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

Optimal. Leaf size=361 \[ \frac{i b d^2 \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{1-c^2 x^2}}-\frac{i b d^2 \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{1-c^2 x^2}}+d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{2 d^2 \sqrt{d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}}+\frac{1}{5} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac{b c^5 d^2 x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+\frac{11 b c^3 d^2 x^3 \sqrt{d-c^2 d x^2}}{45 \sqrt{1-c^2 x^2}}-\frac{23 b c d^2 x \sqrt{d-c^2 d x^2}}{15 \sqrt{1-c^2 x^2}} \]

[Out]

(-23*b*c*d^2*x*Sqrt[d - c^2*d*x^2])/(15*Sqrt[1 - c^2*x^2]) + (11*b*c^3*d^2*x^3*Sqrt[d - c^2*d*x^2])/(45*Sqrt[1
- c^2*x^2]) - (b*c^5*d^2*x^5*Sqrt[d - c^2*d*x^2])/(25*Sqrt[1 - c^2*x^2]) + d^2*Sqrt[d - c^2*d*x^2]*(a + b*Arc
Sin[c*x]) + (d*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/3 + ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/5 -
(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] + (I*b*d^2*Sqrt[d
- c^2*d*x^2]*PolyLog[2, -E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2] - (I*b*d^2*Sqrt[d - c^2*d*x^2]*PolyLog[2, E^(I
*ArcSin[c*x])])/Sqrt[1 - c^2*x^2]

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Rubi [A]  time = 0.462722, antiderivative size = 361, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {4699, 4697, 4709, 4183, 2279, 2391, 8, 194} \[ \frac{i b d^2 \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{1-c^2 x^2}}-\frac{i b d^2 \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{1-c^2 x^2}}+d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{2 d^2 \sqrt{d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}}+\frac{1}{5} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac{b c^5 d^2 x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+\frac{11 b c^3 d^2 x^3 \sqrt{d-c^2 d x^2}}{45 \sqrt{1-c^2 x^2}}-\frac{23 b c d^2 x \sqrt{d-c^2 d x^2}}{15 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-23*b*c*d^2*x*Sqrt[d - c^2*d*x^2])/(15*Sqrt[1 - c^2*x^2]) + (11*b*c^3*d^2*x^3*Sqrt[d - c^2*d*x^2])/(45*Sqrt[1
- c^2*x^2]) - (b*c^5*d^2*x^5*Sqrt[d - c^2*d*x^2])/(25*Sqrt[1 - c^2*x^2]) + d^2*Sqrt[d - c^2*d*x^2]*(a + b*Arc
Sin[c*x]) + (d*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/3 + ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/5 -
(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] + (I*b*d^2*Sqrt[d
- c^2*d*x^2]*PolyLog[2, -E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2] - (I*b*d^2*Sqrt[d - c^2*d*x^2]*PolyLog[2, E^(I
*ArcSin[c*x])])/Sqrt[1 - c^2*x^2]

Rule 4699

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(
f*(m + 2*p + 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, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]
&& (RationalQ[m] || EqQ[n, 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 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && 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} \, dx &=\frac{1}{5} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )+d \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 \left (1-c^2 x^2\right )^2 \, dx}{5 \sqrt{1-c^2 x^2}}\\ &=\frac{1}{3} d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )+d^2 \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 (1-2 c^2 x^2+c^4 x^4\right ) \, dx}{5 \sqrt{1-c^2 x^2}}-\frac{\left (b c d^2 \sqrt{d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right ) \, dx}{3 \sqrt{1-c^2 x^2}}\\ &=-\frac{8 b c d^2 x \sqrt{d-c^2 d x^2}}{15 \sqrt{1-c^2 x^2}}+\frac{11 b c^3 d^2 x^3 \sqrt{d-c^2 d x^2}}{45 \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{\left (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}{\sqrt{1-c^2 x^2}}-\frac{\left (b c d^2 \sqrt{d-c^2 d x^2}\right ) \int 1 \, dx}{\sqrt{1-c^2 x^2}}\\ &=-\frac{23 b c d^2 x \sqrt{d-c^2 d x^2}}{15 \sqrt{1-c^2 x^2}}+\frac{11 b c^3 d^2 x^3 \sqrt{d-c^2 d x^2}}{45 \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{\left (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 )}{\sqrt{1-c^2 x^2}}\\ &=-\frac{23 b c d^2 x \sqrt{d-c^2 d x^2}}{15 \sqrt{1-c^2 x^2}}+\frac{11 b c^3 d^2 x^3 \sqrt{d-c^2 d x^2}}{45 \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )-\frac{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 (b 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 )}{\sqrt{1-c^2 x^2}}+\frac{\left (b 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 )}{\sqrt{1-c^2 x^2}}\\ &=-\frac{23 b c d^2 x \sqrt{d-c^2 d x^2}}{15 \sqrt{1-c^2 x^2}}+\frac{11 b c^3 d^2 x^3 \sqrt{d-c^2 d x^2}}{45 \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )-\frac{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 (i b 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 )}{\sqrt{1-c^2 x^2}}-\frac{\left (i b 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 )}{\sqrt{1-c^2 x^2}}\\ &=-\frac{23 b c d^2 x \sqrt{d-c^2 d x^2}}{15 \sqrt{1-c^2 x^2}}+\frac{11 b c^3 d^2 x^3 \sqrt{d-c^2 d x^2}}{45 \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )-\frac{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{i b d^2 \sqrt{d-c^2 d x^2} \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{1-c^2 x^2}}-\frac{i b d^2 \sqrt{d-c^2 d x^2} \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.7149, size = 394, normalized size = 1.09 \[ \frac{b d^2 \sqrt{d-c^2 d x^2} \left (i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+\sqrt{1-c^2 x^2} \sin ^{-1}(c x)-c x+\sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )}{\sqrt{1-c^2 x^2}}+\frac{1}{15} a d^2 \left (3 c^4 x^4-11 c^2 x^2+23\right ) \sqrt{d-c^2 d x^2}-a d^{5/2} \log \left (\sqrt{d} \sqrt{d-c^2 d x^2}+d\right )+a d^{5/2} \log (x)-\frac{b d^2 \sqrt{d-c^2 d x^2} \left (-3 \sin ^{-1}(c x) \left (3 \sqrt{1-c^2 x^2}+\cos \left (3 \sin ^{-1}(c x)\right )\right )+9 c x+\sin \left (3 \sin ^{-1}(c x)\right )\right )}{18 \sqrt{1-c^2 x^2}}+\frac{b d^2 \sqrt{d-c^2 d x^2} \left (-15 \sin ^{-1}(c x) \left (30 \sqrt{1-c^2 x^2}+5 \cos \left (3 \sin ^{-1}(c x)\right )-3 \cos \left (5 \sin ^{-1}(c x)\right )\right )+450 c x+25 \sin \left (3 \sin ^{-1}(c x)\right )-9 \sin \left (5 \sin ^{-1}(c x)\right )\right )}{3600 \sqrt{1-c^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*d^2*Sqrt[d - c^2*d*x^2]*(23 - 11*c^2*x^2 + 3*c^4*x^4))/15 + a*d^(5/2)*Log[x] - a*d^(5/2)*Log[d + Sqrt[d]*Sq
rt[d - c^2*d*x^2]] + (b*d^2*Sqrt[d - c^2*d*x^2]*(-(c*x) + Sqrt[1 - c^2*x^2]*ArcSin[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*d^2*Sqrt[d - c^2*d*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]]))/(18*Sqrt[1 - c^2*x^2]) + (b*d^2*Sqrt[d - c^2*d*x^2]*(450*c*x -
15*ArcSin[c*x]*(30*Sqrt[1 - c^2*x^2] + 5*Cos[3*ArcSin[c*x]] - 3*Cos[5*ArcSin[c*x]]) + 25*Sin[3*ArcSin[c*x]] -
9*Sin[5*ArcSin[c*x]]))/(3600*Sqrt[1 - c^2*x^2])

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Maple [A]  time = 0.237, size = 652, normalized size = 1.8 \begin{align*}{\frac{a}{5} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{5}{2}}}}+{\frac{ad}{3} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-a{d}^{{\frac{5}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{-{c}^{2}d{x}^{2}+d} \right ) } \right ) +a\sqrt{-{c}^{2}d{x}^{2}+d}{d}^{2}-{\frac{ib{d}^{2}}{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{ib{d}^{2}}{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{b{d}^{2}\arcsin \left ( cx \right ){x}^{6}{c}^{6}}{5\,{c}^{2}{x}^{2}-5}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{14\,{d}^{2}b\arcsin \left ( cx \right ){c}^{4}{x}^{4}}{15\,{c}^{2}{x}^{2}-15}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{34\,{d}^{2}b\arcsin \left ( cx \right ){c}^{2}{x}^{2}}{15\,{c}^{2}{x}^{2}-15}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{b{d}^{2}{x}^{5}{c}^{5}}{25\,{c}^{2}{x}^{2}-25}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{11\,b{d}^{2}{x}^{3}{c}^{3}}{45\,{c}^{2}{x}^{2}-45}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{23\,b{d}^{2}xc}{15\,{c}^{2}{x}^{2}-15}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{b{d}^{2}\arcsin \left ( cx \right ) }{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{b{d}^{2}\arcsin \left ( cx \right ) }{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{23\,b{d}^{2}\arcsin \left ( cx \right ) }{15\,{c}^{2}{x}^{2}-15}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }} \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,x)

[Out]

1/5*(-c^2*d*x^2+d)^(5/2)*a+1/3*a*d*(-c^2*d*x^2+d)^(3/2)-a*d^(5/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)+a
*(-c^2*d*x^2+d)^(1/2)*d^2-I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*d^2*polylog(2,-I*c*x-(-c^2
*x^2+1)^(1/2))+I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*d^2*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2
))+1/5*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(c^2*x^2-1)*arcsin(c*x)*x^6*c^6-14/15*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(c^2*x^
2-1)*arcsin(c*x)*x^4*c^4+34/15*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(c^2*x^2-1)*arcsin(c*x)*x^2*c^2+1/25*b*(-d*(c^2*x^
2-1))^(1/2)*d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^5*c^5-11/45*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(c^2*x^2-1)*(-c^2*x^
2+1)^(1/2)*x^3*c^3+23/15*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x*c+b*(-d*(c^2*x^2-1))^(1
/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*d^2*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-b*(-d*(c^2*x^2-1))^(1/2)*(-c
^2*x^2+1)^(1/2)/(c^2*x^2-1)*d^2*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-23/15*b*(-d*(c^2*x^2-1))^(1/2)*d^2/
(c^2*x^2-1)*arcsin(c*x)

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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,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}, 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,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, 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,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}\,{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,x, algorithm="giac")

[Out]

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