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

Optimal. Leaf size=250 \[ -\frac {x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c^3 \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {b^2 x \left (1-c^2 x^2\right )}{4 c^2 \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{4 c^3 \sqrt {c d x+d} \sqrt {e-c e x}} \]

[Out]

1/4*b^2*x*(-c^2*x^2+1)/c^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/2*x*(-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)-1/4*b^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)/c^3/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/2*b*x^
2*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/6*(a+b*arcsin(c*x))^3*(-c^2*x^2+1)
^(1/2)/b/c^3/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)

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Rubi [A]  time = 0.58, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4739, 4707, 4641, 4627, 321, 216} \[ \frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c^3 \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {b^2 x \left (1-c^2 x^2\right )}{4 c^2 \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{4 c^3 \sqrt {c d x+d} \sqrt {e-c e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x*(1 - c^2*x^2))/(4*c^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - (b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(4*c^3*Sqr
t[d + c*d*x]*Sqrt[e - c*e*x]) + (b*x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(2*c*Sqrt[d + c*d*x]*Sqrt[e - c*
e*x]) - (x*(1 - c^2*x^2)*(a + b*ArcSin[c*x])^2)/(2*c^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (Sqrt[1 - c^2*x^2]*(
a + b*ArcSin[c*x])^3)/(6*b*c^3*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4627

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

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 4707

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

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^2 \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^2 \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 {x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\sqrt {1-c^2 x^2} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}} \, dx}{2 c^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int x \left (a+b \sin ^{-1}(c x)\right ) \, dx}{c \sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c^3 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (b^2 \sqrt {1-c^2 x^2}\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {b^2 x \left (1-c^2 x^2\right )}{4 c^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c^3 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (b^2 \sqrt {1-c^2 x^2}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 c^2 \sqrt {d+c d x} \sqrt {e-c e x}}\\ &=\frac {b^2 x \left (1-c^2 x^2\right )}{4 c^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{4 c^3 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c^3 \sqrt {d+c d x} \sqrt {e-c e x}}\\ \end {align*}

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Mathematica [A]  time = 1.31, size = 326, normalized size = 1.30 \[ \frac {-3 \sqrt {d} \sqrt {e} \left (a^2 \left (4 c x-4 c^3 x^3\right )+a b \sqrt {1-c^2 x^2}+a b \cos \left (3 \sin ^{-1}(c x)\right )+2 b^2 c x \left (c^2 x^2-1\right )\right )-12 a^2 \sqrt {c d x+d} \sqrt {e-c e x} \tan ^{-1}\left (\frac {c x \sqrt {c d x+d} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (c^2 x^2-1\right )}\right )+12 b \sqrt {d} \sqrt {e} \sin ^{-1}(c x)^2 \left (a \sqrt {1-c^2 x^2}+b c x \left (c^2 x^2-1\right )\right )-3 b \sqrt {d} \sqrt {e} \sin ^{-1}(c x) \left (2 a c x+2 a \sin \left (3 \sin ^{-1}(c x)\right )+b \sqrt {1-c^2 x^2}+b \cos \left (3 \sin ^{-1}(c x)\right )\right )+4 b^2 \sqrt {d} \sqrt {e} \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^3}{24 c^3 \sqrt {d} \sqrt {e} \sqrt {c d x+d} \sqrt {e-c e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*b*Sqrt[d]*Sqrt[e]*(a*Sqrt[1 - c^2*x^2] + b*c*x*(-1 + c^2*x^2))*ArcSin[c*x]^2 + 4*b^2*Sqrt[d]*Sqrt[e]*Sqrt[
1 - c^2*x^2]*ArcSin[c*x]^3 - 12*a^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x
])/(Sqrt[d]*Sqrt[e]*(-1 + c^2*x^2))] - 3*Sqrt[d]*Sqrt[e]*(a*b*Sqrt[1 - c^2*x^2] + 2*b^2*c*x*(-1 + c^2*x^2) + a
^2*(4*c*x - 4*c^3*x^3) + a*b*Cos[3*ArcSin[c*x]]) - 3*b*Sqrt[d]*Sqrt[e]*ArcSin[c*x]*(2*a*c*x + b*Sqrt[1 - c^2*x
^2] + b*Cos[3*ArcSin[c*x]] + 2*a*Sin[3*ArcSin[c*x]]))/(24*c^3*Sqrt[d]*Sqrt[e]*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b^{2} x^{2} \arcsin \left (c x\right )^{2} + 2 \, a b x^{2} \arcsin \left (c x\right ) + a^{2} x^{2}\right )} \sqrt {c d x + d} \sqrt {-c e x + e}}{c^{2} d e x^{2} - d e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [F]  time = 0.67, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {c d x +d}\, \sqrt {-c e x +e}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*(a+b*arcsin(c*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 \[ -\frac {1}{2} \, a^{2} {\left (\frac {\sqrt {-c^{2} d e x^{2} + d e} x}{c^{2} d e} - \frac {\arcsin \left (c x\right )}{\sqrt {d e} c^{3}}\right )} - \sqrt {d} \sqrt {e} \int \frac {{\left (b^{2} x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, a b x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{2} d e x^{2} - d e}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^2*(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 \[ \int \frac {x^{2} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\sqrt {d \left (c x + 1\right )} \sqrt {- e \left (c x - 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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