3.165 \(\int x^2 (a+a \cos (x))^{3/2} \, dx\)

Optimal. Leaf size=145 \[ \frac {4}{3} a x^2 \sin \left (\frac {x}{2}\right ) \cos \left (\frac {x}{2}\right ) \sqrt {a \cos (x)+a}+\frac {8}{3} a x^2 \tan \left (\frac {x}{2}\right ) \sqrt {a \cos (x)+a}+\frac {16}{9} a x \cos ^2\left (\frac {x}{2}\right ) \sqrt {a \cos (x)+a}+\frac {32}{3} a x \sqrt {a \cos (x)+a}-\frac {224}{9} a \tan \left (\frac {x}{2}\right ) \sqrt {a \cos (x)+a}+\frac {32}{27} a \sin ^2\left (\frac {x}{2}\right ) \tan \left (\frac {x}{2}\right ) \sqrt {a \cos (x)+a} \]

[Out]

32/3*a*x*(a+a*cos(x))^(1/2)+16/9*a*x*cos(1/2*x)^2*(a+a*cos(x))^(1/2)+4/3*a*x^2*cos(1/2*x)*sin(1/2*x)*(a+a*cos(
x))^(1/2)-224/9*a*(a+a*cos(x))^(1/2)*tan(1/2*x)+8/3*a*x^2*(a+a*cos(x))^(1/2)*tan(1/2*x)+32/27*a*sin(1/2*x)^2*(
a+a*cos(x))^(1/2)*tan(1/2*x)

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Rubi [A]  time = 0.14, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3319, 3311, 3296, 2637, 2633} \[ \frac {4}{3} a x^2 \sin \left (\frac {x}{2}\right ) \cos \left (\frac {x}{2}\right ) \sqrt {a \cos (x)+a}+\frac {8}{3} a x^2 \tan \left (\frac {x}{2}\right ) \sqrt {a \cos (x)+a}+\frac {16}{9} a x \cos ^2\left (\frac {x}{2}\right ) \sqrt {a \cos (x)+a}+\frac {32}{3} a x \sqrt {a \cos (x)+a}-\frac {224}{9} a \tan \left (\frac {x}{2}\right ) \sqrt {a \cos (x)+a}+\frac {32}{27} a \sin ^2\left (\frac {x}{2}\right ) \tan \left (\frac {x}{2}\right ) \sqrt {a \cos (x)+a} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + a*Cos[x])^(3/2),x]

[Out]

(32*a*x*Sqrt[a + a*Cos[x]])/3 + (16*a*x*Cos[x/2]^2*Sqrt[a + a*Cos[x]])/9 + (4*a*x^2*Cos[x/2]*Sqrt[a + a*Cos[x]
]*Sin[x/2])/3 - (224*a*Sqrt[a + a*Cos[x]]*Tan[x/2])/9 + (8*a*x^2*Sqrt[a + a*Cos[x]]*Tan[x/2])/3 + (32*a*Sqrt[a
+ a*Cos[x]]*Sin[x/2]^2*Tan[x/2])/27

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((2*a)^IntPart[n
]*(a + b*Sin[e + f*x])^FracPart[n])/Sin[e/2 + (a*Pi)/(4*b) + (f*x)/2]^(2*FracPart[n]), Int[(c + d*x)^m*Sin[e/2
+ (a*Pi)/(4*b) + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps

\begin {align*} \int x^2 (a+a \cos (x))^{3/2} \, dx &=\left (2 a \sqrt {a+a \cos (x)} \sec \left (\frac {x}{2}\right )\right ) \int x^2 \cos ^3\left (\frac {x}{2}\right ) \, dx\\ &=\frac {16}{9} a x \cos ^2\left (\frac {x}{2}\right ) \sqrt {a+a \cos (x)}+\frac {4}{3} a x^2 \cos \left (\frac {x}{2}\right ) \sqrt {a+a \cos (x)} \sin \left (\frac {x}{2}\right )+\frac {1}{3} \left (4 a \sqrt {a+a \cos (x)} \sec \left (\frac {x}{2}\right )\right ) \int x^2 \cos \left (\frac {x}{2}\right ) \, dx-\frac {1}{9} \left (16 a \sqrt {a+a \cos (x)} \sec \left (\frac {x}{2}\right )\right ) \int \cos ^3\left (\frac {x}{2}\right ) \, dx\\ &=\frac {16}{9} a x \cos ^2\left (\frac {x}{2}\right ) \sqrt {a+a \cos (x)}+\frac {4}{3} a x^2 \cos \left (\frac {x}{2}\right ) \sqrt {a+a \cos (x)} \sin \left (\frac {x}{2}\right )+\frac {8}{3} a x^2 \sqrt {a+a \cos (x)} \tan \left (\frac {x}{2}\right )+\frac {1}{9} \left (32 a \sqrt {a+a \cos (x)} \sec \left (\frac {x}{2}\right )\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin \left (\frac {x}{2}\right )\right )-\frac {1}{3} \left (16 a \sqrt {a+a \cos (x)} \sec \left (\frac {x}{2}\right )\right ) \int x \sin \left (\frac {x}{2}\right ) \, dx\\ &=\frac {32}{3} a x \sqrt {a+a \cos (x)}+\frac {16}{9} a x \cos ^2\left (\frac {x}{2}\right ) \sqrt {a+a \cos (x)}+\frac {4}{3} a x^2 \cos \left (\frac {x}{2}\right ) \sqrt {a+a \cos (x)} \sin \left (\frac {x}{2}\right )-\frac {32}{9} a \sqrt {a+a \cos (x)} \tan \left (\frac {x}{2}\right )+\frac {8}{3} a x^2 \sqrt {a+a \cos (x)} \tan \left (\frac {x}{2}\right )+\frac {32}{27} a \sqrt {a+a \cos (x)} \sin ^2\left (\frac {x}{2}\right ) \tan \left (\frac {x}{2}\right )-\frac {1}{3} \left (32 a \sqrt {a+a \cos (x)} \sec \left (\frac {x}{2}\right )\right ) \int \cos \left (\frac {x}{2}\right ) \, dx\\ &=\frac {32}{3} a x \sqrt {a+a \cos (x)}+\frac {16}{9} a x \cos ^2\left (\frac {x}{2}\right ) \sqrt {a+a \cos (x)}+\frac {4}{3} a x^2 \cos \left (\frac {x}{2}\right ) \sqrt {a+a \cos (x)} \sin \left (\frac {x}{2}\right )-\frac {224}{9} a \sqrt {a+a \cos (x)} \tan \left (\frac {x}{2}\right )+\frac {8}{3} a x^2 \sqrt {a+a \cos (x)} \tan \left (\frac {x}{2}\right )+\frac {32}{27} a \sqrt {a+a \cos (x)} \sin ^2\left (\frac {x}{2}\right ) \tan \left (\frac {x}{2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 54, normalized size = 0.37 \[ \frac {2}{27} a \sqrt {a (\cos (x)+1)} \left (\left (45 x^2-328\right ) \tan \left (\frac {x}{2}\right )+\cos (x) \left (\left (9 x^2-8\right ) \tan \left (\frac {x}{2}\right )+12 x\right )+156 x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + a*Cos[x])^(3/2),x]

[Out]

(2*a*Sqrt[a*(1 + Cos[x])]*(156*x + (-328 + 45*x^2)*Tan[x/2] + Cos[x]*(12*x + (-8 + 9*x^2)*Tan[x/2])))/27

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [A]  time = 0.37, size = 85, normalized size = 0.59 \[ \frac {1}{27} \, \sqrt {2} {\left (12 \, a x \cos \left (\frac {3}{2} \, x\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right ) + 324 \, a x \cos \left (\frac {1}{2} \, x\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right ) + {\left (9 \, a x^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right ) - 8 \, a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right )\right )} \sin \left (\frac {3}{2} \, x\right ) + 81 \, {\left (a x^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right ) - 8 \, a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right )\right )} \sin \left (\frac {1}{2} \, x\right )\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*sqrt(2)*(12*a*x*cos(3/2*x)*sgn(cos(1/2*x)) + 324*a*x*cos(1/2*x)*sgn(cos(1/2*x)) + (9*a*x^2*sgn(cos(1/2*x)
) - 8*a*sgn(cos(1/2*x)))*sin(3/2*x) + 81*(a*x^2*sgn(cos(1/2*x)) - 8*a*sgn(cos(1/2*x)))*sin(1/2*x))*sqrt(a)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a +a \cos \relax (x )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.08, size = 72, normalized size = 0.50 \[ \frac {1}{27} \, {\left (81 \, \sqrt {2} a x^{2} \sin \left (\frac {1}{2} \, x\right ) + 12 \, \sqrt {2} a x \cos \left (\frac {3}{2} \, x\right ) + 324 \, \sqrt {2} a x \cos \left (\frac {1}{2} \, x\right ) - 648 \, \sqrt {2} a \sin \left (\frac {1}{2} \, x\right ) + {\left (9 \, \sqrt {2} a x^{2} - 8 \, \sqrt {2} a\right )} \sin \left (\frac {3}{2} \, x\right )\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*(81*sqrt(2)*a*x^2*sin(1/2*x) + 12*sqrt(2)*a*x*cos(3/2*x) + 324*sqrt(2)*a*x*cos(1/2*x) - 648*sqrt(2)*a*sin
(1/2*x) + (9*sqrt(2)*a*x^2 - 8*sqrt(2)*a)*sin(3/2*x))*sqrt(a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\left (a+a\,\cos \relax (x)\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + a*cos(x))^(3/2),x)

[Out]

int(x^2*(a + a*cos(x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a \left (\cos {\relax (x )} + 1\right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+a*cos(x))**(3/2),x)

[Out]

Integral(x**2*(a*(cos(x) + 1))**(3/2), x)

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