3.132 \(\int \frac{x^3 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=150 \[ -\frac{a+b \sin ^{-1}(c x)}{c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{a+b \sin ^{-1}(c x)}{3 c^4 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b x \sqrt{d-c^2 d x^2}}{6 c^3 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{5 b \sqrt{d-c^2 d x^2} \tanh ^{-1}(c x)}{6 c^4 d^3 \sqrt{1-c^2 x^2}} \]

[Out]

-(b*x*Sqrt[d - c^2*d*x^2])/(6*c^3*d^3*(1 - c^2*x^2)^(3/2)) + (a + b*ArcSin[c*x])/(3*c^4*d*(d - c^2*d*x^2)^(3/2
)) - (a + b*ArcSin[c*x])/(c^4*d^2*Sqrt[d - c^2*d*x^2]) + (5*b*Sqrt[d - c^2*d*x^2]*ArcTanh[c*x])/(6*c^4*d^3*Sqr
t[1 - c^2*x^2])

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Rubi [A]  time = 0.197092, antiderivative size = 155, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {4703, 4677, 206, 288} \[ -\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b x}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{5 b \sqrt{1-c^2 x^2} \tanh ^{-1}(c x)}{6 c^4 d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

-(b*x)/(6*c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (x^2*(a + b*ArcSin[c*x]))/(3*c^2*d*(d - c^2*d*x^2)^
(3/2)) - (2*(a + b*ArcSin[c*x]))/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) + (5*b*Sqrt[1 - c^2*x^2]*ArcTanh[c*x])/(6*c^4
*d^2*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 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])

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 \int \frac{x \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{3 c^2 d}-\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int \frac{x^2}{\left (1-c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b x}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int \frac{1}{1-c^2 x^2} \, dx}{6 c^3 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (2 b \sqrt{1-c^2 x^2}\right ) \int \frac{1}{1-c^2 x^2} \, dx}{3 c^3 d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b x}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{5 b \sqrt{1-c^2 x^2} \tanh ^{-1}(c x)}{6 c^4 d^2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.22869, size = 143, normalized size = 0.95 \[ \frac{\sqrt{d-c^2 d x^2} \left (\sqrt{-c^2} \left (6 a c^2 x^2-4 a-b c x \sqrt{1-c^2 x^2}+2 b \left (3 c^2 x^2-2\right ) \sin ^{-1}(c x)\right )-5 i b c \left (1-c^2 x^2\right )^{3/2} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-c^2} x\right ),1\right )\right )}{6 c^4 \sqrt{-c^2} d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(Sqrt[d - c^2*d*x^2]*(Sqrt[-c^2]*(-4*a + 6*a*c^2*x^2 - b*c*x*Sqrt[1 - c^2*x^2] + 2*b*(-2 + 3*c^2*x^2)*ArcSin[c
*x]) - (5*I)*b*c*(1 - c^2*x^2)^(3/2)*EllipticF[I*ArcSinh[Sqrt[-c^2]*x], 1]))/(6*c^4*Sqrt[-c^2]*d^3*(-1 + c^2*x
^2)^2)

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Maple [C]  time = 0.218, size = 307, normalized size = 2.1 \begin{align*}{\frac{a{x}^{2}}{{c}^{2}d} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,a}{3\,d{c}^{4}} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}+{\frac{b\arcsin \left ( cx \right ){x}^{2}}{{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) ^{2}{c}^{2}}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{bx}{6\,{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) ^{2}{c}^{3}}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{2\,b\arcsin \left ( cx \right ) }{3\,{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) ^{2}{c}^{4}}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{5\,b}{6\,{c}^{4}{d}^{3} \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{5\,b}{6\,{c}^{4}{d}^{3} \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)^(5/2),x)

[Out]

a*x^2/c^2/d/(-c^2*d*x^2+d)^(3/2)-2/3*a/d/c^4/(-c^2*d*x^2+d)^(3/2)+b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c
^2*arcsin(c*x)*x^2-1/6*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^3*(-c^2*x^2+1)^(1/2)*x-2/3*b*(-d*(c^2*x^2-
1))^(1/2)/d^3/(c^2*x^2-1)^2/c^4*arcsin(c*x)-5/6*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^4/d^3/(c^2*x^2-1
)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)+5/6*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^4/d^3/(c^2*x^2-1)*ln(I*c*x+
(-c^2*x^2+1)^(1/2)-I)

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Maxima [A]  time = 1.62098, size = 216, normalized size = 1.44 \begin{align*} \frac{1}{12} \, b c{\left (\frac{2 \, x}{c^{6} d^{\frac{5}{2}} x^{2} - c^{4} d^{\frac{5}{2}}} + \frac{5 \, \log \left (c x + 1\right )}{c^{5} d^{\frac{5}{2}}} - \frac{5 \, \log \left (c x - 1\right )}{c^{5} d^{\frac{5}{2}}}\right )} + \frac{1}{3} \, b{\left (\frac{3 \, x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d} - \frac{2}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{4} d}\right )} \arcsin \left (c x\right ) + \frac{1}{3} \, a{\left (\frac{3 \, x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d} - \frac{2}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{4} d}\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)^(5/2),x, algorithm="maxima")

[Out]

1/12*b*c*(2*x/(c^6*d^(5/2)*x^2 - c^4*d^(5/2)) + 5*log(c*x + 1)/(c^5*d^(5/2)) - 5*log(c*x - 1)/(c^5*d^(5/2))) +
1/3*b*(3*x^2/((-c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d))*arcsin(c*x) + 1/3*a*(3*x^2/((
-c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d))

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Fricas [A]  time = 2.29991, size = 913, normalized size = 6.09 \begin{align*} \left [-\frac{4 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} b c x - 5 \,{\left (b c^{4} x^{4} - 2 \, 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 ) - 8 \,{\left (3 \, a c^{2} x^{2} +{\left (3 \, b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt{-c^{2} d x^{2} + d}}{24 \,{\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}}, -\frac{2 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} b c x - 5 \,{\left (b c^{4} x^{4} - 2 \, 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 ) - 4 \,{\left (3 \, a c^{2} x^{2} +{\left (3 \, b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt{-c^{2} d x^{2} + d}}{12 \,{\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

[-1/24*(4*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x - 5*(b*c^4*x^4 - 2*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)) - 8*(3*a*c^2*x^2 + (3*b*c^2*x^2 - 2*b)*arcsin(c*x) - 2*a)*sqrt(-c^2*d*x^2 +
d))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3), -1/12*(2*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x - 5*(b*c^4
*x^4 - 2*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)
) - 4*(3*a*c^2*x^2 + (3*b*c^2*x^2 - 2*b)*arcsin(c*x) - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2
+ c^4*d^3)]

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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)**(5/2),x)

[Out]

Exception raised: TypeError

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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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)*x^3/(-c^2*d*x^2 + d)^(5/2), x)