3.54 \(\int \frac{a+b \sinh ^{-1}(c x)}{x^4 (d+c^2 d x^2)^3} \, dx\)

Optimal. Leaf size=295 \[ -\frac{35 i b c^3 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}+\frac{35 i b c^3 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (c^2 x^2+1\right )}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (c^2 x^2+1\right )^2}+\frac{7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (c^2 x^2+1\right )^2}-\frac{a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (c^2 x^2+1\right )^2}+\frac{35 c^3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}+\frac{29 b c^3}{24 d^3 \sqrt{c^2 x^2+1}}-\frac{b c^3}{12 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac{b c}{6 d^3 x^2 \left (c^2 x^2+1\right )^{3/2}}+\frac{19 b c^3 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{6 d^3} \]

[Out]

-(b*c^3)/(12*d^3*(1 + c^2*x^2)^(3/2)) - (b*c)/(6*d^3*x^2*(1 + c^2*x^2)^(3/2)) + (29*b*c^3)/(24*d^3*Sqrt[1 + c^
2*x^2]) - (a + b*ArcSinh[c*x])/(3*d^3*x^3*(1 + c^2*x^2)^2) + (7*c^2*(a + b*ArcSinh[c*x]))/(3*d^3*x*(1 + c^2*x^
2)^2) + (35*c^4*x*(a + b*ArcSinh[c*x]))/(12*d^3*(1 + c^2*x^2)^2) + (35*c^4*x*(a + b*ArcSinh[c*x]))/(8*d^3*(1 +
c^2*x^2)) + (35*c^3*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(4*d^3) + (19*b*c^3*ArcTanh[Sqrt[1 + c^2*x^2
]])/(6*d^3) - (((35*I)/8)*b*c^3*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d^3 + (((35*I)/8)*b*c^3*PolyLog[2, I*E^ArcSin
h[c*x]])/d^3

________________________________________________________________________________________

Rubi [A]  time = 0.367184, antiderivative size = 345, normalized size of antiderivative = 1.17, number of steps used = 23, number of rules used = 11, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.458, Rules used = {5747, 5690, 5693, 4180, 2279, 2391, 261, 266, 51, 63, 208} \[ -\frac{35 i b c^3 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}+\frac{35 i b c^3 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (c^2 x^2+1\right )}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (c^2 x^2+1\right )^2}+\frac{7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (c^2 x^2+1\right )^2}-\frac{a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (c^2 x^2+1\right )^2}+\frac{35 c^3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}+\frac{49 b c^3}{24 d^3 \sqrt{c^2 x^2+1}}+\frac{7 b c^3}{36 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac{5 b c \sqrt{c^2 x^2+1}}{6 d^3 x^2}+\frac{5 b c}{9 d^3 x^2 \sqrt{c^2 x^2+1}}+\frac{b c}{9 d^3 x^2 \left (c^2 x^2+1\right )^{3/2}}+\frac{19 b c^3 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{6 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^3),x]

[Out]

(7*b*c^3)/(36*d^3*(1 + c^2*x^2)^(3/2)) + (b*c)/(9*d^3*x^2*(1 + c^2*x^2)^(3/2)) + (49*b*c^3)/(24*d^3*Sqrt[1 + c
^2*x^2]) + (5*b*c)/(9*d^3*x^2*Sqrt[1 + c^2*x^2]) - (5*b*c*Sqrt[1 + c^2*x^2])/(6*d^3*x^2) - (a + b*ArcSinh[c*x]
)/(3*d^3*x^3*(1 + c^2*x^2)^2) + (7*c^2*(a + b*ArcSinh[c*x]))/(3*d^3*x*(1 + c^2*x^2)^2) + (35*c^4*x*(a + b*ArcS
inh[c*x]))/(12*d^3*(1 + c^2*x^2)^2) + (35*c^4*x*(a + b*ArcSinh[c*x]))/(8*d^3*(1 + c^2*x^2)) + (35*c^3*(a + b*A
rcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(4*d^3) + (19*b*c^3*ArcTanh[Sqrt[1 + c^2*x^2]])/(6*d^3) - (((35*I)/8)*b*c
^3*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d^3 + (((35*I)/8)*b*c^3*PolyLog[2, I*E^ArcSinh[c*x]])/d^3

Rule 5747

Int[((a_.) + ArcSinh[(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*ArcSinh[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*ArcSinh[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
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rule 5690

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

Rule 5693

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

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, 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 \sinh ^{-1}(c x)}{x^4 \left (d+c^2 d x^2\right )^3} \, dx &=-\frac{a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}-\frac{1}{3} \left (7 c^2\right ) \int \frac{a+b \sinh ^{-1}(c x)}{x^2 \left (d+c^2 d x^2\right )^3} \, dx+\frac{(b c) \int \frac{1}{x^3 \left (1+c^2 x^2\right )^{5/2}} \, dx}{3 d^3}\\ &=-\frac{a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac{7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac{1}{3} \left (35 c^4\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^3} \, dx+\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+c^2 x\right )^{5/2}} \, dx,x,x^2\right )}{6 d^3}-\frac{\left (7 b c^3\right ) \int \frac{1}{x \left (1+c^2 x^2\right )^{5/2}} \, dx}{3 d^3}\\ &=\frac{b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}-\frac{a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac{7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac{(5 b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{18 d^3}-\frac{\left (7 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )^{5/2}} \, dx,x,x^2\right )}{6 d^3}-\frac{\left (35 b c^5\right ) \int \frac{x}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{12 d^3}+\frac{\left (35 c^4\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\left (d+c^2 d x^2\right )^2} \, dx}{4 d}\\ &=\frac{7 b c^3}{36 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac{5 b c}{9 d^3 x^2 \sqrt{1+c^2 x^2}}-\frac{a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac{7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac{(5 b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+c^2 x}} \, dx,x,x^2\right )}{6 d^3}-\frac{\left (7 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^3}-\frac{\left (35 b c^5\right ) \int \frac{x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{8 d^3}+\frac{\left (35 c^4\right ) \int \frac{a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx}{8 d^2}\\ &=\frac{7 b c^3}{36 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac{49 b c^3}{24 d^3 \sqrt{1+c^2 x^2}}+\frac{5 b c}{9 d^3 x^2 \sqrt{1+c^2 x^2}}-\frac{5 b c \sqrt{1+c^2 x^2}}{6 d^3 x^2}-\frac{a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac{7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac{\left (35 c^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}-\frac{\left (5 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+c^2 x}} \, dx,x,x^2\right )}{12 d^3}-\frac{\left (7 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+c^2 x}} \, dx,x,x^2\right )}{6 d^3}\\ &=\frac{7 b c^3}{36 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac{49 b c^3}{24 d^3 \sqrt{1+c^2 x^2}}+\frac{5 b c}{9 d^3 x^2 \sqrt{1+c^2 x^2}}-\frac{5 b c \sqrt{1+c^2 x^2}}{6 d^3 x^2}-\frac{a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac{7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac{35 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\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 )}{6 d^3}-\frac{(7 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^3}-\frac{\left (35 i b c^3\right ) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}+\frac{\left (35 i b c^3\right ) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}\\ &=\frac{7 b c^3}{36 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac{49 b c^3}{24 d^3 \sqrt{1+c^2 x^2}}+\frac{5 b c}{9 d^3 x^2 \sqrt{1+c^2 x^2}}-\frac{5 b c \sqrt{1+c^2 x^2}}{6 d^3 x^2}-\frac{a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac{7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac{35 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{19 b c^3 \tanh ^{-1}\left (\sqrt{1+c^2 x^2}\right )}{6 d^3}-\frac{\left (35 i b c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{8 d^3}+\frac{\left (35 i b c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{8 d^3}\\ &=\frac{7 b c^3}{36 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b c}{9 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac{49 b c^3}{24 d^3 \sqrt{1+c^2 x^2}}+\frac{5 b c}{9 d^3 x^2 \sqrt{1+c^2 x^2}}-\frac{5 b c \sqrt{1+c^2 x^2}}{6 d^3 x^2}-\frac{a+b \sinh ^{-1}(c x)}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac{7 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac{35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}+\frac{35 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{19 b c^3 \tanh ^{-1}\left (\sqrt{1+c^2 x^2}\right )}{6 d^3}-\frac{35 i b c^3 \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}+\frac{35 i b c^3 \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}\\ \end{align*}

Mathematica [C]  time = 1.04604, size = 380, normalized size = 1.29 \[ \frac{\frac{4 b c^3 \text{Hypergeometric2F1}\left (-\frac{3}{2},2,-\frac{1}{2},c^2 x^2+1\right )}{\left (c^2 x^2+1\right )^{3/2}}+\frac{42 b c^3 \text{Hypergeometric2F1}\left (-\frac{1}{2},2,\frac{1}{2},c^2 x^2+1\right )}{\sqrt{c^2 x^2+1}}+210 b \left (-c^2\right )^{3/2} \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )-210 b \left (-c^2\right )^{3/2} \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+\frac{42 a}{c^2 x^5+x^3}+\frac{12 a}{x^3 \left (c^2 x^2+1\right )^2}+\frac{210 a c^2}{x}+210 a c^3 \tan ^{-1}(c x)-\frac{70 a}{x^3}-\frac{35 b c \sqrt{c^2 x^2+1}}{x^2}+\frac{42 b \sinh ^{-1}(c x)}{c^2 x^5+x^3}+\frac{12 b \sinh ^{-1}(c x)}{x^3 \left (c^2 x^2+1\right )^2}+245 b c^3 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )+\frac{210 b c^2 \sinh ^{-1}(c x)}{x}-210 b \left (-c^2\right )^{3/2} \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+210 b \left (-c^2\right )^{3/2} \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )-\frac{70 b \sinh ^{-1}(c x)}{x^3}}{48 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^3),x]

[Out]

((-70*a)/x^3 + (210*a*c^2)/x + (12*a)/(x^3*(1 + c^2*x^2)^2) - (35*b*c*Sqrt[1 + c^2*x^2])/x^2 + (42*a)/(x^3 + c
^2*x^5) - (70*b*ArcSinh[c*x])/x^3 + (210*b*c^2*ArcSinh[c*x])/x + (12*b*ArcSinh[c*x])/(x^3*(1 + c^2*x^2)^2) + (
42*b*ArcSinh[c*x])/(x^3 + c^2*x^5) + 210*a*c^3*ArcTan[c*x] + 245*b*c^3*ArcTanh[Sqrt[1 + c^2*x^2]] + (4*b*c^3*H
ypergeometric2F1[-3/2, 2, -1/2, 1 + c^2*x^2])/(1 + c^2*x^2)^(3/2) + (42*b*c^3*Hypergeometric2F1[-1/2, 2, 1/2,
1 + c^2*x^2])/Sqrt[1 + c^2*x^2] - 210*b*(-c^2)^(3/2)*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 210
*b*(-c^2)^(3/2)*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] + 210*b*(-c^2)^(3/2)*PolyLog[2, (c*E^ArcSi
nh[c*x])/Sqrt[-c^2]] - 210*b*(-c^2)^(3/2)*PolyLog[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c])/(48*d^3)

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Maple [A]  time = 0.02, size = 425, normalized size = 1.4 \begin{align*} -{\frac{a}{3\,{d}^{3}{x}^{3}}}+3\,{\frac{{c}^{2}a}{{d}^{3}x}}+{\frac{11\,{c}^{6}a{x}^{3}}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{13\,{c}^{4}ax}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{35\,{c}^{3}a\arctan \left ( cx \right ) }{8\,{d}^{3}}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{3\,{d}^{3}{x}^{3}}}+3\,{\frac{{c}^{2}b{\it Arcsinh} \left ( cx \right ) }{{d}^{3}x}}+{\frac{11\,{c}^{6}b{\it Arcsinh} \left ( cx \right ){x}^{3}}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{13\,{c}^{4}b{\it Arcsinh} \left ( cx \right ) x}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{35\,b{c}^{3}{\it Arcsinh} \left ( cx \right ) \arctan \left ( cx \right ) }{8\,{d}^{3}}}+{\frac{35\,{c}^{5}b{x}^{2}}{8\,{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{103\,b{c}^{3}}{24\,{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{\frac{19\,b{c}^{3}}{6\,{d}^{3}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{19\,b{c}^{3}}{6\,{d}^{3}}{\it Artanh} \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}} \right ) }-{\frac{bc}{6\,{d}^{3}{x}^{2}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,b{c}^{3}\arctan \left ( cx \right ) }{8\,{d}^{3}}\ln \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{35\,b{c}^{3}\arctan \left ( cx \right ) }{8\,{d}^{3}}\ln \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{35\,i}{8}}{c}^{3}b}{{d}^{3}}{\it dilog} \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{35\,i}{8}}{c}^{3}b}{{d}^{3}}{\it dilog} \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^3,x)

[Out]

-1/3*a/d^3/x^3+3*c^2*a/d^3/x+11/8*c^6*a/d^3*x^3/(c^2*x^2+1)^2+13/8*c^4*a/d^3*x/(c^2*x^2+1)^2+35/8*c^3*a/d^3*ar
ctan(c*x)-1/3*b/d^3*arcsinh(c*x)/x^3+3*c^2*b/d^3*arcsinh(c*x)/x+11/8*c^6*b/d^3*arcsinh(c*x)*x^3/(c^2*x^2+1)^2+
13/8*c^4*b/d^3*arcsinh(c*x)*x/(c^2*x^2+1)^2+35/8*c^3*b/d^3*arcsinh(c*x)*arctan(c*x)+35/8*c^5*b/d^3*x^2/(c^2*x^
2+1)^(3/2)+103/24*b*c^3/d^3/(c^2*x^2+1)^(3/2)-19/6*b*c^3/d^3/(c^2*x^2+1)^(1/2)+19/6*c^3*b/d^3*arctanh(1/(c^2*x
^2+1)^(1/2))-1/6*b*c/d^3/x^2/(c^2*x^2+1)^(3/2)+35/8*c^3*b/d^3*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-
35/8*c^3*b/d^3*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+35/8*I*c^3*b/d^3*dilog(1-I*(1+I*c*x)/(c^2*x^2+1
)^(1/2))-35/8*I*c^3*b/d^3*dilog(1+I*(1+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}{24} \, a{\left (\frac{105 \, c^{3} \arctan \left (c x\right )}{d^{3}} + \frac{105 \, c^{6} x^{6} + 175 \, c^{4} x^{4} + 56 \, c^{2} x^{2} - 8}{c^{4} d^{3} x^{7} + 2 \, c^{2} d^{3} x^{5} + d^{3} x^{3}}\right )} + b \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{6} d^{3} x^{10} + 3 \, c^{4} d^{3} x^{8} + 3 \, c^{2} d^{3} x^{6} + d^{3} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^3,x, algorithm="maxima")

[Out]

1/24*a*(105*c^3*arctan(c*x)/d^3 + (105*c^6*x^6 + 175*c^4*x^4 + 56*c^2*x^2 - 8)/(c^4*d^3*x^7 + 2*c^2*d^3*x^5 +
d^3*x^3)) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^10 + 3*c^4*d^3*x^8 + 3*c^2*d^3*x^6 + d^3*x^4),
x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arsinh}\left (c x\right ) + a}{c^{6} d^{3} x^{10} + 3 \, c^{4} d^{3} x^{8} + 3 \, c^{2} d^{3} x^{6} + d^{3} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*arcsinh(c*x) + a)/(c^6*d^3*x^10 + 3*c^4*d^3*x^8 + 3*c^2*d^3*x^6 + d^3*x^4), 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*asinh(c*x))/x**4/(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 \operatorname{arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{3} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^3,x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/((c^2*d*x^2 + d)^3*x^4), x)