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

Optimal. Leaf size=355 \[ \frac {5}{2} c^2 d^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )-\frac {5 c^2 d^2 \sqrt {c^2 d x^2+d} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {5}{6} c^2 d \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {5 b c^2 d^2 \sqrt {c^2 d x^2+d} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {c^2 x^2+1}}+\frac {5 b c^2 d^2 \sqrt {c^2 d x^2+d} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {c^2 x^2+1}}-\frac {b c d^2 \sqrt {c^2 d x^2+d}}{2 x \sqrt {c^2 x^2+1}}-\frac {b c^5 d^2 x^3 \sqrt {c^2 d x^2+d}}{9 \sqrt {c^2 x^2+1}}-\frac {7 b c^3 d^2 x \sqrt {c^2 d x^2+d}}{3 \sqrt {c^2 x^2+1}} \]

[Out]

5/6*c^2*d*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))-1/2*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))/x^2+5/2*c^2*d^2*(a
+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)-1/2*b*c*d^2*(c^2*d*x^2+d)^(1/2)/x/(c^2*x^2+1)^(1/2)-7/3*b*c^3*d^2*x*(c^2*
d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-1/9*b*c^5*d^2*x^3*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-5*c^2*d^2*(a+b*arcsin
h(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-5/2*b*c^2*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)+5/2*b*c^2*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 = 0.45, antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {5739, 5744, 5742, 5760, 4182, 2279, 2391, 8, 270} \[ -\frac {5 b c^2 d^2 \sqrt {c^2 d x^2+d} \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {c^2 x^2+1}}+\frac {5 b c^2 d^2 \sqrt {c^2 d x^2+d} \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {c^2 x^2+1}}+\frac {5}{2} c^2 d^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )-\frac {5 c^2 d^2 \sqrt {c^2 d x^2+d} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {5}{6} c^2 d \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {b c^5 d^2 x^3 \sqrt {c^2 d x^2+d}}{9 \sqrt {c^2 x^2+1}}-\frac {7 b c^3 d^2 x \sqrt {c^2 d x^2+d}}{3 \sqrt {c^2 x^2+1}}-\frac {b c d^2 \sqrt {c^2 d x^2+d}}{2 x \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d^2*Sqrt[d + c^2*d*x^2])/(2*x*Sqrt[1 + c^2*x^2]) - (7*b*c^3*d^2*x*Sqrt[d + c^2*d*x^2])/(3*Sqrt[1 + c^2*x
^2]) - (b*c^5*d^2*x^3*Sqrt[d + c^2*d*x^2])/(9*Sqrt[1 + c^2*x^2]) + (5*c^2*d^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSi
nh[c*x]))/2 + (5*c^2*d*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/6 - ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c
*x]))/(2*x^2) - (5*c^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]
- (5*b*c^2*d^2*Sqrt[d + c^2*d*x^2]*PolyLog[2, -E^ArcSinh[c*x]])/(2*Sqrt[1 + c^2*x^2]) + (5*b*c^2*d^2*Sqrt[d +
c^2*d*x^2]*PolyLog[2, E^ArcSinh[c*x]])/(2*Sqrt[1 + c^2*x^2])

Rule 8

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 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 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 5739

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

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^3} \, dx &=-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {1}{2} \left (5 c^2 d\right ) \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 \frac {\left (1+c^2 x^2\right )^2}{x^2} \, dx}{2 \sqrt {1+c^2 x^2}}\\ &=\frac {5}{6} c^2 d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {1}{2} \left (5 c^2 d^2\right ) \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 (2 c^2+\frac {1}{x^2}+c^4 x^2\right ) \, dx}{2 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c^3 d^2 \sqrt {d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right ) \, dx}{6 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d+c^2 d x^2}}{2 x \sqrt {1+c^2 x^2}}+\frac {b c^3 d^2 x \sqrt {d+c^2 d x^2}}{6 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d+c^2 d x^2}}{9 \sqrt {1+c^2 x^2}}+\frac {5}{2} c^2 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{6} c^2 d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {\left (5 c^2 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}{2 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c^3 d^2 \sqrt {d+c^2 d x^2}\right ) \int 1 \, dx}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d+c^2 d x^2}}{2 x \sqrt {1+c^2 x^2}}-\frac {7 b c^3 d^2 x \sqrt {d+c^2 d x^2}}{3 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d+c^2 d x^2}}{9 \sqrt {1+c^2 x^2}}+\frac {5}{2} c^2 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{6} c^2 d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {\left (5 c^2 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 )}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d+c^2 d x^2}}{2 x \sqrt {1+c^2 x^2}}-\frac {7 b c^3 d^2 x \sqrt {d+c^2 d x^2}}{3 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d+c^2 d x^2}}{9 \sqrt {1+c^2 x^2}}+\frac {5}{2} c^2 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{6} c^2 d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {5 c^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 (5 b c^2 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 )}{2 \sqrt {1+c^2 x^2}}+\frac {\left (5 b c^2 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 )}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d+c^2 d x^2}}{2 x \sqrt {1+c^2 x^2}}-\frac {7 b c^3 d^2 x \sqrt {d+c^2 d x^2}}{3 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d+c^2 d x^2}}{9 \sqrt {1+c^2 x^2}}+\frac {5}{2} c^2 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{6} c^2 d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {5 c^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 (5 b c^2 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 )}{2 \sqrt {1+c^2 x^2}}+\frac {\left (5 b c^2 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 )}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c d^2 \sqrt {d+c^2 d x^2}}{2 x \sqrt {1+c^2 x^2}}-\frac {7 b c^3 d^2 x \sqrt {d+c^2 d x^2}}{3 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 x^3 \sqrt {d+c^2 d x^2}}{9 \sqrt {1+c^2 x^2}}+\frac {5}{2} c^2 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{6} c^2 d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {5 c^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 {5 b c^2 d^2 \sqrt {d+c^2 d x^2} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {1+c^2 x^2}}+\frac {5 b c^2 d^2 \sqrt {d+c^2 d x^2} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 6.74, size = 467, normalized size = 1.32 \[ -\frac {5}{2} a c^2 d^{5/2} \log \left (\sqrt {d} \sqrt {d \left (c^2 x^2+1\right )}+d\right )+\frac {5}{2} a c^2 d^{5/2} \log (x)+\sqrt {d \left (c^2 x^2+1\right )} \left (\frac {1}{3} a c^4 d^2 x^2+\frac {7}{3} a c^2 d^2-\frac {a d^2}{2 x^2}\right )+\frac {2 b c^2 d^2 \sqrt {d \left (c^2 x^2+1\right )} \left (\sqrt {c^2 x^2+1} \sinh ^{-1}(c x)+\text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )-c x+\sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )}{\sqrt {c^2 x^2+1}}+\frac {b c^2 d^2 \sqrt {d \left (c^2 x^2+1\right )} \left (4 \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-4 \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+4 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-4 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+2 \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-2 \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\sinh ^{-1}(c x) \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\sinh ^{-1}(c x) \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{8 \sqrt {c^2 x^2+1}}+b c^2 d^2 \left (\frac {1}{3} \left (c^2 x^2+1\right ) \sqrt {d \left (c^2 x^2+1\right )} \sinh ^{-1}(c x)-\frac {c x \left (c^2 x^2+3\right ) \sqrt {d \left (c^2 x^2+1\right )}}{9 \sqrt {c^2 x^2+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[d*(1 + c^2*x^2)]*((7*a*c^2*d^2)/3 - (a*d^2)/(2*x^2) + (a*c^4*d^2*x^2)/3) + b*c^2*d^2*(-1/9*(c*x*Sqrt[d*(1
+ c^2*x^2)]*(3 + c^2*x^2))/Sqrt[1 + c^2*x^2] + ((1 + c^2*x^2)*Sqrt[d*(1 + c^2*x^2)]*ArcSinh[c*x])/3) + (5*a*c
^2*d^(5/2)*Log[x])/2 - (5*a*c^2*d^(5/2)*Log[d + Sqrt[d]*Sqrt[d*(1 + c^2*x^2)]])/2 + (2*b*c^2*d^2*Sqrt[d*(1 + c
^2*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] +
(b*c^2*d^2*Sqrt[d*(1 + c^2*x^2)]*(-2*Coth[ArcSinh[c*x]/2] - ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^2 + 4*ArcSinh[c
*x]*Log[1 - E^(-ArcSinh[c*x])] - 4*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 4*PolyLog[2, -E^(-ArcSinh[c*x])]
- 4*PolyLog[2, E^(-ArcSinh[c*x])] - ArcSinh[c*x]*Sech[ArcSinh[c*x]/2]^2 + 2*Tanh[ArcSinh[c*x]/2]))/(8*Sqrt[1 +
c^2*x^2])

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

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^3,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^3, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.30, size = 588, normalized size = 1.66 \[ -\frac {a \left (c^{2} d \,x^{2}+d \right )^{\frac {7}{2}}}{2 d \,x^{2}}+\frac {a \,c^{2} \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{2}+\frac {5 a \,c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{6}-\frac {5 a \,c^{2} d^{\frac {5}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c^{2} d \,x^{2}+d}}{x}\right )}{2}+\frac {5 a \,c^{2} \sqrt {c^{2} d \,x^{2}+d}\, d^{2}}{2}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right ) c^{2} d^{2}}{2 \sqrt {c^{2} x^{2}+1}}-\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2} d^{2}}{2 \sqrt {c^{2} x^{2}+1}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right ) c^{2} d^{2}}{2 \sqrt {c^{2} x^{2}+1}}-\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right ) c^{2} d^{2}}{2 \sqrt {c^{2} x^{2}+1}}+\frac {11 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, c^{2} d^{2} \arcsinh \left (c x \right )}{6 \left (c^{2} x^{2}+1\right )}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} \arcsinh \left (c x \right )}{2 x^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, c^{6} d^{2} \arcsinh \left (c x \right ) x^{4}}{3 c^{2} x^{2}+3}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, c^{5} d^{2} x^{3}}{9 \sqrt {c^{2} x^{2}+1}}+\frac {8 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, c^{4} d^{2} \arcsinh \left (c x \right ) x^{2}}{3 \left (c^{2} x^{2}+1\right )}-\frac {7 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, c^{3} d^{2} x}{3 \sqrt {c^{2} x^{2}+1}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, d^{2} c}{2 x \sqrt {c^{2} x^{2}+1}} \]

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^3,x)

[Out]

-1/2*a/d/x^2*(c^2*d*x^2+d)^(7/2)+1/2*a*c^2*(c^2*d*x^2+d)^(5/2)+5/6*a*c^2*d*(c^2*d*x^2+d)^(3/2)-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*a*c^2*(c^2*d*x^2+d)^(1/2)*d^2+5/2*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))*c^2*d^2-5/2*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))*c^2*d^2+5/2*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))*c^2*d^2-5/2*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))*c^2*
d^2+11/6*b*(d*(c^2*x^2+1))^(1/2)*c^2*d^2/(c^2*x^2+1)*arcsinh(c*x)-1/2*b*(d*(c^2*x^2+1))^(1/2)*d^2/x^2/(c^2*x^2
+1)*arcsinh(c*x)+1/3*b*(d*(c^2*x^2+1))^(1/2)*c^6*d^2/(c^2*x^2+1)*arcsinh(c*x)*x^4-1/9*b*(d*(c^2*x^2+1))^(1/2)*
c^5*d^2/(c^2*x^2+1)^(1/2)*x^3+8/3*b*(d*(c^2*x^2+1))^(1/2)*c^4*d^2/(c^2*x^2+1)*arcsinh(c*x)*x^2-7/3*b*(d*(c^2*x
^2+1))^(1/2)*c^3*d^2/(c^2*x^2+1)^(1/2)*x-1/2*b*(d*(c^2*x^2+1))^(1/2)*d^2/x/(c^2*x^2+1)^(1/2)*c

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{6} \, {\left (15 \, c^{2} d^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) - 3 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} c^{2} - 5 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d - 15 \, \sqrt {c^{2} d x^{2} + d} c^{2} d^{2} + \frac {3 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {7}{2}}}{d x^{2}}\right )} a + b \int \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{x^{3}}\,{d x} \]

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^3,x, algorithm="maxima")

[Out]

-1/6*(15*c^2*d^(5/2)*arcsinh(1/(c*abs(x))) - 3*(c^2*d*x^2 + d)^(5/2)*c^2 - 5*(c^2*d*x^2 + d)^(3/2)*c^2*d - 15*
sqrt(c^2*d*x^2 + d)*c^2*d^2 + 3*(c^2*d*x^2 + d)^(7/2)/(d*x^2))*a + b*integrate((c^2*d*x^2 + d)^(5/2)*log(c*x +
sqrt(c^2*x^2 + 1))/x^3, x)

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

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^3,x)

[Out]

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

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

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**3,x)

[Out]

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

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