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

Optimal. Leaf size=329 \[ -\frac{b d^2 \sqrt{c^2 d x^2+d} \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 x^2+1}}+\frac{b d^2 \sqrt{c^2 d x^2+d} \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 x^2+1}}+d^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 d^2 \sqrt{c^2 d x^2+d} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 x^2+1}}+\frac{1}{5} \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{b c^5 d^2 x^5 \sqrt{c^2 d x^2+d}}{25 \sqrt{c^2 x^2+1}}-\frac{11 b c^3 d^2 x^3 \sqrt{c^2 d x^2+d}}{45 \sqrt{c^2 x^2+1}}-\frac{23 b c d^2 x \sqrt{c^2 d x^2+d}}{15 \sqrt{c^2 x^2+1}} \]

[Out]

(-23*b*c*d^2*x*Sqrt[d + c^2*d*x^2])/(15*Sqrt[1 + c^2*x^2]) - (11*b*c^3*d^2*x^3*Sqrt[d + c^2*d*x^2])/(45*Sqrt[1
+ c^2*x^2]) - (b*c^5*d^2*x^5*Sqrt[d + c^2*d*x^2])/(25*Sqrt[1 + c^2*x^2]) + d^2*Sqrt[d + c^2*d*x^2]*(a + b*Arc
Sinh[c*x]) + (d*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/3 + ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/5
- (2*d^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] - (b*d^2*Sqrt[d
+ c^2*d*x^2]*PolyLog[2, -E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] + (b*d^2*Sqrt[d + c^2*d*x^2]*PolyLog[2, E^ArcSinh[
c*x]])/Sqrt[1 + c^2*x^2]

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Rubi [A]  time = 0.439084, antiderivative size = 329, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5744, 5742, 5760, 4182, 2279, 2391, 8, 194} \[ -\frac{b d^2 \sqrt{c^2 d x^2+d} \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 x^2+1}}+\frac{b d^2 \sqrt{c^2 d x^2+d} \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 x^2+1}}+d^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 d^2 \sqrt{c^2 d x^2+d} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 x^2+1}}+\frac{1}{5} \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{b c^5 d^2 x^5 \sqrt{c^2 d x^2+d}}{25 \sqrt{c^2 x^2+1}}-\frac{11 b c^3 d^2 x^3 \sqrt{c^2 d x^2+d}}{45 \sqrt{c^2 x^2+1}}-\frac{23 b c d^2 x \sqrt{c^2 d x^2+d}}{15 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-23*b*c*d^2*x*Sqrt[d + c^2*d*x^2])/(15*Sqrt[1 + c^2*x^2]) - (11*b*c^3*d^2*x^3*Sqrt[d + c^2*d*x^2])/(45*Sqrt[1
+ c^2*x^2]) - (b*c^5*d^2*x^5*Sqrt[d + c^2*d*x^2])/(25*Sqrt[1 + c^2*x^2]) + d^2*Sqrt[d + c^2*d*x^2]*(a + b*Arc
Sinh[c*x]) + (d*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/3 + ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/5
- (2*d^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] - (b*d^2*Sqrt[d
+ c^2*d*x^2]*PolyLog[2, -E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] + (b*d^2*Sqrt[d + c^2*d*x^2]*PolyLog[2, E^ArcSinh[
c*x]])/Sqrt[1 + c^2*x^2]

Rule 5744

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*(a + b*ArcSinh[c*x])^n)/(f*(m + 2*p + 1)), x] + (Dist[(2*d*p)/(m + 2*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)^FracPart[p]
)/(f*(m + 2*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] && GtQ[p, 0] &&  !LtQ[m, -1]
&& (RationalQ[m] || EqQ[n, 1])

Rule 5742

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(
(f*x)^(m + 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(f*(m + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1
+ c^2*x^2]), Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(f*
(m + 2)*Sqrt[1 + c^2*x^2]), Int[(f*x)^(m + 1)*(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[m, -1] && (RationalQ[m] || 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 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 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 194

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

Rubi steps

\begin{align*} \int \frac{\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx &=\frac{1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+d \int \frac{\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx-\frac{\left (b c d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^2 \, dx}{5 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+d^2 \int \frac{\sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx-\frac{\left (b c d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (1+2 c^2 x^2+c^4 x^4\right ) \, dx}{5 \sqrt{1+c^2 x^2}}-\frac{\left (b c d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right ) \, dx}{3 \sqrt{1+c^2 x^2}}\\ &=-\frac{8 b c d^2 x \sqrt{d+c^2 d x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{11 b c^3 d^2 x^3 \sqrt{d+c^2 d x^2}}{45 \sqrt{1+c^2 x^2}}-\frac{b c^5 d^2 x^5 \sqrt{d+c^2 d x^2}}{25 \sqrt{1+c^2 x^2}}+d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (d^2 \sqrt{d+c^2 d x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{x \sqrt{1+c^2 x^2}} \, dx}{\sqrt{1+c^2 x^2}}-\frac{\left (b c d^2 \sqrt{d+c^2 d x^2}\right ) \int 1 \, dx}{\sqrt{1+c^2 x^2}}\\ &=-\frac{23 b c d^2 x \sqrt{d+c^2 d x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{11 b c^3 d^2 x^3 \sqrt{d+c^2 d x^2}}{45 \sqrt{1+c^2 x^2}}-\frac{b c^5 d^2 x^5 \sqrt{d+c^2 d x^2}}{25 \sqrt{1+c^2 x^2}}+d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (d^2 \sqrt{d+c^2 d x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}\\ &=-\frac{23 b c d^2 x \sqrt{d+c^2 d x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{11 b c^3 d^2 x^3 \sqrt{d+c^2 d x^2}}{45 \sqrt{1+c^2 x^2}}-\frac{b c^5 d^2 x^5 \sqrt{d+c^2 d x^2}}{25 \sqrt{1+c^2 x^2}}+d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}-\frac{\left (b d^2 \sqrt{d+c^2 d x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}+\frac{\left (b d^2 \sqrt{d+c^2 d x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}\\ &=-\frac{23 b c d^2 x \sqrt{d+c^2 d x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{11 b c^3 d^2 x^3 \sqrt{d+c^2 d x^2}}{45 \sqrt{1+c^2 x^2}}-\frac{b c^5 d^2 x^5 \sqrt{d+c^2 d x^2}}{25 \sqrt{1+c^2 x^2}}+d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}-\frac{\left (b d^2 \sqrt{d+c^2 d x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}+\frac{\left (b d^2 \sqrt{d+c^2 d x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}\\ &=-\frac{23 b c d^2 x \sqrt{d+c^2 d x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{11 b c^3 d^2 x^3 \sqrt{d+c^2 d x^2}}{45 \sqrt{1+c^2 x^2}}-\frac{b c^5 d^2 x^5 \sqrt{d+c^2 d x^2}}{25 \sqrt{1+c^2 x^2}}+d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}-\frac{b d^2 \sqrt{d+c^2 d x^2} \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}+\frac{b d^2 \sqrt{d+c^2 d x^2} \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.23533, size = 361, normalized size = 1.1 \[ \frac{b d^2 \sqrt{c^2 d x^2+d} \left (\text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-\text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+\sqrt{c^2 x^2+1} \sinh ^{-1}(c x)-c x+\sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )}{\sqrt{c^2 x^2+1}}+\frac{1}{15} a d^2 \left (3 c^4 x^4+11 c^2 x^2+23\right ) \sqrt{c^2 d x^2+d}-a d^{5/2} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+d\right )+a d^{5/2} \log (x)-\frac{b c^3 d^2 x^3 \left (3 c^2 x^2+5\right ) \sqrt{c^2 d x^2+d}}{75 \sqrt{c^2 x^2+1}}-\frac{8 b c d^2 x \left (c^2 x^2+3\right ) \sqrt{c^2 d x^2+d}}{45 \sqrt{c^2 x^2+1}}+\frac{2}{3} b d^2 \left (c^2 x^2+1\right ) \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)+\frac{1}{15} b d \left (3 c^2 x^2-2\right ) \left (c^2 d x^2+d\right )^{3/2} \sinh ^{-1}(c x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-8*b*c*d^2*x*(3 + c^2*x^2)*Sqrt[d + c^2*d*x^2])/(45*Sqrt[1 + c^2*x^2]) - (b*c^3*d^2*x^3*(5 + 3*c^2*x^2)*Sqrt[
d + c^2*d*x^2])/(75*Sqrt[1 + c^2*x^2]) + (a*d^2*Sqrt[d + c^2*d*x^2]*(23 + 11*c^2*x^2 + 3*c^4*x^4))/15 + (2*b*d
^2*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x])/3 + (b*d*(-2 + 3*c^2*x^2)*(d + c^2*d*x^2)^(3/2)*ArcSinh[c*x
])/15 + a*d^(5/2)*Log[x] - a*d^(5/2)*Log[d + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + (b*d^2*Sqrt[d + c^2*d*x^2]*(-(c*x)
+ Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - ArcSinh[c*x]*Log[1 + E^(-ArcSinh
[c*x])] + PolyLog[2, -E^(-ArcSinh[c*x])] - PolyLog[2, E^(-ArcSinh[c*x])]))/Sqrt[1 + c^2*x^2]

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Maple [A]  time = 0.202, size = 540, normalized size = 1.6 \begin{align*}{\frac{a}{5} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{5}{2}}}}+{\frac{ad}{3} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-a{d}^{{\frac{5}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{{c}^{2}d{x}^{2}+d} \right ) } \right ) +a\sqrt{{c}^{2}d{x}^{2}+d}{d}^{2}+{\frac{23\,b{d}^{2}{\it Arcsinh} \left ( cx \right ) }{15\,{c}^{2}{x}^{2}+15}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{b{d}^{2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it polylog} \left ( 2,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{b{d}^{2}{\it Arcsinh} \left ( cx \right ) \sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{b{d}^{2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{b{d}^{2}{\it Arcsinh} \left ( cx \right ) \sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{b{d}^{2}{\it Arcsinh} \left ( cx \right ){x}^{6}{c}^{6}}{5\,{c}^{2}{x}^{2}+5}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{b{d}^{2}{c}^{5}{x}^{5}}{25}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{14\,b{d}^{2}{\it Arcsinh} \left ( cx \right ){x}^{4}{c}^{4}}{15\,{c}^{2}{x}^{2}+15}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{11\,b{d}^{2}{c}^{3}{x}^{3}}{45}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{34\,b{d}^{2}{\it Arcsinh} \left ( cx \right ){x}^{2}{c}^{2}}{15\,{c}^{2}{x}^{2}+15}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{23\,b{d}^{2}cx}{15}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(c^2*d*x^2+d)^(5/2)*a+1/3*a*d*(c^2*d*x^2+d)^(3/2)-a*d^(5/2)*ln((2*d+2*d^(1/2)*(c^2*d*x^2+d)^(1/2))/x)+a*(c
^2*d*x^2+d)^(1/2)*d^2+23/15*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)+b*(d*(c^2*x^2+1))^(1/2)/(c^2*
x^2+1)^(1/2)*polylog(2,c*x+(c^2*x^2+1)^(1/2))*d^2-b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*ln(1+
c*x+(c^2*x^2+1)^(1/2))*d^2-b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*d^2+b*(
d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))*d^2+1/5*b*(d*(c^2*x^2+1))^(1/2
)*d^2/(c^2*x^2+1)*arcsinh(c*x)*x^6*c^6-1/25*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)^(1/2)*c^5*x^5+14/15*b*(d*(
c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)*x^4*c^4-11/45*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)^(1/2)*c^3
*x^3+34/15*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)*x^2*c^2-23/15*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2
*x^2+1)^(1/2)*c*x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/x,x, algorithm="fricas")

[Out]

integral((a*c^4*d^2*x^4 + 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 + 2*b*c^2*d^2*x^2 + b*d^2)*arcsinh(c*x))*sq
rt(c^2*d*x^2 + d)/x, 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((c**2*d*x**2+d)**(5/2)*(a+b*asinh(c*x))/x,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} + d\right )}^{\frac{5}{2}}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/x,x, algorithm="giac")

[Out]

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