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

Optimal. Leaf size=244 \[ \frac {4 i b \sqrt {1-c^2 x^2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d e \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d e \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {c d x+d} \sqrt {e-c e x}} \]

[Out]

(a+b*arcsin(c*x))^2/c^2/d/e/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+4*I*b*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)
^(1/2))*(-c^2*x^2+1)^(1/2)/c^2/d/e/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-2*I*b^2*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(
1/2)))*(-c^2*x^2+1)^(1/2)/c^2/d/e/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+2*I*b^2*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/
2)))*(-c^2*x^2+1)^(1/2)/c^2/d/e/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)

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Rubi [A]  time = 0.49, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4739, 4677, 4657, 4181, 2279, 2391} \[ -\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {4 i b \sqrt {1-c^2 x^2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d e \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d e \sqrt {c d x+d} \sqrt {e-c e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*ArcSin[c*x])^2/(c^2*d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + ((4*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x
])*ArcTan[E^(I*ArcSin[c*x])])/(c^2*d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - ((2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog
[2, (-I)*E^(I*ArcSin[c*x])])/(c^2*d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + ((2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[
2, I*E^(I*ArcSin[c*x])])/(c^2*d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

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 4181

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

Rule 4657

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

Rule 4677

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

Rule 4739

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((h_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(
q_), x_Symbol] :> Dist[((-((d^2*g)/e))^IntPart[q]*(d + e*x)^FracPart[q]*(f + g*x)^FracPart[q])/(1 - c^2*x^2)^F
racPart[q], Int[(h*x)^m*(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]

Rubi steps

\begin {align*} \int \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{d e \sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{1-c^2 x^2} \, dx}{c d e \sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {4 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {4 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 i b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (2 i b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {4 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}\\ \end {align*}

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Mathematica [A]  time = 1.35, size = 453, normalized size = 1.86 \[ \frac {a^2+2 a b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )-2 a b \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+2 a b \sin ^{-1}(c x)-2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )+2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+i \pi b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)-2 b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-\pi b^2 \sqrt {1-c^2 x^2} \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+2 b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-\pi b^2 \sqrt {1-c^2 x^2} \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+\pi b^2 \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+\pi b^2 \sqrt {1-c^2 x^2} \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+b^2 \sin ^{-1}(c x)^2}{c^2 d e \sqrt {c d x+d} \sqrt {e-c e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2 + 2*a*b*ArcSin[c*x] + I*b^2*Pi*Sqrt[1 - c^2*x^2]*ArcSin[c*x] + b^2*ArcSin[c*x]^2 - b^2*Pi*Sqrt[1 - c^2*x^
2]*Log[1 - I*E^(I*ArcSin[c*x])] - 2*b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - b^2*Pi*Sq
rt[1 - c^2*x^2]*Log[1 + I*E^(I*ArcSin[c*x])] + 2*b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])
] + b^2*Pi*Sqrt[1 - c^2*x^2]*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + 2*a*b*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2
] - Sin[ArcSin[c*x]/2]] - 2*a*b*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] + b^2*Pi*Sqrt[1
- c^2*x^2]*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])]
+ (2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^2*d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x \arcsin \left (c x\right )^{2} + 2 \, a b x \arcsin \left (c x\right ) + a^{2} x\right )} \sqrt {c d x + d} \sqrt {-c e x + e}}{c^{4} d^{2} e^{2} x^{4} - 2 \, c^{2} d^{2} e^{2} x^{2} + d^{2} e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.85, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (c d x +d \right )^{\frac {3}{2}} \left (-c e x +e \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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