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]

-1/8*a*(a+b*sin(x)^2)^4/b^2+1/10*(a+b*sin(x)^2)^5/b^2

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Rubi [A]  time = 0.08, antiderivative size = 36, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [B]  time = 0.35, 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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fricas [B]  time = 1.19, size = 103, normalized size = 2.86 \[ -\frac {1}{10} \, b^{3} \cos \relax (x)^{10} + \frac {1}{8} \, {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \relax (x)^{8} - \frac {1}{2} \, {\left (a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \relax (x)^{6} + \frac {1}{4} \, {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \cos \relax (x)^{4} - \frac {1}{2} \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \relax (x)^{2} \]

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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giac [A]  time = 0.12, size = 39, normalized size = 1.08 \[ \frac {1}{10} \, b^{3} \sin \relax (x)^{10} + \frac {3}{8} \, a b^{2} \sin \relax (x)^{8} + \frac {1}{2} \, a^{2} b \sin \relax (x)^{6} + \frac {1}{4} \, a^{3} \sin \relax (x)^{4} \]

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

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maple [A]  time = 0.01, size = 40, normalized size = 1.11 \[ \frac {b^{3} \left (\sin ^{10}\relax (x )\right )}{10}+\frac {3 a \,b^{2} \left (\sin ^{8}\relax (x )\right )}{8}+\frac {a^{2} b \left (\sin ^{6}\relax (x )\right )}{2}+\frac {a^{3} \left (\sin ^{4}\relax (x )\right )}{4} \]

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*a^3*sin(x)^4

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maxima [A]  time = 0.33, size = 39, normalized size = 1.08 \[ \frac {1}{10} \, b^{3} \sin \relax (x)^{10} + \frac {3}{8} \, a b^{2} \sin \relax (x)^{8} + \frac {1}{2} \, a^{2} b \sin \relax (x)^{6} + \frac {1}{4} \, a^{3} \sin \relax (x)^{4} \]

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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mupad [B]  time = 0.07, size = 73, normalized size = 2.03 \[ \frac {b^2\,{\cos \relax (x)}^8\,\left (3\,a+4\,b\right )}{8}-\frac {b^3\,{\cos \relax (x)}^{10}}{10}-\frac {{\cos \relax (x)}^2\,{\left (a+b\right )}^3}{2}-\frac {b\,{\cos \relax (x)}^6\,\left (a^2+3\,a\,b+2\,b^2\right )}{2}+\frac {{\cos \relax (x)}^4\,{\left (a+b\right )}^2\,\left (a+4\,b\right )}{4} \]

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]

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

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sympy [A]  time = 11.38, size = 44, normalized size = 1.22 \[ \frac {a^{3} \sin ^{4}{\relax (x )}}{4} + \frac {a^{2} b \sin ^{6}{\relax (x )}}{2} + \frac {3 a b^{2} \sin ^{8}{\relax (x )}}{8} + \frac {b^{3} \sin ^{10}{\relax (x )}}{10} \]

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)**10/10

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