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

Optimal. Leaf size=196 \[ \frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3}-\frac {3 b}{8 c d^3 \sqrt {1-c^2 x^2}}-\frac {b}{12 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {3 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{8 c d^3}-\frac {3 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{8 c d^3} \]

[Out]

-1/12*b/c/d^3/(-c^2*x^2+1)^(3/2)+1/4*x*(a+b*arcsin(c*x))/d^3/(-c^2*x^2+1)^2+3/8*x*(a+b*arcsin(c*x))/d^3/(-c^2*
x^2+1)-3/4*I*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/c/d^3+3/8*I*b*polylog(2,-I*(I*c*x+(-c^2*x^2+1)
^(1/2)))/c/d^3-3/8*I*b*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/d^3-3/8*b/c/d^3/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4655, 4657, 4181, 2279, 2391, 261} \[ \frac {3 i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{8 c d^3}-\frac {3 i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{8 c d^3}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3}-\frac {3 b}{8 c d^3 \sqrt {1-c^2 x^2}}-\frac {b}{12 c d^3 \left (1-c^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 4655

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSin[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a + b*
ArcSin[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 - c^2*x^2)^FracPart[p
]), Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

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]

Rubi steps

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

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Mathematica [B]  time = 1.61, size = 501, normalized size = 2.56 \[ -\frac {\frac {6 a x}{c^2 x^2-1}-\frac {4 a x}{\left (c^2 x^2-1\right )^2}+\frac {3 a \log (1-c x)}{c}-\frac {3 a \log (c x+1)}{c}+\frac {3 b \sqrt {1-c^2 x^2}}{c-c^2 x}+\frac {3 b \sqrt {1-c^2 x^2}}{c^2 x+c}-\frac {b x \sqrt {1-c^2 x^2}}{3 (c x-1)^2}+\frac {2 b \sqrt {1-c^2 x^2}}{3 c (c x-1)^2}+\frac {b x \sqrt {1-c^2 x^2}}{3 (c x+1)^2}+\frac {2 b \sqrt {1-c^2 x^2}}{3 c (c x+1)^2}-\frac {3 b \sin ^{-1}(c x)}{c-c^2 x}+\frac {3 b \sin ^{-1}(c x)}{c^2 x+c}-\frac {6 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c}+\frac {6 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac {b \sin ^{-1}(c x)}{c (c x-1)^2}+\frac {b \sin ^{-1}(c x)}{c (c x+1)^2}+\frac {3 i \pi b \sin ^{-1}(c x)}{c}-\frac {6 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac {3 \pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )}{c}+\frac {6 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac {3 \pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )}{c}+\frac {3 \pi b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{c}+\frac {3 \pi b \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{c}}{16 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/16*((2*b*Sqrt[1 - c^2*x^2])/(3*c*(-1 + c*x)^2) - (b*x*Sqrt[1 - c^2*x^2])/(3*(-1 + c*x)^2) + (2*b*Sqrt[1 - c
^2*x^2])/(3*c*(1 + c*x)^2) + (b*x*Sqrt[1 - c^2*x^2])/(3*(1 + c*x)^2) + (3*b*Sqrt[1 - c^2*x^2])/(c - c^2*x) + (
3*b*Sqrt[1 - c^2*x^2])/(c + c^2*x) - (4*a*x)/(-1 + c^2*x^2)^2 + (6*a*x)/(-1 + c^2*x^2) + ((3*I)*b*Pi*ArcSin[c*
x])/c - (b*ArcSin[c*x])/(c*(-1 + c*x)^2) + (b*ArcSin[c*x])/(c*(1 + c*x)^2) - (3*b*ArcSin[c*x])/(c - c^2*x) + (
3*b*ArcSin[c*x])/(c + c^2*x) - (3*b*Pi*Log[1 - I*E^(I*ArcSin[c*x])])/c - (6*b*ArcSin[c*x]*Log[1 - I*E^(I*ArcSi
n[c*x])])/c - (3*b*Pi*Log[1 + I*E^(I*ArcSin[c*x])])/c + (6*b*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])])/c + (3*
a*Log[1 - c*x])/c - (3*a*Log[1 + c*x])/c + (3*b*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]])/c + (3*b*Pi*Log[Sin[(Pi
+ 2*ArcSin[c*x])/4]])/c - ((6*I)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/c + ((6*I)*b*PolyLog[2, I*E^(I*ArcSin[c
*x])])/c)/d^3

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \arcsin \left (c x\right ) + a}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(-c^2*d*x^2+d)^3,x, algorithm="fricas")

[Out]

integral(-(b*arcsin(c*x) + a)/(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {b \arcsin \left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(-c^2*d*x^2+d)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.19, size = 384, normalized size = 1.96 \[ -\frac {a}{16 c \,d^{3} \left (c x +1\right )^{2}}-\frac {3 a}{16 c \,d^{3} \left (c x +1\right )}+\frac {3 a \ln \left (c x +1\right )}{16 c \,d^{3}}+\frac {a}{16 c \,d^{3} \left (c x -1\right )^{2}}-\frac {3 a}{16 c \,d^{3} \left (c x -1\right )}-\frac {3 a \ln \left (c x -1\right )}{16 c \,d^{3}}-\frac {3 c^{2} b \arcsin \left (c x \right ) x^{3}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {3 c b \,x^{2} \sqrt {-c^{2} x^{2}+1}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {5 b \arcsin \left (c x \right ) x}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {11 b \sqrt {-c^{2} x^{2}+1}}{24 c \,d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 c \,d^{3}}+\frac {3 b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 c \,d^{3}}+\frac {3 i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 c \,d^{3}}-\frac {3 i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 c \,d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/c*a/d^3/(c*x+1)^2-3/16/c*a/d^3/(c*x+1)+3/16/c*a/d^3*ln(c*x+1)+1/16/c*a/d^3/(c*x-1)^2-3/16/c*a/d^3/(c*x-1
)-3/16/c*a/d^3*ln(c*x-1)-3/8*c^2*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*x^3+3/8*c*b/d^3/(c^4*x^4-2*c^2*x^2+1)
*x^2*(-c^2*x^2+1)^(1/2)+5/8*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*x-11/24/c*b/d^3/(c^4*x^4-2*c^2*x^2+1)*(-c^
2*x^2+1)^(1/2)-3/8/c*b/d^3*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+3/8/c*b/d^3*arcsin(c*x)*ln(1-I*(I*c*
x+(-c^2*x^2+1)^(1/2)))+3/8*I/c*b/d^3*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3/8*I/c*b/d^3*dilog(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 \[ -\frac {1}{16} \, a {\left (\frac {2 \, {\left (3 \, c^{2} x^{3} - 5 \, x\right )}}{c^{4} d^{3} x^{4} - 2 \, c^{2} d^{3} x^{2} + d^{3}} - \frac {3 \, \log \left (c x + 1\right )}{c d^{3}} + \frac {3 \, \log \left (c x - 1\right )}{c d^{3}}\right )} + \frac {{\left (3 \, {\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) - 3 \, {\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \, {\left (3 \, c^{3} x^{3} - 5 \, c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (c^{5} d^{3} x^{4} - 2 \, c^{3} d^{3} x^{2} + c d^{3}\right )} \int \frac {{\left (6 \, c^{3} x^{3} - 10 \, c x - 3 \, {\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \log \left (c x + 1\right ) + 3 \, {\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \log \left (-c x + 1\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}\,{d x}\right )} b}{16 \, {\left (c^{5} d^{3} x^{4} - 2 \, c^{3} d^{3} x^{2} + c d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(-c^2*d*x^2+d)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d-c^2\,d\,x^2\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))/(d - c^2*d*x^2)^3,x)

[Out]

int((a + b*asin(c*x))/(d - c^2*d*x^2)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))/(-c**2*d*x**2+d)**3,x)

[Out]

Timed out

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