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

Optimal. Leaf size=257 \[ -\frac {c \sqrt {c d x+d} \sqrt {e-c e x} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b \sqrt {1-c^2 x^2}}-\frac {i c \sqrt {c d x+d} \sqrt {e-c e x} \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}}+\frac {2 b c \sqrt {c d x+d} \sqrt {e-c e x} \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (a+b \sin ^{-1}(c x)\right )^2}{x}-\frac {i b^2 c \sqrt {c d x+d} \sqrt {e-c e x} \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.59, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4739, 4693, 4625, 3717, 2190, 2279, 2391, 4641} \[ -\frac {i b^2 c \sqrt {c d x+d} \sqrt {e-c e x} \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {c \sqrt {c d x+d} \sqrt {e-c e x} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b \sqrt {1-c^2 x^2}}-\frac {i c \sqrt {c d x+d} \sqrt {e-c e x} \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}}+\frac {2 b c \sqrt {c d x+d} \sqrt {e-c e x} \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (a+b \sin ^{-1}(c x)\right )^2}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(a + b*ArcSin[c*x])^2)/x) - (I*c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(a + b*Arc
Sin[c*x])^2)/Sqrt[1 - c^2*x^2] - (c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(a + b*ArcSin[c*x])^3)/(3*b*Sqrt[1 - c^2*x
^2]) + (2*b*c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(a + b*ArcSin[c*x])*Log[1 - E^((2*I)*ArcSin[c*x])])/Sqrt[1 - c^2
*x^2] - (I*b^2*c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*PolyLog[2, E^((2*I)*ArcSin[c*x])])/Sqrt[1 - c^2*x^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 3717

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 4625

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

Rule 4641

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

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

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Mathematica [A]  time = 1.25, size = 374, normalized size = 1.46 \[ \frac {-3 a^2 \sqrt {1-c^2 x^2} \sqrt {c d x+d} \sqrt {e-c e x}+3 a^2 c \sqrt {d} \sqrt {e} x \sqrt {1-c^2 x^2} \tan ^{-1}\left (\frac {c x \sqrt {c d x+d} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (c^2 x^2-1\right )}\right )-3 i b \sqrt {c d x+d} \sqrt {e-c e x} \sin ^{-1}(c x)^2 \left (-i a c x-i b \sqrt {1-c^2 x^2}+b c x\right )+6 b \sqrt {c d x+d} \sqrt {e-c e x} \sin ^{-1}(c x) \left (-a \sqrt {1-c^2 x^2}+b c x \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )\right )+6 a b c x \sqrt {c d x+d} \sqrt {e-c e x} \log (c x)-3 i b^2 c x \sqrt {c d x+d} \sqrt {e-c e x} \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-b^2 c x \sqrt {c d x+d} \sqrt {e-c e x} \sin ^{-1}(c x)^3}{3 x \sqrt {1-c^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*a^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[1 - c^2*x^2] - (3*I)*b*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*((-I)*a*c*
x + b*c*x - I*b*Sqrt[1 - c^2*x^2])*ArcSin[c*x]^2 - b^2*c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 + 3*a
^2*c*Sqrt[d]*Sqrt[e]*x*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sqrt[d]*Sqrt[e]*(-1 + c
^2*x^2))] + 6*b*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]*(-(a*Sqrt[1 - c^2*x^2]) + b*c*x*Log[1 - E^((2*I)*A
rcSin[c*x])]) + 6*a*b*c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Log[c*x] - (3*I)*b^2*c*x*Sqrt[d + c*d*x]*Sqrt[e - c*
e*x]*PolyLog[2, E^((2*I)*ArcSin[c*x])])/(3*x*Sqrt[1 - c^2*x^2])

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \sqrt {c d x + d} \sqrt {-c e x + e}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c d x +d}\, \sqrt {-c e x +e}\, \left (a +b \arcsin \left (c x \right )\right )^{2}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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