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

Optimal. Leaf size=177 \[ \frac {2 a b x \sqrt {1-c^2 x^2}}{c \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {\left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 b^2 \left (1-c^2 x^2\right )}{c^2 \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c \sqrt {c d x+d} \sqrt {e-c e x}} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4739, 4677, 4619, 261} \[ \frac {2 a b x \sqrt {1-c^2 x^2}}{c \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {\left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 b^2 \left (1-c^2 x^2\right )}{c^2 \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c \sqrt {c d x+d} \sqrt {e-c e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*x*Sqrt[1 - c^2*x^2])/(c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (2*b^2*(1 - c^2*x^2))/(c^2*Sqrt[d + c*d*x]*S
qrt[e - c*e*x]) + (2*b^2*x*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - ((1 - c^2*x^2)
*(a + b*ArcSin[c*x])^2)/(c^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 4619

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSin[c*x])^n, x] - Dist[b*c*n, Int[
(x*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[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}{\sqrt {d+c d x} \sqrt {e-c e x}} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {x \left (a+b \sin ^{-1}(c x)\right )^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}{c^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int \left (a+b \sin ^{-1}(c x)\right ) \, dx}{c \sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {2 a b x \sqrt {1-c^2 x^2}}{c \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}{c^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \int \sin ^{-1}(c x) \, dx}{c \sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {2 a b x \sqrt {1-c^2 x^2}}{c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c \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}{c^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \int \frac {x}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {2 a b x \sqrt {1-c^2 x^2}}{c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b^2 \left (1-c^2 x^2\right )}{c^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c \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}{c^2 \sqrt {d+c d x} \sqrt {e-c e x}}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 150, normalized size = 0.85 \[ -\frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (a^2 \left (c^2 x^2-1\right )+2 a b c x \sqrt {1-c^2 x^2}+2 b \sin ^{-1}(c x) \left (a \left (c^2 x^2-1\right )+b c x \sqrt {1-c^2 x^2}\right )-2 b^2 \left (c^2 x^2-1\right )+b^2 \left (c^2 x^2-1\right ) \sin ^{-1}(c x)^2\right )}{c^2 d e (c x-1) (c x+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(2*a*b*c*x*Sqrt[1 - c^2*x^2] + a^2*(-1 + c^2*x^2) - 2*b^2*(-1 + c^2*x^2) +
2*b*(b*c*x*Sqrt[1 - c^2*x^2] + a*(-1 + c^2*x^2))*ArcSin[c*x] + b^2*(-1 + c^2*x^2)*ArcSin[c*x]^2))/(c^2*d*e*(-1
+ c*x)*(1 + c*x)))

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fricas [A]  time = 0.53, size = 138, normalized size = 0.78 \[ -\frac {{\left ({\left (a^{2} - 2 \, b^{2}\right )} c^{2} x^{2} + {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arcsin \left (c x\right )^{2} - a^{2} + 2 \, b^{2} + 2 \, {\left (a b c^{2} x^{2} - a b\right )} \arcsin \left (c x\right ) + 2 \, {\left (b^{2} c x \arcsin \left (c x\right ) + a b c x\right )} \sqrt {-c^{2} x^{2} + 1}\right )} \sqrt {c d x + d} \sqrt {-c e x + e}}{c^{4} d e x^{2} - c^{2} d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.80, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {c d x +d}\, \sqrt {-c e x +e}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 157, normalized size = 0.89 \[ 2 \, b^{2} {\left (\frac {x \arcsin \left (c x\right )}{c \sqrt {d} \sqrt {e}} + \frac {\sqrt {-c^{2} x^{2} + 1}}{c^{2} \sqrt {d} \sqrt {e}}\right )} + \frac {2 \, a b x}{c \sqrt {d} \sqrt {e}} - \frac {\sqrt {-c^{2} d e x^{2} + d e} b^{2} \arcsin \left (c x\right )^{2}}{c^{2} d e} - \frac {2 \, \sqrt {-c^{2} d e x^{2} + d e} a b \arcsin \left (c x\right )}{c^{2} d e} - \frac {\sqrt {-c^{2} d e x^{2} + d e} a^{2}}{c^{2} d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{\sqrt {d+c\,d\,x}\,\sqrt {e-c\,e\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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