3.83 \(\int \frac {(d-c^2 d x^2)^{3/2} (a+b \sin ^{-1}(c x))}{x^5} \, dx\)

Optimal. Leaf size=307 \[ \frac {3 c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac {3 c^4 d \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 \sqrt {1-c^2 x^2}}-\frac {b c d \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {3 i b c^4 d \sqrt {d-c^2 d x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}-\frac {3 i b c^4 d \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 d \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}} \]

[Out]

-1/4*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^4+3/8*c^2*d*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/x^2-1/12*b*c*
d*(-c^2*d*x^2+d)^(1/2)/x^3/(-c^2*x^2+1)^(1/2)+5/8*b*c^3*d*(-c^2*d*x^2+d)^(1/2)/x/(-c^2*x^2+1)^(1/2)-3/4*c^4*d*
(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)+3/8*I*b*c^4*d*poly
log(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)-3/8*I*b*c^4*d*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 = 0.32, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {4695, 4693, 30, 4709, 4183, 2279, 2391, 14} \[ \frac {3 i b c^4 d \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}-\frac {3 i b c^4 d \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}+\frac {3 c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac {3 c^4 d \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 d \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}}-\frac {b c d \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 4693

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

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 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 )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x^5} \, dx &=-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac {1}{4} \left (3 c^2 d\right ) \int \frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x^3} \, dx+\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int \frac {1-c^2 x^2}{x^4} \, dx}{4 \sqrt {1-c^2 x^2}}\\ &=\frac {3 c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int \left (\frac {1}{x^4}-\frac {c^2}{x^2}\right ) \, dx}{4 \sqrt {1-c^2 x^2}}-\frac {\left (3 b c^3 d \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{x^2} \, dx}{8 \sqrt {1-c^2 x^2}}+\frac {\left (3 c^4 d \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 {b c d \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 d \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}}+\frac {3 c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac {\left (3 c^4 d \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 \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 d \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}}+\frac {3 c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac {3 c^4 d \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 (3 b c^4 d \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 (3 b c^4 d \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 \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 d \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}}+\frac {3 c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac {3 c^4 d \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 (3 i b c^4 d \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 (3 i b c^4 d \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 \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 d \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}}+\frac {3 c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}-\frac {3 c^4 d \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 {3 i b c^4 d \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 {3 i b c^4 d \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 = 5.97, size = 494, normalized size = 1.61 \[ \frac {3}{8} a c^4 d^{3/2} \log (x)+\frac {a d \left (5 c^2 x^2-2\right ) \sqrt {d-c^2 d x^2}}{8 x^4}-\frac {3}{8} a c^4 d^{3/2} \log \left (\sqrt {d} \sqrt {d-c^2 d x^2}+d\right )-\frac {b c^4 d^2 \sqrt {1-c^2 x^2} \left (-4 i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )+4 i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-4 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+4 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )-2 \tan \left (\frac {1}{2} \sin ^{-1}(c x)\right )-2 \cot \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin ^{-1}(c x) \csc ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )+\sin ^{-1}(c x) \sec ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{8 \sqrt {d-c^2 d x^2}}+\frac {b c^4 d \sqrt {d-c^2 d x^2} \left (-\frac {16 \sin ^4\left (\frac {1}{2} \sin ^{-1}(c x)\right )}{c^3 x^3}-24 i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )+24 i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-24 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+24 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )+8 \tan \left (\frac {1}{2} \sin ^{-1}(c x)\right )+8 \cot \left (\frac {1}{2} \sin ^{-1}(c x)\right )-c x \csc ^4\left (\frac {1}{2} \sin ^{-1}(c x)\right )-3 \sin ^{-1}(c x) \csc ^4\left (\frac {1}{2} \sin ^{-1}(c x)\right )+6 \sin ^{-1}(c x) \csc ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )+3 \sin ^{-1}(c x) \sec ^4\left (\frac {1}{2} \sin ^{-1}(c x)\right )-6 \sin ^{-1}(c x) \sec ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{192 \sqrt {1-c^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*d*(-2 + 5*c^2*x^2)*Sqrt[d - c^2*d*x^2])/(8*x^4) + (3*a*c^4*d^(3/2)*Log[x])/8 - (3*a*c^4*d^(3/2)*Log[d + Sqr
t[d]*Sqrt[d - c^2*d*x^2]])/8 - (b*c^4*d^2*Sqrt[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]))/(8*Sqrt[d - c^2*d*x^2]) + (b*c^4*d*Sqrt[d - c^2*d*x^2]*(8*Cot[ArcSin[c*x]/2] + 6*ArcSin[c*x]*Csc[Ar
cSin[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*A
rcSin[c*x])] + 24*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] - (24*I)*PolyLog[2, -E^(I*ArcSin[c*x])] + (24*I)*Poly
Log[2, E^(I*ArcSin[c*x])] - 6*ArcSin[c*x]*Sec[ArcSin[c*x]/2]^2 + 3*ArcSin[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 = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a c^{2} d x^{2} - a d + {\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^5,x, algorithm="fricas")

[Out]

integral(-(a*c^2*d*x^2 - a*d + (b*c^2*d*x^2 - b*d)*arcsin(c*x))*sqrt(-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)^(3/2)*(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 [B]  time = 0.44, size = 601, normalized size = 1.96 \[ -\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{4 d \,x^{4}}+\frac {a \,c^{2} \left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{8 d \,x^{2}}+\frac {a \,c^{4} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{8}-\frac {3 a \,c^{4} d^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{8}+\frac {3 a \,c^{4} \sqrt {-c^{2} d \,x^{2}+d}\, d}{8}+\frac {5 b d \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c^{4}}{8 \left (c^{2} x^{2}-1\right )}-\frac {5 b d \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{3}}{8 x \left (c^{2} x^{2}-1\right )}-\frac {7 b d \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c^{2}}{8 x^{2} \left (c^{2} x^{2}-1\right )}+\frac {b d \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c}{12 x^{3} \left (c^{2} x^{2}-1\right )}+\frac {b d \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{4 x^{4} \left (c^{2} x^{2}-1\right )}-\frac {3 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d \,c^{4} \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right ) \arcsin \left (c x \right )}{8 c^{2} x^{2}-8}+\frac {3 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d \,c^{4} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right ) \arcsin \left (c x \right )}{8 c^{2} x^{2}-8}+\frac {3 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d \,c^{4} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{8 c^{2} x^{2}-8}-\frac {3 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, d \,c^{4} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{8 c^{2} x^{2}-8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a/d/x^4*(-c^2*d*x^2+d)^(5/2)+1/8*a*c^2/d/x^2*(-c^2*d*x^2+d)^(5/2)+1/8*a*c^4*(-c^2*d*x^2+d)^(3/2)-3/8*a*c^
4*d^(3/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)+3/8*a*c^4*(-c^2*d*x^2+d)^(1/2)*d+5/8*b*d*(-d*(c^2*x^2-1))
^(1/2)/(c^2*x^2-1)*arcsin(c*x)*c^4-5/8*b*d*(-d*(c^2*x^2-1))^(1/2)/x/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c^3-7/8*b*d
*(-d*(c^2*x^2-1))^(1/2)/x^2/(c^2*x^2-1)*arcsin(c*x)*c^2+1/12*b*d*(-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*(-d*(c^2*x^2-1))^(1/2)/x^4/(c^2*x^2-1)*arcsin(c*x)-3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2
+1)^(1/2)*d*c^4/(8*c^2*x^2-8)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))*arcsin(c*x)+3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+
1)^(1/2)*d*c^4/(8*c^2*x^2-8)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))*arcsin(c*x)+3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2
+1)^(1/2)*d*c^4/(8*c^2*x^2-8)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))-3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1
/2)*d*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 \[ -b \sqrt {d} \int \frac {{\left (c^{2} d x^{2} - d\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x^{5}}\,{d x} - \frac {1}{8} \, {\left (3 \, c^{4} d^{\frac {3}{2}} \log \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {d}}{{\left | x \right |}} + \frac {2 \, d}{{\left | x \right |}}\right ) - {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} - 3 \, \sqrt {-c^{2} d x^{2} + d} c^{4} d - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} c^{2}}{d x^{2}} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{d x^{4}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^5,x, algorithm="maxima")

[Out]

-b*sqrt(d)*integrate((c^2*d*x^2 - d)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/x
^5, x) - 1/8*(3*c^4*d^(3/2)*log(2*sqrt(-c^2*d*x^2 + d)*sqrt(d)/abs(x) + 2*d/abs(x)) - (-c^2*d*x^2 + d)^(3/2)*c
^4 - 3*sqrt(-c^2*d*x^2 + d)*c^4*d - (-c^2*d*x^2 + d)^(5/2)*c^2/(d*x^2) + 2*(-c^2*d*x^2 + d)^(5/2)/(d*x^4))*a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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