1/96*(2*a*b^3 + b^4)*sin(6*x) + 1/128*(24*a^2*b^2 + 24*a*b^ 
3 + 7*b^4)*sin(4*x) - 1/32*(32*a^3*b + 48*a^2*b^2 + 30*a*b^3 + 7*b^4)*sin( 
2*x)
 
3.1.77.9 Mupad [B] (verification not implemented)

Time = 13.57 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.05 \[ \int \left (a+b \sin ^2(x)\right )^4 \, dx=x\,a^4-2\,\sin \left (x\right )\,a^3\,b\,\cos \left (x\right )+2\,x\,a^3\,b+\frac {3\,\sin \left (x\right )\,a^2\,b^2\,{\cos \left (x\right )}^3}{2}-\frac {15\,\sin \left (x\right )\,a^2\,b^2\,\cos \left (x\right )}{4}+\frac {9\,x\,a^2\,b^2}{4}-\frac {2\,\sin \left (x\right )\,a\,b^3\,{\cos \left (x\right )}^5}{3}+\frac {13\,\sin \left (x\right )\,a\,b^3\,{\cos \left (x\right )}^3}{6}-\frac {11\,\sin \left (x\right )\,a\,b^3\,\cos \left (x\right )}{4}+\frac {5\,x\,a\,b^3}{4}+\frac {\sin \left (x\right )\,b^4\,{\cos \left (x\right )}^7}{8}-\frac {25\,\sin \left (x\right )\,b^4\,{\cos \left (x\right )}^5}{48}+\frac {163\,\sin \left (x\right )\,b^4\,{\cos \left (x\right )}^3}{192}-\frac {93\,\sin \left (x\right )\,b^4\,\cos \left (x\right )}{128}+\frac {35\,x\,b^4}{128} \]

input
int((a + b*sin(x)^2)^4,x)
 
output
a^4*x + (35*b^4*x)/128 + (163*b^4*cos(x)^3*sin(x))/192 - (25*b^4*cos(x)^5* 
sin(x))/48 + (b^4*cos(x)^7*sin(x))/8 + (9*a^2*b^2*x)/4 - (93*b^4*cos(x)*si 
n(x))/128 + (5*a*b^3*x)/4 + 2*a^3*b*x + (3*a^2*b^2*cos(x)^3*sin(x))/2 - (1 
1*a*b^3*cos(x)*sin(x))/4 - 2*a^3*b*cos(x)*sin(x) - (15*a^2*b^2*cos(x)*sin( 
x))/4 + (13*a*b^3*cos(x)^3*sin(x))/6 - (2*a*b^3*cos(x)^5*sin(x))/3