3.6.90 \(\int \frac {(a+b \text {ArcSin}(c x))^2}{x^2 \sqrt {d+c d x} \sqrt {e-c e x}} \, dx\) [590]

Optimal. Leaf size=214 \[ -\frac {i c \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))^2}{x \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b c \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {i b^2 c \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}} \]

[Out]

-(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/x/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-I*c*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2
)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+2*b*c*(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/
2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-I*b^2*c*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^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 = 0.42, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4823, 4771, 4721, 3798, 2221, 2317, 2438} \begin {gather*} -\frac {\left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))^2}{x \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {i c \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{\sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 b c \sqrt {1-c^2 x^2} \log \left (1-e^{2 i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{\sqrt {c d x+d} \sqrt {e-c e x}}-\frac {i b^2 c \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{2 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^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]),x]

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x))))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4721

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n*Cot[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 4771

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/(d*f*(m + 1))), x] - Dist[b*c*(n/(f*(m + 1)))*Simp[(d
+ e*x^2)^p/(1 - c^2*x^2)^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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x^2 \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^2 \sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=-\frac {\left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (2 b c \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{x} \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=-\frac {\left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (2 b c \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int (a+b x) \cot (x) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=-\frac {i c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (4 i b c \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=-\frac {i c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 b^2 c \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=-\frac {i c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (i b^2 c \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=-\frac {i c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {i b^2 c \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{2 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 = 0.70, size = 189, normalized size = 0.88 \begin {gather*} \frac {b^2 \left (-1+c^2 x^2-i c x \sqrt {1-c^2 x^2}\right ) \text {ArcSin}(c x)^2+2 b \text {ArcSin}(c x) \left (-a+a c^2 x^2+b c x \sqrt {1-c^2 x^2} \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )\right )+a \left (-a+a c^2 x^2+2 b c x \sqrt {1-c^2 x^2} \log (c x)\right )-i b^2 c x \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c x)}\right )}{x \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^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]),x]

[Out]

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

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Maple [F]
time = 1.36, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{x^{2} \sqrt {c d x +d}\, \sqrt {-c e x +e}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{x^{2} \sqrt {d \left (c x + 1\right )} \sqrt {- e \left (c x - 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2/(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^2), 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^2\,\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^2*(d + c*d*x)^(1/2)*(e - c*e*x)^(1/2)),x)

[Out]

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

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