3.1.97 \(\int \frac {(d-c^2 d x^2)^{5/2} (a+b \text {ArcSin}(c x))}{x^3} \, dx\) [97]

Optimal. Leaf size=386 \[ -\frac {b c d^2 \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}+\frac {7 b c^3 d^2 x \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d-c^2 d x^2}}{9 \sqrt {1-c^2 x^2}}-\frac {5}{2} c^2 d^2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))-\frac {5}{6} c^2 d \left (d-c^2 d x^2\right )^{3/2} (a+b \text {ArcSin}(c x))-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \text {ArcSin}(c x))}{2 x^2}+\frac {5 c^2 d^2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x)) \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {5 i b c^2 d^2 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}+\frac {5 i b c^2 d^2 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )}{2 \sqrt {1-c^2 x^2}} \]

[Out]

-5/6*c^2*d*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))-1/2*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^2-5/2*c^2*d^2*(
a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)-1/2*b*c*d^2*(-c^2*d*x^2+d)^(1/2)/x/(-c^2*x^2+1)^(1/2)+7/3*b*c^3*d^2*x*(-
c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-1/9*b*c^5*d^2*x^3*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+5*c^2*d^2*(a+b
*arcsin(c*x))*arctanh(I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-5/2*I*b*c^2*d^2*polylo
g(2,-I*c*x-(-c^2*x^2+1)^(1/2))*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+5/2*I*b*c^2*d^2*polylog(2,I*c*x+(-c^2*x
^2+1)^(1/2))*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.30, antiderivative size = 386, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4785, 4787, 4783, 4803, 4268, 2317, 2438, 8, 276} \begin {gather*} -\frac {5}{2} c^2 d^2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))+\frac {5 c^2 d^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{\sqrt {1-c^2 x^2}}-\frac {5}{6} c^2 d \left (d-c^2 d x^2\right )^{3/2} (a+b \text {ArcSin}(c x))-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \text {ArcSin}(c x))}{2 x^2}-\frac {5 i b c^2 d^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (-e^{i \text {ArcSin}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}+\frac {5 i b c^2 d^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{i \text {ArcSin}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d-c^2 d x^2}}{9 \sqrt {1-c^2 x^2}}+\frac {7 b c^3 d^2 x \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*c*d^2*Sqrt[d - c^2*d*x^2])/(x*Sqrt[1 - c^2*x^2]) + (7*b*c^3*d^2*x*Sqrt[d - c^2*d*x^2])/(3*Sqrt[1 - c^2
*x^2]) - (b*c^5*d^2*x^3*Sqrt[d - c^2*d*x^2])/(9*Sqrt[1 - c^2*x^2]) - (5*c^2*d^2*Sqrt[d - c^2*d*x^2]*(a + b*Arc
Sin[c*x]))/2 - (5*c^2*d*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/6 - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*
x]))/(2*x^2) + (5*c^2*d^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2
] - (((5*I)/2)*b*c^2*d^2*Sqrt[d - c^2*d*x^2]*PolyLog[2, -E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2] + (((5*I)/2)*b*
c^2*d^2*Sqrt[d - c^2*d*x^2]*PolyLog[2, E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4268

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*
x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[d*(m/f), Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4783

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(f
*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcSin[c*x])^n/(f*(m + 2))), x] + (Dist[(1/(m + 2))*Simp[Sqrt[d + e*x^2]/S
qrt[1 - c^2*x^2]], Int[(f*x)^m*((a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2]), x], x] - Dist[b*c*(n/(f*(m + 2)))*Si
mp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(f*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b,
c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && (IGtQ[m, -2] || EqQ[n, 1])

Rule 4785

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/(f*(m + 1)))*Simp[(d + e*x^2)^p/(1 - c
^2*x^2)^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]

Rule 4787

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

Rule 4803

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[(1/c^(m
+ 1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]], Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; Free
Q[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[m]

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^3} \, dx &=-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}-\frac {1}{2} \left (5 c^2 d\right ) \int \frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x} \, 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^2} \, dx}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {5}{6} c^2 d \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 )}{2 x^2}-\frac {1}{2} \left (5 c^2 d^2\right ) \int \frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x} \, dx+\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \int \left (-2 c^2+\frac {1}{x^2}+c^4 x^2\right ) \, dx}{2 \sqrt {1-c^2 x^2}}+\frac {\left (5 b c^3 d^2 \sqrt {d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right ) \, dx}{6 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}-\frac {b c^3 d^2 x \sqrt {d-c^2 d x^2}}{6 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d-c^2 d x^2}}{9 \sqrt {1-c^2 x^2}}-\frac {5}{2} c^2 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {5}{6} c^2 d \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 )}{2 x^2}-\frac {\left (5 c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {1-c^2 x^2}}+\frac {\left (5 b c^3 d^2 \sqrt {d-c^2 d x^2}\right ) \int 1 \, dx}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}+\frac {7 b c^3 d^2 x \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d-c^2 d x^2}}{9 \sqrt {1-c^2 x^2}}-\frac {5}{2} c^2 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {5}{6} c^2 d \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 )}{2 x^2}-\frac {\left (5 c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}+\frac {7 b c^3 d^2 x \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d-c^2 d x^2}}{9 \sqrt {1-c^2 x^2}}-\frac {5}{2} c^2 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {5}{6} c^2 d \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 )}{2 x^2}+\frac {5 c^2 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {\left (5 b c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}-\frac {\left (5 b c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}+\frac {7 b c^3 d^2 x \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d-c^2 d x^2}}{9 \sqrt {1-c^2 x^2}}-\frac {5}{2} c^2 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {5}{6} c^2 d \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 )}{2 x^2}+\frac {5 c^2 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {\left (5 i b c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}+\frac {\left (5 i b c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}+\frac {7 b c^3 d^2 x \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d-c^2 d x^2}}{9 \sqrt {1-c^2 x^2}}-\frac {5}{2} c^2 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {5}{6} c^2 d \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 )}{2 x^2}+\frac {5 c^2 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {5 i b c^2 d^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}+\frac {5 i b c^2 d^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{2 \sqrt {1-c^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 3.02, size = 484, normalized size = 1.25 \begin {gather*} \frac {-12 a d^3 \left (-1+c^2 x^2\right ) \left (-3-14 c^2 x^2+2 c^4 x^4\right )-180 a c^2 d^{5/2} x^2 \sqrt {d-c^2 d x^2} \log (x)+180 a c^2 d^{5/2} x^2 \sqrt {d-c^2 d x^2} \log \left (d+\sqrt {d} \sqrt {d-c^2 d x^2}\right )+144 b c^2 d^3 x^2 \sqrt {1-c^2 x^2} \left (c x-\sqrt {1-c^2 x^2} \text {ArcSin}(c x)-\text {ArcSin}(c x) \left (\log \left (1-e^{i \text {ArcSin}(c x)}\right )-\log \left (1+e^{i \text {ArcSin}(c x)}\right )\right )-i \left (\text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )-\text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )\right )\right )+2 b c^2 d^3 x^2 \sqrt {1-c^2 x^2} \left (9 c x-3 \text {ArcSin}(c x) \left (3 \sqrt {1-c^2 x^2}+\cos (3 \text {ArcSin}(c x))\right )+\sin (3 \text {ArcSin}(c x))\right )-9 b c^2 d^3 x^2 \sqrt {1-c^2 x^2} \left (2 \cot \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\text {ArcSin}(c x) \csc ^2\left (\frac {1}{2} \text {ArcSin}(c x)\right )+4 \text {ArcSin}(c x) \log \left (1-e^{i \text {ArcSin}(c x)}\right )-4 \text {ArcSin}(c x) \log \left (1+e^{i \text {ArcSin}(c x)}\right )+4 i \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )-4 i \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )-\text {ArcSin}(c x) \sec ^2\left (\frac {1}{2} \text {ArcSin}(c x)\right )+2 \tan \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}{72 x^2 \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*a*d^3*(-1 + c^2*x^2)*(-3 - 14*c^2*x^2 + 2*c^4*x^4) - 180*a*c^2*d^(5/2)*x^2*Sqrt[d - c^2*d*x^2]*Log[x] + 1
80*a*c^2*d^(5/2)*x^2*Sqrt[d - c^2*d*x^2]*Log[d + Sqrt[d]*Sqrt[d - c^2*d*x^2]] + 144*b*c^2*d^3*x^2*Sqrt[1 - c^2
*x^2]*(c*x - Sqrt[1 - c^2*x^2]*ArcSin[c*x] - ArcSin[c*x]*(Log[1 - E^(I*ArcSin[c*x])] - Log[1 + E^(I*ArcSin[c*x
])]) - I*(PolyLog[2, -E^(I*ArcSin[c*x])] - PolyLog[2, E^(I*ArcSin[c*x])])) + 2*b*c^2*d^3*x^2*Sqrt[1 - c^2*x^2]
*(9*c*x - 3*ArcSin[c*x]*(3*Sqrt[1 - c^2*x^2] + Cos[3*ArcSin[c*x]]) + Sin[3*ArcSin[c*x]]) - 9*b*c^2*d^3*x^2*Sqr
t[1 - c^2*x^2]*(2*Cot[ArcSin[c*x]/2] + ArcSin[c*x]*Csc[ArcSin[c*x]/2]^2 + 4*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*
x])] - 4*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] + (4*I)*PolyLog[2, -E^(I*ArcSin[c*x])] - (4*I)*PolyLog[2, E^(I
*ArcSin[c*x])] - ArcSin[c*x]*Sec[ArcSin[c*x]/2]^2 + 2*Tan[ArcSin[c*x]/2]))/(72*x^2*Sqrt[d - c^2*d*x^2])

________________________________________________________________________________________

Maple [A]
time = 0.26, size = 704, normalized size = 1.82

method	result	size
default	\(-\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {7}{2}}}{2 d \,x^{2}}-\frac {a \,c^{2} \left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{2}-\frac {5 a \,c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{6}+\frac {5 a \,c^{2} d^{\frac {5}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{2}-\frac {5 a \,c^{2} \sqrt {-c^{2} d \,x^{2}+d}\, d^{2}}{2}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{5} d^{2} \sqrt {-c^{2} x^{2}+1}\, x^{3}}{9 c^{2} x^{2}-9}-\frac {7 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{3} d^{2} \sqrt {-c^{2} x^{2}+1}\, x}{3 \left (c^{2} x^{2}-1\right )}+\frac {5 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} d^{2} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2}-\frac {5 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} d^{2} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2}+\frac {11 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} d^{2} \arcsin \left (c x \right )}{6 \left (c^{2} x^{2}-1\right )}+\frac {b \,d^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{2 x^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{6} d^{2} \arcsin \left (c x \right ) x^{4}}{3 c^{2} x^{2}-3}-\frac {8 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{4} d^{2} \arcsin \left (c x \right ) x^{2}}{3 \left (c^{2} x^{2}-1\right )}+\frac {b \,d^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c}{2 x \left (c^{2} x^{2}-1\right )}-\frac {5 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} d^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2}+\frac {5 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} d^{2} \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2}\)	\(704\)

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^3,x,method=_RETURNVERBOSE)

[Out]

-1/2*a/d/x^2*(-c^2*d*x^2+d)^(7/2)-1/2*a*c^2*(-c^2*d*x^2+d)^(5/2)-5/6*a*c^2*d*(-c^2*d*x^2+d)^(3/2)+5/2*a*c^2*d^
(5/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)-5/2*a*c^2*(-c^2*d*x^2+d)^(1/2)*d^2+1/9*b*(-d*(c^2*x^2-1))^(1/
2)*c^5*d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^3-7/3*b*(-d*(c^2*x^2-1))^(1/2)*c^3*d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1
/2)*x-5*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^2*d^2/(2*c^2*x^2-2)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2)
)+5*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^2*d^2/(2*c^2*x^2-2)*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))+1
1/6*b*(-d*(c^2*x^2-1))^(1/2)*c^2*d^2/(c^2*x^2-1)*arcsin(c*x)+1/2*b*d^2*(-d*(c^2*x^2-1))^(1/2)/x^2/(c^2*x^2-1)*
arcsin(c*x)+1/3*b*(-d*(c^2*x^2-1))^(1/2)*c^6*d^2/(c^2*x^2-1)*arcsin(c*x)*x^4-8/3*b*(-d*(c^2*x^2-1))^(1/2)*c^4*
d^2/(c^2*x^2-1)*arcsin(c*x)*x^2+1/2*b*d^2*(-d*(c^2*x^2-1))^(1/2)/x/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c-5*b*(-d*(c
^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^2*d^2/(2*c^2*x^2-2)*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+5*b*(-d*(
c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^2*d^2/(2*c^2*x^2-2)*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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^3,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^3, x) + 1/6*(15*c^2*d^(5/2)*log(2*sqrt(-c^2*d*x^2 + d)*sqrt(d)/abs(x) + 2*d/abs(x)) - 3*(-
c^2*d*x^2 + d)^(5/2)*c^2 - 5*(-c^2*d*x^2 + d)^(3/2)*c^2*d - 15*sqrt(-c^2*d*x^2 + d)*c^2*d^2 - 3*(-c^2*d*x^2 +
d)^(7/2)/(d*x^2))*a

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^3,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^3, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{x^{3}}\, dx \end {gather*}

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**3,x)

[Out]

Integral((-d*(c*x - 1)*(c*x + 1))**(5/2)*(a + b*asin(c*x))/x**3, x)

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{5/2}}{x^3} \,d x \end {gather*}

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^3,x)

[Out]

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

________________________________________________________________________________________