3.77 \(\int (a+b \sin ^2(x))^4 \, dx\)

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

[Out]

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*
cos(x)*sin(x)*(a+b*sin(x)^2)^3

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Rubi [A]  time = 0.17, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3180, 3170, 3169} \[ \frac {1}{128} x \left (288 a^2 b^2+256 a^3 b+128 a^4+160 a b^3+35 b^4\right )-\frac {1}{192} b^2 \left (104 a^2+104 a b+35 b^2\right ) \sin ^3(x) \cos (x)-\frac {1}{384} b \left (808 a^2 b+608 a^3+480 a b^2+105 b^3\right ) \sin (x) \cos (x)-\frac {1}{8} b \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^3-\frac {7}{48} b (2 a+b) \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x]^2)^4,x]

[Out]

((128*a^4 + 256*a^3*b + 288*a^2*b^2 + 160*a*b^3 + 35*b^4)*x)/128 - (b*(608*a^3 + 808*a^2*b + 480*a*b^2 + 105*b
^3)*Cos[x]*Sin[x])/384 - (b^2*(104*a^2 + 104*a*b + 35*b^2)*Cos[x]*Sin[x]^3)/192 - (7*b*(2*a + b)*Cos[x]*Sin[x]
*(a + b*Sin[x]^2)^2)/48 - (b*Cos[x]*Sin[x]*(a + b*Sin[x]^2)^3)/8

Rule 3169

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 3170

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[(B*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(2*f*(p + 1)), x] + Dist[1/(2*(p + 1)), Int[(a + b*Si
n[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] && GtQ[p, 0]

Rule 3180

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
e + f*x]^2)^(p - 1))/(2*f*p), x] + Dist[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]

Rubi steps

\begin {align*} \int \left (a+b \sin ^2(x)\right )^4 \, dx &=-\frac {1}{8} b \cos (x) \sin (x) \left (a+b \sin ^2(x)\right )^3+\frac {1}{8} \int \left (a+b \sin ^2(x)\right )^2 \left (a (8 a+b)+7 b (2 a+b) \sin ^2(x)\right ) \, dx\\ &=-\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+\frac {1}{48} \int \left (a+b \sin ^2(x)\right ) \left (a \left (48 a^2+20 a b+7 b^2\right )+b \left (104 a^2+104 a b+35 b^2\right ) \sin ^2(x)\right ) \, 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\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 113, normalized size = 0.81 \[ \frac {24 b^2 \left (24 a^2+24 a b+7 b^2\right ) \sin (4 x)-96 b (2 a+b) \left (16 a^2+16 a b+7 b^2\right ) \sin (2 x)+24 x \left (128 a^4+256 a^3 b+288 a^2 b^2+160 a b^3+35 b^4\right )-32 b^3 (2 a+b) \sin (6 x)+3 b^4 \sin (8 x)}{3072} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[x]^2)^4,x]

[Out]

(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

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fricas [A]  time = 0.44, size = 123, normalized size = 0.88 \[ \frac {1}{128} \, {\left (128 \, a^{4} + 256 \, a^{3} b + 288 \, a^{2} b^{2} + 160 \, a b^{3} + 35 \, b^{4}\right )} x + \frac {1}{384} \, {\left (48 \, b^{4} \cos \relax (x)^{7} - 8 \, {\left (32 \, a b^{3} + 25 \, b^{4}\right )} \cos \relax (x)^{5} + 2 \, {\left (288 \, a^{2} b^{2} + 416 \, a b^{3} + 163 \, b^{4}\right )} \cos \relax (x)^{3} - 3 \, {\left (256 \, a^{3} b + 480 \, a^{2} b^{2} + 352 \, a b^{3} + 93 \, b^{4}\right )} \cos \relax (x)\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(x)^2)^4,x, algorithm="fricas")

[Out]

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)

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giac [A]  time = 0.14, size = 118, normalized size = 0.84 \[ \frac {1}{1024} \, b^{4} \sin \left (8 \, x\right ) + \frac {1}{128} \, {\left (128 \, a^{4} + 256 \, a^{3} b + 288 \, a^{2} b^{2} + 160 \, a b^{3} + 35 \, b^{4}\right )} x - \frac {1}{96} \, {\left (2 \, a b^{3} + b^{4}\right )} \sin \left (6 \, x\right ) + \frac {1}{128} \, {\left (24 \, a^{2} b^{2} + 24 \, a b^{3} + 7 \, b^{4}\right )} \sin \left (4 \, x\right ) - \frac {1}{32} \, {\left (32 \, a^{3} b + 48 \, a^{2} b^{2} + 30 \, a b^{3} + 7 \, b^{4}\right )} \sin \left (2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(x)^2)^4,x, algorithm="giac")

[Out]

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)*si
n(2*x)

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maple [A]  time = 0.56, size = 110, normalized size = 0.79 \[ b^{4} \left (-\frac {\left (\sin ^{7}\relax (x )+\frac {7 \left (\sin ^{5}\relax (x )\right )}{6}+\frac {35 \left (\sin ^{3}\relax (x )\right )}{24}+\frac {35 \sin \relax (x )}{16}\right ) \cos \relax (x )}{8}+\frac {35 x}{128}\right )+4 a \,b^{3} \left (-\frac {\left (\sin ^{5}\relax (x )+\frac {5 \left (\sin ^{3}\relax (x )\right )}{4}+\frac {15 \sin \relax (x )}{8}\right ) \cos \relax (x )}{6}+\frac {5 x}{16}\right )+6 a^{2} b^{2} \left (-\frac {\left (\sin ^{3}\relax (x )+\frac {3 \sin \relax (x )}{2}\right ) \cos \relax (x )}{4}+\frac {3 x}{8}\right )+4 a^{3} b \left (-\frac {\sin \relax (x ) \cos \relax (x )}{2}+\frac {x}{2}\right )+a^{4} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(x)^2)^4,x)

[Out]

b^4*(-1/8*(sin(x)^7+7/6*sin(x)^5+35/24*sin(x)^3+35/16*sin(x))*cos(x)+35/128*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*sin(x)*cos
(x)+1/2*x)+a^4*x

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maxima [A]  time = 0.34, size = 108, normalized size = 0.77 \[ \frac {1}{48} \, {\left (4 \, \sin \left (2 \, x\right )^{3} + 60 \, x + 9 \, \sin \left (4 \, x\right ) - 48 \, \sin \left (2 \, x\right )\right )} a b^{3} + \frac {1}{3072} \, {\left (128 \, \sin \left (2 \, x\right )^{3} + 840 \, x + 3 \, \sin \left (8 \, x\right ) + 168 \, \sin \left (4 \, x\right ) - 768 \, \sin \left (2 \, x\right )\right )} b^{4} + \frac {3}{16} \, a^{2} b^{2} {\left (12 \, x + \sin \left (4 \, x\right ) - 8 \, \sin \left (2 \, x\right )\right )} + a^{3} b {\left (2 \, x - \sin \left (2 \, x\right )\right )} + a^{4} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(x)^2)^4,x, algorithm="maxima")

[Out]

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) + 16
8*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

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mupad [B]  time = 13.63, size = 147, normalized size = 1.05 \[ x\,a^4-2\,\sin \relax (x)\,a^3\,b\,\cos \relax (x)+2\,x\,a^3\,b+\frac {3\,\sin \relax (x)\,a^2\,b^2\,{\cos \relax (x)}^3}{2}-\frac {15\,\sin \relax (x)\,a^2\,b^2\,\cos \relax (x)}{4}+\frac {9\,x\,a^2\,b^2}{4}-\frac {2\,\sin \relax (x)\,a\,b^3\,{\cos \relax (x)}^5}{3}+\frac {13\,\sin \relax (x)\,a\,b^3\,{\cos \relax (x)}^3}{6}-\frac {11\,\sin \relax (x)\,a\,b^3\,\cos \relax (x)}{4}+\frac {5\,x\,a\,b^3}{4}+\frac {\sin \relax (x)\,b^4\,{\cos \relax (x)}^7}{8}-\frac {25\,\sin \relax (x)\,b^4\,{\cos \relax (x)}^5}{48}+\frac {163\,\sin \relax (x)\,b^4\,{\cos \relax (x)}^3}{192}-\frac {93\,\sin \relax (x)\,b^4\,\cos \relax (x)}{128}+\frac {35\,x\,b^4}{128} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(x)^2)^4,x)

[Out]

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)*sin(x))/128 + (5*a*b^3*x)/4 + 2*a^3*b*x + (3*a^2*b^2*cos(x)^3*sin(x))/2 -
(11*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

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sympy [B]  time = 7.52, size = 410, normalized size = 2.93 \[ a^{4} x + 2 a^{3} b x \sin ^{2}{\relax (x )} + 2 a^{3} b x \cos ^{2}{\relax (x )} - 2 a^{3} b \sin {\relax (x )} \cos {\relax (x )} + \frac {9 a^{2} b^{2} x \sin ^{4}{\relax (x )}}{4} + \frac {9 a^{2} b^{2} x \sin ^{2}{\relax (x )} \cos ^{2}{\relax (x )}}{2} + \frac {9 a^{2} b^{2} x \cos ^{4}{\relax (x )}}{4} - \frac {15 a^{2} b^{2} \sin ^{3}{\relax (x )} \cos {\relax (x )}}{4} - \frac {9 a^{2} b^{2} \sin {\relax (x )} \cos ^{3}{\relax (x )}}{4} + \frac {5 a b^{3} x \sin ^{6}{\relax (x )}}{4} + \frac {15 a b^{3} x \sin ^{4}{\relax (x )} \cos ^{2}{\relax (x )}}{4} + \frac {15 a b^{3} x \sin ^{2}{\relax (x )} \cos ^{4}{\relax (x )}}{4} + \frac {5 a b^{3} x \cos ^{6}{\relax (x )}}{4} - \frac {11 a b^{3} \sin ^{5}{\relax (x )} \cos {\relax (x )}}{4} - \frac {10 a b^{3} \sin ^{3}{\relax (x )} \cos ^{3}{\relax (x )}}{3} - \frac {5 a b^{3} \sin {\relax (x )} \cos ^{5}{\relax (x )}}{4} + \frac {35 b^{4} x \sin ^{8}{\relax (x )}}{128} + \frac {35 b^{4} x \sin ^{6}{\relax (x )} \cos ^{2}{\relax (x )}}{32} + \frac {105 b^{4} x \sin ^{4}{\relax (x )} \cos ^{4}{\relax (x )}}{64} + \frac {35 b^{4} x \sin ^{2}{\relax (x )} \cos ^{6}{\relax (x )}}{32} + \frac {35 b^{4} x \cos ^{8}{\relax (x )}}{128} - \frac {93 b^{4} \sin ^{7}{\relax (x )} \cos {\relax (x )}}{128} - \frac {511 b^{4} \sin ^{5}{\relax (x )} \cos ^{3}{\relax (x )}}{384} - \frac {385 b^{4} \sin ^{3}{\relax (x )} \cos ^{5}{\relax (x )}}{384} - \frac {35 b^{4} \sin {\relax (x )} \cos ^{7}{\relax (x )}}{128} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(x)**2)**4,x)

[Out]

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*
sin(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

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