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

Optimal. Leaf size=220 \[ \frac {a+b \sin ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}+\frac {i b \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {i b \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4705, 4713, 4709, 4183, 2279, 2391, 206} \[ \frac {i b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {i b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {a+b \sin ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 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 4705

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)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p +
1)), Int[(f*x)^m*(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)^Frac
Part[p])/(2*f*(p + 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, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

Rule 4709

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

Rule 4713

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[
Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2], Int[((f*x)^m*(a + b*ArcSin[c*x])^n)/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] &&  !GtQ[d, 0] && (IntegerQ[m] || EqQ[n, 1])

Rubi steps

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

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Mathematica [A]  time = 1.16, size = 300, normalized size = 1.36 \[ \frac {-\frac {a \sqrt {d-c^2 d x^2}}{c^2 x^2-1}-a \sqrt {d} \log \left (\sqrt {d} \sqrt {d-c^2 d x^2}+d\right )+a \sqrt {d} \log (x)+\frac {b d \left (i \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-i \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )+\sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )+\sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )-\sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+\sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{4} d^{2} x^{5} - 2 \, c^{2} d^{2} x^{3} + d^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^4*d^2*x^5 - 2*c^2*d^2*x^3 + d^2*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.22, size = 344, normalized size = 1.56 \[ \frac {a}{d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {a \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{d^{\frac {3}{2}}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \dilog \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \dilog \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))/(x*(-d*(c*x - 1)*(c*x + 1))**(3/2)), x)

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