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

Optimal. Leaf size=36 \[ \frac{\left (a+b \sin ^2(x)\right )^5}{10 b^2}-\frac{a \left (a+b \sin ^2(x)\right )^4}{8 b^2} \]

[Out]

-(a*(a + b*Sin[x]^2)^4)/(8*b^2) + (a + b*Sin[x]^2)^5/(10*b^2)

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Rubi [A]  time = 0.076912, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3198, 266, 43} \[ \frac{\left (a+b \sin ^2(x)\right )^5}{10 b^2}-\frac{a \left (a+b \sin ^2(x)\right )^4}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]*Sin[x]^3*(a + b*Sin[x]^2)^3,x]

[Out]

-(a*(a + b*Sin[x]^2)^4)/(8*b^2) + (a + b*Sin[x]^2)^5/(10*b^2)

Rule 3198

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^
2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^n*(1 - ff^2*x^2
)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[
(m - 1)/2]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \cos (x) \sin ^3(x) \left (a+b \sin ^2(x)\right )^3 \, dx &=\operatorname{Subst}\left (\int x^3 \left (a+b x^2\right )^3 \, dx,x,\sin (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int x (a+b x)^3 \, dx,x,\sin ^2(x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{a (a+b x)^3}{b}+\frac{(a+b x)^4}{b}\right ) \, dx,x,\sin ^2(x)\right )\\ &=-\frac{a \left (a+b \sin ^2(x)\right )^4}{8 b^2}+\frac{\left (a+b \sin ^2(x)\right )^5}{10 b^2}\\ \end{align*}

Mathematica [B]  time = 0.42499, size = 128, normalized size = 3.56 \[ \frac{-20 \left (64 a^3+24 a b^2+7 b^3\right ) \cos (2 x)+20 \left (16 a^3+18 a b^2+5 b^3\right ) \cos (4 x)+b \left (3840 a^2 \sin ^4(x)-1280 a^2 \sin (3 x) \sin ^3(x)+2560 a b \sin ^6(x)-10 b (16 a+5 b) \cos (6 x)+15 b (2 a+b) \cos (8 x)+640 b^2 \sin ^8(x)-2 b^2 \cos (10 x)\right )}{10240} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]*Sin[x]^3*(a + b*Sin[x]^2)^3,x]

[Out]

(-20*(64*a^3 + 24*a*b^2 + 7*b^3)*Cos[2*x] + 20*(16*a^3 + 18*a*b^2 + 5*b^3)*Cos[4*x] + b*(-10*b*(16*a + 5*b)*Co
s[6*x] + 15*b*(2*a + b)*Cos[8*x] - 2*b^2*Cos[10*x] + 3840*a^2*Sin[x]^4 + 2560*a*b*Sin[x]^6 + 640*b^2*Sin[x]^8
- 1280*a^2*Sin[x]^3*Sin[3*x]))/10240

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Maple [A]  time = 0.011, size = 40, normalized size = 1.1 \begin{align*}{\frac{{b}^{3} \left ( \sin \left ( x \right ) \right ) ^{10}}{10}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( x \right ) \right ) ^{8}}{8}}+{\frac{{a}^{2}b \left ( \sin \left ( x \right ) \right ) ^{6}}{2}}+{\frac{ \left ( \sin \left ( x \right ) \right ) ^{4}{a}^{3}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^3*sin(x)^10+3/8*a*b^2*sin(x)^8+1/2*a^2*b*sin(x)^6+1/4*sin(x)^4*a^3

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Maxima [A]  time = 0.982705, size = 53, normalized size = 1.47 \begin{align*} \frac{1}{10} \, b^{3} \sin \left (x\right )^{10} + \frac{3}{8} \, a b^{2} \sin \left (x\right )^{8} + \frac{1}{2} \, a^{2} b \sin \left (x\right )^{6} + \frac{1}{4} \, a^{3} \sin \left (x\right )^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^3*sin(x)^10 + 3/8*a*b^2*sin(x)^8 + 1/2*a^2*b*sin(x)^6 + 1/4*a^3*sin(x)^4

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Fricas [B]  time = 2.22598, size = 258, normalized size = 7.17 \begin{align*} -\frac{1}{10} \, b^{3} \cos \left (x\right )^{10} + \frac{1}{8} \,{\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (x\right )^{8} - \frac{1}{2} \,{\left (a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (x\right )^{6} + \frac{1}{4} \,{\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (x\right )^{4} - \frac{1}{2} \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (x\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*b^3*cos(x)^10 + 1/8*(3*a*b^2 + 4*b^3)*cos(x)^8 - 1/2*(a^2*b + 3*a*b^2 + 2*b^3)*cos(x)^6 + 1/4*(a^3 + 6*a
^2*b + 9*a*b^2 + 4*b^3)*cos(x)^4 - 1/2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(x)^2

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Sympy [B]  time = 16.575, size = 102, normalized size = 2.83 \begin{align*} \frac{a^{3} \sin ^{4}{\left (x \right )}}{4} + \frac{a^{2} b \sin ^{6}{\left (x \right )}}{2} + \frac{3 a b^{2} \sin ^{8}{\left (x \right )}}{8} - \frac{b^{3} \sin ^{8}{\left (x \right )} \cos ^{2}{\left (x \right )}}{2} - b^{3} \sin ^{6}{\left (x \right )} \cos ^{4}{\left (x \right )} - b^{3} \sin ^{4}{\left (x \right )} \cos ^{6}{\left (x \right )} - \frac{b^{3} \sin ^{2}{\left (x \right )} \cos ^{8}{\left (x \right )}}{2} - \frac{b^{3} \cos ^{10}{\left (x \right )}}{10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*sin(x)**4/4 + a**2*b*sin(x)**6/2 + 3*a*b**2*sin(x)**8/8 - b**3*sin(x)**8*cos(x)**2/2 - b**3*sin(x)**6*cos
(x)**4 - b**3*sin(x)**4*cos(x)**6 - b**3*sin(x)**2*cos(x)**8/2 - b**3*cos(x)**10/10

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Giac [A]  time = 1.13995, size = 53, normalized size = 1.47 \begin{align*} \frac{1}{10} \, b^{3} \sin \left (x\right )^{10} + \frac{3}{8} \, a b^{2} \sin \left (x\right )^{8} + \frac{1}{2} \, a^{2} b \sin \left (x\right )^{6} + \frac{1}{4} \, a^{3} \sin \left (x\right )^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^3*sin(x)^10 + 3/8*a*b^2*sin(x)^8 + 1/2*a^2*b*sin(x)^6 + 1/4*a^3*sin(x)^4