3.2.40 \(\int \frac {(d+c^2 d x^2)^{5/2} (a+b \sinh ^{-1}(c x))}{x} \, dx\) [140]

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

[Out]

1/3*d*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))+1/5*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))+d^2*(a+b*arcsinh(c*x))
*(c^2*d*x^2+d)^(1/2)-23/15*b*c*d^2*x*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-11/45*b*c^3*d^2*x^3*(c^2*d*x^2+d)^(
1/2)/(c^2*x^2+1)^(1/2)-1/25*b*c^5*d^2*x^5*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-2*d^2*(a+b*arcsinh(c*x))*arcta
nh(c*x+(c^2*x^2+1)^(1/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-b*d^2*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*(c^2*d
*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+b*d^2*polylog(2,c*x+(c^2*x^2+1)^(1/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 329, normalized size of antiderivative = 1.00, 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 = {5808, 5806, 5816, 4267, 2317, 2438, 8, 200} \begin {gather*} 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 d^2 \sqrt {c^2 d x^2+d} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}+\frac {b d^2 \sqrt {c^2 d x^2+d} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\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}}-\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}} \end {gather*}

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 8

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

Rule 200

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]

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 5806

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[(1/(m + 2))*Simp[Sqrt[d + e*x^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/(f*(m + 2)))
*Simp[Sqrt[d + e*x^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] && IGtQ[n, 0] && (IGtQ[m, -2] || EqQ[n, 1])

Rule 5808

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/(f*(m + 2*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] && GtQ[p, 0] &&  !LtQ[m, -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 {\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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.91, size = 353, normalized size = 1.07 \begin {gather*} \frac {1}{225} d^2 \left (-\frac {40 b c x \left (3+c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}-\frac {3 b c^3 x^3 \left (5+3 c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}+15 a \sqrt {d+c^2 d x^2} \left (23+11 c^2 x^2+3 c^4 x^4\right )+150 b \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)+15 b \left (1+c^2 x^2\right ) \left (-2+3 c^2 x^2\right ) \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)+225 a \sqrt {d} \log (x)-225 a \sqrt {d} \log \left (d+\sqrt {d} \sqrt {d+c^2 d x^2}\right )+\frac {225 b \sqrt {d+c^2 d x^2} \left (-c x+\sqrt {1+c^2 x^2} \sinh ^{-1}(c x)+\sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\sinh ^{-1}(c x) \log \left (1+e^{-\sinh ^{-1}(c x)}\right )+\text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-\text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )\right )}{\sqrt {1+c^2 x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*((-40*b*c*x*(3 + c^2*x^2)*Sqrt[d + c^2*d*x^2])/Sqrt[1 + c^2*x^2] - (3*b*c^3*x^3*(5 + 3*c^2*x^2)*Sqrt[d +
c^2*d*x^2])/Sqrt[1 + c^2*x^2] + 15*a*Sqrt[d + c^2*d*x^2]*(23 + 11*c^2*x^2 + 3*c^4*x^4) + 150*b*(1 + c^2*x^2)*S
qrt[d + c^2*d*x^2]*ArcSinh[c*x] + 15*b*(1 + c^2*x^2)*(-2 + 3*c^2*x^2)*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x] + 225*a
*Sqrt[d]*Log[x] - 225*a*Sqrt[d]*Log[d + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + (225*b*Sqrt[d + c^2*d*x^2]*(-(c*x) + Sq
rt[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]))/225

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Maple [A]
time = 1.67, size = 540, normalized size = 1.64

method	result	size
default	\(\frac {\left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}} a}{5}+\frac {a d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{3}-a \,d^{\frac {5}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c^{2} d \,x^{2}+d}}{x}\right )+a \sqrt {c^{2} d \,x^{2}+d}\, d^{2}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right ) d^{2}}{\sqrt {c^{2} x^{2}+1}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right ) d^{2}}{\sqrt {c^{2} x^{2}+1}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right ) d^{2}}{\sqrt {c^{2} x^{2}+1}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right ) d^{2}}{\sqrt {c^{2} x^{2}+1}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} \arcsinh \left (c x \right ) x^{6} c^{6}}{5 c^{2} x^{2}+5}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} x^{5} c^{5}}{25 \sqrt {c^{2} x^{2}+1}}+\frac {14 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} \arcsinh \left (c x \right ) x^{4} c^{4}}{15 \left (c^{2} x^{2}+1\right )}-\frac {11 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} x^{3} c^{3}}{45 \sqrt {c^{2} x^{2}+1}}+\frac {34 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} \arcsinh \left (c x \right ) x^{2} c^{2}}{15 \left (c^{2} x^{2}+1\right )}-\frac {23 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} x c}{15 \sqrt {c^{2} x^{2}+1}}+\frac {23 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} \arcsinh \left (c x \right )}{15 \left (c^{2} x^{2}+1\right )}\)	\(540\)

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,method=_RETURNVERBOSE)

[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-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)*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)*polylog(2,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)*x^5*c^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^4-11/4
5*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)^(1/2)*x^3*c^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)*x*c+23/15*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^
2+1)*arcsinh(c*x)

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

[Out]

-1/15*(15*d^(5/2)*arcsinh(1/(c*abs(x))) - 3*(c^2*d*x^2 + d)^(5/2) - 5*(c^2*d*x^2 + d)^(3/2)*d - 15*sqrt(c^2*d*
x^2 + d)*d^2)*a + b*integrate((c^2*d*x^2 + d)^(5/2)*log(c*x + sqrt(c^2*x^2 + 1))/x, 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((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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{x}\, dx \end {gather*}

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]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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