3.174 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^3 (d+c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=400 \[ -\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {c^2 d x^2+d}}+\frac {5 c^2 \sqrt {c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt {c^2 d x^2+d}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \left (c^2 d x^2+d\right )^{3/2}}+\frac {5 b c^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt {c^2 d x^2+d}}-\frac {5 b c^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt {c^2 d x^2+d}}-\frac {3 b c \sqrt {c^2 x^2+1}}{4 d^2 x \sqrt {c^2 d x^2+d}}+\frac {b c}{4 d^2 x \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}+\frac {13 b c^2 \sqrt {c^2 x^2+1} \tan ^{-1}(c x)}{6 d^2 \sqrt {c^2 d x^2+d}}+\frac {5 b c^3 x}{12 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}} \]

[Out]

-5/6*c^2*(a+b*arcsinh(c*x))/d/(c^2*d*x^2+d)^(3/2)+1/2*(-a-b*arcsinh(c*x))/d/x^2/(c^2*d*x^2+d)^(3/2)-5/2*c^2*(a
+b*arcsinh(c*x))/d^2/(c^2*d*x^2+d)^(1/2)+1/4*b*c/d^2/x/(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)+5/12*b*c^3*x/d^2/
(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)-3/4*b*c*(c^2*x^2+1)^(1/2)/d^2/x/(c^2*d*x^2+d)^(1/2)+13/6*b*c^2*arctan(c*
x)*(c^2*x^2+1)^(1/2)/d^2/(c^2*d*x^2+d)^(1/2)+5*c^2*(a+b*arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))*(c^2*x^2+
1)^(1/2)/d^2/(c^2*d*x^2+d)^(1/2)+5/2*b*c^2*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*(c^2*x^2+1)^(1/2)/d^2/(c^2*d*x^2+
d)^(1/2)-5/2*b*c^2*polylog(2,c*x+(c^2*x^2+1)^(1/2))*(c^2*x^2+1)^(1/2)/d^2/(c^2*d*x^2+d)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.56, antiderivative size = 400, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {5747, 5755, 5764, 5760, 4182, 2279, 2391, 203, 199, 290, 325} \[ \frac {5 b c^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt {c^2 d x^2+d}}-\frac {5 b c^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt {c^2 d x^2+d}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {c^2 d x^2+d}}+\frac {5 c^2 \sqrt {c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt {c^2 d x^2+d}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \left (c^2 d x^2+d\right )^{3/2}}+\frac {5 b c^3 x}{12 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}-\frac {3 b c \sqrt {c^2 x^2+1}}{4 d^2 x \sqrt {c^2 d x^2+d}}+\frac {b c}{4 d^2 x \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}+\frac {13 b c^2 \sqrt {c^2 x^2+1} \tan ^{-1}(c x)}{6 d^2 \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x^3*(d + c^2*d*x^2)^(5/2)),x]

[Out]

(b*c)/(4*d^2*x*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]) + (5*b*c^3*x)/(12*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x
^2]) - (3*b*c*Sqrt[1 + c^2*x^2])/(4*d^2*x*Sqrt[d + c^2*d*x^2]) - (5*c^2*(a + b*ArcSinh[c*x]))/(6*d*(d + c^2*d*
x^2)^(3/2)) - (a + b*ArcSinh[c*x])/(2*d*x^2*(d + c^2*d*x^2)^(3/2)) - (5*c^2*(a + b*ArcSinh[c*x]))/(2*d^2*Sqrt[
d + c^2*d*x^2]) + (13*b*c^2*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(6*d^2*Sqrt[d + c^2*d*x^2]) + (5*c^2*Sqrt[1 + c^2*x
^2]*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/(d^2*Sqrt[d + c^2*d*x^2]) + (5*b*c^2*Sqrt[1 + c^2*x^2]*PolyL
og[2, -E^ArcSinh[c*x]])/(2*d^2*Sqrt[d + c^2*d*x^2]) - (5*b*c^2*Sqrt[1 + c^2*x^2]*PolyLog[2, E^ArcSinh[c*x]])/(
2*d^2*Sqrt[d + c^2*d*x^2])

Rule 199

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}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 290

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 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])

Rule 5760

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

Rule 5764

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist
[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2], Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a
, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &&  !GtQ[d, 0] && (IntegerQ[m] || 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 )^{5/2}} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \left (d+c^2 d x^2\right )^{3/2}}-\frac {1}{2} \left (5 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^{5/2}} \, dx+\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )^2} \, dx}{2 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b c}{4 d^2 x \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \left (d+c^2 d x^2\right )^{3/2}}-\frac {\left (5 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^{3/2}} \, dx}{2 d}+\frac {\left (3 b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx}{4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (5 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{\left (1+c^2 x^2\right )^2} \, dx}{6 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b c}{4 d^2 x \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {5 b c^3 x}{12 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 d^2 x \sqrt {d+c^2 d x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (5 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {d+c^2 d x^2}} \, dx}{2 d^2}+\frac {\left (5 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{12 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (3 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (5 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b c}{4 d^2 x \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {5 b c^3 x}{12 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 d^2 x \sqrt {d+c^2 d x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {d+c^2 d x^2}}+\frac {13 b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (5 c^2 \sqrt {1+c^2 x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{2 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b c}{4 d^2 x \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {5 b c^3 x}{12 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 d^2 x \sqrt {d+c^2 d x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {d+c^2 d x^2}}+\frac {13 b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (5 c^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b c}{4 d^2 x \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {5 b c^3 x}{12 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 d^2 x \sqrt {d+c^2 d x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {d+c^2 d x^2}}+\frac {13 b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 c^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (5 b c^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (5 b c^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b c}{4 d^2 x \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {5 b c^3 x}{12 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 d^2 x \sqrt {d+c^2 d x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {d+c^2 d x^2}}+\frac {13 b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 c^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (5 b c^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (5 b c^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b c}{4 d^2 x \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {5 b c^3 x}{12 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 d^2 x \sqrt {d+c^2 d x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 d x^2 \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \sqrt {d+c^2 d x^2}}+\frac {13 b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 c^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt {d+c^2 d x^2}}+\frac {5 b c^2 \sqrt {1+c^2 x^2} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt {d+c^2 d x^2}}-\frac {5 b c^2 \sqrt {1+c^2 x^2} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt {d+c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 6.56, size = 437, normalized size = 1.09 \[ \frac {5 a c^2 \log \left (\sqrt {d} \sqrt {d \left (c^2 x^2+1\right )}+d\right )}{2 d^{5/2}}-\frac {5 a c^2 \log (x)}{2 d^{5/2}}+\sqrt {d \left (c^2 x^2+1\right )} \left (-\frac {2 a c^2}{d^3 \left (c^2 x^2+1\right )}-\frac {a c^2}{3 d^3 \left (c^2 x^2+1\right )^2}-\frac {a}{2 d^3 x^2}\right )+\frac {b c^2 \left (-60 \sqrt {c^2 x^2+1} \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )+60 \sqrt {c^2 x^2+1} \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+\frac {4 c x}{\sqrt {c^2 x^2+1}}-\frac {8 \sinh ^{-1}(c x)}{c^2 x^2+1}-60 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+60 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+6 \sqrt {c^2 x^2+1} \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-6 \sqrt {c^2 x^2+1} \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-3 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )-3 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )+104 \sqrt {c^2 x^2+1} \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-48 \sinh ^{-1}(c x)\right )}{24 d^2 \sqrt {d \left (c^2 x^2+1\right )}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x^3*(d + c^2*d*x^2)^(5/2)),x]

[Out]

Sqrt[d*(1 + c^2*x^2)]*(-1/2*a/(d^3*x^2) - (a*c^2)/(3*d^3*(1 + c^2*x^2)^2) - (2*a*c^2)/(d^3*(1 + c^2*x^2))) - (
5*a*c^2*Log[x])/(2*d^(5/2)) + (5*a*c^2*Log[d + Sqrt[d]*Sqrt[d*(1 + c^2*x^2)]])/(2*d^(5/2)) + (b*c^2*((4*c*x)/S
qrt[1 + c^2*x^2] - 48*ArcSinh[c*x] - (8*ArcSinh[c*x])/(1 + c^2*x^2) + 104*Sqrt[1 + c^2*x^2]*ArcTan[Tanh[ArcSin
h[c*x]/2]] - 6*Sqrt[1 + c^2*x^2]*Coth[ArcSinh[c*x]/2] - 3*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^
2 - 60*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] + 60*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 + E
^(-ArcSinh[c*x])] - 60*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^(-ArcSinh[c*x])] + 60*Sqrt[1 + c^2*x^2]*PolyLog[2, E^(-
ArcSinh[c*x])] - 3*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Sech[ArcSinh[c*x]/2]^2 + 6*Sqrt[1 + c^2*x^2]*Tanh[ArcSinh[c*
x]/2]))/(24*d^2*Sqrt[d*(1 + c^2*x^2)])

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c^{2} d x^{2} + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{c^{6} d^{3} x^{9} + 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} + d^{3} x^{3}}, x\right ) \]

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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*d*x^2 + d)*(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 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{3}}\,{d x} \]

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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/((c^2*d*x^2 + d)^(5/2)*x^3), x)

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maple [A]  time = 0.31, size = 546, normalized size = 1.36 \[ -\frac {a}{2 d \,x^{2} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {5 a \,c^{2}}{6 d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {5 a \,c^{2}}{2 d^{2} \sqrt {c^{2} d \,x^{2}+d}}+\frac {5 a \,c^{2} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c^{2} d \,x^{2}+d}}{x}\right )}{2 d^{\frac {5}{2}}}-\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x^{2} \arcsinh \left (c x \right ) c^{4}}{2 \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) d^{3}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x \,c^{3} \sqrt {c^{2} x^{2}+1}}{3 \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) d^{3}}-\frac {10 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) c^{2}}{3 \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) d^{3}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, c \sqrt {c^{2} x^{2}+1}}{2 \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) d^{3} x}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{2 \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) d^{3} x^{2}}+\frac {13 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arctan \left (c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{3 \sqrt {c^{2} x^{2}+1}\, d^{3}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{2 \sqrt {c^{2} x^{2}+1}\, d^{3}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (1+c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{2 \sqrt {c^{2} x^{2}+1}\, d^{3}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{2 \sqrt {c^{2} x^{2}+1}\, d^{3}} \]

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)^(5/2),x)

[Out]

-1/2*a/d/x^2/(c^2*d*x^2+d)^(3/2)-5/6*a*c^2/d/(c^2*d*x^2+d)^(3/2)-5/2*a*c^2/d^2/(c^2*d*x^2+d)^(1/2)+5/2*a*c^2/d
^(5/2)*ln((2*d+2*d^(1/2)*(c^2*d*x^2+d)^(1/2))/x)-5/2*b*(d*(c^2*x^2+1))^(1/2)/(c^4*x^4+2*c^2*x^2+1)/d^3*x^2*arc
sinh(c*x)*c^4-1/3*b*(d*(c^2*x^2+1))^(1/2)/(c^4*x^4+2*c^2*x^2+1)/d^3*x*c^3*(c^2*x^2+1)^(1/2)-10/3*b*(d*(c^2*x^2
+1))^(1/2)/(c^4*x^4+2*c^2*x^2+1)/d^3*arcsinh(c*x)*c^2-1/2*b*(d*(c^2*x^2+1))^(1/2)/(c^4*x^4+2*c^2*x^2+1)/d^3/x*
c*(c^2*x^2+1)^(1/2)-1/2*b*(d*(c^2*x^2+1))^(1/2)/(c^4*x^4+2*c^2*x^2+1)/d^3/x^2*arcsinh(c*x)+13/3*b*(d*(c^2*x^2+
1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3*arctan(c*x+(c^2*x^2+1)^(1/2))*c^2+5/2*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2
)/d^3*dilog(c*x+(c^2*x^2+1)^(1/2))*c^2+5/2*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3*dilog(1+c*x+(c^2*x^2+
1)^(1/2))*c^2+5/2*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))*c^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, a {\left (\frac {15 \, c^{2} \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right )}{d^{\frac {5}{2}}} - \frac {15 \, c^{2}}{\sqrt {c^{2} d x^{2} + d} d^{2}} - \frac {5 \, c^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d} - \frac {3}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x^{2}}\right )} + b \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{3}}\,{d x} \]

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)^(5/2),x, algorithm="maxima")

[Out]

1/6*a*(15*c^2*arcsinh(1/(c*abs(x)))/d^(5/2) - 15*c^2/(sqrt(c^2*d*x^2 + d)*d^2) - 5*c^2/((c^2*d*x^2 + d)^(3/2)*
d) - 3/((c^2*d*x^2 + d)^(3/2)*d*x^2)) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/((c^2*d*x^2 + d)^(5/2)*x^3),
x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x^3\,{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \]

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)^(5/2)),x)

[Out]

int((a + b*asinh(c*x))/(x^3*(d + c^2*d*x^2)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{x^{3} \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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)**(5/2),x)

[Out]

Integral((a + b*asinh(c*x))/(x**3*(d*(c**2*x**2 + 1))**(5/2)), x)

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