3.6.89 \(\int \frac {(a+b \text {ArcSin}(c x))^2}{x \sqrt {d+c d x} \sqrt {e-c e x}} \, dx\) [589]

Optimal. Leaf size=287 \[ -\frac {2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2 \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 i b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 i b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (3,-e^{i \text {ArcSin}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (3,e^{i \text {ArcSin}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}} \]

[Out]

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

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Rubi [A]
time = 0.41, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4823, 4803, 4268, 2611, 2320, 6724} \begin {gather*} \frac {2 i b \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{\sqrt {c d x+d} \sqrt {e-c e x}}-\frac {2 i b \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{\sqrt {c d x+d} \sqrt {e-c e x}}-\frac {2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))^2}{\sqrt {c d x+d} \sqrt {e-c e x}}-\frac {2 b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (-e^{i \text {ArcSin}(c x)}\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (e^{i \text {ArcSin}(c x)}\right )}{\sqrt {c d x+d} \sqrt {e-c e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4268

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 4803

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

Rule 4823

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)^Fr
acPart[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]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt {d+c d x} \sqrt {e-c e x}} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {\sqrt {1-c^2 x^2} \text {Subst}\left (\int (a+b x)^2 \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=-\frac {2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int (a+b x) \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int (a+b x) \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=-\frac {2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 i b^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (2 i b^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \text {Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=-\frac {2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=-\frac {2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ \end {align*}

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Mathematica [A]
time = 0.84, size = 336, normalized size = 1.17 \begin {gather*} \frac {a^2 \log (c x)}{\sqrt {d} \sqrt {e}}-\frac {a^2 \log \left (d e+\sqrt {d} \sqrt {e} \sqrt {d+c d x} \sqrt {e-c e x}\right )}{\sqrt {d} \sqrt {e}}+\frac {2 a b \sqrt {1-c^2 x^2} \left (\text {ArcSin}(c x) \left (\log \left (1-e^{i \text {ArcSin}(c x)}\right )-\log \left (1+e^{i \text {ArcSin}(c x)}\right )\right )+i \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )-i \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b^2 \sqrt {1-c^2 x^2} \left (\text {ArcSin}(c x)^2 \log \left (1-e^{i \text {ArcSin}(c x)}\right )-\text {ArcSin}(c x)^2 \log \left (1+e^{i \text {ArcSin}(c x)}\right )+2 i \text {ArcSin}(c x) \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )-2 i \text {ArcSin}(c x) \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )-2 \text {PolyLog}\left (3,-e^{i \text {ArcSin}(c x)}\right )+2 \text {PolyLog}\left (3,e^{i \text {ArcSin}(c x)}\right )\right )}{\sqrt {d+c d x} \sqrt {e-c e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Log[c*x])/(Sqrt[d]*Sqrt[e]) - (a^2*Log[d*e + Sqrt[d]*Sqrt[e]*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]])/(Sqrt[d]*S
qrt[e]) + (2*a*b*Sqrt[1 - c^2*x^2]*(ArcSin[c*x]*(Log[1 - E^(I*ArcSin[c*x])] - Log[1 + E^(I*ArcSin[c*x])]) + I*
PolyLog[2, -E^(I*ArcSin[c*x])] - I*PolyLog[2, E^(I*ArcSin[c*x])]))/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (b^2*Sq
rt[1 - c^2*x^2]*(ArcSin[c*x]^2*Log[1 - E^(I*ArcSin[c*x])] - ArcSin[c*x]^2*Log[1 + E^(I*ArcSin[c*x])] + (2*I)*A
rcSin[c*x]*PolyLog[2, -E^(I*ArcSin[c*x])] - (2*I)*ArcSin[c*x]*PolyLog[2, E^(I*ArcSin[c*x])] - 2*PolyLog[3, -E^
(I*ArcSin[c*x])] + 2*PolyLog[3, E^(I*ArcSin[c*x])]))/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{x \sqrt {c d x +d}\, \sqrt {-c e x +e}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{x\,\sqrt {d+c\,d\,x}\,\sqrt {e-c\,e\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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