3.139 \(\int \frac{a+b \sin ^{-1}(c x)}{x^4 (d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=310 \[ \frac{16 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac{b c^3 \sqrt{d-c^2 d x^2}}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{b c \sqrt{d-c^2 d x^2}}{6 d^3 x^2 \sqrt{1-c^2 x^2}}+\frac{8 b c^3 \log (x) \sqrt{d-c^2 d x^2}}{3 d^3 \sqrt{1-c^2 x^2}}+\frac{4 b c^3 \sqrt{d-c^2 d x^2} \log \left (1-c^2 x^2\right )}{3 d^3 \sqrt{1-c^2 x^2}} \]

[Out]

-(b*c^3*Sqrt[d - c^2*d*x^2])/(6*d^3*(1 - c^2*x^2)^(3/2)) - (b*c*Sqrt[d - c^2*d*x^2])/(6*d^3*x^2*Sqrt[1 - c^2*x
^2]) - (a + b*ArcSin[c*x])/(3*d*x^3*(d - c^2*d*x^2)^(3/2)) - (2*c^2*(a + b*ArcSin[c*x]))/(d*x*(d - c^2*d*x^2)^
(3/2)) + (8*c^4*x*(a + b*ArcSin[c*x]))/(3*d*(d - c^2*d*x^2)^(3/2)) + (16*c^4*x*(a + b*ArcSin[c*x]))/(3*d^2*Sqr
t[d - c^2*d*x^2]) + (8*b*c^3*Sqrt[d - c^2*d*x^2]*Log[x])/(3*d^3*Sqrt[1 - c^2*x^2]) + (4*b*c^3*Sqrt[d - c^2*d*x
^2]*Log[1 - c^2*x^2])/(3*d^3*Sqrt[1 - c^2*x^2])

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Rubi [A]  time = 0.387039, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {4701, 4655, 4653, 260, 261, 266, 44} \[ \frac{16 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac{b c^3}{6 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{6 d^2 x^2 \sqrt{d-c^2 d x^2}}+\frac{8 b c^3 \sqrt{1-c^2 x^2} \log (x)}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{4 b c^3 \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(x^4*(d - c^2*d*x^2)^(5/2)),x]

[Out]

-(b*c^3)/(6*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) - (b*c*Sqrt[1 - c^2*x^2])/(6*d^2*x^2*Sqrt[d - c^2*d*x^2
]) - (a + b*ArcSin[c*x])/(3*d*x^3*(d - c^2*d*x^2)^(3/2)) - (2*c^2*(a + b*ArcSin[c*x]))/(d*x*(d - c^2*d*x^2)^(3
/2)) + (8*c^4*x*(a + b*ArcSin[c*x]))/(3*d*(d - c^2*d*x^2)^(3/2)) + (16*c^4*x*(a + b*ArcSin[c*x]))/(3*d^2*Sqrt[
d - c^2*d*x^2]) + (8*b*c^3*Sqrt[1 - c^2*x^2]*Log[x])/(3*d^2*Sqrt[d - c^2*d*x^2]) + (4*b*c^3*Sqrt[1 - c^2*x^2]*
Log[1 - c^2*x^2])/(3*d^2*Sqrt[d - c^2*d*x^2])

Rule 4701

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

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]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}(c x)}{x^4 \left (d-c^2 d x^2\right )^{5/2}} \, dx &=-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}+\left (2 c^2\right ) \int \frac{a+b \sin ^{-1}(c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx+\frac{\left (b c \sqrt{1-c^2 x^2}\right ) \int \frac{1}{x^3 \left (1-c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}+\left (8 c^4\right ) \int \frac{a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx+\frac{\left (b c \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-c^2 x\right )^2} \, dx,x,x^2\right )}{6 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (2 b c^3 \sqrt{1-c^2 x^2}\right ) \int \frac{1}{x \left (1-c^2 x^2\right )^2} \, dx}{d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac{8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac{\left (16 c^4\right ) \int \frac{a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{3 d}+\frac{\left (b c \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^2}+\frac{2 c^2}{x}+\frac{c^4}{\left (-1+c^2 x\right )^2}-\frac{2 c^4}{-1+c^2 x}\right ) \, dx,x,x^2\right )}{6 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (b c^3 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )^2} \, dx,x,x^2\right )}{d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (8 b c^5 \sqrt{1-c^2 x^2}\right ) \int \frac{x}{\left (1-c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{7 b c^3}{6 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{6 d^2 x^2 \sqrt{d-c^2 d x^2}}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac{8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac{16 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{2 b c^3 \sqrt{1-c^2 x^2} \log (x)}{3 d^2 \sqrt{d-c^2 d x^2}}-\frac{b c^3 \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (b c^3 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x}+\frac{c^2}{\left (-1+c^2 x\right )^2}-\frac{c^2}{-1+c^2 x}\right ) \, dx,x,x^2\right )}{d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (16 b c^5 \sqrt{1-c^2 x^2}\right ) \int \frac{x}{1-c^2 x^2} \, dx}{3 d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b c^3}{6 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{6 d^2 x^2 \sqrt{d-c^2 d x^2}}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac{8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac{16 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{8 b c^3 \sqrt{1-c^2 x^2} \log (x)}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{4 b c^3 \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 d^2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.345004, size = 213, normalized size = 0.69 \[ -\frac{\sqrt{d-c^2 d x^2} \left (32 a c^6 x^6-48 a c^4 x^4+12 a c^2 x^2+2 a+b c x \sqrt{1-c^2 x^2}+8 b c^5 x^5 \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )-8 b c^3 x^3 \left (1-c^2 x^2\right )^{3/2} \log \left (x^2\right )-8 b c^3 x^3 \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )+2 b \left (16 c^6 x^6-24 c^4 x^4+6 c^2 x^2+1\right ) \sin ^{-1}(c x)\right )}{6 d^3 x^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])/(x^4*(d - c^2*d*x^2)^(5/2)),x]

[Out]

-(Sqrt[d - c^2*d*x^2]*(2*a + 12*a*c^2*x^2 - 48*a*c^4*x^4 + 32*a*c^6*x^6 + b*c*x*Sqrt[1 - c^2*x^2] + 2*b*(1 + 6
*c^2*x^2 - 24*c^4*x^4 + 16*c^6*x^6)*ArcSin[c*x] - 8*b*c^3*x^3*(1 - c^2*x^2)^(3/2)*Log[x^2] - 8*b*c^3*x^3*Sqrt[
1 - c^2*x^2]*Log[1 - c^2*x^2] + 8*b*c^5*x^5*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2]))/(6*d^3*x^3*(-1 + c^2*x^2)^2)

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Maple [C]  time = 0.261, size = 1875, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^(5/2),x)

[Out]

2*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*c^3*(-c^2*x^2+1)^(1/2)+1/3*b*(-
d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3/x^3*arcsin(c*x)+8/3*a*c^4/d*x/(-c^2*d
*x^2+d)^(3/2)+16/3*a*c^4/d^2*x/(-c^2*d*x^2+d)^(1/2)-2*a*c^2/d/x/(-c^2*d*x^2+d)^(3/2)+6*b*(-d*(c^2*x^2-1))^(1/2
)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3/x*arcsin(c*x)*c^2+1/6*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x
^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3/x^2*(-c^2*x^2+1)^(1/2)*c-8/3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^
(1/2)/d^3/(c^2*x^2-1)*ln((I*c*x+(-c^2*x^2+1)^(1/2))^4-1)*c^3-64*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^
6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^7*arcsin(c*x)*c^10+160*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4
*x^4-10*c^2*x^2-1)/d^3*x^5*arcsin(c*x)*c^8-344/3*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10
*c^2*x^2-1)/d^3*x^3*arcsin(c*x)*c^6+8/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^
2-1)/d^3*x*c^4-2*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^2*c^5*(-c^2*x^
2+1)^(1/2)+12*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x*arcsin(c*x)*c^4+1
28/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^11*c^14-448/3*I*b*(-d*(c
^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^9*c^12+560/3*I*b*(-d*(c^2*x^2-1))^(1/2)
/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^7*c^10-280/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*
c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^5*c^8+32/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4
-10*c^2*x^2-1)/d^3*x^3*c^6+80*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x
^5*(-c^2*x^2+1)*c^8+32/3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/(c^2*x^2-1)*arcsin(c*x)*c^3-16/3*I*
b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^
3-40/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^3*(-c^2*x^2+1)*c^6-8/3
*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x*(-c^2*x^2+1)*c^4+128/3*I*b*(
-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^9*(-c^2*x^2+1)*c^12-320/3*I*b*(-d*
(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^7*(-c^2*x^2+1)*c^10-1/3*a/d/x^3/(-c^2
*d*x^2+d)^(3/2)+128*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^4*arcsin(
c*x)*(-c^2*x^2+1)^(1/2)*c^7-64*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*
x^6*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^9-176/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c
^2*x^2-1)/d^3*x^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^5

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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((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}}{c^{6} d^{3} x^{10} - 3 \, c^{4} d^{3} x^{8} + 3 \, c^{2} d^{3} x^{6} - d^{3} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^6*d^3*x^10 - 3*c^4*d^3*x^8 + 3*c^2*d^3*x^6 - d^3*x^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((a+b*asin(c*x))/x**4/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{5}{2}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)/((-c^2*d*x^2 + d)^(5/2)*x^4), x)