3.45 \(\int \frac{a+b \sin ^{-1}(c x)}{x^4 (d-c^2 d x^2)^2} \, dx\)

Optimal. Leaf size=259 \[ \frac{5 i b c^3 \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac{5 i b c^3 \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}+\frac{5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac{5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}-\frac{a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac{5 i c^3 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{b c^3}{3 d^2 \sqrt{1-c^2 x^2}}-\frac{b c}{6 d^2 x^2 \sqrt{1-c^2 x^2}}-\frac{13 b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{6 d^2} \]

[Out]

-(b*c^3)/(3*d^2*Sqrt[1 - c^2*x^2]) - (b*c)/(6*d^2*x^2*Sqrt[1 - c^2*x^2]) - (a + b*ArcSin[c*x])/(3*d^2*x^3*(1 -
c^2*x^2)) - (5*c^2*(a + b*ArcSin[c*x]))/(3*d^2*x*(1 - c^2*x^2)) + (5*c^4*x*(a + b*ArcSin[c*x]))/(2*d^2*(1 - c
^2*x^2)) - ((5*I)*c^3*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/d^2 - (13*b*c^3*ArcTanh[Sqrt[1 - c^2*x^2]
])/(6*d^2) + (((5*I)/2)*b*c^3*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d^2 - (((5*I)/2)*b*c^3*PolyLog[2, I*E^(I*Arc
Sin[c*x])])/d^2

________________________________________________________________________________________

Rubi [A]  time = 0.307726, antiderivative size = 285, normalized size of antiderivative = 1.1, number of steps used = 19, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {4701, 4655, 4657, 4181, 2279, 2391, 261, 266, 51, 63, 208} \[ \frac{5 i b c^3 \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac{5 i b c^3 \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}+\frac{5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac{5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}-\frac{a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac{5 i c^3 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{5 b c^3}{6 d^2 \sqrt{1-c^2 x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{2 d^2 x^2}+\frac{b c}{3 d^2 x^2 \sqrt{1-c^2 x^2}}-\frac{13 b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{6 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(x^4*(d - c^2*d*x^2)^2),x]

[Out]

(-5*b*c^3)/(6*d^2*Sqrt[1 - c^2*x^2]) + (b*c)/(3*d^2*x^2*Sqrt[1 - c^2*x^2]) - (b*c*Sqrt[1 - c^2*x^2])/(2*d^2*x^
2) - (a + b*ArcSin[c*x])/(3*d^2*x^3*(1 - c^2*x^2)) - (5*c^2*(a + b*ArcSin[c*x]))/(3*d^2*x*(1 - c^2*x^2)) + (5*
c^4*x*(a + b*ArcSin[c*x]))/(2*d^2*(1 - c^2*x^2)) - ((5*I)*c^3*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/d
^2 - (13*b*c^3*ArcTanh[Sqrt[1 - c^2*x^2]])/(6*d^2) + (((5*I)/2)*b*c^3*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d^2
- (((5*I)/2)*b*c^3*PolyLog[2, I*E^(I*ArcSin[c*x])])/d^2

Rule 4701

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 + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

Rule 4655

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSin[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(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])/(2*(p + 1)*(1 - c^2*x^2)^FracPart[p
]), Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 4657

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

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}(c x)}{x^4 \left (d-c^2 d x^2\right )^2} \, dx &=-\frac{a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}+\frac{1}{3} \left (5 c^2\right ) \int \frac{a+b \sin ^{-1}(c x)}{x^2 \left (d-c^2 d x^2\right )^2} \, dx+\frac{(b c) \int \frac{1}{x^3 \left (1-c^2 x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=-\frac{a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac{5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\left (5 c^4\right ) \int \frac{a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^2} \, dx+\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{6 d^2}+\frac{\left (5 b c^3\right ) \int \frac{1}{x \left (1-c^2 x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=\frac{b c}{3 d^2 x^2 \sqrt{1-c^2 x^2}}-\frac{a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac{5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\frac{5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{2 d^2}+\frac{\left (5 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{6 d^2}-\frac{\left (5 b c^5\right ) \int \frac{x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{2 d^2}+\frac{\left (5 c^4\right ) \int \frac{a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{2 d}\\ &=-\frac{5 b c^3}{6 d^2 \sqrt{1-c^2 x^2}}+\frac{b c}{3 d^2 x^2 \sqrt{1-c^2 x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{2 d^2 x^2}-\frac{a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac{5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\frac{5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac{\left (5 c^3\right ) \operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 d^2}+\frac{\left (b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{4 d^2}+\frac{\left (5 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{6 d^2}\\ &=-\frac{5 b c^3}{6 d^2 \sqrt{1-c^2 x^2}}+\frac{b c}{3 d^2 x^2 \sqrt{1-c^2 x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{2 d^2 x^2}-\frac{a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac{5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\frac{5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac{5 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^2}\right )}{2 d^2}-\frac{(5 b c) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^2}\right )}{3 d^2}-\frac{\left (5 b c^3\right ) \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 d^2}+\frac{\left (5 b c^3\right ) \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 d^2}\\ &=-\frac{5 b c^3}{6 d^2 \sqrt{1-c^2 x^2}}+\frac{b c}{3 d^2 x^2 \sqrt{1-c^2 x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{2 d^2 x^2}-\frac{a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac{5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\frac{5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac{5 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac{13 b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{6 d^2}+\frac{\left (5 i b c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac{\left (5 i b c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 d^2}\\ &=-\frac{5 b c^3}{6 d^2 \sqrt{1-c^2 x^2}}+\frac{b c}{3 d^2 x^2 \sqrt{1-c^2 x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{2 d^2 x^2}-\frac{a+b \sin ^{-1}(c x)}{3 d^2 x^3 \left (1-c^2 x^2\right )}-\frac{5 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 x \left (1-c^2 x^2\right )}+\frac{5 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}-\frac{5 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2}-\frac{13 b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{6 d^2}+\frac{5 i b c^3 \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac{5 i b c^3 \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{2 d^2}\\ \end{align*}

Mathematica [A]  time = 0.929428, size = 426, normalized size = 1.64 \[ -\frac{-30 i b c^3 \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+30 i b c^3 \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )+\frac{6 a c^4 x}{c^2 x^2-1}+\frac{24 a c^2}{x}+15 a c^3 \log (1-c x)-15 a c^3 \log (c x+1)+\frac{4 a}{x^3}-\frac{3 b c^3 \sqrt{1-c^2 x^2}}{c x-1}+\frac{3 b c^3 \sqrt{1-c^2 x^2}}{c x+1}+\frac{2 b c \sqrt{1-c^2 x^2}}{x^2}+26 b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )+\frac{3 b c^3 \sin ^{-1}(c x)}{c x-1}+\frac{3 b c^3 \sin ^{-1}(c x)}{c x+1}+15 i \pi b c^3 \sin ^{-1}(c x)+\frac{24 b c^2 \sin ^{-1}(c x)}{x}-30 b c^3 \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-15 \pi b c^3 \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+30 b c^3 \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-15 \pi b c^3 \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+15 \pi b c^3 \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+15 \pi b c^3 \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+\frac{4 b \sin ^{-1}(c x)}{x^3}}{12 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c*x])/(x^4*(d - c^2*d*x^2)^2),x]

[Out]

-((4*a)/x^3 + (24*a*c^2)/x + (2*b*c*Sqrt[1 - c^2*x^2])/x^2 - (3*b*c^3*Sqrt[1 - c^2*x^2])/(-1 + c*x) + (3*b*c^3
*Sqrt[1 - c^2*x^2])/(1 + c*x) + (6*a*c^4*x)/(-1 + c^2*x^2) + (15*I)*b*c^3*Pi*ArcSin[c*x] + (4*b*ArcSin[c*x])/x
^3 + (24*b*c^2*ArcSin[c*x])/x + (3*b*c^3*ArcSin[c*x])/(-1 + c*x) + (3*b*c^3*ArcSin[c*x])/(1 + c*x) + 26*b*c^3*
ArcTanh[Sqrt[1 - c^2*x^2]] - 15*b*c^3*Pi*Log[1 - I*E^(I*ArcSin[c*x])] - 30*b*c^3*ArcSin[c*x]*Log[1 - I*E^(I*Ar
cSin[c*x])] - 15*b*c^3*Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 30*b*c^3*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + 1
5*a*c^3*Log[1 - c*x] - 15*a*c^3*Log[1 + c*x] + 15*b*c^3*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + 15*b*c^3*Pi*Log
[Sin[(Pi + 2*ArcSin[c*x])/4]] - (30*I)*b*c^3*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (30*I)*b*c^3*PolyLog[2, I*E^
(I*ArcSin[c*x])])/(12*d^2)

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Maple [A]  time = 0.23, size = 426, normalized size = 1.6 \begin{align*} -{\frac{{c}^{3}a}{4\,{d}^{2} \left ( cx-1 \right ) }}-{\frac{5\,{c}^{3}a\ln \left ( cx-1 \right ) }{4\,{d}^{2}}}-{\frac{{c}^{3}a}{4\,{d}^{2} \left ( cx+1 \right ) }}+{\frac{5\,{c}^{3}a\ln \left ( cx+1 \right ) }{4\,{d}^{2}}}-{\frac{a}{3\,{d}^{2}{x}^{3}}}-2\,{\frac{{c}^{2}a}{{d}^{2}x}}-{\frac{5\,{c}^{4}b\arcsin \left ( cx \right ) x}{2\,{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{b{c}^{3}}{3\,{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{5\,{c}^{2}b\arcsin \left ( cx \right ) }{3\,{d}^{2}x \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{bc}{6\,{d}^{2}{x}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{b\arcsin \left ( cx \right ) }{3\,{d}^{2}{x}^{3} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{13\,b{c}^{3}}{6\,{d}^{2}}\ln \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1}-1 \right ) }-{\frac{13\,b{c}^{3}}{6\,{d}^{2}}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{5\,b{c}^{3}\arcsin \left ( cx \right ) }{2\,{d}^{2}}\ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{{\frac{5\,i}{2}}{c}^{3}b}{{d}^{2}}{\it dilog} \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{{\frac{5\,i}{2}}{c}^{3}b}{{d}^{2}}{\it dilog} \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{5\,b{c}^{3}\arcsin \left ( cx \right ) }{2\,{d}^{2}}\ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^2,x)

[Out]

-1/4*c^3*a/d^2/(c*x-1)-5/4*c^3*a/d^2*ln(c*x-1)-1/4*c^3*a/d^2/(c*x+1)+5/4*c^3*a/d^2*ln(c*x+1)-1/3*a/d^2/x^3-2*c
^2*a/d^2/x-5/2*c^4*b/d^2/(c^2*x^2-1)*arcsin(c*x)*x+1/3*c^3*b/d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)+5/3*c^2*b/d^2/
x/(c^2*x^2-1)*arcsin(c*x)+1/6*c*b/d^2/x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)+1/3*b/d^2/x^3/(c^2*x^2-1)*arcsin(c*x)
+13/6*c^3*b/d^2*ln(I*c*x+(-c^2*x^2+1)^(1/2)-1)-13/6*c^3*b/d^2*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-5/2*c^3*b/d^2*arc
sin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+5/2*I*c^3*b/d^2*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-5/2*I*c^3*b/
d^2*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+5/2*c^3*b/d^2*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{12} \,{\left (\frac{15 \, c^{3} \log \left (c x + 1\right )}{d^{2}} - \frac{15 \, c^{3} \log \left (c x - 1\right )}{d^{2}} - \frac{2 \,{\left (15 \, c^{4} x^{4} - 10 \, c^{2} x^{2} - 2\right )}}{c^{2} d^{2} x^{5} - d^{2} x^{3}}\right )} a + \frac{{\left (15 \,{\left (c^{5} x^{5} - c^{3} x^{3}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) - 15 \,{\left (c^{5} x^{5} - c^{3} x^{3}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \,{\left (15 \, c^{4} x^{4} - 10 \, c^{2} x^{2} - 2\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) -{\left (c^{2} d^{2} x^{5} - d^{2} x^{3}\right )} \int \frac{{\left (30 \, c^{5} x^{4} - 20 \, c^{3} x^{2} - 15 \,{\left (c^{6} x^{5} - c^{4} x^{3}\right )} \log \left (c x + 1\right ) + 15 \,{\left (c^{6} x^{5} - c^{4} x^{3}\right )} \log \left (-c x + 1\right ) - 4 \, c\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{4} d^{2} x^{7} - 2 \, c^{2} d^{2} x^{5} + d^{2} x^{3}}\,{d x}\right )} b}{12 \,{\left (c^{2} d^{2} x^{5} - d^{2} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

1/12*(15*c^3*log(c*x + 1)/d^2 - 15*c^3*log(c*x - 1)/d^2 - 2*(15*c^4*x^4 - 10*c^2*x^2 - 2)/(c^2*d^2*x^5 - d^2*x
^3))*a + 1/12*(15*(c^5*x^5 - c^3*x^3)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - 15*(c^5*x^5 -
c^3*x^3)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) - 2*(15*c^4*x^4 - 10*c^2*x^2 - 2)*arctan2(c*
x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + 12*(c^2*d^2*x^5 - d^2*x^3)*integrate(-1/12*(30*c^5*x^4 - 20*c^3*x^2 - 15*(c
^6*x^5 - c^4*x^3)*log(c*x + 1) + 15*(c^6*x^5 - c^4*x^3)*log(-c*x + 1) - 4*c)*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^4
*d^2*x^7 - 2*c^2*d^2*x^5 + d^2*x^3), x))*b/(c^2*d^2*x^5 - d^2*x^3)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \arcsin \left (c x\right ) + a}{c^{4} d^{2} x^{8} - 2 \, c^{2} d^{2} x^{6} + d^{2} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*arcsin(c*x) + a)/(c^4*d^2*x^8 - 2*c^2*d^2*x^6 + d^2*x^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{4} x^{8} - 2 c^{2} x^{6} + x^{4}}\, dx + \int \frac{b \operatorname{asin}{\left (c x \right )}}{c^{4} x^{8} - 2 c^{2} x^{6} + x^{4}}\, dx}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))/x**4/(-c**2*d*x**2+d)**2,x)

[Out]

(Integral(a/(c**4*x**8 - 2*c**2*x**6 + x**4), x) + Integral(b*asin(c*x)/(c**4*x**8 - 2*c**2*x**6 + x**4), x))/
d**2

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Giac [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((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^2,x, algorithm="giac")

[Out]

Timed out