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

Optimal. Leaf size=84 \[ -\frac {2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d}+\frac {i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4657, 4181, 2279, 2391} \[ \frac {i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac {2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

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Mathematica [B]  time = 0.24, size = 207, normalized size = 2.46 \[ \frac {-a \log (1-c x)+a \log (c x+1)+2 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )-2 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )-i \pi b \sin ^{-1}(c x)+2 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-2 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+\pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+\pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-\pi b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-\pi b \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{2 c d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 170, normalized size = 2.02 \[ \frac {a \arctanh \left (c x \right )}{c d}-\frac {i b \arctanh \left (c x \right ) \ln \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d}+\frac {i b \arctanh \left (c x \right ) \ln \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d}+\frac {b \arctanh \left (c x \right ) \arcsin \left (c x \right )}{c d}-\frac {i b \dilog \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d}+\frac {i b \dilog \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*a/d*arctanh(c*x)-I/c*b/d*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+I/c*b/d*arctanh(c*x)*ln(1+I*(c*x+
1)/(-c^2*x^2+1)^(1/2))+1/c*b/d*arctanh(c*x)*arcsin(c*x)-I/c*b/d*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+I/c*b/d*
dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(log(c*x + 1)/(c*d) - log(c*x - 1)/(c*d)) + 1/2*(2*c*d*integrate(1/2*sqrt(c*x + 1)*sqrt(-c*x + 1)*(log(c
*x + 1) - log(-c*x + 1))/(c^2*d*x^2 - d), x) + arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - arcta
n2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1))*b/(c*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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