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

Optimal. Leaf size=124 \[ -\frac {2 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {a+b \sin ^{-1}(c x)}{2 d x^2}+\frac {i b c^2 \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d}-\frac {i b c^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d}-\frac {b c \sqrt {1-c^2 x^2}}{2 d x} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4701, 4679, 4419, 4183, 2279, 2391, 264} \[ \frac {i b c^2 \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d}-\frac {i b c^2 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}-\frac {2 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {a+b \sin ^{-1}(c x)}{2 d x^2}-\frac {b c \sqrt {1-c^2 x^2}}{2 d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -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 4183

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

Rule 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]

Rule 4679

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Dist[1/d, Subst[Int[(a
+ b*x)^n/(Cos[x]*Sin[x]), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n
, 0]

Rule 4701

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

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 149, normalized size = 1.20 \[ -\frac {a c^2 \log \left (1-c^2 x^2\right )-2 a c^2 \log (x)+\frac {a}{x^2}+b c^2 \left (\frac {\sqrt {1-c^2 x^2}}{c x}+\frac {\sin ^{-1}(c x)}{c^2 x^2}-i \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )+i \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+2 \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \arcsin \left (c x\right ) + a}{c^{2} d x^{5} - d x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*arcsin(c*x) + a)/(c^2*d*x^5 - d*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.33, size = 296, normalized size = 2.39 \[ -\frac {c^{2} a \ln \left (c x +1\right )}{2 d}-\frac {a}{2 d \,x^{2}}+\frac {c^{2} a \ln \left (c x \right )}{d}-\frac {c^{2} a \ln \left (c x -1\right )}{2 d}+\frac {i c^{2} b}{2 d}-\frac {b c \sqrt {-c^{2} x^{2}+1}}{2 d x}-\frac {b \arcsin \left (c x \right )}{2 d \,x^{2}}+\frac {c^{2} b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {i c^{2} b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {c^{2} b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i b \,c^{2} \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 d}+\frac {c^{2} b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {i c^{2} b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^2*a/d*ln(c*x+1)-1/2*a/d/x^2+c^2*a/d*ln(c*x)-1/2*c^2*a/d*ln(c*x-1)+1/2*I*c^2*b/d-1/2*b*c*(-c^2*x^2+1)^(1
/2)/d/x-1/2*b/d*arcsin(c*x)/x^2+c^2*b/d*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-I*c^2*b/d*polylog(2,I*c*x+(
-c^2*x^2+1)^(1/2))-c^2*b/d*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+1/2*I*b*c^2*polylog(2,-(I*c*x+(-c^2*
x^2+1)^(1/2))^2)/d+c^2*b/d*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-I*c^2*b/d*polylog(2,-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 \[ -\frac {1}{2} \, {\left (\frac {c^{2} \log \left (c x + 1\right )}{d} + \frac {c^{2} \log \left (c x - 1\right )}{d} - \frac {2 \, c^{2} \log \relax (x)}{d} + \frac {1}{d x^{2}}\right )} a - b \int \frac {\arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{2} d x^{5} - d x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a}{c^{2} x^{5} - x^{3}}\, dx + \int \frac {b \operatorname {asin}{\left (c x \right )}}{c^{2} x^{5} - x^{3}}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a/(c**2*x**5 - x**3), x) + Integral(b*asin(c*x)/(c**2*x**5 - x**3), x))/d

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