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

Optimal. Leaf size=141 \[ \frac {x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac {i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d^2}-\frac {b}{2 c d^2 \sqrt {1-c^2 x^2}}+\frac {i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{2 c d^2}-\frac {i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{2 c d^2} \]

[Out]

1/2*x*(a+b*arcsin(c*x))/d^2/(-c^2*x^2+1)-I*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/c/d^2+1/2*I*b*po
lylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/d^2-1/2*I*b*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/d^2-1/2*b/c/d^2
/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{2 c d^2}-\frac {i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{2 c d^2}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac {i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d^2}-\frac {b}{2 c d^2 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [B]  time = 0.84, size = 334, normalized size = 2.37 \[ -\frac {\frac {2 a x}{c^2 x^2-1}+\frac {a \log (1-c x)}{c}-\frac {a \log (c x+1)}{c}+\frac {b \sqrt {1-c^2 x^2}}{c-c^2 x}+\frac {b \sqrt {1-c^2 x^2}}{c^2 x+c}+\frac {b \sin ^{-1}(c x)}{c^2 x+c}-\frac {2 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c}+\frac {2 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c}+\frac {b \sin ^{-1}(c x)}{c (c x-1)}+\frac {i \pi b \sin ^{-1}(c x)}{c}-\frac {2 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac {\pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )}{c}+\frac {2 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac {\pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )}{c}+\frac {\pi b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{c}+\frac {\pi b \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{c}}{4 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsin(c*x) + a)/(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 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 260, normalized size = 1.84 \[ -\frac {a}{4 c \,d^{2} \left (c x +1\right )}+\frac {a \ln \left (c x +1\right )}{4 c \,d^{2}}-\frac {a}{4 c \,d^{2} \left (c x -1\right )}-\frac {a \ln \left (c x -1\right )}{4 c \,d^{2}}-\frac {b \arcsin \left (c x \right ) x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{2 c \,d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c \,d^{2}}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c \,d^{2}}+\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c \,d^{2}}-\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/c*a/d^2/(c*x+1)+1/4/c*a/d^2*ln(c*x+1)-1/4/c*a/d^2/(c*x-1)-1/4/c*a/d^2*ln(c*x-1)-1/2*b/d^2/(c^2*x^2-1)*arc
sin(c*x)*x+1/2/c*b/d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-1/2/c*b/d^2*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)
))+1/2/c*b/d^2*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+1/2*I/c*b/d^2*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2
)))-1/2*I/c*b/d^2*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 \[ -\frac {1}{4} \, a {\left (\frac {2 \, x}{c^{2} d^{2} x^{2} - d^{2}} - \frac {\log \left (c x + 1\right )}{c d^{2}} + \frac {\log \left (c x - 1\right )}{c d^{2}}\right )} - \frac {{\left (2 \, c x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (c^{2} x^{2} - 1\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) + {\left (c^{2} x^{2} - 1\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right ) + {\left (c^{3} d^{2} x^{2} - c d^{2}\right )} \int \frac {{\left (2 \, c x - {\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) + {\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}\,{d x}\right )} b}{4 \, {\left (c^{3} d^{2} x^{2} - c d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a*(2*x/(c^2*d^2*x^2 - d^2) - log(c*x + 1)/(c*d^2) + log(c*x - 1)/(c*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^3*d^2*x^2 - c*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^4*d^2*x^4 - 2*c^2*d^
2*x^2 + d^2), x))*b/(c^3*d^2*x^2 - c*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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