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

Optimal. Leaf size=389 \[ \frac{15 i b c^4 d^2 \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt{1-c^2 x^2}}-\frac{15 i b c^4 d^2 \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt{1-c^2 x^2}}+\frac{15}{8} c^4 d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{15 c^4 d^2 \sqrt{d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 \sqrt{1-c^2 x^2}}+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac{b c^5 d^2 x \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}}+\frac{9 b c^3 d^2 \sqrt{d-c^2 d x^2}}{8 x \sqrt{1-c^2 x^2}}-\frac{b c d^2 \sqrt{d-c^2 d x^2}}{12 x^3 \sqrt{1-c^2 x^2}} \]

[Out]

-(b*c*d^2*Sqrt[d - c^2*d*x^2])/(12*x^3*Sqrt[1 - c^2*x^2]) + (9*b*c^3*d^2*Sqrt[d - c^2*d*x^2])/(8*x*Sqrt[1 - c^
2*x^2]) - (b*c^5*d^2*x*Sqrt[d - c^2*d*x^2])/Sqrt[1 - c^2*x^2] + (15*c^4*d^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[
c*x]))/8 + (5*c^2*d*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(8*x^2) - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[
c*x]))/(4*x^4) - (15*c^4*d^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])])/(4*Sqrt[1 - c
^2*x^2]) + (((15*I)/8)*b*c^4*d^2*Sqrt[d - c^2*d*x^2]*PolyLog[2, -E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2] - (((15
*I)/8)*b*c^4*d^2*Sqrt[d - c^2*d*x^2]*PolyLog[2, E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2]

________________________________________________________________________________________

Rubi [A]  time = 0.4552, antiderivative size = 389, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4695, 4697, 4709, 4183, 2279, 2391, 8, 14, 270} \[ \frac{15 i b c^4 d^2 \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt{1-c^2 x^2}}-\frac{15 i b c^4 d^2 \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt{1-c^2 x^2}}+\frac{15}{8} c^4 d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{15 c^4 d^2 \sqrt{d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 \sqrt{1-c^2 x^2}}+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac{b c^5 d^2 x \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}}+\frac{9 b c^3 d^2 \sqrt{d-c^2 d x^2}}{8 x \sqrt{1-c^2 x^2}}-\frac{b c d^2 \sqrt{d-c^2 d x^2}}{12 x^3 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d^2*Sqrt[d - c^2*d*x^2])/(12*x^3*Sqrt[1 - c^2*x^2]) + (9*b*c^3*d^2*Sqrt[d - c^2*d*x^2])/(8*x*Sqrt[1 - c^
2*x^2]) - (b*c^5*d^2*x*Sqrt[d - c^2*d*x^2])/Sqrt[1 - c^2*x^2] + (15*c^4*d^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[
c*x]))/8 + (5*c^2*d*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(8*x^2) - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[
c*x]))/(4*x^4) - (15*c^4*d^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])])/(4*Sqrt[1 - c
^2*x^2]) + (((15*I)/8)*b*c^4*d^2*Sqrt[d - c^2*d*x^2]*PolyLog[2, -E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2] - (((15
*I)/8)*b*c^4*d^2*Sqrt[d - c^2*d*x^2]*PolyLog[2, E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2]

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]

Rule 4697

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[Sqrt[d + e*x^2]/((m + 2)*Sqrt[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*Sqrt[d + e*x^2])/(f*(m
+ 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] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 1])

Rule 4709

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

Rule 4183

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 2279

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 2391

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 8

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

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 270

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]

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^5} \, dx &=-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac{1}{4} \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^3} \, 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^4} \, dx}{4 \sqrt{1-c^2 x^2}}\\ &=\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac{1}{8} \left (15 c^4 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 (c^4+\frac{1}{x^4}-\frac{2 c^2}{x^2}\right ) \, dx}{4 \sqrt{1-c^2 x^2}}-\frac{\left (5 b c^3 d^2 \sqrt{d-c^2 d x^2}\right ) \int \frac{1-c^2 x^2}{x^2} \, dx}{8 \sqrt{1-c^2 x^2}}\\ &=-\frac{b c d^2 \sqrt{d-c^2 d x^2}}{12 x^3 \sqrt{1-c^2 x^2}}+\frac{b c^3 d^2 \sqrt{d-c^2 d x^2}}{2 x \sqrt{1-c^2 x^2}}+\frac{b c^5 d^2 x \sqrt{d-c^2 d x^2}}{4 \sqrt{1-c^2 x^2}}+\frac{15}{8} c^4 d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac{\left (5 b c^3 d^2 \sqrt{d-c^2 d x^2}\right ) \int \left (-c^2+\frac{1}{x^2}\right ) \, dx}{8 \sqrt{1-c^2 x^2}}+\frac{\left (15 c^4 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}{8 \sqrt{1-c^2 x^2}}-\frac{\left (15 b c^5 d^2 \sqrt{d-c^2 d x^2}\right ) \int 1 \, dx}{8 \sqrt{1-c^2 x^2}}\\ &=-\frac{b c d^2 \sqrt{d-c^2 d x^2}}{12 x^3 \sqrt{1-c^2 x^2}}+\frac{9 b c^3 d^2 \sqrt{d-c^2 d x^2}}{8 x \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}}+\frac{15}{8} c^4 d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac{\left (15 c^4 d^2 \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 \sqrt{1-c^2 x^2}}\\ &=-\frac{b c d^2 \sqrt{d-c^2 d x^2}}{12 x^3 \sqrt{1-c^2 x^2}}+\frac{9 b c^3 d^2 \sqrt{d-c^2 d x^2}}{8 x \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}}+\frac{15}{8} c^4 d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac{15 c^4 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 )}{4 \sqrt{1-c^2 x^2}}-\frac{\left (15 b c^4 d^2 \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 \sqrt{1-c^2 x^2}}+\frac{\left (15 b c^4 d^2 \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 \sqrt{1-c^2 x^2}}\\ &=-\frac{b c d^2 \sqrt{d-c^2 d x^2}}{12 x^3 \sqrt{1-c^2 x^2}}+\frac{9 b c^3 d^2 \sqrt{d-c^2 d x^2}}{8 x \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}}+\frac{15}{8} c^4 d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac{15 c^4 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 )}{4 \sqrt{1-c^2 x^2}}+\frac{\left (15 i b c^4 d^2 \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt{1-c^2 x^2}}-\frac{\left (15 i b c^4 d^2 \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt{1-c^2 x^2}}\\ &=-\frac{b c d^2 \sqrt{d-c^2 d x^2}}{12 x^3 \sqrt{1-c^2 x^2}}+\frac{9 b c^3 d^2 \sqrt{d-c^2 d x^2}}{8 x \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}}+\frac{15}{8} c^4 d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac{15 c^4 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 )}{4 \sqrt{1-c^2 x^2}}+\frac{15 i b c^4 d^2 \sqrt{d-c^2 d x^2} \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt{1-c^2 x^2}}-\frac{15 i b c^4 d^2 \sqrt{d-c^2 d x^2} \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 6.13157, size = 640, normalized size = 1.65 \[ \frac{b c^4 d^2 \sqrt{d-c^2 d x^2} \left (i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+\sqrt{1-c^2 x^2} \sin ^{-1}(c x)-c x+\sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )}{\sqrt{1-c^2 x^2}}-\frac{b c^4 d^3 \sqrt{1-c^2 x^2} \left (-4 i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )+4 i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-4 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+4 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )-2 \tan \left (\frac{1}{2} \sin ^{-1}(c x)\right )-2 \cot \left (\frac{1}{2} \sin ^{-1}(c x)\right )-\sin ^{-1}(c x) \csc ^2\left (\frac{1}{2} \sin ^{-1}(c x)\right )+\sin ^{-1}(c x) \sec ^2\left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )}{4 \sqrt{d-c^2 d x^2}}+\frac{b c^4 d^2 \sqrt{d-c^2 d x^2} \left (-24 i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )+24 i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-\frac{16 \sin ^4\left (\frac{1}{2} \sin ^{-1}(c x)\right )}{c^3 x^3}-24 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+24 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )+8 \tan \left (\frac{1}{2} \sin ^{-1}(c x)\right )+8 \cot \left (\frac{1}{2} \sin ^{-1}(c x)\right )-c x \csc ^4\left (\frac{1}{2} \sin ^{-1}(c x)\right )-3 \sin ^{-1}(c x) \csc ^4\left (\frac{1}{2} \sin ^{-1}(c x)\right )+6 \sin ^{-1}(c x) \csc ^2\left (\frac{1}{2} \sin ^{-1}(c x)\right )+3 \sin ^{-1}(c x) \sec ^4\left (\frac{1}{2} \sin ^{-1}(c x)\right )-6 \sin ^{-1}(c x) \sec ^2\left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )}{192 \sqrt{1-c^2 x^2}}+\frac{a d^2 \left (8 c^4 x^4+9 c^2 x^2-2\right ) \sqrt{d-c^2 d x^2}}{8 x^4}-\frac{15}{8} a c^4 d^{5/2} \log \left (\sqrt{d} \sqrt{d-c^2 d x^2}+d\right )+\frac{15}{8} a c^4 d^{5/2} \log (x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*d^2*Sqrt[d - c^2*d*x^2]*(-2 + 9*c^2*x^2 + 8*c^4*x^4))/(8*x^4) + (15*a*c^4*d^(5/2)*Log[x])/8 - (15*a*c^4*d^(
5/2)*Log[d + Sqrt[d]*Sqrt[d - c^2*d*x^2]])/8 + (b*c^4*d^2*Sqrt[d - c^2*d*x^2]*(-(c*x) + Sqrt[1 - c^2*x^2]*ArcS
in[c*x] + ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])] - ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] + I*PolyLog[2, -E^(I
*ArcSin[c*x])] - I*PolyLog[2, E^(I*ArcSin[c*x])]))/Sqrt[1 - c^2*x^2] - (b*c^4*d^3*Sqrt[1 - c^2*x^2]*(-2*Cot[Ar
cSin[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]))/(4*Sqrt[d - c^2*d*x^2]) + (b*c^4*d^2*Sqrt[d - c^2*d*x^2]*(8
*Cot[ArcSin[c*x]/2] + 6*ArcSin[c*x]*Csc[ArcSin[c*x]/2]^2 - c*x*Csc[ArcSin[c*x]/2]^4 - 3*ArcSin[c*x]*Csc[ArcSin
[c*x]/2]^4 - 24*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])] + 24*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] - (24*I)*Po
lyLog[2, -E^(I*ArcSin[c*x])] + (24*I)*PolyLog[2, E^(I*ArcSin[c*x])] - 6*ArcSin[c*x]*Sec[ArcSin[c*x]/2]^2 + 3*A
rcSin[c*x]*Sec[ArcSin[c*x]/2]^4 - (16*Sin[ArcSin[c*x]/2]^4)/(c^3*x^3) + 8*Tan[ArcSin[c*x]/2]))/(192*Sqrt[1 - c
^2*x^2])

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Maple [A]  time = 0.312, size = 727, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

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^5,x)

[Out]

-1/4*a/d/x^4*(-c^2*d*x^2+d)^(7/2)+3/8*a*c^2/d/x^2*(-c^2*d*x^2+d)^(7/2)+3/8*a*c^4*(-c^2*d*x^2+d)^(5/2)+5/8*a*c^
4*d*(-c^2*d*x^2+d)^(3/2)-15/8*a*c^4*d^(5/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)+15/8*a*c^4*(-c^2*d*x^2+
d)^(1/2)*d^2+b*(-d*(c^2*x^2-1))^(1/2)*c^6*d^2/(c^2*x^2-1)*arcsin(c*x)*x^2+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+1/8*b*(-d*(c^2*x^2-1))^(1/2)*c^4*d^2/(c^2*x^2-1)*arcsin(c*x)-9/8*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^3-11/8*b*d^2*(-d*(c^2*x^2-1))^(1/2)/x^2/(c^2*x^2-1)*arcsin(c*
x)*c^2+1/12*b*d^2*(-d*(c^2*x^2-1))^(1/2)/x^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c+1/4*b*d^2*(-d*(c^2*x^2-1))^(1/2)
/x^4/(c^2*x^2-1)*arcsin(c*x)+15*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^4*d^2/(8*c^2*x^2-8)*arcsin(c*x)*
ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-15*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^4*d^2/(8*c^2*x^2-8)*arcsin(c*x
)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))+15*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^4*d^2/(8*c^2*x^2-8)*polylo
g(2,I*c*x+(-c^2*x^2+1)^(1/2))-15*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^4*d^2/(8*c^2*x^2-8)*polylog(2
,-I*c*x-(-c^2*x^2+1)^(1/2))

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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((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^5,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{{\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^{5}}, x\right ) \end{align*}

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

[Out]

Timed out

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

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^5,x, algorithm="giac")

[Out]

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