3.76 \(\int (a+b \sin ^2(x))^3 \, dx\)

Optimal. Leaf size=87 \[ \frac {1}{16} x (2 a+b) \left (8 a^2+8 a b+5 b^2\right )-\frac {1}{48} b \left (64 a^2+54 a b+15 b^2\right ) \sin (x) \cos (x)-\frac {5}{24} b^2 (2 a+b) \sin ^3(x) \cos (x)-\frac {1}{6} b \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^2 \]

[Out]

1/16*(2*a+b)*(8*a^2+8*a*b+5*b^2)*x-1/48*b*(64*a^2+54*a*b+15*b^2)*cos(x)*sin(x)-5/24*b^2*(2*a+b)*cos(x)*sin(x)^
3-1/6*b*cos(x)*sin(x)*(a+b*sin(x)^2)^2

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Rubi [A]  time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3180, 3169} \[ \frac {1}{16} x (2 a+b) \left (8 a^2+8 a b+5 b^2\right )-\frac {1}{48} b \left (64 a^2+54 a b+15 b^2\right ) \sin (x) \cos (x)-\frac {5}{24} b^2 (2 a+b) \sin ^3(x) \cos (x)-\frac {1}{6} b \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x]^2)^3,x]

[Out]

((2*a + b)*(8*a^2 + 8*a*b + 5*b^2)*x)/16 - (b*(64*a^2 + 54*a*b + 15*b^2)*Cos[x]*Sin[x])/48 - (5*b^2*(2*a + b)*
Cos[x]*Sin[x]^3)/24 - (b*Cos[x]*Sin[x]*(a + b*Sin[x]^2)^2)/6

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 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 )^3 \, dx &=-\frac {1}{6} b \cos (x) \sin (x) \left (a+b \sin ^2(x)\right )^2+\frac {1}{6} \int \left (a+b \sin ^2(x)\right ) \left (a (6 a+b)+5 b (2 a+b) \sin ^2(x)\right ) \, dx\\ &=\frac {1}{16} (2 a+b) \left (8 a^2+8 a b+5 b^2\right ) x-\frac {1}{48} b \left (64 a^2+54 a b+15 b^2\right ) \cos (x) \sin (x)-\frac {5}{24} b^2 (2 a+b) \cos (x) \sin ^3(x)-\frac {1}{6} b \cos (x) \sin (x) \left (a+b \sin ^2(x)\right )^2\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 80, normalized size = 0.92 \[ \frac {1}{192} \left (12 x (2 a+b) \left (8 a^2+8 a b+5 b^2\right )+9 b^2 (2 a+b) \sin (4 x)+9 i b (4 i a+(1+2 i) b) (4 a+(2+i) b) \sin (2 x)+b^3 (-\sin (6 x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[x]^2)^3,x]

[Out]

(12*(2*a + b)*(8*a^2 + 8*a*b + 5*b^2)*x + (9*I)*b*((4*I)*a + (1 + 2*I)*b)*(4*a + (2 + I)*b)*Sin[2*x] + 9*b^2*(
2*a + b)*Sin[4*x] - b^3*Sin[6*x])/192

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fricas [A]  time = 0.43, size = 81, normalized size = 0.93 \[ \frac {1}{16} \, {\left (16 \, a^{3} + 24 \, a^{2} b + 18 \, a b^{2} + 5 \, b^{3}\right )} x - \frac {1}{48} \, {\left (8 \, b^{3} \cos \relax (x)^{5} - 2 \, {\left (18 \, a b^{2} + 13 \, b^{3}\right )} \cos \relax (x)^{3} + 3 \, {\left (24 \, a^{2} b + 30 \, a b^{2} + 11 \, b^{3}\right )} \cos \relax (x)\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(x)^2)^3,x, algorithm="fricas")

[Out]

1/16*(16*a^3 + 24*a^2*b + 18*a*b^2 + 5*b^3)*x - 1/48*(8*b^3*cos(x)^5 - 2*(18*a*b^2 + 13*b^3)*cos(x)^3 + 3*(24*
a^2*b + 30*a*b^2 + 11*b^3)*cos(x))*sin(x)

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giac [A]  time = 0.15, size = 76, normalized size = 0.87 \[ -\frac {1}{192} \, b^{3} \sin \left (6 \, x\right ) + \frac {1}{16} \, {\left (16 \, a^{3} + 24 \, a^{2} b + 18 \, a b^{2} + 5 \, b^{3}\right )} x + \frac {3}{64} \, {\left (2 \, a b^{2} + b^{3}\right )} \sin \left (4 \, x\right ) - \frac {3}{64} \, {\left (16 \, a^{2} b + 16 \, a b^{2} + 5 \, b^{3}\right )} \sin \left (2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(x)^2)^3,x, algorithm="giac")

[Out]

-1/192*b^3*sin(6*x) + 1/16*(16*a^3 + 24*a^2*b + 18*a*b^2 + 5*b^3)*x + 3/64*(2*a*b^2 + b^3)*sin(4*x) - 3/64*(16
*a^2*b + 16*a*b^2 + 5*b^3)*sin(2*x)

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maple [A]  time = 0.47, size = 73, normalized size = 0.84 \[ b^{3} \left (-\frac {\left (\sin ^{5}\relax (x )+\frac {5 \left (\sin ^{3}\relax (x )\right )}{4}+\frac {15 \sin \relax (x )}{8}\right ) \cos \relax (x )}{6}+\frac {5 x}{16}\right )+3 a \,b^{2} \left (-\frac {\left (\sin ^{3}\relax (x )+\frac {3 \sin \relax (x )}{2}\right ) \cos \relax (x )}{4}+\frac {3 x}{8}\right )+3 a^{2} b \left (-\frac {\sin \relax (x ) \cos \relax (x )}{2}+\frac {x}{2}\right )+a^{3} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(x)^2)^3,x)

[Out]

b^3*(-1/6*(sin(x)^5+5/4*sin(x)^3+15/8*sin(x))*cos(x)+5/16*x)+3*a*b^2*(-1/4*(sin(x)^3+3/2*sin(x))*cos(x)+3/8*x)
+3*a^2*b*(-1/2*sin(x)*cos(x)+1/2*x)+a^3*x

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maxima [A]  time = 0.35, size = 71, normalized size = 0.82 \[ \frac {1}{192} \, {\left (4 \, \sin \left (2 \, x\right )^{3} + 60 \, x + 9 \, \sin \left (4 \, x\right ) - 48 \, \sin \left (2 \, x\right )\right )} b^{3} + \frac {3}{32} \, a b^{2} {\left (12 \, x + \sin \left (4 \, x\right ) - 8 \, \sin \left (2 \, x\right )\right )} + \frac {3}{4} \, a^{2} b {\left (2 \, x - \sin \left (2 \, x\right )\right )} + a^{3} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(x)^2)^3,x, algorithm="maxima")

[Out]

1/192*(4*sin(2*x)^3 + 60*x + 9*sin(4*x) - 48*sin(2*x))*b^3 + 3/32*a*b^2*(12*x + sin(4*x) - 8*sin(2*x)) + 3/4*a
^2*b*(2*x - sin(2*x)) + a^3*x

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mupad [B]  time = 14.13, size = 118, normalized size = 1.36 \[ a^3\,x+\frac {5\,b^3\,x}{16}-\frac {\left (72\,a^2\,b+90\,a\,b^2+33\,b^3\right )\,{\mathrm {tan}\relax (x)}^5+\left (144\,a^2\,b+144\,a\,b^2+40\,b^3\right )\,{\mathrm {tan}\relax (x)}^3+\left (72\,a^2\,b+54\,a\,b^2+15\,b^3\right )\,\mathrm {tan}\relax (x)}{48\,{\mathrm {tan}\relax (x)}^6+144\,{\mathrm {tan}\relax (x)}^4+144\,{\mathrm {tan}\relax (x)}^2+48}+\frac {9\,a\,b^2\,x}{8}+\frac {3\,a^2\,b\,x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(x)^2)^3,x)

[Out]

a^3*x + (5*b^3*x)/16 - (tan(x)^5*(90*a*b^2 + 72*a^2*b + 33*b^3) + tan(x)^3*(144*a*b^2 + 144*a^2*b + 40*b^3) +
tan(x)*(54*a*b^2 + 72*a^2*b + 15*b^3))/(144*tan(x)^2 + 144*tan(x)^4 + 48*tan(x)^6 + 48) + (9*a*b^2*x)/8 + (3*a
^2*b*x)/2

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sympy [B]  time = 2.76, size = 246, normalized size = 2.83 \[ a^{3} x + \frac {3 a^{2} b x \sin ^{2}{\relax (x )}}{2} + \frac {3 a^{2} b x \cos ^{2}{\relax (x )}}{2} - \frac {3 a^{2} b \sin {\relax (x )} \cos {\relax (x )}}{2} + \frac {9 a b^{2} x \sin ^{4}{\relax (x )}}{8} + \frac {9 a b^{2} x \sin ^{2}{\relax (x )} \cos ^{2}{\relax (x )}}{4} + \frac {9 a b^{2} x \cos ^{4}{\relax (x )}}{8} - \frac {15 a b^{2} \sin ^{3}{\relax (x )} \cos {\relax (x )}}{8} - \frac {9 a b^{2} \sin {\relax (x )} \cos ^{3}{\relax (x )}}{8} + \frac {5 b^{3} x \sin ^{6}{\relax (x )}}{16} + \frac {15 b^{3} x \sin ^{4}{\relax (x )} \cos ^{2}{\relax (x )}}{16} + \frac {15 b^{3} x \sin ^{2}{\relax (x )} \cos ^{4}{\relax (x )}}{16} + \frac {5 b^{3} x \cos ^{6}{\relax (x )}}{16} - \frac {11 b^{3} \sin ^{5}{\relax (x )} \cos {\relax (x )}}{16} - \frac {5 b^{3} \sin ^{3}{\relax (x )} \cos ^{3}{\relax (x )}}{6} - \frac {5 b^{3} \sin {\relax (x )} \cos ^{5}{\relax (x )}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(x)**2)**3,x)

[Out]

a**3*x + 3*a**2*b*x*sin(x)**2/2 + 3*a**2*b*x*cos(x)**2/2 - 3*a**2*b*sin(x)*cos(x)/2 + 9*a*b**2*x*sin(x)**4/8 +
9*a*b**2*x*sin(x)**2*cos(x)**2/4 + 9*a*b**2*x*cos(x)**4/8 - 15*a*b**2*sin(x)**3*cos(x)/8 - 9*a*b**2*sin(x)*co
s(x)**3/8 + 5*b**3*x*sin(x)**6/16 + 15*b**3*x*sin(x)**4*cos(x)**2/16 + 15*b**3*x*sin(x)**2*cos(x)**4/16 + 5*b*
*3*x*cos(x)**6/16 - 11*b**3*sin(x)**5*cos(x)/16 - 5*b**3*sin(x)**3*cos(x)**3/6 - 5*b**3*sin(x)*cos(x)**5/16

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