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

Optimal. Leaf size=119 \[ \frac{a+b \sin ^{-1}(c x)}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b x}{6 c d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{b \sqrt{1-c^2 x^2} \tanh ^{-1}(c x)}{6 c^2 d^2 \sqrt{d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.080653, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4677, 199, 206} \[ \frac{a+b \sin ^{-1}(c x)}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b x}{6 c d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{b \sqrt{1-c^2 x^2} \tanh ^{-1}(c x)}{6 c^2 d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

Mathematica [A]  time = 0.0463611, size = 85, normalized size = 0.71 \[ \frac{-2 a+b c x \sqrt{1-c^2 x^2}+b \left (1-c^2 x^2\right )^{3/2} \tanh ^{-1}(c x)-2 b \sin ^{-1}(c x)}{6 c^2 d^2 \left (c^2 x^2-1\right ) \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.13, size = 259, normalized size = 2.2 \begin{align*}{\frac{a}{3\,{c}^{2}d} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}-{\frac{bx}{6\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) c}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{b\arcsin \left ( cx \right ) }{3\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ){c}^{2}}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{b}{6\,{c}^{2}{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{b}{6\,{c}^{2}{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*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x)

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.1984, size = 815, normalized size = 6.85 \begin{align*} \left [-\frac{4 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} b c x -{\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 \, \sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}}{24 \,{\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}}, -\frac{2 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} b c x +{\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 \, \sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}}{12 \,{\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(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 - (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*sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a))/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c
^2*d^3), -1/12*(2*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x + (b*c^4*x^4 - 2*b*c^2*x^2 + b)*sqrt(-d)*arcta
n(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*sqrt(-c^2*d*x^2 + d)*(b*arcsin(c
*x) + a))/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (a + b \operatorname{asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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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}{{\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*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

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