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

Optimal. Leaf size=187 \[ \frac {3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{2 c^4 d^2}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {3 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{2 c^5 d^2}+\frac {3 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{2 c^5 d^2}+\frac {b \sqrt {1-c^2 x^2}}{c^5 d^2}-\frac {b}{2 c^5 d^2 \sqrt {1-c^2 x^2}} \]

[Out]

3/2*x*(a+b*arcsin(c*x))/c^4/d^2+1/2*x^3*(a+b*arcsin(c*x))/c^2/d^2/(-c^2*x^2+1)+3*I*(a+b*arcsin(c*x))*arctan(I*
c*x+(-c^2*x^2+1)^(1/2))/c^5/d^2-3/2*I*b*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^5/d^2+3/2*I*b*polylog(2,I*(
I*c*x+(-c^2*x^2+1)^(1/2)))/c^5/d^2-1/2*b/c^5/d^2/(-c^2*x^2+1)^(1/2)+b*(-c^2*x^2+1)^(1/2)/c^5/d^2

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Rubi [A]  time = 0.24, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4703, 4715, 4657, 4181, 2279, 2391, 261, 266, 43} \[ -\frac {3 i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{2 c^5 d^2}+\frac {3 i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{2 c^5 d^2}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{2 c^4 d^2}+\frac {3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac {b \sqrt {1-c^2 x^2}}{c^5 d^2}-\frac {b}{2 c^5 d^2 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.44, size = 332, normalized size = 1.78 \[ \frac {-\frac {2 a c x}{c^2 x^2-1}+4 a c x+3 a \log (1-c x)-3 a \log (c x+1)+\frac {b \sqrt {1-c^2 x^2}}{c x-1}-\frac {b \sqrt {1-c^2 x^2}}{c x+1}+4 b \sqrt {1-c^2 x^2}-6 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )+6 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+4 b c x \sin ^{-1}(c x)+\frac {b \sin ^{-1}(c x)}{1-c x}-\frac {b \sin ^{-1}(c x)}{c x+1}+3 i \pi b \sin ^{-1}(c x)-6 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-3 \pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+6 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-3 \pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+3 \pi b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+3 \pi b \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{4 c^5 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{4} \arcsin \left (c x\right ) + a x^{4}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

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)^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} - d\right )}^{2}}\,{d x} \]

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)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.40, size = 305, normalized size = 1.63 \[ \frac {a x}{c^{4} d^{2}}-\frac {a}{4 c^{5} d^{2} \left (c x +1\right )}-\frac {3 a \ln \left (c x +1\right )}{4 c^{5} d^{2}}-\frac {a}{4 c^{5} d^{2} \left (c x -1\right )}+\frac {3 a \ln \left (c x -1\right )}{4 c^{5} d^{2}}+\frac {b \sqrt {-c^{2} x^{2}+1}}{c^{5} d^{2}}+\frac {b \arcsin \left (c x \right ) x}{c^{4} d^{2}}-\frac {b \arcsin \left (c x \right ) x}{2 c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{2 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {3 b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c^{5} d^{2}}-\frac {3 b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c^{5} d^{2}}-\frac {3 i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c^{5} d^{2}}+\frac {3 i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c^{5} d^{2}} \]

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)^2,x)

[Out]

1/c^4*a/d^2*x-1/4/c^5*a/d^2/(c*x+1)-3/4/c^5*a/d^2*ln(c*x+1)-1/4/c^5*a/d^2/(c*x-1)+3/4/c^5*a/d^2*ln(c*x-1)+b*(-
c^2*x^2+1)^(1/2)/c^5/d^2+1/c^4*b/d^2*arcsin(c*x)*x-1/2/c^4*b/d^2/(c^2*x^2-1)*arcsin(c*x)*x+1/2/c^5*b/d^2/(c^2*
x^2-1)*(-c^2*x^2+1)^(1/2)+3/2/c^5*b/d^2*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3/2/c^5*b/d^2*arcsin(c*
x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3/2*I/c^5*b/d^2*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+3/2*I/c^5*b/d^2*di
log(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{4} \, a {\left (\frac {2 \, x}{c^{6} d^{2} x^{2} - c^{4} d^{2}} - \frac {4 \, x}{c^{4} d^{2}} + \frac {3 \, \log \left (c x + 1\right )}{c^{5} d^{2}} - \frac {3 \, \log \left (c x - 1\right )}{c^{5} d^{2}}\right )} - \frac {{\left (3 \, {\left (c^{2} x^{2} - 1\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) - 3 \, {\left (c^{2} x^{2} - 1\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \, {\left (2 \, c^{3} x^{3} - 3 \, c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (c^{7} d^{2} x^{2} - c^{5} d^{2}\right )} \int \frac {{\left (4 \, c^{3} x^{3} - 6 \, c x - 3 \, {\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) + 3 \, {\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{8} d^{2} x^{4} - 2 \, c^{6} d^{2} x^{2} + c^{4} d^{2}}\,{d x}\right )} b}{4 \, {\left (c^{7} d^{2} x^{2} - c^{5} d^{2}\right )}} \]

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)^2,x, algorithm="maxima")

[Out]

-1/4*a*(2*x/(c^6*d^2*x^2 - c^4*d^2) - 4*x/(c^4*d^2) + 3*log(c*x + 1)/(c^5*d^2) - 3*log(c*x - 1)/(c^5*d^2)) - 1
/4*(3*(c^2*x^2 - 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - 3*(c^2*x^2 - 1)*arctan2(c*x, sqr
t(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) - 2*(2*c^3*x^3 - 3*c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) +
4*(c^7*d^2*x^2 - c^5*d^2)*integrate(-1/4*(4*c^3*x^3 - 6*c*x - 3*(c^2*x^2 - 1)*log(c*x + 1) + 3*(c^2*x^2 - 1)*l
og(-c*x + 1))*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^8*d^2*x^4 - 2*c^6*d^2*x^2 + c^4*d^2), x))*b/(c^7*d^2*x^2 - c^5*d
^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a x^{4}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{4} \operatorname {asin}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \]

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)**2,x)

[Out]

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

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