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

Optimal. Leaf size=232 \[ -\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^3 \left (c^2 x^2+1\right )}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{2 d^3 x^2 \left (c^2 x^2+1\right )^2}+\frac {6 c^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}+\frac {3 b c^2 \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}-\frac {3 b c^2 \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}-\frac {b c}{2 d^3 x \left (c^2 x^2+1\right )^{3/2}}+\frac {2 b c^3 x}{3 d^3 \sqrt {c^2 x^2+1}}-\frac {5 b c^3 x}{12 d^3 \left (c^2 x^2+1\right )^{3/2}} \]

[Out]

-1/2*b*c/d^3/x/(c^2*x^2+1)^(3/2)-5/12*b*c^3*x/d^3/(c^2*x^2+1)^(3/2)-3/4*c^2*(a+b*arcsinh(c*x))/d^3/(c^2*x^2+1)
^2+1/2*(-a-b*arcsinh(c*x))/d^3/x^2/(c^2*x^2+1)^2-3/2*c^2*(a+b*arcsinh(c*x))/d^3/(c^2*x^2+1)+6*c^2*(a+b*arcsinh
(c*x))*arctanh((c*x+(c^2*x^2+1)^(1/2))^2)/d^3+3/2*b*c^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^3-3/2*b*c^2*po
lylog(2,(c*x+(c^2*x^2+1)^(1/2))^2)/d^3+2/3*b*c^3*x/d^3/(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5747, 5755, 5720, 5461, 4182, 2279, 2391, 191, 192, 271} \[ \frac {3 b c^2 \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}-\frac {3 b c^2 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^3 \left (c^2 x^2+1\right )}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{2 d^3 x^2 \left (c^2 x^2+1\right )^2}+\frac {6 c^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}+\frac {2 b c^3 x}{3 d^3 \sqrt {c^2 x^2+1}}-\frac {5 b c^3 x}{12 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {b c}{2 d^3 x \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c)/(2*d^3*x*(1 + c^2*x^2)^(3/2)) - (5*b*c^3*x)/(12*d^3*(1 + c^2*x^2)^(3/2)) + (2*b*c^3*x)/(3*d^3*Sqrt[1 +
c^2*x^2]) - (3*c^2*(a + b*ArcSinh[c*x]))/(4*d^3*(1 + c^2*x^2)^2) - (a + b*ArcSinh[c*x])/(2*d^3*x^2*(1 + c^2*x^
2)^2) - (3*c^2*(a + b*ArcSinh[c*x]))/(2*d^3*(1 + c^2*x^2)) + (6*c^2*(a + b*ArcSinh[c*x])*ArcTanh[E^(2*ArcSinh[
c*x])])/d^3 + (3*b*c^2*PolyLog[2, -E^(2*ArcSinh[c*x])])/(2*d^3) - (3*b*c^2*PolyLog[2, E^(2*ArcSinh[c*x])])/(2*
d^3)

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

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

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && 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 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 5720

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

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 5755

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)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(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)^F
racPart[p])/(2*f*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[
c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x^3 \left (d+c^2 d x^2\right )^3} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\left (3 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^3} \, dx+\frac {(b c) \int \frac {1}{x^2 \left (1+c^2 x^2\right )^{5/2}} \, dx}{2 d^3}\\ &=-\frac {b c}{2 d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}+\frac {\left (3 b c^3\right ) \int \frac {1}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{4 d^3}-\frac {\left (2 b c^3\right ) \int \frac {1}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{d^3}-\frac {\left (3 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^2} \, dx}{d}\\ &=-\frac {b c}{2 d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^3 \left (1+c^2 x^2\right )}+\frac {\left (b c^3\right ) \int \frac {1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 d^3}-\frac {\left (4 b c^3\right ) \int \frac {1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^3}+\frac {\left (3 b c^3\right ) \int \frac {1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 d^3}-\frac {\left (3 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )} \, dx}{d^2}\\ &=-\frac {b c}{2 d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {2 b c^3 x}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^3 \left (1+c^2 x^2\right )}-\frac {\left (3 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b c}{2 d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {2 b c^3 x}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^3 \left (1+c^2 x^2\right )}-\frac {\left (6 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \text {csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b c}{2 d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {2 b c^3 x}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}-\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b c}{2 d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {2 b c^3 x}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}-\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}\\ &=-\frac {b c}{2 d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {2 b c^3 x}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {a+b \sinh ^{-1}(c x)}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {3 b c^2 \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}-\frac {3 b c^2 \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.83, size = 353, normalized size = 1.52 \[ \frac {-18 c^2 \left (2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )+b \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )\right )+\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{\left (c^2 x^3+x\right )^2}+\frac {9 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 x^4+x^2}+\frac {18 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{b}-\frac {18 \left (a+b \sinh ^{-1}(c x)\right )}{x^2}+18 a c^2 \log \left (c^2 x^2+1\right )+36 b c^2 \text {Li}_2\left (\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+36 b c^2 \text {Li}_2\left (\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+\frac {9 b c \left (2 c^2 x^2+1\right )}{x \sqrt {c^2 x^2+1}}-\frac {18 b c \sqrt {c^2 x^2+1}}{x}-18 b c^2 \sinh ^{-1}(c x)^2+36 b c^2 \sinh ^{-1}(c x) \log \left (\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}+1\right )+36 b c^2 \sinh ^{-1}(c x) \log \left (\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )+\frac {b c \left (8 c^4 x^4+12 c^2 x^2+3\right )}{x \left (c^2 x^2+1\right )^{3/2}}}{12 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-18*b*c*Sqrt[1 + c^2*x^2])/x + (9*b*c*(1 + 2*c^2*x^2))/(x*Sqrt[1 + c^2*x^2]) + (b*c*(3 + 12*c^2*x^2 + 8*c^4*
x^4))/(x*(1 + c^2*x^2)^(3/2)) - 18*b*c^2*ArcSinh[c*x]^2 - (18*(a + b*ArcSinh[c*x]))/x^2 + (3*(a + b*ArcSinh[c*
x]))/(x + c^2*x^3)^2 + (9*(a + b*ArcSinh[c*x]))/(x^2 + c^2*x^4) + (18*c^2*(a + b*ArcSinh[c*x])^2)/b + 36*b*c^2
*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 36*b*c^2*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x
])/c] + 18*a*c^2*Log[1 + c^2*x^2] + 36*b*c^2*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 36*b*c^2*PolyLog[2, (
Sqrt[-c^2]*E^ArcSinh[c*x])/c] - 18*c^2*(2*(a + b*ArcSinh[c*x])*Log[1 - E^(2*ArcSinh[c*x])] + b*PolyLog[2, E^(2
*ArcSinh[c*x])]))/(12*d^3)

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arsinh}\left (c x\right ) + a}{c^{6} d^{3} x^{9} + 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} + d^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsinh(c*x) + a)/(c^6*d^3*x^9 + 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5 + d^3*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.26, size = 575, normalized size = 2.48 \[ -\frac {a}{2 d^{3} x^{2}}-\frac {3 c^{2} a \ln \left (c x \right )}{d^{3}}-\frac {c^{2} a}{d^{3} \left (c^{2} x^{2}+1\right )}-\frac {c^{2} a}{4 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {3 c^{2} a \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}+\frac {2 c^{5} b \,x^{3} \sqrt {c^{2} x^{2}+1}}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}-\frac {2 c^{6} b \,x^{4}}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}-\frac {3 c^{4} b \arcsinh \left (c x \right ) x^{2}}{2 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {c^{3} b x \sqrt {c^{2} x^{2}+1}}{4 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}-\frac {4 c^{4} b \,x^{2}}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}-\frac {9 c^{2} b \arcsinh \left (c x \right )}{4 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}-\frac {c b \sqrt {c^{2} x^{2}+1}}{2 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) x}-\frac {2 c^{2} b}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}-\frac {b \arcsinh \left (c x \right )}{2 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) x^{2}}+\frac {3 c^{2} b \arcsinh \left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{3}}+\frac {3 b \,c^{2} \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d^{3}}-\frac {3 c^{2} b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}-\frac {3 c^{2} b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}-\frac {3 c^{2} b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}-\frac {3 c^{2} b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d^3/x^2-3*c^2*a/d^3*ln(c*x)-c^2*a/d^3/(c^2*x^2+1)-1/4*c^2*a/d^3/(c^2*x^2+1)^2+3/2*c^2*a/d^3*ln(c^2*x^2+
1)+2/3*c^5*b/d^3/(c^4*x^4+2*c^2*x^2+1)*x^3*(c^2*x^2+1)^(1/2)-2/3*c^6*b/d^3/(c^4*x^4+2*c^2*x^2+1)*x^4-3/2*c^4*b
/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)*x^2+1/4*c^3*b/d^3/(c^4*x^4+2*c^2*x^2+1)*x*(c^2*x^2+1)^(1/2)-4/3*c^4*b/
d^3/(c^4*x^4+2*c^2*x^2+1)*x^2-9/4*c^2*b/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)-1/2*c*b/d^3/(c^4*x^4+2*c^2*x^2+
1)/x*(c^2*x^2+1)^(1/2)-2/3*c^2*b/d^3/(c^4*x^4+2*c^2*x^2+1)-1/2*b/d^3/(c^4*x^4+2*c^2*x^2+1)/x^2*arcsinh(c*x)+3*
c^2*b/d^3*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+3/2*b*c^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^3-3*c
^2*b/d^3*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))-3*c^2*b/d^3*polylog(2,-c*x-(c^2*x^2+1)^(1/2))-3*c^2*b/d^3*ar
csinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))-3*c^2*b/d^3*polylog(2,c*x+(c^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{4} \, a {\left (\frac {6 \, c^{4} x^{4} + 9 \, c^{2} x^{2} + 2}{c^{4} d^{3} x^{6} + 2 \, c^{2} d^{3} x^{4} + d^{3} x^{2}} - \frac {6 \, c^{2} \log \left (c^{2} x^{2} + 1\right )}{d^{3}} + \frac {12 \, c^{2} \log \relax (x)}{d^{3}}\right )} + b \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{6} d^{3} x^{9} + 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} + d^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a*((6*c^4*x^4 + 9*c^2*x^2 + 2)/(c^4*d^3*x^6 + 2*c^2*d^3*x^4 + d^3*x^2) - 6*c^2*log(c^2*x^2 + 1)/d^3 + 12*
c^2*log(x)/d^3) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^9 + 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5 + d^3*
x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{6} x^{9} + 3 c^{4} x^{7} + 3 c^{2} x^{5} + x^{3}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{9} + 3 c^{4} x^{7} + 3 c^{2} x^{5} + x^{3}}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**6*x**9 + 3*c**4*x**7 + 3*c**2*x**5 + x**3), x) + Integral(b*asinh(c*x)/(c**6*x**9 + 3*c**4*x**
7 + 3*c**2*x**5 + x**3), x))/d**3

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