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 \[ \frac {5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac {15}{8} c^4 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {15 c^4 d^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 \sqrt {1-c^2 x^2}}-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 x \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}+\frac {15 i b c^4 d^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}-\frac {15 i b c^4 d^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}+\frac {9 b c^3 d^2 \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}} \]

[Out]

5/8*c^2*d*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^2-1/4*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^4+15/8*c^4*d
^2*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)-1/12*b*c*d^2*(-c^2*d*x^2+d)^(1/2)/x^3/(-c^2*x^2+1)^(1/2)+9/8*b*c^3*d
^2*(-c^2*d*x^2+d)^(1/2)/x/(-c^2*x^2+1)^(1/2)-b*c^5*d^2*x*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-15/4*c^4*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)+15/8*I*b*c^4*d^2*p
olylog(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)-15/8*I*b*c^4*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 = 389, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 6.46, size = 640, normalized size = 1.65 \[ \frac {15}{8} a c^4 d^{5/2} \log (x)-\frac {15}{8} a c^4 d^{5/2} \log \left (\sqrt {d} \sqrt {d-c^2 d x^2}+d\right )+\frac {a d^2 \left (8 c^4 x^4+9 c^2 x^2-2\right ) \sqrt {d-c^2 d x^2}}{8 x^4}-\frac {b c^4 d^3 \sqrt {1-c^2 x^2} \left (-4 i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )+4 i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-4 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+4 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )-2 \tan \left (\frac {1}{2} \sin ^{-1}(c x)\right )-2 \cot \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin ^{-1}(c x) \csc ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )+\sin ^{-1}(c x) \sec ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{4 \sqrt {d-c^2 d x^2}}+\frac {b c^4 d^2 \sqrt {d-c^2 d x^2} \left (\sqrt {1-c^2 x^2} \sin ^{-1}(c x)+i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-c x+\sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )}{\sqrt {1-c^2 x^2}}+\frac {b c^4 d^2 \sqrt {d-c^2 d x^2} \left (-\frac {16 \sin ^4\left (\frac {1}{2} \sin ^{-1}(c x)\right )}{c^3 x^3}-24 i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )+24 i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-24 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+24 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )+8 \tan \left (\frac {1}{2} \sin ^{-1}(c x)\right )+8 \cot \left (\frac {1}{2} \sin ^{-1}(c x)\right )-c x \csc ^4\left (\frac {1}{2} \sin ^{-1}(c x)\right )-3 \sin ^{-1}(c x) \csc ^4\left (\frac {1}{2} \sin ^{-1}(c x)\right )+6 \sin ^{-1}(c x) \csc ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )+3 \sin ^{-1}(c x) \sec ^4\left (\frac {1}{2} \sin ^{-1}(c x)\right )-6 \sin ^{-1}(c x) \sec ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{192 \sqrt {1-c^2 x^2}} \]

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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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a c^{4} d^{2} x^{4} - 2 \, a c^{2} d^{2} x^{2} + a d^{2} + {\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{x^{5}}, x\right ) \]

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^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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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^5,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 = 0.48, size = 727, normalized size = 1.87 \[ -\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {7}{2}}}{4 d \,x^{4}}+\frac {3 a \,c^{2} \left (-c^{2} d \,x^{2}+d \right )^{\frac {7}{2}}}{8 d \,x^{2}}+\frac {3 a \,c^{4} \left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{8}+\frac {5 a \,c^{4} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{8}-\frac {15 a \,c^{4} d^{\frac {5}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{8}+\frac {15 a \,c^{4} \sqrt {-c^{2} d \,x^{2}+d}\, d^{2}}{8}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d^{2} c^{5} \sqrt {-c^{2} x^{2}+1}\, x}{c^{2} x^{2}-1}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d^{2} c^{6} \arcsin \left (c x \right ) x^{2}}{c^{2} x^{2}-1}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d^{2} c^{4} \arcsin \left (c x \right )}{8 c^{2} x^{2}-8}-\frac {9 b \,d^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{3}}{8 x \left (c^{2} x^{2}-1\right )}-\frac {11 b \,d^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c^{2}}{8 x^{2} \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}{12 x^{3} \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 )}{4 x^{4} \left (c^{2} x^{2}-1\right )}-\frac {15 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d^{2} c^{4} \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{8 c^{2} x^{2}-8}+\frac {15 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d^{2} c^{4} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{8 c^{2} x^{2}-8}+\frac {15 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d^{2} c^{4} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{8 c^{2} x^{2}-8}-\frac {15 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d^{2} c^{4} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{8 c^{2} x^{2}-8} \]

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)*d^2*c^5/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x+b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^
6/(c^2*x^2-1)*arcsin(c*x)*x^2+1/8*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^4/(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)*d^2*c^4/(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)*d^2*c^4/(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)*d^2*c^4/(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)*d^2*c^4/(8*c^2*x^2-8)*polylog(
2,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 \[ b \sqrt {d} \int \frac {{\left (c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x^{5}}\,{d x} - \frac {1}{8} \, {\left (15 \, c^{4} d^{\frac {5}{2}} \log \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {d}}{{\left | x \right |}} + \frac {2 \, d}{{\left | x \right |}}\right ) - 3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} c^{4} - 5 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d - 15 \, \sqrt {-c^{2} d x^{2} + d} c^{4} d^{2} - \frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}} c^{2}}{d x^{2}} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}}}{d x^{4}}\right )} a \]

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^5,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^5, x) - 1/8*(15*c^4*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^4 - 5*(-c^2*d*x^2 + d)^(3/2)*c^4*d - 15*sqrt(-c^2*d*x^2 + d)*c^4*d^2 - 3*(-c^2*d*x^2 +
d)^(7/2)*c^2/(d*x^2) + 2*(-c^2*d*x^2 + d)^(7/2)/(d*x^4))*a

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

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^5,x)

[Out]

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

[Out]

Timed out

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