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

Optimal. Leaf size=222 \[ \frac{15 i b c \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac{15 i b c \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (c^2 x^2+1\right )}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac{a+b \sinh ^{-1}(c x)}{d^3 x \left (c^2 x^2+1\right )^2}-\frac{15 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac{7 b c}{8 d^3 \sqrt{c^2 x^2+1}}-\frac{b c}{12 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac{b c \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{d^3} \]

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

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Rubi [A]  time = 0.238339, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 16, 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} \[ \frac{15 i b c \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac{15 i b c \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (c^2 x^2+1\right )}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac{a+b \sinh ^{-1}(c x)}{d^3 x \left (c^2 x^2+1\right )^2}-\frac{15 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac{7 b c}{8 d^3 \sqrt{c^2 x^2+1}}-\frac{b c}{12 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac{b c \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{d^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[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*ArcSinh[c*x])/(d^3*x*(1 + c^2
*x^2)^2) - (5*c^2*x*(a + b*ArcSinh[c*x]))/(4*d^3*(1 + c^2*x^2)^2) - (15*c^2*x*(a + b*ArcSinh[c*x]))/(8*d^3*(1
+ c^2*x^2)) - (15*c*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(4*d^3) - (b*c*ArcTanh[Sqrt[1 + c^2*x^2]])/d^
3 + (((15*I)/8)*b*c*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d^3 - (((15*I)/8)*b*c*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^2 \left (d+c^2 d x^2\right )^3} \, dx &=-\frac{a+b \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\left (5 c^2\right ) \int \frac{a+b \sinh ^{-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 \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-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 \sinh ^{-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 \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-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 \sinh ^{-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 \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}-\frac{(15 c) \operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-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 \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}-\frac{15 c \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-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 i b c) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}-\frac{(15 i b c) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-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 \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}-\frac{15 c \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-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^{\sinh ^{-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^{\sinh ^{-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 \sinh ^{-1}(c x)}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}-\frac{15 c \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-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^{\sinh ^{-1}(c x)}\right )}{8 d^3}-\frac{15 i b c \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{8 d^3}\\ \end{align*}

Mathematica [C]  time = 1.21853, size = 298, normalized size = 1.34 \[ -\frac{\frac{2 b c \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},c^2 x^2+1\right )}{\left (c^2 x^2+1\right )^{3/2}}+\frac{15 b c \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},c^2 x^2+1\right )}{\sqrt{c^2 x^2+1}}-45 b \sqrt{-c^2} \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )+45 b \sqrt{-c^2} \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-\frac{15 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 x^3+x}-\frac{6 \left (a+b \sinh ^{-1}(c x)\right )}{x \left (c^2 x^2+1\right )^2}+\frac{45 \left (a+b \sinh ^{-1}(c x)\right )}{x}+45 a c \tan ^{-1}(c x)+45 b c \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )+45 b \sqrt{-c^2} \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )-45 b \sqrt{-c^2} \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )}{24 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((45*(a + b*ArcSinh[c*x]))/x - (6*(a + b*ArcSinh[c*x]))/(x*(1 + c^2*x^2)^2) - (15*(a + b*ArcSinh[c*x]))/(x +
c^2*x^3) + 45*a*c*ArcTan[c*x] + 45*b*c*ArcTanh[Sqrt[1 + c^2*x^2]] + (2*b*c*Hypergeometric2F1[-3/2, 1, -1/2, 1
+ c^2*x^2])/(1 + c^2*x^2)^(3/2) + (15*b*c*Hypergeometric2F1[-1/2, 1, 1/2, 1 + c^2*x^2])/Sqrt[1 + c^2*x^2] + 45
*b*Sqrt[-c^2]*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 45*b*Sqrt[-c^2]*ArcSinh[c*x]*Log[1 + (Sqrt
[-c^2]*E^ArcSinh[c*x])/c] - 45*b*Sqrt[-c^2]*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 45*b*Sqrt[-c^2]*PolyLo
g[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c])/(24*d^3)

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Maple [A]  time = 0.016, size = 357, normalized size = 1.6 \begin{align*} -{\frac{a}{{d}^{3}x}}-{\frac{7\,a{c}^{4}{x}^{3}}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{9\,a{c}^{2}x}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{15\,ca\arctan \left ( cx \right ) }{8\,{d}^{3}}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{{d}^{3}x}}-{\frac{7\,b{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{3}}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{9\,b{\it Arcsinh} \left ( cx \right ){c}^{2}x}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{15\,bc{\it Arcsinh} \left ( cx \right ) \arctan \left ( cx \right ) }{8\,{d}^{3}}}-{\frac{15\,bc\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{15\,bc\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{15\,i}{8}}cb}{{d}^{3}}{\it dilog} \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{15\,i}{8}}cb}{{d}^{3}}{\it dilog} \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{15\,b{c}^{3}{x}^{2}}{8\,{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{\frac{47\,bc}{24\,{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{bc}{{d}^{3}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{bc}{{d}^{3}}{\it Artanh} \left ({\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^2/(c^2*d*x^2+d)^3,x)

[Out]

-a/d^3/x-7/8*a/d^3*c^4*x^3/(c^2*x^2+1)^2-9/8*a/d^3*c^2*x/(c^2*x^2+1)^2-15/8*c*a/d^3*arctan(c*x)-b/d^3*arcsinh(
c*x)/x-7/8*b/d^3*arcsinh(c*x)*c^4*x^3/(c^2*x^2+1)^2-9/8*b/d^3*arcsinh(c*x)*c^2*x/(c^2*x^2+1)^2-15/8*c*b/d^3*ar
csinh(c*x)*arctan(c*x)-15/8*c*b/d^3*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+15/8*c*b/d^3*arctan(c*x)*l
n(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+15/8*I*c*b/d^3*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-15/8*I*c*b/d^3*dilog(
1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-15/8*b/d^3*c^3*x^2/(c^2*x^2+1)^(3/2)-47/24*b*c/d^3/(c^2*x^2+1)^(3/2)+b*c/d^3/
(c^2*x^2+1)^(1/2)-c*b/d^3*arctanh(1/(c^2*x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{8} \, a{\left (\frac{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} + \frac{15 \, c \arctan \left (c x\right )}{d^{3}}\right )} + b \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{6} d^{3} x^{8} + 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} + d^{3} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a*((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*arctan(c*x)/d^3) + b*integr
ate(log(c*x + sqrt(c^2*x^2 + 1))/(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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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^{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*arcsinh(c*x))/x^2/(c^2*d*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*arcsinh(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{6} x^{8} + 3 c^{4} x^{6} + 3 c^{2} x^{4} + x^{2}}\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{c^{6} x^{8} + 3 c^{4} x^{6} + 3 c^{2} x^{4} + x^{2}}\, dx}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/x**2/(c**2*d*x**2+d)**3,x)

[Out]

(Integral(a/(c**6*x**8 + 3*c**4*x**6 + 3*c**2*x**4 + x**2), x) + Integral(b*asinh(c*x)/(c**6*x**8 + 3*c**4*x**
6 + 3*c**2*x**4 + x**2), x))/d**3

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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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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