3.46 \(\int \frac{x^4 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^3} \, dx\)

Optimal. Leaf size=204 \[ \frac{3 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}-\frac{3 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}-\frac{3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c^5 d^3}+\frac{5 b}{8 c^5 d^3 \sqrt{1-c^2 x^2}}-\frac{b}{12 c^5 d^3 \left (1-c^2 x^2\right )^{3/2}} \]

[Out]

-b/(12*c^5*d^3*(1 - c^2*x^2)^(3/2)) + (5*b)/(8*c^5*d^3*Sqrt[1 - c^2*x^2]) + (x^3*(a + b*ArcSin[c*x]))/(4*c^2*d
^3*(1 - c^2*x^2)^2) - (3*x*(a + b*ArcSin[c*x]))/(8*c^4*d^3*(1 - c^2*x^2)) - (((3*I)/4)*(a + b*ArcSin[c*x])*Arc
Tan[E^(I*ArcSin[c*x])])/(c^5*d^3) + (((3*I)/8)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^5*d^3) - (((3*I)/8)*b*
PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^5*d^3)

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Rubi [A]  time = 0.240689, antiderivative size = 204, 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 = {4703, 4657, 4181, 2279, 2391, 261, 266, 43} \[ \frac{3 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}-\frac{3 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}-\frac{3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c^5 d^3}+\frac{5 b}{8 c^5 d^3 \sqrt{1-c^2 x^2}}-\frac{b}{12 c^5 d^3 \left (1-c^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^3,x]

[Out]

-b/(12*c^5*d^3*(1 - c^2*x^2)^(3/2)) + (5*b)/(8*c^5*d^3*Sqrt[1 - c^2*x^2]) + (x^3*(a + b*ArcSin[c*x]))/(4*c^2*d
^3*(1 - c^2*x^2)^2) - (3*x*(a + b*ArcSin[c*x]))/(8*c^4*d^3*(1 - c^2*x^2)) - (((3*I)/4)*(a + b*ArcSin[c*x])*Arc
Tan[E^(I*ArcSin[c*x])])/(c^5*d^3) + (((3*I)/8)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^5*d^3) - (((3*I)/8)*b*
PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^5*d^3)

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]

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 )}{\left (d-c^2 d x^2\right )^3} \, dx &=\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{b \int \frac{x^3}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{4 c d^3}-\frac{3 \int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^2} \, dx}{4 c^2 d}\\ &=\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}+\frac{(3 b) \int \frac{x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{8 c^3 d^3}-\frac{b \operatorname{Subst}\left (\int \frac{x}{\left (1-c^2 x\right )^{5/2}} \, dx,x,x^2\right )}{8 c d^3}+\frac{3 \int \frac{a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{8 c^4 d^2}\\ &=\frac{3 b}{8 c^5 d^3 \sqrt{1-c^2 x^2}}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}+\frac{3 \operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^5 d^3}-\frac{b \operatorname{Subst}\left (\int \left (\frac{1}{c^2 \left (1-c^2 x\right )^{5/2}}-\frac{1}{c^2 \left (1-c^2 x\right )^{3/2}}\right ) \, dx,x,x^2\right )}{8 c d^3}\\ &=-\frac{b}{12 c^5 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{5 b}{8 c^5 d^3 \sqrt{1-c^2 x^2}}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}-\frac{3 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac{(3 b) \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^5 d^3}+\frac{(3 b) \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^5 d^3}\\ &=-\frac{b}{12 c^5 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{5 b}{8 c^5 d^3 \sqrt{1-c^2 x^2}}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}-\frac{3 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac{(3 i b) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}-\frac{(3 i b) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}\\ &=-\frac{b}{12 c^5 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{5 b}{8 c^5 d^3 \sqrt{1-c^2 x^2}}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{3 x \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d^3 \left (1-c^2 x^2\right )}-\frac{3 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac{3 i b \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}-\frac{3 i b \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{8 c^5 d^3}\\ \end{align*}

Mathematica [B]  time = 1.07112, size = 445, normalized size = 2.18 \[ \frac{18 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )-18 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )+\frac{30 a c x}{c^2 x^2-1}+\frac{12 a c x}{\left (c^2 x^2-1\right )^2}-9 a \log (1-c x)+9 a \log (c x+1)-\frac{15 b \sqrt{1-c^2 x^2}}{c x-1}+\frac{15 b \sqrt{1-c^2 x^2}}{c x+1}+\frac{b c x \sqrt{1-c^2 x^2}}{(c x-1)^2}-\frac{2 b \sqrt{1-c^2 x^2}}{(c x-1)^2}-\frac{b c x \sqrt{1-c^2 x^2}}{(c x+1)^2}-\frac{2 b \sqrt{1-c^2 x^2}}{(c x+1)^2}+\frac{15 b \sin ^{-1}(c x)}{c x-1}+\frac{15 b \sin ^{-1}(c x)}{c x+1}+\frac{3 b \sin ^{-1}(c x)}{(c x-1)^2}-\frac{3 b \sin ^{-1}(c x)}{(c x+1)^2}-9 i \pi b \sin ^{-1}(c x)+18 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+9 \pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-18 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+9 \pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-9 \pi b \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-9 \pi b \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{48 c^5 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^3,x]

[Out]

((-2*b*Sqrt[1 - c^2*x^2])/(-1 + c*x)^2 + (b*c*x*Sqrt[1 - c^2*x^2])/(-1 + c*x)^2 - (15*b*Sqrt[1 - c^2*x^2])/(-1
+ c*x) - (2*b*Sqrt[1 - c^2*x^2])/(1 + c*x)^2 - (b*c*x*Sqrt[1 - c^2*x^2])/(1 + c*x)^2 + (15*b*Sqrt[1 - c^2*x^2
])/(1 + c*x) + (12*a*c*x)/(-1 + c^2*x^2)^2 + (30*a*c*x)/(-1 + c^2*x^2) - (9*I)*b*Pi*ArcSin[c*x] + (3*b*ArcSin[
c*x])/(-1 + c*x)^2 + (15*b*ArcSin[c*x])/(-1 + c*x) - (3*b*ArcSin[c*x])/(1 + c*x)^2 + (15*b*ArcSin[c*x])/(1 + 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])])/(48*c^5*d^3)

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Maple [A]  time = 0.378, size = 389, normalized size = 1.9 \begin{align*}{\frac{a}{16\,{c}^{5}{d}^{3} \left ( cx-1 \right ) ^{2}}}+{\frac{5\,a}{16\,{c}^{5}{d}^{3} \left ( cx-1 \right ) }}-{\frac{3\,a\ln \left ( cx-1 \right ) }{16\,{c}^{5}{d}^{3}}}-{\frac{a}{16\,{c}^{5}{d}^{3} \left ( cx+1 \right ) ^{2}}}+{\frac{5\,a}{16\,{c}^{5}{d}^{3} \left ( cx+1 \right ) }}+{\frac{3\,a\ln \left ( cx+1 \right ) }{16\,{c}^{5}{d}^{3}}}+{\frac{5\,b\arcsin \left ( cx \right ){x}^{3}}{8\,{c}^{2}{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}-{\frac{5\,b{x}^{2}}{8\,{c}^{3}{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{3\,b\arcsin \left ( cx \right ) x}{8\,{c}^{4}{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{13\,b}{24\,{c}^{5}{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{3\,b\arcsin \left ( cx \right ) }{8\,{c}^{5}{d}^{3}}\ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{3\,b\arcsin \left ( cx \right ) }{8\,{c}^{5}{d}^{3}}\ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{{\frac{3\,i}{8}}b}{{c}^{5}{d}^{3}}{\it dilog} \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{{\frac{3\,i}{8}}b}{{c}^{5}{d}^{3}}{\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)^3,x)

[Out]

1/16/c^5*a/d^3/(c*x-1)^2+5/16/c^5*a/d^3/(c*x-1)-3/16/c^5*a/d^3*ln(c*x-1)-1/16/c^5*a/d^3/(c*x+1)^2+5/16/c^5*a/d
^3/(c*x+1)+3/16/c^5*a/d^3*ln(c*x+1)+5/8/c^2*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*x^3-5/8/c^3*b/d^3/(c^4*x^4
-2*c^2*x^2+1)*x^2*(-c^2*x^2+1)^(1/2)-3/8/c^4*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*x+13/24/c^5*b/d^3/(c^4*x^
4-2*c^2*x^2+1)*(-c^2*x^2+1)^(1/2)-3/8/c^5*b/d^3*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+3/8/c^5*b/d^3*a
rcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+3/8*I/c^5*b/d^3*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3/8*I/c^5*
b/d^3*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}{16} \, a{\left (\frac{2 \,{\left (5 \, c^{2} x^{3} - 3 \, x\right )}}{c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}} + \frac{3 \, \log \left (c x + 1\right )}{c^{5} d^{3}} - \frac{3 \, \log \left (c x - 1\right )}{c^{5} d^{3}}\right )} + \frac{{\left (3 \,{\left (c^{4} x^{4} - 2 \, 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 ) - 3 \,{\left (c^{4} x^{4} - 2 \, 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 ) + 2 \,{\left (5 \, c^{3} x^{3} - 3 \, c x\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) +{\left (c^{9} d^{3} x^{4} - 2 \, c^{7} d^{3} x^{2} + c^{5} d^{3}\right )} \int \frac{{\left (10 \, c^{3} x^{3} - 6 \, c x + 3 \,{\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \log \left (c x + 1\right ) - 3 \,{\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \log \left (-c x + 1\right )\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{10} d^{3} x^{6} - 3 \, c^{8} d^{3} x^{4} + 3 \, c^{6} d^{3} x^{2} - c^{4} d^{3}}\,{d x}\right )} b}{16 \,{\left (c^{9} d^{3} x^{4} - 2 \, c^{7} d^{3} x^{2} + c^{5} d^{3}\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)^3,x, algorithm="maxima")

[Out]

1/16*a*(2*(5*c^2*x^3 - 3*x)/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3) + 3*log(c*x + 1)/(c^5*d^3) - 3*log(c*x - 1
)/(c^5*d^3)) + 1/16*(3*(c^4*x^4 - 2*c^2*x^2 + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - 3*(
c^4*x^4 - 2*c^2*x^2 + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) + 2*(5*c^3*x^3 - 3*c*x)*arct
an2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + 16*(c^9*d^3*x^4 - 2*c^7*d^3*x^2 + c^5*d^3)*integrate(1/16*(10*c^3*x^3
- 6*c*x + 3*(c^4*x^4 - 2*c^2*x^2 + 1)*log(c*x + 1) - 3*(c^4*x^4 - 2*c^2*x^2 + 1)*log(-c*x + 1))*sqrt(c*x + 1)
*sqrt(-c*x + 1)/(c^10*d^3*x^6 - 3*c^8*d^3*x^4 + 3*c^6*d^3*x^2 - c^4*d^3), x))*b/(c^9*d^3*x^4 - 2*c^7*d^3*x^2 +
c^5*d^3)

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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^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, 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)^3,x, algorithm="fricas")

[Out]

integral(-(b*x^4*arcsin(c*x) + a*x^4)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a x^{4}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac{b x^{4} \operatorname{asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \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)**3,x)

[Out]

-(Integral(a*x**4/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Integral(b*x**4*asin(c*x)/(c**6*x**6 - 3*c
**4*x**4 + 3*c**2*x**2 - 1), x))/d**3

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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}}{{\left (c^{2} d x^{2} - d\right )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

integrate(-(b*arcsin(c*x) + a)*x^4/(c^2*d*x^2 - d)^3, x)