3.28 \(\int \frac{x^4 (a+b \sin ^{-1}(c x))}{d-c^2 d x^2} \, dx\)

Optimal. Leaf size=172 \[ \frac{i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac{2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d}+\frac{b \left (1-c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac{4 b \sqrt{1-c^2 x^2}}{3 c^5 d} \]

[Out]

(-4*b*Sqrt[1 - c^2*x^2])/(3*c^5*d) + (b*(1 - c^2*x^2)^(3/2))/(9*c^5*d) - (x*(a + b*ArcSin[c*x]))/(c^4*d) - (x^
3*(a + b*ArcSin[c*x]))/(3*c^2*d) - ((2*I)*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/(c^5*d) + (I*b*PolyLo
g[2, (-I)*E^(I*ArcSin[c*x])])/(c^5*d) - (I*b*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^5*d)

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Rubi [A]  time = 0.237931, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4715, 4657, 4181, 2279, 2391, 261, 266, 43} \[ \frac{i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac{2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d}+\frac{b \left (1-c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac{4 b \sqrt{1-c^2 x^2}}{3 c^5 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2),x]

[Out]

(-4*b*Sqrt[1 - c^2*x^2])/(3*c^5*d) + (b*(1 - c^2*x^2)^(3/2))/(9*c^5*d) - (x*(a + b*ArcSin[c*x]))/(c^4*d) - (x^
3*(a + b*ArcSin[c*x]))/(3*c^2*d) - ((2*I)*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/(c^5*d) + (I*b*PolyLo
g[2, (-I)*E^(I*ArcSin[c*x])])/(c^5*d) - (I*b*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^5*d)

Rule 4715

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)/(e*(m + 2*p + 1)), x] + (Dist[(f^2*(m - 1))/(c^2*(m
+ 2*p + 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*f*n*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(c*(m + 2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a +
b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[m,
1] && NeQ[m + 2*p + 1, 0] && IntegerQ[m]

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]

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 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])

Rubi steps

\begin{align*} \int \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx &=-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}+\frac{\int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx}{c^2}+\frac{b \int \frac{x^3}{\sqrt{1-c^2 x^2}} \, dx}{3 c d}\\ &=-\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}+\frac{\int \frac{a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{c^4}+\frac{b \int \frac{x}{\sqrt{1-c^2 x^2}} \, dx}{c^3 d}+\frac{b \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-c^2 x}} \, dx,x,x^2\right )}{6 c d}\\ &=-\frac{b \sqrt{1-c^2 x^2}}{c^5 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}+\frac{\operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}+\frac{b \operatorname{Subst}\left (\int \left (\frac{1}{c^2 \sqrt{1-c^2 x}}-\frac{\sqrt{1-c^2 x}}{c^2}\right ) \, dx,x,x^2\right )}{6 c d}\\ &=-\frac{4 b \sqrt{1-c^2 x^2}}{3 c^5 d}+\frac{b \left (1-c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}-\frac{2 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{b \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}+\frac{b \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}\\ &=-\frac{4 b \sqrt{1-c^2 x^2}}{3 c^5 d}+\frac{b \left (1-c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}-\frac{2 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^5 d}\\ &=-\frac{4 b \sqrt{1-c^2 x^2}}{3 c^5 d}+\frac{b \left (1-c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}-\frac{2 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac{i b \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{i b \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}\\ \end{align*}

Mathematica [A]  time = 0.304937, size = 286, normalized size = 1.66 \[ -\frac{-18 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+18 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )+6 a c^3 x^3+18 a c x+9 a \log (1-c x)-9 a \log (c x+1)+2 b c^2 x^2 \sqrt{1-c^2 x^2}+22 b \sqrt{1-c^2 x^2}+6 b c^3 x^3 \sin ^{-1}(c x)+18 b c x \sin ^{-1}(c x)+9 i \pi b \sin ^{-1}(c x)-18 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-9 \pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+18 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-9 \pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+9 \pi b \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+9 \pi b \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{18 c^5 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2),x]

[Out]

-(18*a*c*x + 6*a*c^3*x^3 + 22*b*Sqrt[1 - c^2*x^2] + 2*b*c^2*x^2*Sqrt[1 - c^2*x^2] + (9*I)*b*Pi*ArcSin[c*x] + 1
8*b*c*x*ArcSin[c*x] + 6*b*c^3*x^3*ArcSin[c*x] - 9*b*Pi*Log[1 - I*E^(I*ArcSin[c*x])] - 18*b*ArcSin[c*x]*Log[1 -
I*E^(I*ArcSin[c*x])] - 9*b*Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 18*b*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] +
9*a*Log[1 - c*x] - 9*a*Log[1 + c*x] + 9*b*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + 9*b*Pi*Log[Sin[(Pi + 2*ArcSin
[c*x])/4]] - (18*I)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (18*I)*b*PolyLog[2, I*E^(I*ArcSin[c*x])])/(18*c^5*d
)

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Maple [A]  time = 0.253, size = 270, normalized size = 1.6 \begin{align*} -{\frac{a{x}^{3}}{3\,{c}^{2}d}}-{\frac{ax}{{c}^{4}d}}-{\frac{a\ln \left ( cx-1 \right ) }{2\,{c}^{5}d}}+{\frac{a\ln \left ( cx+1 \right ) }{2\,{c}^{5}d}}-{\frac{b\arcsin \left ( cx \right ){x}^{3}}{3\,{c}^{2}d}}-{\frac{b\arcsin \left ( cx \right ) x}{{c}^{4}d}}-{\frac{ib}{{c}^{5}d}{\it dilog} \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{b{x}^{2}}{9\,{c}^{3}d}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{11\,b}{9\,{c}^{5}d}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arcsin \left ( cx \right ) }{{c}^{5}d}\ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{b\arcsin \left ( cx \right ) }{{c}^{5}d}\ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{ib}{{c}^{5}d}{\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^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d),x)

[Out]

-1/3/c^2*a/d*x^3-1/c^4*a/d*x-1/2/c^5*a/d*ln(c*x-1)+1/2/c^5*a/d*ln(c*x+1)-1/3/c^2*b/d*arcsin(c*x)*x^3-1/c^4*b/d
*arcsin(c*x)*x-I/c^5*b/d*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-1/9/c^3*b/d*(-c^2*x^2+1)^(1/2)*x^2-11/9*b*(-c^2
*x^2+1)^(1/2)/c^5/d-1/c^5*b/d*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+1/c^5*b/d*arcsin(c*x)*ln(1-I*(I*c
*x+(-c^2*x^2+1)^(1/2)))+I/c^5*b/d*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}{6} \, a{\left (\frac{2 \,{\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4} d} - \frac{3 \, \log \left (c x + 1\right )}{c^{5} d} + \frac{3 \, \log \left (c x - 1\right )}{c^{5} d}\right )} + \frac{-\frac{1}{3} \,{\left (c^{5} d{\left (\frac{2 \,{\left (c^{2} x^{2} + 2\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{5} d} + \frac{18 \, \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{5} d} + 3 \, \int -\frac{3 \, \sqrt{c x + 1} \sqrt{-c x + 1}{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{c^{6} d x^{2} - c^{4} d}\,{d x}\right )} + 6 \,{\left (c^{3} x^{3} + 3 \, c x\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) - 9 \, \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) + 9 \, \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right )\right )} b}{6 \, c^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{4}}{c^{2} d x^{2} - d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d),x, algorithm="giac")

[Out]

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