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

Optimal. Leaf size=139 \[ \frac{1}{2} i b c^2 d \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac{i c^2 d \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}-c^2 d \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac{b c d \sqrt{1-c^2 x^2}}{2 x}-\frac{1}{2} b c^2 d \sin ^{-1}(c x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.120644, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4685, 277, 216, 4625, 3717, 2190, 2279, 2391} \[ \frac{1}{2} i b c^2 d \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac{i c^2 d \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}-c^2 d \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac{b c d \sqrt{1-c^2 x^2}}{2 x}-\frac{1}{2} b c^2 d \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4685

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((f*x)
^(m + 1)*(d + e*x^2)^p*(a + b*ArcSin[c*x]))/(f*(m + 1)), x] + (-Dist[(b*c*d^p)/(f*(m + 1)), Int[(f*x)^(m + 1)*
(1 - c^2*x^2)^(p - 1/2), x], x] - Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^(p - 1)*(a + b*Arc
Sin[c*x]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0] && ILtQ[(m + 1)/2, 0]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 4625

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tan[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

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]

Rubi steps

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

Mathematica [A]  time = 0.105223, size = 110, normalized size = 0.79 \[ -\frac{d \left (-i b c^2 x^2 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+2 a c^2 x^2 \log (x)+a+b c x \sqrt{1-c^2 x^2}-i b c^2 x^2 \sin ^{-1}(c x)^2+b \sin ^{-1}(c x) \left (1+2 c^2 x^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )\right )\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.23, size = 195, normalized size = 1.4 \begin{align*} -{\frac{da}{2\,{x}^{2}}}-{c}^{2}da\ln \left ( cx \right ) +{\frac{i}{2}}{c}^{2}db \left ( \arcsin \left ( cx \right ) \right ) ^{2}+{\frac{i}{2}}{c}^{2}db-{\frac{bcd}{2\,x}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{bd\arcsin \left ( cx \right ) }{2\,{x}^{2}}}-{c}^{2}db\arcsin \left ( cx \right ) \ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) -{c}^{2}db\arcsin \left ( cx \right ) \ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) +i{c}^{2}db{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) +i{c}^{2}db{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -b c^{2} d \int \frac{\arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )}{x}\,{d x} - a c^{2} d \log \left (x\right ) - \frac{1}{2} \, b d{\left (\frac{\sqrt{-c^{2} x^{2} + 1} c}{x} + \frac{\arcsin \left (c x\right )}{x^{2}}\right )} - \frac{a d}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{a c^{2} d x^{2} - a d +{\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - d \left (\int - \frac{a}{x^{3}}\, dx + \int \frac{a c^{2}}{x}\, dx + \int - \frac{b \operatorname{asin}{\left (c x \right )}}{x^{3}}\, dx + \int \frac{b c^{2} \operatorname{asin}{\left (c x \right )}}{x}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (c^{2} d x^{2} - d\right )}{\left (b \arcsin \left (c x\right ) + a\right )}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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