3.140 \(\int \frac{\sin ^{-1}(a x)}{(c-a^2 c x^2)^{7/2}} \, dx\)

Optimal. Leaf size=210 \[ -\frac{2}{15 a c^3 \sqrt{1-a^2 x^2} \sqrt{c-a^2 c x^2}}-\frac{1}{20 a c^3 \left (1-a^2 x^2\right )^{3/2} \sqrt{c-a^2 c x^2}}+\frac{4 \sqrt{1-a^2 x^2} \log \left (1-a^2 x^2\right )}{15 a c^3 \sqrt{c-a^2 c x^2}}+\frac{8 x \sin ^{-1}(a x)}{15 c^3 \sqrt{c-a^2 c x^2}}+\frac{4 x \sin ^{-1}(a x)}{15 c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{x \sin ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

-1/(20*a*c^3*(1 - a^2*x^2)^(3/2)*Sqrt[c - a^2*c*x^2]) - 2/(15*a*c^3*Sqrt[1 - a^2*x^2]*Sqrt[c - a^2*c*x^2]) + (
x*ArcSin[a*x])/(5*c*(c - a^2*c*x^2)^(5/2)) + (4*x*ArcSin[a*x])/(15*c^2*(c - a^2*c*x^2)^(3/2)) + (8*x*ArcSin[a*
x])/(15*c^3*Sqrt[c - a^2*c*x^2]) + (4*Sqrt[1 - a^2*x^2]*Log[1 - a^2*x^2])/(15*a*c^3*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.115313, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4655, 4653, 260, 261} \[ -\frac{2}{15 a c^3 \sqrt{1-a^2 x^2} \sqrt{c-a^2 c x^2}}-\frac{1}{20 a c^3 \left (1-a^2 x^2\right )^{3/2} \sqrt{c-a^2 c x^2}}+\frac{4 \sqrt{1-a^2 x^2} \log \left (1-a^2 x^2\right )}{15 a c^3 \sqrt{c-a^2 c x^2}}+\frac{8 x \sin ^{-1}(a x)}{15 c^3 \sqrt{c-a^2 c x^2}}+\frac{4 x \sin ^{-1}(a x)}{15 c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{x \sin ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]/(c - a^2*c*x^2)^(7/2),x]

[Out]

-1/(20*a*c^3*(1 - a^2*x^2)^(3/2)*Sqrt[c - a^2*c*x^2]) - 2/(15*a*c^3*Sqrt[1 - a^2*x^2]*Sqrt[c - a^2*c*x^2]) + (
x*ArcSin[a*x])/(5*c*(c - a^2*c*x^2)^(5/2)) + (4*x*ArcSin[a*x])/(15*c^2*(c - a^2*c*x^2)^(3/2)) + (8*x*ArcSin[a*
x])/(15*c^3*Sqrt[c - a^2*c*x^2]) + (4*Sqrt[1 - a^2*x^2]*Log[1 - a^2*x^2])/(15*a*c^3*Sqrt[c - a^2*c*x^2])

Rule 4655

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSin[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a + b*
ArcSin[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 - c^2*x^2)^FracPart[p
]), Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 4653

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 - c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSin[c*x
])^(n - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 260

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

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]

Rubi steps

\begin{align*} \int \frac{\sin ^{-1}(a x)}{\left (c-a^2 c x^2\right )^{7/2}} \, dx &=\frac{x \sin ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}}+\frac{4 \int \frac{\sin ^{-1}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx}{5 c}-\frac{\left (a \sqrt{1-a^2 x^2}\right ) \int \frac{x}{\left (1-a^2 x^2\right )^3} \, dx}{5 c^3 \sqrt{c-a^2 c x^2}}\\ &=-\frac{1}{20 a c^3 \left (1-a^2 x^2\right )^{3/2} \sqrt{c-a^2 c x^2}}+\frac{x \sin ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}}+\frac{4 x \sin ^{-1}(a x)}{15 c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{8 \int \frac{\sin ^{-1}(a x)}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{15 c^2}-\frac{\left (4 a \sqrt{1-a^2 x^2}\right ) \int \frac{x}{\left (1-a^2 x^2\right )^2} \, dx}{15 c^3 \sqrt{c-a^2 c x^2}}\\ &=-\frac{1}{20 a c^3 \left (1-a^2 x^2\right )^{3/2} \sqrt{c-a^2 c x^2}}-\frac{2}{15 a c^3 \sqrt{1-a^2 x^2} \sqrt{c-a^2 c x^2}}+\frac{x \sin ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}}+\frac{4 x \sin ^{-1}(a x)}{15 c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{8 x \sin ^{-1}(a x)}{15 c^3 \sqrt{c-a^2 c x^2}}-\frac{\left (8 a \sqrt{1-a^2 x^2}\right ) \int \frac{x}{1-a^2 x^2} \, dx}{15 c^3 \sqrt{c-a^2 c x^2}}\\ &=-\frac{1}{20 a c^3 \left (1-a^2 x^2\right )^{3/2} \sqrt{c-a^2 c x^2}}-\frac{2}{15 a c^3 \sqrt{1-a^2 x^2} \sqrt{c-a^2 c x^2}}+\frac{x \sin ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}}+\frac{4 x \sin ^{-1}(a x)}{15 c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{8 x \sin ^{-1}(a x)}{15 c^3 \sqrt{c-a^2 c x^2}}+\frac{4 \sqrt{1-a^2 x^2} \log \left (1-a^2 x^2\right )}{15 a c^3 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.216464, size = 111, normalized size = 0.53 \[ -\frac{\sqrt{c-a^2 c x^2} \left (\sqrt{1-a^2 x^2} \left (8 a^2 x^2+16 \left (a^2 x^2-1\right )^2 \log \left (a^2 x^2-1\right )-11\right )+4 a x \left (8 a^4 x^4-20 a^2 x^2+15\right ) \sin ^{-1}(a x)\right )}{60 a c^4 \left (a^2 x^2-1\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a*x]/(c - a^2*c*x^2)^(7/2),x]

[Out]

-(Sqrt[c - a^2*c*x^2]*(4*a*x*(15 - 20*a^2*x^2 + 8*a^4*x^4)*ArcSin[a*x] + Sqrt[1 - a^2*x^2]*(-11 + 8*a^2*x^2 +
16*(-1 + a^2*x^2)^2*Log[-1 + a^2*x^2])))/(60*a*c^4*(-1 + a^2*x^2)^3)

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Maple [C]  time = 0.217, size = 409, normalized size = 2. \begin{align*}{\frac{{\frac{16\,i}{15}}\arcsin \left ( ax \right ) }{a{c}^{4} \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{1}{60\,{c}^{4} \left ( 40\,{a}^{10}{x}^{10}-215\,{x}^{8}{a}^{8}+469\,{a}^{6}{x}^{6}-517\,{a}^{4}{x}^{4}+287\,{a}^{2}{x}^{2}-64 \right ) a}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) } \left ( 8\,{a}^{5}{x}^{5}-20\,{a}^{3}{x}^{3}+8\,i\sqrt{-{a}^{2}{x}^{2}+1}{x}^{4}{a}^{4}+15\,ax-16\,i\sqrt{-{a}^{2}{x}^{2}+1}{x}^{2}{a}^{2}+8\,i\sqrt{-{a}^{2}{x}^{2}+1} \right ) \left ( 64\,i{x}^{8}{a}^{8}+64\,\sqrt{-{a}^{2}{x}^{2}+1}{x}^{7}{a}^{7}-280\,i{x}^{6}{a}^{6}-248\,\sqrt{-{a}^{2}{x}^{2}+1}{a}^{5}{x}^{5}+160\,{a}^{4}{x}^{4}\arcsin \left ( ax \right ) +456\,i{x}^{4}{a}^{4}+340\,{a}^{3}{x}^{3}\sqrt{-{a}^{2}{x}^{2}+1}-380\,{a}^{2}{x}^{2}\arcsin \left ( ax \right ) -328\,i{a}^{2}{x}^{2}-165\,ax\sqrt{-{a}^{2}{x}^{2}+1}+256\,\arcsin \left ( ax \right ) +88\,i \right ) }-{\frac{8}{15\,a{c}^{4} \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}\ln \left ( 1+ \left ( iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(a*x)/(-a^2*c*x^2+c)^(7/2),x)

[Out]

16/15*I*(-c*(a^2*x^2-1))^(1/2)*(-a^2*x^2+1)^(1/2)/a/c^4/(a^2*x^2-1)*arcsin(a*x)-1/60*(-c*(a^2*x^2-1))^(1/2)*(8
*a^5*x^5-20*a^3*x^3+8*I*(-a^2*x^2+1)^(1/2)*x^4*a^4+15*a*x-16*I*(-a^2*x^2+1)^(1/2)*x^2*a^2+8*I*(-a^2*x^2+1)^(1/
2))*(64*I*x^8*a^8+64*(-a^2*x^2+1)^(1/2)*x^7*a^7-280*I*x^6*a^6-248*(-a^2*x^2+1)^(1/2)*a^5*x^5+160*a^4*x^4*arcsi
n(a*x)+456*I*x^4*a^4+340*a^3*x^3*(-a^2*x^2+1)^(1/2)-380*a^2*x^2*arcsin(a*x)-328*I*a^2*x^2-165*a*x*(-a^2*x^2+1)
^(1/2)+256*arcsin(a*x)+88*I)/c^4/(40*a^10*x^10-215*a^8*x^8+469*a^6*x^6-517*a^4*x^4+287*a^2*x^2-64)/a-8/15*(-c*
(a^2*x^2-1))^(1/2)*(-a^2*x^2+1)^(1/2)/a/c^4/(a^2*x^2-1)*ln(1+(I*a*x+(-a^2*x^2+1)^(1/2))^2)

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Maxima [A]  time = 1.70594, size = 209, normalized size = 1. \begin{align*} -\frac{1}{60} \, a{\left (\frac{16 \, \sqrt{\frac{1}{a^{4} c}} \log \left (x^{2} - \frac{1}{a^{2}}\right )}{c^{3}} + \frac{3}{{\left (a^{6} c^{\frac{5}{2}} x^{4} - 2 \, a^{4} c^{\frac{5}{2}} x^{2} + a^{2} c^{\frac{5}{2}}\right )} c} - \frac{8}{{\left (a^{4} c^{\frac{3}{2}} x^{2} - a^{2} c^{\frac{3}{2}}\right )} c^{2}}\right )} + \frac{1}{15} \,{\left (\frac{8 \, x}{\sqrt{-a^{2} c x^{2} + c} c^{3}} + \frac{4 \, x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}} c^{2}} + \frac{3 \, x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}} c}\right )} \arcsin \left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/(-a^2*c*x^2+c)^(7/2),x, algorithm="maxima")

[Out]

-1/60*a*(16*sqrt(1/(a^4*c))*log(x^2 - 1/a^2)/c^3 + 3/((a^6*c^(5/2)*x^4 - 2*a^4*c^(5/2)*x^2 + a^2*c^(5/2))*c) -
8/((a^4*c^(3/2)*x^2 - a^2*c^(3/2))*c^2)) + 1/15*(8*x/(sqrt(-a^2*c*x^2 + c)*c^3) + 4*x/((-a^2*c*x^2 + c)^(3/2)
*c^2) + 3*x/((-a^2*c*x^2 + c)^(5/2)*c))*arcsin(a*x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} c x^{2} + c} \arcsin \left (a x\right )}{a^{8} c^{4} x^{8} - 4 \, a^{6} c^{4} x^{6} + 6 \, a^{4} c^{4} x^{4} - 4 \, a^{2} c^{4} x^{2} + c^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/(-a^2*c*x^2+c)^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*c*x^2 + c)*arcsin(a*x)/(a^8*c^4*x^8 - 4*a^6*c^4*x^6 + 6*a^4*c^4*x^4 - 4*a^2*c^4*x^2 + c^4),
x)

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Sympy [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(asin(a*x)/(-a**2*c*x**2+c)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.43052, size = 173, normalized size = 0.82 \begin{align*} -\frac{1}{60} \, \sqrt{c}{\left (\frac{16 \, \log \left ({\left | a^{2} x^{2} - 1 \right |}\right )}{a c^{4}} - \frac{24 \, a^{4} x^{4} - 56 \, a^{2} x^{2} + 35}{{\left (a^{2} x^{2} - 1\right )}^{2} a c^{4}}\right )} - \frac{\sqrt{-a^{2} c x^{2} + c}{\left (4 \,{\left (\frac{2 \, a^{4} x^{2}}{c} - \frac{5 \, a^{2}}{c}\right )} x^{2} + \frac{15}{c}\right )} x \arcsin \left (a x\right )}{15 \,{\left (a^{2} c x^{2} - c\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/(-a^2*c*x^2+c)^(7/2),x, algorithm="giac")

[Out]

-1/60*sqrt(c)*(16*log(abs(a^2*x^2 - 1))/(a*c^4) - (24*a^4*x^4 - 56*a^2*x^2 + 35)/((a^2*x^2 - 1)^2*a*c^4)) - 1/
15*sqrt(-a^2*c*x^2 + c)*(4*(2*a^4*x^2/c - 5*a^2/c)*x^2 + 15/c)*x*arcsin(a*x)/(a^2*c*x^2 - c)^3