3.121 \(\int \frac{x^3 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=142 \[ \frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2}+\frac{a+b \sin ^{-1}(c x)}{c^4 d \sqrt{d-c^2 d x^2}}-\frac{b x \sqrt{d-c^2 d x^2}}{c^3 d^2 \sqrt{1-c^2 x^2}}-\frac{b \sqrt{d-c^2 d x^2} \tanh ^{-1}(c x)}{c^4 d^2 \sqrt{1-c^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.180346, antiderivative size = 146, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {4703, 4677, 8, 321, 206} \[ \frac{2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}}-\frac{b x \sqrt{1-c^2 x^2}}{c^3 d \sqrt{d-c^2 d x^2}}-\frac{b \sqrt{1-c^2 x^2} \tanh ^{-1}(c x)}{c^4 d \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(3/2),x]

[Out]

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

Rule 4703

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p +
1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*f*n*d^IntPart[p]*(d + e*x^2
)^FracPart[p])/(2*c*(p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSi
n[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 1]

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 206

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

Rubi steps

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

Mathematica [C]  time = 0.2074, size = 136, normalized size = 0.96 \[ \frac{\sqrt{d-c^2 d x^2} \left (\sqrt{-c^2} \left (a c^2 x^2-2 a+b c x \sqrt{1-c^2 x^2}+b \left (c^2 x^2-2\right ) \sin ^{-1}(c x)\right )-i b c \sqrt{1-c^2 x^2} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-c^2} x\right ),1\right )\right )}{c^4 \sqrt{-c^2} d^2 \left (c^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(3/2),x]

[Out]

(Sqrt[d - c^2*d*x^2]*(Sqrt[-c^2]*(-2*a + a*c^2*x^2 + b*c*x*Sqrt[1 - c^2*x^2] + b*(-2 + c^2*x^2)*ArcSin[c*x]) -
I*b*c*Sqrt[1 - c^2*x^2]*EllipticF[I*ArcSinh[Sqrt[-c^2]*x], 1]))/(c^4*Sqrt[-c^2]*d^2*(-1 + c^2*x^2))

________________________________________________________________________________________

Maple [C]  time = 0.214, size = 306, normalized size = 2.2 \begin{align*} -{\frac{a{x}^{2}}{{c}^{2}d}{\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}}+2\,{\frac{a}{d{c}^{4}\sqrt{-{c}^{2}d{x}^{2}+d}}}+{\frac{bx}{{c}^{3}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{b\arcsin \left ( cx \right ){x}^{2}}{{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-2\,{\frac{b\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\arcsin \left ( cx \right ) }{{d}^{2}{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{b}{{d}^{2}{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1}+i \right ) }-{\frac{b}{{d}^{2}{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1}-i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.11859, size = 810, normalized size = 5.7 \begin{align*} \left [\frac{4 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} b c x +{\left (b c^{2} x^{2} - b\right )} \sqrt{d} \log \left (-\frac{c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} + 4 \,{\left (c^{3} x^{3} + c x\right )} \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} \sqrt{d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) + 4 \,{\left (a c^{2} x^{2} +{\left (b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt{-c^{2} d x^{2} + d}}{4 \,{\left (c^{6} d^{2} x^{2} - c^{4} d^{2}\right )}}, \frac{2 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} b c x -{\left (b c^{2} x^{2} - b\right )} \sqrt{-d} \arctan \left (\frac{2 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} c \sqrt{-d} x}{c^{4} d x^{4} - d}\right ) + 2 \,{\left (a c^{2} x^{2} +{\left (b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt{-c^{2} d x^{2} + d}}{2 \,{\left (c^{6} d^{2} x^{2} - c^{4} d^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(4*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x + (b*c^2*x^2 - b)*sqrt(d)*log(-(c^6*d*x^6 + 5*c^4*d*x^4
- 5*c^2*d*x^2 + 4*(c^3*x^3 + c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*sqrt(d) - d)/(c^6*x^6 - 3*c^4*x^4 +
3*c^2*x^2 - 1)) + 4*(a*c^2*x^2 + (b*c^2*x^2 - 2*b)*arcsin(c*x) - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^6*d^2*x^2 - c^4
*d^2), 1/2*(2*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x - (b*c^2*x^2 - b)*sqrt(-d)*arctan(2*sqrt(-c^2*d*x^
2 + d)*sqrt(-c^2*x^2 + 1)*c*sqrt(-d)*x/(c^4*d*x^4 - d)) + 2*(a*c^2*x^2 + (b*c^2*x^2 - 2*b)*arcsin(c*x) - 2*a)*
sqrt(-c^2*d*x^2 + d))/(c^6*d^2*x^2 - c^4*d^2)]

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(3/2),x)

[Out]

Exception raised: TypeError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (c x\right ) + a\right )} x^{3}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="giac")

[Out]

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