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

Optimal. Leaf size=242 \[ \frac{15 i b c \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}-\frac{15 i b c \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}-\frac{15 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 d^3}-\frac{7 b c}{8 d^3 \sqrt{1-c^2 x^2}}-\frac{b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{b c \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{d^3} \]

[Out]

-(b*c)/(12*d^3*(1 - c^2*x^2)^(3/2)) - (7*b*c)/(8*d^3*Sqrt[1 - c^2*x^2]) - (a + b*ArcSin[c*x])/(d^3*x*(1 - c^2*
x^2)^2) + (5*c^2*x*(a + b*ArcSin[c*x]))/(4*d^3*(1 - c^2*x^2)^2) + (15*c^2*x*(a + b*ArcSin[c*x]))/(8*d^3*(1 - c
^2*x^2)) - (((15*I)/4)*c*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/d^3 - (b*c*ArcTanh[Sqrt[1 - c^2*x^2]])
/d^3 + (((15*I)/8)*b*c*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d^3 - (((15*I)/8)*b*c*PolyLog[2, I*E^(I*ArcSin[c*x]
)])/d^3

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Rubi [A]  time = 0.24238, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 16, 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} \[ \frac{15 i b c \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}-\frac{15 i b c \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}-\frac{15 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 d^3}-\frac{7 b c}{8 d^3 \sqrt{1-c^2 x^2}}-\frac{b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{b c \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c)/(12*d^3*(1 - c^2*x^2)^(3/2)) - (7*b*c)/(8*d^3*Sqrt[1 - c^2*x^2]) - (a + b*ArcSin[c*x])/(d^3*x*(1 - c^2*
x^2)^2) + (5*c^2*x*(a + b*ArcSin[c*x]))/(4*d^3*(1 - c^2*x^2)^2) + (15*c^2*x*(a + b*ArcSin[c*x]))/(8*d^3*(1 - c
^2*x^2)) - (((15*I)/4)*c*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/d^3 - (b*c*ArcTanh[Sqrt[1 - c^2*x^2]])
/d^3 + (((15*I)/8)*b*c*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d^3 - (((15*I)/8)*b*c*PolyLog[2, I*E^(I*ArcSin[c*x]
)])/d^3

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^2 \left (d-c^2 d x^2\right )^3} \, dx &=-\frac{a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\left (5 c^2\right ) \int \frac{a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^3} \, dx+\frac{(b c) \int \frac{1}{x \left (1-c^2 x^2\right )^{5/2}} \, dx}{d^3}\\ &=-\frac{a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )^{5/2}} \, dx,x,x^2\right )}{2 d^3}-\frac{\left (5 b c^3\right ) \int \frac{x}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{4 d^3}+\frac{\left (15 c^2\right ) \int \frac{a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d}\\ &=-\frac{b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{2 d^3}-\frac{\left (15 b c^3\right ) \int \frac{x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{8 d^3}+\frac{\left (15 c^2\right ) \int \frac{a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{8 d^2}\\ &=-\frac{b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{7 b c}{8 d^3 \sqrt{1-c^2 x^2}}-\frac{a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac{(15 c) \operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 d^3}+\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{2 d^3}\\ &=-\frac{b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{7 b c}{8 d^3 \sqrt{1-c^2 x^2}}-\frac{a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}-\frac{15 i c \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{b \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^2}\right )}{c d^3}-\frac{(15 b c) \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 d^3}+\frac{(15 b c) \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 d^3}\\ &=-\frac{b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{7 b c}{8 d^3 \sqrt{1-c^2 x^2}}-\frac{a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}-\frac{15 i c \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{b c \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{d^3}+\frac{(15 i b c) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 d^3}-\frac{(15 i b c) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 d^3}\\ &=-\frac{b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{7 b c}{8 d^3 \sqrt{1-c^2 x^2}}-\frac{a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}-\frac{15 i c \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{b c \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{d^3}+\frac{15 i b c \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}-\frac{15 i b c \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}\\ \end{align*}

Mathematica [B]  time = 1.51178, size = 512, normalized size = 2.12 \[ -\frac{-30 i b c \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+30 i b c \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )+\frac{14 a c^2 x}{c^2 x^2-1}-\frac{4 a c^2 x}{\left (c^2 x^2-1\right )^2}+15 a c \log (1-c x)-15 a c \log (c x+1)+\frac{16 a}{x}-\frac{b c^2 x \sqrt{1-c^2 x^2}}{3 (c x-1)^2}+\frac{b c^2 x \sqrt{1-c^2 x^2}}{3 (c x+1)^2}-\frac{7 b c \sqrt{1-c^2 x^2}}{c x-1}+\frac{7 b c \sqrt{1-c^2 x^2}}{c x+1}+\frac{2 b c \sqrt{1-c^2 x^2}}{3 (c x-1)^2}+\frac{2 b c \sqrt{1-c^2 x^2}}{3 (c x+1)^2}+16 b c \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )+\frac{7 b c \sin ^{-1}(c x)}{c x-1}+\frac{7 b c \sin ^{-1}(c x)}{c x+1}-\frac{b c \sin ^{-1}(c x)}{(c x-1)^2}+\frac{b c \sin ^{-1}(c x)}{(c x+1)^2}+15 i \pi b c \sin ^{-1}(c x)+\frac{16 b \sin ^{-1}(c x)}{x}-30 b c \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-15 \pi b c \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+30 b c \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-15 \pi b c \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+15 \pi b c \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+15 \pi b c \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{16 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((16*a)/x + (2*b*c*Sqrt[1 - c^2*x^2])/(3*(-1 + c*x)^2) - (b*c^2*x*Sqrt[1 - c^2*x^2])/(3*(-1 + c*x)^2) - (7*b*
c*Sqrt[1 - c^2*x^2])/(-1 + c*x) + (2*b*c*Sqrt[1 - c^2*x^2])/(3*(1 + c*x)^2) + (b*c^2*x*Sqrt[1 - c^2*x^2])/(3*(
1 + c*x)^2) + (7*b*c*Sqrt[1 - c^2*x^2])/(1 + c*x) - (4*a*c^2*x)/(-1 + c^2*x^2)^2 + (14*a*c^2*x)/(-1 + c^2*x^2)
+ (15*I)*b*c*Pi*ArcSin[c*x] + (16*b*ArcSin[c*x])/x - (b*c*ArcSin[c*x])/(-1 + c*x)^2 + (7*b*c*ArcSin[c*x])/(-1
+ c*x) + (b*c*ArcSin[c*x])/(1 + c*x)^2 + (7*b*c*ArcSin[c*x])/(1 + c*x) + 16*b*c*ArcTanh[Sqrt[1 - c^2*x^2]] -
15*b*c*Pi*Log[1 - I*E^(I*ArcSin[c*x])] - 30*b*c*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 15*b*c*Pi*Log[1 + I
*E^(I*ArcSin[c*x])] + 30*b*c*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + 15*a*c*Log[1 - c*x] - 15*a*c*Log[1 + c
*x] + 15*b*c*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + 15*b*c*Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (30*I)*b*c*Po
lyLog[2, (-I)*E^(I*ArcSin[c*x])] + (30*I)*b*c*PolyLog[2, I*E^(I*ArcSin[c*x])])/(16*d^3)

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Maple [A]  time = 0.223, size = 461, normalized size = 1.9 \begin{align*}{\frac{ca}{16\,{d}^{3} \left ( cx-1 \right ) ^{2}}}-{\frac{7\,ca}{16\,{d}^{3} \left ( cx-1 \right ) }}-{\frac{15\,ca\ln \left ( cx-1 \right ) }{16\,{d}^{3}}}-{\frac{ca}{16\,{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{7\,ca}{16\,{d}^{3} \left ( cx+1 \right ) }}+{\frac{15\,ca\ln \left ( cx+1 \right ) }{16\,{d}^{3}}}-{\frac{a}{{d}^{3}x}}-{\frac{15\,b\arcsin \left ( cx \right ){c}^{4}{x}^{3}}{8\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{7\,b{c}^{3}{x}^{2}}{8\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{25\,b\arcsin \left ( cx \right ){c}^{2}x}{8\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}-{\frac{23\,bc}{24\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arcsin \left ( cx \right ) }{{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) x}}+{\frac{bc}{{d}^{3}}\ln \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1}-1 \right ) }-{\frac{bc}{{d}^{3}}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{15\,bc\arcsin \left ( cx \right ) }{8\,{d}^{3}}\ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{15\,bc\arcsin \left ( cx \right ) }{8\,{d}^{3}}\ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{{\frac{15\,i}{8}}cb}{{d}^{3}}{\it dilog} \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{{\frac{15\,i}{8}}cb}{{d}^{3}}{\it dilog} \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^2/(-c^2*d*x^2+d)^3,x)

[Out]

1/16*c*a/d^3/(c*x-1)^2-7/16*c*a/d^3/(c*x-1)-15/16*c*a/d^3*ln(c*x-1)-1/16*c*a/d^3/(c*x+1)^2-7/16*c*a/d^3/(c*x+1
)+15/16*c*a/d^3*ln(c*x+1)-a/d^3/x-15/8*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*c^4*x^3+7/8*b/d^3/(c^4*x^4-2*c^
2*x^2+1)*c^3*x^2*(-c^2*x^2+1)^(1/2)+25/8*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*c^2*x-23/24*c*b/d^3/(c^4*x^4-
2*c^2*x^2+1)*(-c^2*x^2+1)^(1/2)-b/d^3/(c^4*x^4-2*c^2*x^2+1)/x*arcsin(c*x)+c*b/d^3*ln(I*c*x+(-c^2*x^2+1)^(1/2)-
1)-c*b/d^3*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-15/8*c*b/d^3*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+15/8*c*b
/d^3*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-15/8*I*c*b/d^3*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+15/8*
I*c*b/d^3*dilog(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}{16} \, a{\left (\frac{2 \,{\left (15 \, c^{4} x^{4} - 25 \, c^{2} x^{2} + 8\right )}}{c^{4} d^{3} x^{5} - 2 \, c^{2} d^{3} x^{3} + d^{3} x} - \frac{15 \, c \log \left (c x + 1\right )}{d^{3}} + \frac{15 \, c \log \left (c x - 1\right )}{d^{3}}\right )} + \frac{{\left (15 \,{\left (c^{5} x^{5} - 2 \, c^{3} x^{3} + c x\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} - 2 \, c^{3} x^{3} + c x\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} - 25 \, c^{2} x^{2} + 8\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) -{\left (c^{4} d^{3} x^{5} - 2 \, c^{2} d^{3} x^{3} + d^{3} x\right )} \int \frac{{\left (30 \, c^{5} x^{4} - 50 \, c^{3} x^{2} - 15 \,{\left (c^{6} x^{5} - 2 \, c^{4} x^{3} + c^{2} x\right )} \log \left (c x + 1\right ) + 15 \,{\left (c^{6} x^{5} - 2 \, c^{4} x^{3} + c^{2} x\right )} \log \left (-c x + 1\right ) + 16 \, c\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{6} d^{3} x^{7} - 3 \, c^{4} d^{3} x^{5} + 3 \, c^{2} d^{3} x^{3} - d^{3} x}\,{d x}\right )} b}{16 \,{\left (c^{4} d^{3} x^{5} - 2 \, c^{2} d^{3} x^{3} + d^{3} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*a*(2*(15*c^4*x^4 - 25*c^2*x^2 + 8)/(c^4*d^3*x^5 - 2*c^2*d^3*x^3 + d^3*x) - 15*c*log(c*x + 1)/d^3 + 15*c*
log(c*x - 1)/d^3) + 1/16*(15*(c^5*x^5 - 2*c^3*x^3 + c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x +
1) - 15*(c^5*x^5 - 2*c^3*x^3 + c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) - 2*(15*c^4*x^4 -
25*c^2*x^2 + 8)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + 16*(c^4*d^3*x^5 - 2*c^2*d^3*x^3 + d^3*x)*integra
te(-1/16*(30*c^5*x^4 - 50*c^3*x^2 - 15*(c^6*x^5 - 2*c^4*x^3 + c^2*x)*log(c*x + 1) + 15*(c^6*x^5 - 2*c^4*x^3 +
c^2*x)*log(-c*x + 1) + 16*c)*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^6*d^3*x^7 - 3*c^4*d^3*x^5 + 3*c^2*d^3*x^3 - d^3*x
), x))*b/(c^4*d^3*x^5 - 2*c^2*d^3*x^3 + d^3*x)

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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^{6} d^{3} x^{8} - 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} - d^{3} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*arcsin(c*x) + a)/(c^6*d^3*x^8 - 3*c^4*d^3*x^6 + 3*c^2*d^3*x^4 - d^3*x^2), 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((a+b*asin(c*x))/x**2/(-c**2*d*x**2+d)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{b \arcsin \left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )}^{3} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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