3.2.74 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^3 (d+c^2 d x^2)^{5/2}} \, dx\) [174]

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

[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)

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Rubi [A]
time = 0.35, antiderivative size = 400, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5809, 5811, 5816, 4267, 2317, 2438, 209, 205, 296, 331} \begin {gather*} -\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 {13 b c^2 \sqrt {c^2 x^2+1} \text {ArcTan}(c x)}{6 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 {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 {5 b c^3 x}{12 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}} \end {gather*}

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 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 209

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

Rule 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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 2317

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 2438

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 4267

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 5809

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/(f*(m + 1)))*Simp[(d +
e*x^2)^p/(1 + c^2*x^2)^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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && ILtQ[m, -1]

Rule 5811

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/(2*f*(p + 1)))*Simp[(d +
e*x^2)^p/(1 + c^2*x^2)^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 5816

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

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 4.69, size = 409, normalized size = 1.02 \begin {gather*} \frac {-\frac {4 a \sqrt {d+c^2 d x^2} \left (3+20 c^2 x^2+15 c^4 x^4\right )}{\left (x+c^2 x^3\right )^2}-60 a c^2 \sqrt {d} \log (x)+60 a c^2 \sqrt {d} \log \left (d+\sqrt {d} \sqrt {d+c^2 d x^2}\right )+\frac {b c^2 d \left (\frac {4 c x}{\sqrt {1+c^2 x^2}}-48 \sinh ^{-1}(c x)-\frac {8 \sinh ^{-1}(c x)}{1+c^2 x^2}+104 \sqrt {1+c^2 x^2} \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-6 \sqrt {1+c^2 x^2} \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-3 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )-60 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+60 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \log \left (1+e^{-\sinh ^{-1}(c x)}\right )-60 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )+60 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-3 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )+6 \sqrt {1+c^2 x^2} \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{\sqrt {d+c^2 d x^2}}}{24 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*a*Sqrt[d + c^2*d*x^2]*(3 + 20*c^2*x^2 + 15*c^4*x^4))/(x + c^2*x^3)^2 - 60*a*c^2*Sqrt[d]*Log[x] + 60*a*c^2
*Sqrt[d]*Log[d + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + (b*c^2*d*((4*c*x)/Sqrt[1 + c^2*x^2] - 48*ArcSinh[c*x] - (8*Arc
Sinh[c*x])/(1 + c^2*x^2) + 104*Sqrt[1 + c^2*x^2]*ArcTan[Tanh[ArcSinh[c*x]/2]] - 6*Sqrt[1 + c^2*x^2]*Coth[ArcSi
nh[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]*Po
lyLog[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]*ArcSin
h[c*x]*Sech[ArcSinh[c*x]/2]^2 + 6*Sqrt[1 + c^2*x^2]*Tanh[ArcSinh[c*x]/2]))/Sqrt[d + c^2*d*x^2])/(24*d^3)

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Maple [A]
time = 3.32, size = 546, normalized size = 1.36

method	result	size
default	\(-\frac {a}{2 d \,x^{2} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {5 a \,c^{2}}{6 d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {5 a \,c^{2}}{2 d^{2} \sqrt {c^{2} d \,x^{2}+d}}+\frac {5 a \,c^{2} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c^{2} d \,x^{2}+d}}{x}\right )}{2 d^{\frac {5}{2}}}-\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x^{2} \arcsinh \left (c x \right ) c^{4}}{2 \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) d^{3}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x \sqrt {c^{2} x^{2}+1}\, c^{3}}{3 \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) d^{3}}-\frac {10 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) c^{2}}{3 \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) d^{3}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \sqrt {c^{2} x^{2}+1}\, c}{2 \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) d^{3} x}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{2 \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right ) d^{3} x^{2}}+\frac {13 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arctan \left (c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{3 \sqrt {c^{2} x^{2}+1}\, d^{3}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{2 \sqrt {c^{2} x^{2}+1}\, d^{3}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \dilog \left (1+c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{2 \sqrt {c^{2} x^{2}+1}\, d^{3}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2}}{2 \sqrt {c^{2} x^{2}+1}\, d^{3}}\)	\(546\)

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,method=_RETURNVERBOSE)

[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^2*x^2+1)^(1/2)*c^3-10/3*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^2*x^2+1)^(1/2)*c-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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