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

Optimal. Leaf size=268 \[ -\frac {15}{8} c^2 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {15 c d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b \sqrt {1-c^2 x^2}}-\frac {5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac {b c d^2 \log (x) \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}+\frac {9 b c^3 d^2 x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}} \]

[Out]

-5/4*c^2*d*x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))-(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x-15/8*c^2*d^2*x*(a
+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)+9/16*b*c^3*d^2*x^2*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-1/16*b*c^5*d^2
*x^4*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-15/16*c*d^2*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/b/(-c^2*x^2+
1)^(1/2)+b*c*d^2*ln(x)*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {4695, 4649, 4647, 4641, 30, 14, 266, 43} \[ -\frac {15}{8} c^2 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {15 c d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b \sqrt {1-c^2 x^2}}-\frac {5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {b c^5 d^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}+\frac {9 b c^3 d^2 x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}+\frac {b c d^2 \log (x) \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*b*c^3*d^2*x^2*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1 - c^2*x^2]) - (b*c^5*d^2*x^4*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1
- c^2*x^2]) - (15*c^2*d^2*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/8 - (5*c^2*d*x*(d - c^2*d*x^2)^(3/2)*(a +
b*ArcSin[c*x]))/4 - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/x - (15*c*d^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcS
in[c*x])^2)/(16*b*Sqrt[1 - c^2*x^2]) + (b*c*d^2*Sqrt[d - c^2*d*x^2]*Log[x])/Sqrt[1 - c^2*x^2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 4641

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

Rule 4647

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*(
a + b*ArcSin[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 - c^2*x^2]), Int[x*(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]

Rule 4649

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(x*(d + e*x^2)^p*(
a + b*ArcSin[c*x])^n)/(2*p + 1), x] + (Dist[(2*d*p)/(2*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] && Gt
Q[n, 0] && GtQ[p, 0]

Rule 4695

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*(a + b*ArcSin[c*x])^n)/(f*(m + 1)), x] + (-Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x)^
(m + 2)*(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])/
(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}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x^2} \, dx &=-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\left (5 c^2 d\right ) \int \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx+\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {\left (1-c^2 x^2\right )^2}{x} \, dx}{\sqrt {1-c^2 x^2}}\\ &=-\frac {5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{4} \left (15 c^2 d^2\right ) \int \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx+\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {\left (1-c^2 x\right )^2}{x} \, dx,x,x^2\right )}{2 \sqrt {1-c^2 x^2}}+\frac {\left (5 b c^3 d^2 \sqrt {d-c^2 d x^2}\right ) \int x \left (1-c^2 x^2\right ) \, dx}{4 \sqrt {1-c^2 x^2}}\\ &=-\frac {15}{8} c^2 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (-2 c^2+\frac {1}{x}+c^4 x\right ) \, dx,x,x^2\right )}{2 \sqrt {1-c^2 x^2}}-\frac {\left (15 c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{8 \sqrt {1-c^2 x^2}}+\frac {\left (5 b c^3 d^2 \sqrt {d-c^2 d x^2}\right ) \int \left (x-c^2 x^3\right ) \, dx}{4 \sqrt {1-c^2 x^2}}+\frac {\left (15 b c^3 d^2 \sqrt {d-c^2 d x^2}\right ) \int x \, dx}{8 \sqrt {1-c^2 x^2}}\\ &=\frac {9 b c^3 d^2 x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}-\frac {15}{8} c^2 d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {15 c d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b \sqrt {1-c^2 x^2}}+\frac {b c d^2 \sqrt {d-c^2 d x^2} \log (x)}{\sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.36, size = 257, normalized size = 0.96 \[ \frac {d^2 \left (\sqrt {d-c^2 d x^2} \left (16 \left (a \sqrt {1-c^2 x^2} \left (2 c^4 x^4-9 c^2 x^2-8\right )+8 b c x \log (c x)\right )-32 b c x \cos \left (2 \sin ^{-1}(c x)\right )-b c x \cos \left (4 \sin ^{-1}(c x)\right )\right )+240 a c \sqrt {d} x \sqrt {1-c^2 x^2} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )-120 b c x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)^2-4 b \sqrt {d-c^2 d x^2} \left (32 \sqrt {1-c^2 x^2}+16 c x \sin \left (2 \sin ^{-1}(c x)\right )+c x \sin \left (4 \sin ^{-1}(c x)\right )\right ) \sin ^{-1}(c x)\right )}{128 x \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(-120*b*c*x*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]^2 + 240*a*c*Sqrt[d]*x*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d -
c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))] + Sqrt[d - c^2*d*x^2]*(-32*b*c*x*Cos[2*ArcSin[c*x]] - b*c*x*Cos[4*ArcSin
[c*x]] + 16*(a*Sqrt[1 - c^2*x^2]*(-8 - 9*c^2*x^2 + 2*c^4*x^4) + 8*b*c*x*Log[c*x])) - 4*b*Sqrt[d - c^2*d*x^2]*A
rcSin[c*x]*(32*Sqrt[1 - c^2*x^2] + 16*c*x*Sin[2*ArcSin[c*x]] + c*x*Sin[4*ArcSin[c*x]])))/(128*x*Sqrt[1 - c^2*x
^2])

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a c^{4} d^{2} x^{4} - 2 \, a c^{2} d^{2} x^{2} + a d^{2} + {\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*c^4*d^2*x^4 - 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 - 2*b*c^2*d^2*x^2 + b*d^2)*arcsin(c*x))*sqr
t(-c^2*d*x^2 + d)/x^2, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.35, size = 1391, normalized size = 5.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)*d^2/(c^2*x^2-1)/x+33/256*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c/(c^2*x^2-1)*(-c^2
*x^2+1)^(1/2)+33/256*b*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin(c*x))*d^2*c/(c^2*x^2-1)-15/64*b*(-d*(c^2*x^2-1))^(1
/2)*cos(3*arcsin(c*x))*d^2*c^2/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x+15/64*I*b*(-d*(c^2*x^2-1))^(1/2)*c
os(3*arcsin(c*x))*d^2*c^3/(c^2*x^2-1)*arcsin(c*x)*x^2-33/256*I*b*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin(c*x))*d^2
*c^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x-1/8*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^5/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2
+1)^(1/2)*x^4-3/8*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^3/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^2-a/d/x*(-c^
2*d*x^2+d)^(7/2)-a*c^2*x*(-c^2*d*x^2+d)^(5/2)-5/4*a*c^2*d*x*(-c^2*d*x^2+d)^(3/2)-15/8*a*c^2*d^2*x*(-c^2*d*x^2+
d)^(1/2)-b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*ln((I*c*x+(-c^2*x^2+1)^(1/2))^2-1)*d^2*c+17/6
4*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d^2*c/(c^2*x^2-1)*arcsin(c*x)+1/8*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c
^6/(c^2*x^2-1)*arcsin(c*x)*x^5-11/16*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^4/(c^2*x^2-1)*arcsin(c*x)*x^3-9/32*b*(-d*(
c^2*x^2-1))^(1/2)*d^2*c^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2-33/256*b*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin(c*x)
)*d^2*c^3/(c^2*x^2-1)*x^2+1/32*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^5/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^4+31/256*I*b*
(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d^2*c/(c^2*x^2-1)+1/32*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^6/(c^2*x^2-1
)*x^5+13/64*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^4/(c^2*x^2-1)*x^3-15/64*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^2/(c^2*x
^2-1)*x+15/16*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*arcsin(c*x)^2*d^2*c-7/16*b*(-d*(c^2*x^2-
1))^(1/2)*d^2*c^2/(c^2*x^2-1)*arcsin(c*x)*x-31/256*I*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d^2*c^3/(c^2*
x^2-1)*x^2+79/64*I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)*arcsin(c*x)*d^2*c+31/256*b*(-d*(c^2
*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d^2*c^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x-17/64*b*(-d*(c^2*x^2-1))^(1/2)*sin(
3*arcsin(c*x))*d^2*c^3/(c^2*x^2-1)*arcsin(c*x)*x^2-15/64*I*b*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin(c*x))*d^2*c/(
c^2*x^2-1)*arcsin(c*x)-17/64*I*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d^2*c^2/(c^2*x^2-1)*arcsin(c*x)*(-c
^2*x^2+1)^(1/2)*x-15/8*a*c^2*d^3/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \sqrt {d} \int \frac {{\left (c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x^{2}}\,{d x} - \frac {1}{8} \, {\left (10 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d x + 15 \, \sqrt {-c^{2} d x^{2} + d} c^{2} d^{2} x + 15 \, c d^{\frac {5}{2}} \arcsin \left (c x\right ) + \frac {8 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{x}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*sqrt(d)*integrate((c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1
)*sqrt(-c*x + 1))/x^2, x) - 1/8*(10*(-c^2*d*x^2 + d)^(3/2)*c^2*d*x + 15*sqrt(-c^2*d*x^2 + d)*c^2*d^2*x + 15*c*
d^(5/2)*arcsin(c*x) + 8*(-c^2*d*x^2 + d)^(5/2)/x)*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{5/2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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