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

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

[Out]

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

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Rubi [A]  time = 0.09, 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 = {4335, 266, 43} \[ \frac {a \left (a+b \cos ^2(x)\right )^4}{8 b^2}-\frac {\left (a+b \cos ^2(x)\right )^5}{10 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a + b*Cos[x]^2)^4)/(8*b^2) - (a + b*Cos[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 4335

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[d/(
b*c), Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

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

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Mathematica [B]  time = 0.28, size = 137, normalized size = 3.81 \[ \frac {1}{32} \left (-4 a^3 \cos (2 x)-a^3 \cos (4 x)-12 a^2 b \cos ^4(x)-4 a^2 b \cos (3 x) \cos ^3(x)-8 a b^2 \cos ^6(x)-\frac {1}{32} a b^2 (48 \cos (2 x)+36 \cos (4 x)+16 \cos (6 x)+3 \cos (8 x))-2 b^3 \cos ^8(x)-\frac {1}{320} b^3 (140 \cos (2 x)+100 \cos (4 x)+50 \cos (6 x)+15 \cos (8 x)+2 \cos (10 x))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*a^2*b*Cos[x]^4 - 8*a*b^2*Cos[x]^6 - 2*b^3*Cos[x]^8 - 4*a^3*Cos[2*x] - 4*a^2*b*Cos[x]^3*Cos[3*x] - a^3*Cos
[4*x] - (a*b^2*(48*Cos[2*x] + 36*Cos[4*x] + 16*Cos[6*x] + 3*Cos[8*x]))/32 - (b^3*(140*Cos[2*x] + 100*Cos[4*x]
+ 50*Cos[6*x] + 15*Cos[8*x] + 2*Cos[10*x]))/320)/32

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fricas [A]  time = 0.92, size = 39, normalized size = 1.08 \[ -\frac {1}{10} \, b^{3} \cos \relax (x)^{10} - \frac {3}{8} \, a b^{2} \cos \relax (x)^{8} - \frac {1}{2} \, a^{2} b \cos \relax (x)^{6} - \frac {1}{4} \, a^{3} \cos \relax (x)^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 39, normalized size = 1.08 \[ -\frac {1}{10} \, b^{3} \cos \relax (x)^{10} - \frac {3}{8} \, a b^{2} \cos \relax (x)^{8} - \frac {1}{2} \, a^{2} b \cos \relax (x)^{6} - \frac {1}{4} \, a^{3} \cos \relax (x)^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.39, size = 103, normalized size = 2.86 \[ \frac {1}{10} \, b^{3} \sin \relax (x)^{10} - \frac {1}{8} \, {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \sin \relax (x)^{8} + \frac {1}{2} \, {\left (a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} \sin \relax (x)^{6} - \frac {1}{4} \, {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \sin \relax (x)^{4} + \frac {1}{2} \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sin \relax (x)^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.09, size = 39, normalized size = 1.08 \[ -\frac {a^3\,{\cos \relax (x)}^4}{4}-\frac {a^2\,b\,{\cos \relax (x)}^6}{2}-\frac {3\,a\,b^2\,{\cos \relax (x)}^8}{8}-\frac {b^3\,{\cos \relax (x)}^{10}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 11.65, size = 46, normalized size = 1.28 \[ - \frac {a^{3} \cos ^{4}{\relax (x )}}{4} - \frac {a^{2} b \cos ^{6}{\relax (x )}}{2} - \frac {3 a b^{2} \cos ^{8}{\relax (x )}}{8} - \frac {b^{3} \cos ^{10}{\relax (x )}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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