3.135 \(\int \frac {x^2}{\sqrt {a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=293 \[ -\frac {16 \text {Li}_3\left (-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d^3 \sqrt {a \sin (c+d x)+a}}+\frac {16 \text {Li}_3\left (e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d^3 \sqrt {a \sin (c+d x)+a}}+\frac {8 i x \text {Li}_2\left (-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d^2 \sqrt {a \sin (c+d x)+a}}-\frac {8 i x \text {Li}_2\left (e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d^2 \sqrt {a \sin (c+d x)+a}}-\frac {4 x^2 \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-4*x^2*arctanh(exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d/(a+a*sin(d*x+c))^(1/2)+8*I*x*polylog(2,-
exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^2/(a+a*sin(d*x+c))^(1/2)-8*I*x*polylog(2,exp(1/4*I*(2*d
*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^2/(a+a*sin(d*x+c))^(1/2)-16*polylog(3,-exp(1/4*I*(2*d*x+Pi+2*c)))*sin
(1/2*c+1/4*Pi+1/2*d*x)/d^3/(a+a*sin(d*x+c))^(1/2)+16*polylog(3,exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2
*d*x)/d^3/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3319, 4183, 2531, 2282, 6589} \[ \frac {8 i x \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \text {PolyLog}\left (2,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt {a \sin (c+d x)+a}}-\frac {8 i x \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \text {PolyLog}\left (2,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^2 \sqrt {a \sin (c+d x)+a}}-\frac {16 \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \text {PolyLog}\left (3,-e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^3 \sqrt {a \sin (c+d x)+a}}+\frac {16 \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \text {PolyLog}\left (3,e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d^3 \sqrt {a \sin (c+d x)+a}}-\frac {4 x^2 \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+2 d x+\pi )}\right )}{d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-4*x^2*ArcTanh[E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d*Sqrt[a + a*Sin[c + d*x]]) + ((8*I)
*x*PolyLog[2, -E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^2*Sqrt[a + a*Sin[c + d*x]]) - ((8*I
)*x*PolyLog[2, E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^2*Sqrt[a + a*Sin[c + d*x]]) - (16*P
olyLog[3, -E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^3*Sqrt[a + a*Sin[c + d*x]]) + (16*PolyL
og[3, E^((I/4)*(2*c + Pi + 2*d*x))]*Sin[c/2 + Pi/4 + (d*x)/2])/(d^3*Sqrt[a + a*Sin[c + d*x]])

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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])

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt {a+a \sin (c+d x)}} \, dx &=\frac {\sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \int x^2 \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx}{\sqrt {a+a \sin (c+d x)}}\\ &=-\frac {4 x^2 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}-\frac {\left (4 \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int x \log \left (1-e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right ) \, dx}{d \sqrt {a+a \sin (c+d x)}}+\frac {\left (4 \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int x \log \left (1+e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right ) \, dx}{d \sqrt {a+a \sin (c+d x)}}\\ &=-\frac {4 x^2 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}+\frac {8 i x \text {Li}_2\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {8 i x \text {Li}_2\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {\left (8 i \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int \text {Li}_2\left (-e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right ) \, dx}{d^2 \sqrt {a+a \sin (c+d x)}}+\frac {\left (8 i \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int \text {Li}_2\left (e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right ) \, dx}{d^2 \sqrt {a+a \sin (c+d x)}}\\ &=-\frac {4 x^2 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}+\frac {8 i x \text {Li}_2\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {8 i x \text {Li}_2\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {\left (16 \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}+\frac {\left (16 \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}\\ &=-\frac {4 x^2 \tanh ^{-1}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}+\frac {8 i x \text {Li}_2\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {8 i x \text {Li}_2\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {16 \text {Li}_3\left (-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}+\frac {16 \text {Li}_3\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^3 \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 245, normalized size = 0.84 \[ \frac {\sqrt [4]{-1} \sqrt {2} e^{-\frac {1}{2} i (c+d x)} \left (e^{i (c+d x)}+i\right ) \left (4 d x \text {Li}_2\left (-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )-i \left (d^2 x^2 \log \left (1-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )-d^2 x^2 \log \left (1+\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )-4 i d x \text {Li}_2\left (\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )-8 \text {Li}_3\left (-\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )+8 \text {Li}_3\left (\sqrt [4]{-1} e^{\frac {1}{2} i (c+d x)}\right )\right )\right )}{d^3 \sqrt {-i a e^{-i (c+d x)} \left (e^{i (c+d x)}+i\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((-1)^(1/4)*Sqrt[2]*(I + E^(I*(c + d*x)))*(4*d*x*PolyLog[2, -((-1)^(1/4)*E^((I/2)*(c + d*x)))] - I*(d^2*x^2*Lo
g[1 - (-1)^(1/4)*E^((I/2)*(c + d*x))] - d^2*x^2*Log[1 + (-1)^(1/4)*E^((I/2)*(c + d*x))] - (4*I)*d*x*PolyLog[2,
(-1)^(1/4)*E^((I/2)*(c + d*x))] - 8*PolyLog[3, -((-1)^(1/4)*E^((I/2)*(c + d*x)))] + 8*PolyLog[3, (-1)^(1/4)*E
^((I/2)*(c + d*x))])))/(d^3*E^((I/2)*(c + d*x))*Sqrt[((-I)*a*(I + E^(I*(c + d*x)))^2)/E^(I*(c + d*x))])

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{\sqrt {a \sin \left (d x + c\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(x^2/sqrt(a*sin(d*x + c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {a \sin \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(x^2/sqrt(a*sin(d*x + c) + a), x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {a +a \sin \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+a*sin(d*x+c))^(1/2),x)

[Out]

int(x^2/(a+a*sin(d*x+c))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {a \sin \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/sqrt(a*sin(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + a*sin(c + d*x))^(1/2),x)

[Out]

int(x^2/(a + a*sin(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(x**2/sqrt(a*(sin(c + d*x) + 1)), x)

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