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

Optimal. Leaf size=512 \[ -\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {8 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^6 d^3}-\frac {22 b \sqrt {c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {16 a b x \sqrt {c^2 x^2+1}}{3 c^5 d^2 \sqrt {c^2 d x^2+d}}+\frac {11 b x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {c^2 d x^2+d}}-\frac {4 x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {c^2 d x^2+d}}-\frac {b x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}+\frac {11 i b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {11 i b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {c^2 d x^2+d}}+\frac {2 b^2 \left (c^2 x^2+1\right )}{c^6 d^2 \sqrt {c^2 d x^2+d}}+\frac {b^2}{3 c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {16 b^2 x \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{3 c^5 d^2 \sqrt {c^2 d x^2+d}} \]

[Out]

-1/3*x^4*(a+b*arcsinh(c*x))^2/c^2/d/(c^2*d*x^2+d)^(3/2)+1/3*b^2/c^6/d^2/(c^2*d*x^2+d)^(1/2)+2*b^2*(c^2*x^2+1)/
c^6/d^2/(c^2*d*x^2+d)^(1/2)-4/3*x^2*(a+b*arcsinh(c*x))^2/c^4/d^2/(c^2*d*x^2+d)^(1/2)-1/3*b*x^3*(a+b*arcsinh(c*
x))/c^3/d^2/(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)-16/3*a*b*x*(c^2*x^2+1)^(1/2)/c^5/d^2/(c^2*d*x^2+d)^(1/2)-16/
3*b^2*x*arcsinh(c*x)*(c^2*x^2+1)^(1/2)/c^5/d^2/(c^2*d*x^2+d)^(1/2)+11/3*b*x*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/
2)/c^5/d^2/(c^2*d*x^2+d)^(1/2)-22/3*b*(a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/2))*(c^2*x^2+1)^(1/2)/c^6/d
^2/(c^2*d*x^2+d)^(1/2)+11/3*I*b^2*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))*(c^2*x^2+1)^(1/2)/c^6/d^2/(c^2*d*x^2+d
)^(1/2)-11/3*I*b^2*polylog(2,I*(c*x+(c^2*x^2+1)^(1/2)))*(c^2*x^2+1)^(1/2)/c^6/d^2/(c^2*d*x^2+d)^(1/2)+8/3*(a+b
*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/c^6/d^3

________________________________________________________________________________________

Rubi [A]  time = 0.88, antiderivative size = 512, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 11, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {5751, 5717, 5653, 261, 5767, 5693, 4180, 2279, 2391, 266, 43} \[ \frac {11 i b^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {11 i b^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {16 a b x \sqrt {c^2 x^2+1}}{3 c^5 d^2 \sqrt {c^2 d x^2+d}}-\frac {b x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}-\frac {4 x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {c^2 d x^2+d}}+\frac {11 b x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {c^2 d x^2+d}}+\frac {8 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^6 d^3}-\frac {22 b \sqrt {c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}+\frac {2 b^2 \left (c^2 x^2+1\right )}{c^6 d^2 \sqrt {c^2 d x^2+d}}+\frac {b^2}{3 c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {16 b^2 x \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{3 c^5 d^2 \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^2/(3*c^6*d^2*Sqrt[d + c^2*d*x^2]) - (16*a*b*x*Sqrt[1 + c^2*x^2])/(3*c^5*d^2*Sqrt[d + c^2*d*x^2]) + (2*b^2*(1
+ c^2*x^2))/(c^6*d^2*Sqrt[d + c^2*d*x^2]) - (16*b^2*x*Sqrt[1 + c^2*x^2]*ArcSinh[c*x])/(3*c^5*d^2*Sqrt[d + c^2
*d*x^2]) - (b*x^3*(a + b*ArcSinh[c*x]))/(3*c^3*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]) + (11*b*x*Sqrt[1 + c
^2*x^2]*(a + b*ArcSinh[c*x]))/(3*c^5*d^2*Sqrt[d + c^2*d*x^2]) - (x^4*(a + b*ArcSinh[c*x])^2)/(3*c^2*d*(d + c^2
*d*x^2)^(3/2)) - (4*x^2*(a + b*ArcSinh[c*x])^2)/(3*c^4*d^2*Sqrt[d + c^2*d*x^2]) + (8*Sqrt[d + c^2*d*x^2]*(a +
b*ArcSinh[c*x])^2)/(3*c^6*d^3) - (22*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(3*c^6*d
^2*Sqrt[d + c^2*d*x^2]) + (((11*I)/3)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c^6*d^2*Sqrt[d +
c^2*d*x^2]) - (((11*I)/3)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, I*E^ArcSinh[c*x]])/(c^6*d^2*Sqrt[d + c^2*d*x^2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5693

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

Rule 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 5751

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p
+ 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(2*c*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Ar
cSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 1]

Rule 5767

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(e*(m + 2*p + 1)), x] + (-Dist[(f^2*(m - 1))/(c^2
*(m + 2*p + 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d
+ e*x^2)^FracPart[p])/(c*(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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[m
, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^{5/2}} \, dx &=-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {4 \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^{3/2}} \, dx}{3 c^2 d}+\frac {\left (2 b \sqrt {1+c^2 x^2}\right ) \int \frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {4 x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {8 \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {d+c^2 d x^2}} \, dx}{3 c^4 d^2}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c^3 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (8 b \sqrt {1+c^2 x^2}\right ) \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 c^3 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (b^2 \sqrt {1+c^2 x^2}\right ) \int \frac {x^3}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 c^2 d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {11 b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {4 x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {8 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^6 d^3}-\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{1+c^2 x^2} \, dx}{c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (8 b \sqrt {1+c^2 x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{1+c^2 x^2} \, dx}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (16 b \sqrt {1+c^2 x^2}\right ) \int \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (b^2 \sqrt {1+c^2 x^2}\right ) \int \frac {x}{\sqrt {1+c^2 x^2}} \, dx}{c^4 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (8 b^2 \sqrt {1+c^2 x^2}\right ) \int \frac {x}{\sqrt {1+c^2 x^2}} \, dx}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (b^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{6 c^2 d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {16 a b x \sqrt {1+c^2 x^2}}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {11 b^2 \left (1+c^2 x^2\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {b x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {11 b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {4 x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {8 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^6 d^3}-\frac {\left (b \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (8 b \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (16 b^2 \sqrt {1+c^2 x^2}\right ) \int \sinh ^{-1}(c x) \, dx}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (b^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{c^2 \left (1+c^2 x\right )^{3/2}}+\frac {1}{c^2 \sqrt {1+c^2 x}}\right ) \, dx,x,x^2\right )}{6 c^2 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b^2}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {16 a b x \sqrt {1+c^2 x^2}}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {10 b^2 \left (1+c^2 x^2\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {16 b^2 x \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {b x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {11 b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {4 x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {8 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^6 d^3}-\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (i b^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (i b^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^6 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (8 i b^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (8 i b^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (16 b^2 \sqrt {1+c^2 x^2}\right ) \int \frac {x}{\sqrt {1+c^2 x^2}} \, dx}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b^2}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {16 a b x \sqrt {1+c^2 x^2}}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}+\frac {2 b^2 \left (1+c^2 x^2\right )}{c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {16 b^2 x \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {b x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {11 b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {4 x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {8 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^6 d^3}-\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (i b^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (i b^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c^6 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (8 i b^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (8 i b^2 \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {b^2}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {16 a b x \sqrt {1+c^2 x^2}}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}+\frac {2 b^2 \left (1+c^2 x^2\right )}{c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {16 b^2 x \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {b x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {11 b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {4 x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {8 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^6 d^3}-\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}+\frac {11 i b^2 \sqrt {1+c^2 x^2} \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}-\frac {11 i b^2 \sqrt {1+c^2 x^2} \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 c^6 d^2 \sqrt {d+c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.77, size = 333, normalized size = 0.65 \[ \frac {\sqrt {c^2 d x^2+d} \left (a^2 \left (3 c^4 x^4+12 c^2 x^2+8\right )+a b \left (2 \left (3 c^4 x^4+12 c^2 x^2+8\right ) \sinh ^{-1}(c x)-\sqrt {c^2 x^2+1} \left (c x \left (6 c^2 x^2+5\right )+22 \left (c^2 x^2+1\right ) \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )\right )+b^2 \left (11 i \left (c^2 x^2+1\right )^{3/2} \left (\text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )\right )+3 \left (c^2 x^2+1\right )^2 \left (\sinh ^{-1}(c x)^2+2\right )-6 c x \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)+\left (c^2 x^2+1\right ) \left (6 \sinh ^{-1}(c x)^2+1\right )+c x \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)+11 i \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x) \left (\log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right )-\sinh ^{-1}(c x)^2\right )\right )}{3 c^6 d^3 \left (c^2 x^2+1\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[d + c^2*d*x^2]*(a^2*(8 + 12*c^2*x^2 + 3*c^4*x^4) + a*b*(2*(8 + 12*c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x] - Sq
rt[1 + c^2*x^2]*(c*x*(5 + 6*c^2*x^2) + 22*(1 + c^2*x^2)*ArcTan[Tanh[ArcSinh[c*x]/2]])) + b^2*(c*x*Sqrt[1 + c^2
*x^2]*ArcSinh[c*x] - 6*c*x*(1 + c^2*x^2)^(3/2)*ArcSinh[c*x] - ArcSinh[c*x]^2 + 3*(1 + c^2*x^2)^2*(2 + ArcSinh[
c*x]^2) + (1 + c^2*x^2)*(1 + 6*ArcSinh[c*x]^2) + (11*I)*(1 + c^2*x^2)^(3/2)*ArcSinh[c*x]*(Log[1 - I/E^ArcSinh[
c*x]] - Log[1 + I/E^ArcSinh[c*x]]) + (11*I)*(1 + c^2*x^2)^(3/2)*(PolyLog[2, (-I)/E^ArcSinh[c*x]] - PolyLog[2,
I/E^ArcSinh[c*x]]))))/(3*c^6*d^3*(1 + c^2*x^2)^2)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{5} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b x^{5} \operatorname {arsinh}\left (c x\right ) + a^{2} x^{5}\right )} \sqrt {c^{2} d x^{2} + d}}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x^5*arcsinh(c*x)^2 + 2*a*b*x^5*arcsinh(c*x) + a^2*x^5)*sqrt(c^2*d*x^2 + d)/(c^6*d^3*x^6 + 3*c^4*
d^3*x^4 + 3*c^2*d^3*x^2 + d^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(x^5*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(5/2),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 [B]  time = 0.50, size = 1040, normalized size = 2.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^6+2*b^2*(d*(c^2*x^2+1))^(1/2)/c^4/d^3/(c^2*x^2+1)*x^2+b^2*(d
*(c^2*x^2+1))^(1/2)/c^6/d^3/(c^2*x^2+1)*arcsinh(c*x)^2+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^4*x^2
+5/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^6*arcsinh(c*x)^2+2*b^2*(d*(c^2*x^2+1))^(1/2)/c^6/d^3/(c^2*x
^2+1)+a^2*x^4/c^2/d/(c^2*d*x^2+d)^(3/2)+4*a^2/c^4*x^2/d/(c^2*d*x^2+d)^(3/2)-2*b^2*(d*(c^2*x^2+1))^(1/2)/c^5/d^
3/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)/c^5*arcsinh(c*x)*x+b^2*
(d*(c^2*x^2+1))^(1/2)/c^4/d^3/(c^2*x^2+1)*arcsinh(c*x)^2*x^2+2*b^2*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^4
*arcsinh(c*x)^2*x^2+11/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^3*dilog(1+I*(c*x+(c^2*x^2+1)^(1/2
)))-11/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^3*dilog(1-I*(c*x+(c^2*x^2+1)^(1/2)))-2*a*b*(d*(c^
2*x^2+1))^(1/2)/c^5/d^3/(c^2*x^2+1)^(1/2)*x+1/3*a*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)/c^5*x+2*a*b*(d
*(c^2*x^2+1))^(1/2)/c^6/d^3/(c^2*x^2+1)*arcsinh(c*x)+10/3*a*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^6*arcs
inh(c*x)+8/3*a^2/c^6/d/(c^2*d*x^2+d)^(3/2)+11/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^3*arcsinh(
c*x)*ln(1+I*(c*x+(c^2*x^2+1)^(1/2)))-11/3*I*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^3*arcsinh(c*x)*l
n(1-I*(c*x+(c^2*x^2+1)^(1/2)))+2*a*b*(d*(c^2*x^2+1))^(1/2)/c^4/d^3/(c^2*x^2+1)*arcsinh(c*x)*x^2+4*a*b*(d*(c^2*
x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^4*arcsinh(c*x)*x^2+11/3*I*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^
3*ln(c*x+(c^2*x^2+1)^(1/2)-I)-11/3*I*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^3*ln(c*x+(c^2*x^2+1)^(1
/2)+I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, a^{2} {\left (\frac {3 \, x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} + \frac {12 \, x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d} + \frac {8}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{6} d}\right )} + \frac {{\left (3 \, b^{2} c^{4} \sqrt {d} x^{4} + 12 \, b^{2} c^{2} \sqrt {d} x^{2} + 8 \, b^{2} \sqrt {d}\right )} \sqrt {c^{2} x^{2} + 1} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{3 \, {\left (c^{10} d^{3} x^{4} + 2 \, c^{8} d^{3} x^{2} + c^{6} d^{3}\right )}} + \int -\frac {2 \, {\left ({\left (12 \, b^{2} c^{3} x^{3} - 3 \, {\left (a b c^{5} - b^{2} c^{5}\right )} x^{5} + 8 \, b^{2} c x\right )} {\left (c^{2} x^{2} + 1\right )} + {\left (15 \, b^{2} c^{4} x^{4} - 3 \, {\left (a b c^{6} - b^{2} c^{6}\right )} x^{6} + 20 \, b^{2} c^{2} x^{2} + 8 \, b^{2}\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{3 \, {\left (c^{12} d^{\frac {5}{2}} x^{7} + 3 \, c^{10} d^{\frac {5}{2}} x^{5} + 3 \, c^{8} d^{\frac {5}{2}} x^{3} + c^{6} d^{\frac {5}{2}} x + {\left (c^{11} d^{\frac {5}{2}} x^{6} + 3 \, c^{9} d^{\frac {5}{2}} x^{4} + 3 \, c^{7} d^{\frac {5}{2}} x^{2} + c^{5} d^{\frac {5}{2}}\right )} \sqrt {c^{2} x^{2} + 1}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*(3*x^4/((c^2*d*x^2 + d)^(3/2)*c^2*d) + 12*x^2/((c^2*d*x^2 + d)^(3/2)*c^4*d) + 8/((c^2*d*x^2 + d)^(3/2)
*c^6*d)) + 1/3*(3*b^2*c^4*sqrt(d)*x^4 + 12*b^2*c^2*sqrt(d)*x^2 + 8*b^2*sqrt(d))*sqrt(c^2*x^2 + 1)*log(c*x + sq
rt(c^2*x^2 + 1))^2/(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^6*d^3) + integrate(-2/3*((12*b^2*c^3*x^3 - 3*(a*b*c^5 - b
^2*c^5)*x^5 + 8*b^2*c*x)*(c^2*x^2 + 1) + (15*b^2*c^4*x^4 - 3*(a*b*c^6 - b^2*c^6)*x^6 + 20*b^2*c^2*x^2 + 8*b^2)
*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1))/(c^12*d^(5/2)*x^7 + 3*c^10*d^(5/2)*x^5 + 3*c^8*d^(5/2)*x^3 +
c^6*d^(5/2)*x + (c^11*d^(5/2)*x^6 + 3*c^9*d^(5/2)*x^4 + 3*c^7*d^(5/2)*x^2 + c^5*d^(5/2))*sqrt(c^2*x^2 + 1)), x
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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