3.39 \(\int \frac{x^2 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^2} \, dx\)

Optimal. Leaf size=144 \[ -\frac{i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac{i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2}-\frac{b}{2 c^3 d^2 \sqrt{1-c^2 x^2}} \]

[Out]

-b/(2*c^3*d^2*Sqrt[1 - c^2*x^2]) + (x*(a + b*ArcSin[c*x]))/(2*c^2*d^2*(1 - c^2*x^2)) + (I*(a + b*ArcSin[c*x])*
ArcTan[E^(I*ArcSin[c*x])])/(c^3*d^2) - ((I/2)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^3*d^2) + ((I/2)*b*PolyL
og[2, I*E^(I*ArcSin[c*x])])/(c^3*d^2)

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Rubi [A]  time = 0.136193, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4703, 4657, 4181, 2279, 2391, 261} \[ -\frac{i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac{i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2}-\frac{b}{2 c^3 d^2 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^2,x]

[Out]

-b/(2*c^3*d^2*Sqrt[1 - c^2*x^2]) + (x*(a + b*ArcSin[c*x]))/(2*c^2*d^2*(1 - c^2*x^2)) + (I*(a + b*ArcSin[c*x])*
ArcTan[E^(I*ArcSin[c*x])])/(c^3*d^2) - ((I/2)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^3*d^2) + ((I/2)*b*PolyL
og[2, I*E^(I*ArcSin[c*x])])/(c^3*d^2)

Rule 4703

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p +
1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*f*n*d^IntPart[p]*(d + e*x^2
)^FracPart[p])/(2*c*(p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSi
n[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 1]

Rule 4657

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(c*d), Subst[Int[(a +
b*x)^n*Sec[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rule 4181

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 261

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

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^2} \, dx &=\frac{x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{b \int \frac{x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{2 c d^2}-\frac{\int \frac{a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{2 c^2 d}\\ &=-\frac{b}{2 c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{\operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 c^3 d^2}\\ &=-\frac{b}{2 c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{b \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 c^3 d^2}-\frac{b \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 c^3 d^2}\\ &=-\frac{b}{2 c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}\\ &=-\frac{b}{2 c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac{i b \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac{i b \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}\\ \end{align*}

Mathematica [B]  time = 0.173277, size = 463, normalized size = 3.22 \[ \frac{b \left (\frac{-\frac{2 i \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac{i \sin ^{-1}(c x)^2}{2 c}+\frac{3 i \pi \sin ^{-1}(c x)}{2 c}+\frac{2 \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )}{c}+\frac{2 \pi \log \left (1+e^{-i \sin ^{-1}(c x)}\right )}{c}-\frac{\pi \log \left (1+i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac{2 \pi \log \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )}{c}+\frac{\pi \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{c}}{4 c^2}-\frac{-\frac{2 i \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac{i \sin ^{-1}(c x)^2}{2 c}+\frac{i \pi \sin ^{-1}(c x)}{2 c}+\frac{2 \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )}{c}+\frac{2 \pi \log \left (1+e^{-i \sin ^{-1}(c x)}\right )}{c}+\frac{\pi \log \left (1-i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac{\pi \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{c}-\frac{2 \pi \log \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )}{c}}{4 c^2}+\frac{\sqrt{1-c^2 x^2}-\sin ^{-1}(c x)}{4 c^3 (c x-1)}-\frac{\sqrt{1-c^2 x^2}+\sin ^{-1}(c x)}{4 c^2 \left (c^2 x+c\right )}\right )}{d^2}-\frac{a x}{2 c^2 d^2 \left (c^2 x^2-1\right )}+\frac{a \log (1-c x)}{4 c^3 d^2}-\frac{a \log (c x+1)}{4 c^3 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^2*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^2,x]

[Out]

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

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Maple [A]  time = 0.171, size = 263, normalized size = 1.8 \begin{align*} -{\frac{a}{4\,{c}^{3}{d}^{2} \left ( cx-1 \right ) }}+{\frac{a\ln \left ( cx-1 \right ) }{4\,{c}^{3}{d}^{2}}}-{\frac{a}{4\,{c}^{3}{d}^{2} \left ( cx+1 \right ) }}-{\frac{a\ln \left ( cx+1 \right ) }{4\,{c}^{3}{d}^{2}}}-{\frac{b\arcsin \left ( cx \right ) x}{2\,{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{b}{2\,{c}^{3}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{b\arcsin \left ( cx \right ) }{2\,{c}^{3}{d}^{2}}\ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{b\arcsin \left ( cx \right ) }{2\,{c}^{3}{d}^{2}}\ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{{\frac{i}{2}}b}{{c}^{3}{d}^{2}}{\it dilog} \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{{\frac{i}{2}}b}{{c}^{3}{d}^{2}}{\it dilog} \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x)

[Out]

-1/4/c^3*a/d^2/(c*x-1)+1/4/c^3*a/d^2*ln(c*x-1)-1/4/c^3*a/d^2/(c*x+1)-1/4/c^3*a/d^2*ln(c*x+1)-1/2/c^2*b/d^2/(c^
2*x^2-1)*arcsin(c*x)*x+1/2/c^3*b/d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)+1/2/c^3*b/d^2*arcsin(c*x)*ln(1+I*(I*c*x+(-
c^2*x^2+1)^(1/2)))-1/2/c^3*b/d^2*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-1/2*I/c^3*b/d^2*dilog(1+I*(I*c
*x+(-c^2*x^2+1)^(1/2)))+1/2*I/c^3*b/d^2*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{4} \, a{\left (\frac{2 \, x}{c^{4} d^{2} x^{2} - c^{2} d^{2}} + \frac{\log \left (c x + 1\right )}{c^{3} d^{2}} - \frac{\log \left (c x - 1\right )}{c^{3} d^{2}}\right )} - \frac{{\left (2 \, c x \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) +{\left (c^{2} x^{2} - 1\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) -{\left (c^{2} x^{2} - 1\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right ) +{\left (c^{5} d^{2} x^{2} - c^{3} d^{2}\right )} \int \frac{{\left (2 \, c x +{\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) -{\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{6} d^{2} x^{4} - 2 \, c^{4} d^{2} x^{2} + c^{2} d^{2}}\,{d x}\right )} b}{4 \,{\left (c^{5} d^{2} x^{2} - c^{3} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

-1/4*a*(2*x/(c^4*d^2*x^2 - c^2*d^2) + log(c*x + 1)/(c^3*d^2) - log(c*x - 1)/(c^3*d^2)) - 1/4*(2*c*x*arctan2(c*
x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + (c^2*x^2 - 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - (c^
2*x^2 - 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) + 4*(c^5*d^2*x^2 - c^3*d^2)*integrate(1/4*
(2*c*x + (c^2*x^2 - 1)*log(c*x + 1) - (c^2*x^2 - 1)*log(-c*x + 1))*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^6*d^2*x^4 -
2*c^4*d^2*x^2 + c^2*d^2), x))*b/(c^5*d^2*x^2 - c^3*d^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{2} \arcsin \left (c x\right ) + a x^{2}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*x^2*arcsin(c*x) + a*x^2)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x^{2}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac{b x^{2} \operatorname{asin}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*asin(c*x))/(-c**2*d*x**2+d)**2,x)

[Out]

(Integral(a*x**2/(c**4*x**4 - 2*c**2*x**2 + 1), x) + Integral(b*x**2*asin(c*x)/(c**4*x**4 - 2*c**2*x**2 + 1),
x))/d**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}{{\left (c^{2} d x^{2} - d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)*x^2/(c^2*d*x^2 - d)^2, x)