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

Optimal. Leaf size=217 \[ -\frac {i \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{(c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {2 b \left (1-c^2 x^2\right )^{3/2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}-\frac {i b^2 \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {4673, 4651, 4675, 3719, 2190, 2279, 2391} \[ -\frac {i b^2 \left (1-c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}-\frac {i \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{(c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {2 b \left (1-c^2 x^2\right )^{3/2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(1 - c^2*x^2)*(a + b*ArcSin[c*x])^2)/((d + c*d*x)^(3/2)*(e - c*e*x)^(3/2)) - (I*(1 - c^2*x^2)^(3/2)*(a + b*
ArcSin[c*x])^2)/(c*(d + c*d*x)^(3/2)*(e - c*e*x)^(3/2)) + (2*b*(1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x])*Log[1 +
E^((2*I)*ArcSin[c*x])])/(c*(d + c*d*x)^(3/2)*(e - c*e*x)^(3/2)) - (I*b^2*(1 - c^2*x^2)^(3/2)*PolyLog[2, -E^((
2*I)*ArcSin[c*x])])/(c*(d + c*d*x)^(3/2)*(e - c*e*x)^(3/2))

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]

Rule 3719

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

Rule 4651

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n)/Sqrt[d], Int[(x*(a + b*ArcSin[c*x])^(n - 1))/(d + e*x^2), x], x
] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[d, 0]

Rule 4673

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> D
ist[((d + e*x)^q*(f + g*x)^q)/(1 - c^2*x^2)^q, Int[(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, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q]
&& GeQ[p - q, 0]

Rule 4675

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

Rubi steps

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

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Mathematica [B]  time = 0.76, size = 550, normalized size = 2.53 \[ \frac {a^2 c x+2 a b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+2 a b \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+2 a b c x \sin ^{-1}(c x)-2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )-2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )-i b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2+2 i \pi b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)+4 \pi b^2 \sqrt {1-c^2 x^2} \log \left (1+e^{-i \sin ^{-1}(c x)}\right )+2 b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+\pi b^2 \sqrt {1-c^2 x^2} \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+2 b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-\pi b^2 \sqrt {1-c^2 x^2} \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-\pi b^2 \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-4 \pi b^2 \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+\pi b^2 \sqrt {1-c^2 x^2} \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+b^2 c x \sin ^{-1}(c x)^2}{c d e \sqrt {c d x+d} \sqrt {e-c e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*x + 2*a*b*c*x*ArcSin[c*x] + (2*I)*b^2*Pi*Sqrt[1 - c^2*x^2]*ArcSin[c*x] + b^2*c*x*ArcSin[c*x]^2 - I*b^2*
Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2 + 4*b^2*Pi*Sqrt[1 - c^2*x^2]*Log[1 + E^((-I)*ArcSin[c*x])] + 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*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])] - 4*b^2*Pi*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2]] + 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]*Lo
g[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*ArcS
in[c*x])])/(c*d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \sqrt {c d x + d} \sqrt {-c e x + e}}{c^{4} d^{2} e^{2} x^{4} - 2 \, c^{2} d^{2} e^{2} x^{2} + d^{2} e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*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 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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((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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