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

Optimal. Leaf size=225 \[ -\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {c^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {i b c^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}+\frac {i b c^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}-\frac {b c \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4693, 30, 4709, 4183, 2279, 2391} \[ -\frac {i b c^2 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}+\frac {i b c^2 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {c^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {b c \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && 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 4693

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 2.36, size = 239, normalized size = 1.06 \[ \frac {1}{8} \left (-\frac {4 a \sqrt {d-c^2 d x^2}}{x^2}+4 a c^2 \sqrt {d} \log \left (\sqrt {d} \sqrt {d-c^2 d x^2}+d\right )-4 a c^2 \sqrt {d} \log (x)+\frac {b c^2 d \sqrt {1-c^2 x^2} \left (-4 i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )+4 i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-4 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+4 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )-2 \tan \left (\frac {1}{2} \sin ^{-1}(c x)\right )-2 \cot \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin ^{-1}(c x) \csc ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )+\sin ^{-1}(c x) \sec ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{\sqrt {d-c^2 d x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*a*Sqrt[d - c^2*d*x^2])/x^2 - 4*a*c^2*Sqrt[d]*Log[x] + 4*a*c^2*Sqrt[d]*Log[d + Sqrt[d]*Sqrt[d - c^2*d*x^2]
] + (b*c^2*d*Sqrt[1 - c^2*x^2]*(-2*Cot[ArcSin[c*x]/2] - ArcSin[c*x]*Csc[ArcSin[c*x]/2]^2 - 4*ArcSin[c*x]*Log[1
- E^(I*ArcSin[c*x])] + 4*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] - (4*I)*PolyLog[2, -E^(I*ArcSin[c*x])] + (4*I
)*PolyLog[2, E^(I*ArcSin[c*x])] + ArcSin[c*x]*Sec[ArcSin[c*x]/2]^2 - 2*Tan[ArcSin[c*x]/2]))/Sqrt[d - c^2*d*x^2
])/8

________________________________________________________________________________________

fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [B]  time = 0.36, size = 462, normalized size = 2.05 \[ -\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{2 d \,x^{2}}+\frac {a \sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right ) c^{2}}{2}-\frac {a \sqrt {-c^{2} d \,x^{2}+d}\, c^{2}}{2}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c^{2}}{2 \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c}{2 \left (c^{2} x^{2}-1\right ) x}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{2 \left (c^{2} x^{2}-1\right ) x^{2}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2}-\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^3,x)

[Out]

-1/2*a/d/x^2*(-c^2*d*x^2+d)^(3/2)+1/2*a*d^(1/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)*c^2-1/2*a*(-c^2*d*x
^2+d)^(1/2)*c^2-1/2*b*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)*arcsin(c*x)*c^2+1/2*b*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2
-1)/x*(-c^2*x^2+1)^(1/2)*c+1/2*b*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)/x^2*arcsin(c*x)+b*(-d*(c^2*x^2-1))^(1/2)*(
-c^2*x^2+1)^(1/2)*c^2/(2*c^2*x^2-2)*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*
x^2+1)^(1/2)*c^2/(2*c^2*x^2-2)*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2
+1)^(1/2)*c^2/(2*c^2*x^2-2)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*
c^2/(2*c^2*x^2-2)*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________