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

Optimal. Leaf size=361 \[ d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {2 d^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {1}{5} \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {i b d^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {i b d^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {23 b c d^2 x \sqrt {d-c^2 d x^2}}{15 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}+\frac {11 b c^3 d^2 x^3 \sqrt {d-c^2 d x^2}}{45 \sqrt {1-c^2 x^2}} \]

[Out]

1/3*d*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))+1/5*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))+d^2*(a+b*arcsin(c*x))*
(-c^2*d*x^2+d)^(1/2)-23/15*b*c*d^2*x*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+11/45*b*c^3*d^2*x^3*(-c^2*d*x^2+d
)^(1/2)/(-c^2*x^2+1)^(1/2)-1/25*b*c^5*d^2*x^5*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-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)+I*b*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)-I*b*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)

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Rubi [A]  time = 0.46, antiderivative size = 361, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 2.03, size = 394, normalized size = 1.09 \[ -a d^{5/2} \log \left (\sqrt {d} \sqrt {d-c^2 d x^2}+d\right )+\frac {1}{15} a d^2 \left (3 c^4 x^4-11 c^2 x^2+23\right ) \sqrt {d-c^2 d x^2}+a d^{5/2} \log (x)+\frac {b d^2 \sqrt {d-c^2 d x^2} \left (\sqrt {1-c^2 x^2} \sin ^{-1}(c x)+i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-c x+\sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )}{\sqrt {1-c^2 x^2}}-\frac {b d^2 \sqrt {d-c^2 d x^2} \left (-3 \sin ^{-1}(c x) \left (3 \sqrt {1-c^2 x^2}+\cos \left (3 \sin ^{-1}(c x)\right )\right )+9 c x+\sin \left (3 \sin ^{-1}(c x)\right )\right )}{18 \sqrt {1-c^2 x^2}}+\frac {b d^2 \sqrt {d-c^2 d x^2} \left (-15 \sin ^{-1}(c x) \left (30 \sqrt {1-c^2 x^2}+5 \cos \left (3 \sin ^{-1}(c x)\right )-3 \cos \left (5 \sin ^{-1}(c x)\right )\right )+450 c x+25 \sin \left (3 \sin ^{-1}(c x)\right )-9 \sin \left (5 \sin ^{-1}(c x)\right )\right )}{3600 \sqrt {1-c^2 x^2}} \]

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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fricas [F]  time = 1.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a c^{4} d^{2} x^{4} - 2 \, a c^{2} d^{2} x^{2} + a d^{2} + {\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{x}, x\right ) \]

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,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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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,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 [A]  time = 0.34, size = 652, normalized size = 1.81 \[ \frac {\left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}} a}{5}+\frac {a d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{3}-a \,d^{\frac {5}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )+a \sqrt {-c^{2} d \,x^{2}+d}\, d^{2}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d^{2} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}-\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d^{2} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}-\frac {23 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d^{2} \arcsin \left (c x \right )}{15 \left (c^{2} x^{2}-1\right )}+\frac {34 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d^{2} \arcsin \left (c x \right ) x^{2} c^{2}}{15 \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d^{2} \arcsin \left (c x \right ) x^{6} c^{6}}{5 c^{2} x^{2}-5}-\frac {14 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d^{2} \arcsin \left (c x \right ) x^{4} c^{4}}{15 \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d^{2} \sqrt {-c^{2} x^{2}+1}\, x^{5} c^{5}}{25 c^{2} x^{2}-25}-\frac {11 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d^{2} \sqrt {-c^{2} x^{2}+1}\, x^{3} c^{3}}{45 \left (c^{2} x^{2}-1\right )}+\frac {23 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d^{2} \sqrt {-c^{2} x^{2}+1}\, x c}{15 \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d^{2} \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1} \]

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)*d^2/(c^2*x^2-1)*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)*d^2/(c^2*x^2-1)*polylog(2,-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)+34/15*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(c^2*x^2-1)*a
rcsin(c*x)*x^2*c^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+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)*d^2/(c^2*x^2-1)*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)*d^2/(c^2*x^2-1)*arcsin(c*x)*ln(1+I*
c*x+(-c^2*x^2+1)^(1/2))

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

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

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

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,x)

[Out]

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

[Out]

Timed out

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