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

Optimal. Leaf size=82 \[ \frac {i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^2 d}-\frac {\log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d}+\frac {i b \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4675, 3719, 2190, 2279, 2391} \[ \frac {i b \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^2 d}+\frac {i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^2 d}-\frac {\log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && 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 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4675

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[e^(-1), Subst[In
t[(a + b*x)^n*Tan[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 {x \left (a+b \sin ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx &=\frac {\operatorname {Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d}\\ &=\frac {i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^2 d}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d}\\ &=\frac {i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}+\frac {b \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d}\\ &=\frac {i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 c^2 d}\\ &=\frac {i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}+\frac {i b \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.09, size = 244, normalized size = 2.98 \[ -\frac {a \log \left (1-c^2 x^2\right )-2 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )-2 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )-i b \sin ^{-1}(c x)^2+2 i \pi b \sin ^{-1}(c x)+2 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+2 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+4 \pi b \log \left (1+e^{-i \sin ^{-1}(c x)}\right )+\pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-\pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-\pi b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-4 \pi b \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+\pi b \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{2 c^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x \arcsin \left (c x\right ) + a x}{c^{2} d x^{2} - d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.07, size = 118, normalized size = 1.44 \[ -\frac {a \ln \left (c x -1\right )}{2 c^{2} d}-\frac {a \ln \left (c x +1\right )}{2 c^{2} d}+\frac {i b \arcsin \left (c x \right )^{2}}{2 c^{2} d}-\frac {b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{c^{2} d}+\frac {i b \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________