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]

((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

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Rubi [A]  time = 0.0830855, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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]

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]

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*}

Mathematica [C]  time = 0.10195, 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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Maple [A]  time = 0.024, size = 73, normalized size = 0.8 \begin{align*}{b}^{3} \left ( -{\frac{\cos \left ( x \right ) }{6} \left ( \left ( \sin \left ( x \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( x \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( x \right ) }{8}} \right ) }+{\frac{5\,x}{16}} \right ) +3\,a{b}^{2} \left ( -1/4\, \left ( \left ( \sin \left ( x \right ) \right ) ^{3}+3/2\,\sin \left ( x \right ) \right ) \cos \left ( x \right ) +3/8\,x \right ) +3\,{a}^{2}b \left ( -1/2\,\sin \left ( x \right ) \cos \left ( x \right ) +x/2 \right ) +{a}^{3}x \end{align*}

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 = 0.93779, size = 96, normalized size = 1.1 \begin{align*} \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 \end{align*}

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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Fricas [A]  time = 1.66068, size = 207, normalized size = 2.38 \begin{align*} \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 \left (x\right )^{5} - 2 \,{\left (18 \, a b^{2} + 13 \, b^{3}\right )} \cos \left (x\right )^{3} + 3 \,{\left (24 \, a^{2} b + 30 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (x\right )\right )} \sin \left (x\right ) \end{align*}

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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Sympy [B]  time = 5.62157, size = 246, normalized size = 2.83 \begin{align*} a^{3} x + \frac{3 a^{2} b x \sin ^{2}{\left (x \right )}}{2} + \frac{3 a^{2} b x \cos ^{2}{\left (x \right )}}{2} - \frac{3 a^{2} b \sin{\left (x \right )} \cos{\left (x \right )}}{2} + \frac{9 a b^{2} x \sin ^{4}{\left (x \right )}}{8} + \frac{9 a b^{2} x \sin ^{2}{\left (x \right )} \cos ^{2}{\left (x \right )}}{4} + \frac{9 a b^{2} x \cos ^{4}{\left (x \right )}}{8} - \frac{15 a b^{2} \sin ^{3}{\left (x \right )} \cos{\left (x \right )}}{8} - \frac{9 a b^{2} \sin{\left (x \right )} \cos ^{3}{\left (x \right )}}{8} + \frac{5 b^{3} x \sin ^{6}{\left (x \right )}}{16} + \frac{15 b^{3} x \sin ^{4}{\left (x \right )} \cos ^{2}{\left (x \right )}}{16} + \frac{15 b^{3} x \sin ^{2}{\left (x \right )} \cos ^{4}{\left (x \right )}}{16} + \frac{5 b^{3} x \cos ^{6}{\left (x \right )}}{16} - \frac{11 b^{3} \sin ^{5}{\left (x \right )} \cos{\left (x \right )}}{16} - \frac{5 b^{3} \sin ^{3}{\left (x \right )} \cos ^{3}{\left (x \right )}}{6} - \frac{5 b^{3} \sin{\left (x \right )} \cos ^{5}{\left (x \right )}}{16} \end{align*}

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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Giac [A]  time = 1.10996, size = 103, normalized size = 1.18 \begin{align*} -\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 ) \end{align*}

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)