3.174 \(\int \frac {(c e+d e x)^4}{(a+b \sinh ^{-1}(c+d x))^4} \, dx\)

Optimal. Leaf size=410 \[ -\frac {e^4 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{48 b^4 d}+\frac {27 e^4 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^4 d}-\frac {125 e^4 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{96 b^4 d}+\frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{48 b^4 d}-\frac {27 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^4 d}+\frac {125 e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{96 b^4 d}-\frac {25 e^4 \sqrt {(c+d x)^2+1} (c+d x)^4}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {2 e^4 \sqrt {(c+d x)^2+1} (c+d x)^2}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {5 e^4 (c+d x)^5}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e^4 \sqrt {(c+d x)^2+1} (c+d x)^4}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]

[Out]

-2/3*e^4*(d*x+c)^3/b^2/d/(a+b*arcsinh(d*x+c))^2-5/6*e^4*(d*x+c)^5/b^2/d/(a+b*arcsinh(d*x+c))^2+1/48*e^4*cosh(a
/b)*Shi((a+b*arcsinh(d*x+c))/b)/b^4/d-27/32*e^4*cosh(3*a/b)*Shi(3*(a+b*arcsinh(d*x+c))/b)/b^4/d+125/96*e^4*cos
h(5*a/b)*Shi(5*(a+b*arcsinh(d*x+c))/b)/b^4/d-1/48*e^4*Chi((a+b*arcsinh(d*x+c))/b)*sinh(a/b)/b^4/d+27/32*e^4*Ch
i(3*(a+b*arcsinh(d*x+c))/b)*sinh(3*a/b)/b^4/d-125/96*e^4*Chi(5*(a+b*arcsinh(d*x+c))/b)*sinh(5*a/b)/b^4/d-1/3*e
^4*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcsinh(d*x+c))^3-2*e^4*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)/b^3/d/(a+b*arc
sinh(d*x+c))-25/6*e^4*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)/b^3/d/(a+b*arcsinh(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.88, antiderivative size = 406, normalized size of antiderivative = 0.99, number of steps used = 24, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5865, 12, 5667, 5774, 5665, 3303, 3298, 3301} \[ -\frac {e^4 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{48 b^4 d}+\frac {27 e^4 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{32 b^4 d}-\frac {125 e^4 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{96 b^4 d}+\frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{48 b^4 d}-\frac {27 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{32 b^4 d}+\frac {125 e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{96 b^4 d}-\frac {5 e^4 (c+d x)^5}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {25 e^4 \sqrt {(c+d x)^2+1} (c+d x)^4}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {2 e^4 (c+d x)^3}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e^4 \sqrt {(c+d x)^2+1} (c+d x)^2}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e^4 \sqrt {(c+d x)^2+1} (c+d x)^4}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4/(a + b*ArcSinh[c + d*x])^4,x]

[Out]

-(e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^3) - (2*e^4*(c + d*x)^3)/(3*b^2*d*(a
+ b*ArcSinh[c + d*x])^2) - (5*e^4*(c + d*x)^5)/(6*b^2*d*(a + b*ArcSinh[c + d*x])^2) - (2*e^4*(c + d*x)^2*Sqrt[
1 + (c + d*x)^2])/(b^3*d*(a + b*ArcSinh[c + d*x])) - (25*e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(6*b^3*d*(a +
b*ArcSinh[c + d*x])) - (e^4*CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b])/(48*b^4*d) + (27*e^4*CoshIntegral[
(3*a)/b + 3*ArcSinh[c + d*x]]*Sinh[(3*a)/b])/(32*b^4*d) - (125*e^4*CoshIntegral[(5*a)/b + 5*ArcSinh[c + d*x]]*
Sinh[(5*a)/b])/(96*b^4*d) + (e^4*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]])/(48*b^4*d) - (27*e^4*Cosh[(3*
a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[c + d*x]])/(32*b^4*d) + (125*e^4*Cosh[(5*a)/b]*SinhIntegral[(5*a)/b + 5
*ArcSinh[c + d*x]])/(96*b^4*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 5665

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]

Rule 5667

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

Rule 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), 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] && LtQ[n, -
1] && GtQ[d, 0]

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

\begin {align*} \int \frac {(c e+d e x)^4}{\left (a+b \sinh ^{-1}(c+d x)\right )^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^4 x^4}{\left (a+b \sinh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b \sinh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac {\left (4 e^4\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{3 b d}+\frac {\left (5 e^4\right ) \operatorname {Subst}\left (\int \frac {x^5}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {2 e^4 (c+d x)^3}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {5 e^4 (c+d x)^5}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac {\left (2 e^4\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b^2 d}+\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{6 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {2 e^4 (c+d x)^3}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {5 e^4 (c+d x)^5}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^2 \sqrt {1+(c+d x)^2}}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {25 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {\left (2 e^4\right ) \operatorname {Subst}\left (\int \left (-\frac {\sinh (x)}{4 (a+b x)}+\frac {3 \sinh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^3 d}+\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \left (\frac {\sinh (x)}{8 (a+b x)}-\frac {9 \sinh (3 x)}{16 (a+b x)}+\frac {5 \sinh (5 x)}{16 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{6 b^3 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {2 e^4 (c+d x)^3}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {5 e^4 (c+d x)^5}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^2 \sqrt {1+(c+d x)^2}}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {25 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e^4 \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^3 d}+\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{48 b^3 d}+\frac {\left (125 e^4\right ) \operatorname {Subst}\left (\int \frac {\sinh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 b^3 d}+\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^3 d}-\frac {\left (75 e^4\right ) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 b^3 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {2 e^4 (c+d x)^3}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {5 e^4 (c+d x)^5}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^2 \sqrt {1+(c+d x)^2}}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {25 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {\left (e^4 \cosh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^3 d}+\frac {\left (25 e^4 \cosh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{48 b^3 d}+\frac {\left (3 e^4 \cosh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^3 d}-\frac {\left (75 e^4 \cosh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 b^3 d}+\frac {\left (125 e^4 \cosh \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 b^3 d}+\frac {\left (e^4 \sinh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^3 d}-\frac {\left (25 e^4 \sinh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{48 b^3 d}-\frac {\left (3 e^4 \sinh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^3 d}+\frac {\left (75 e^4 \sinh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {\left (125 e^4 \sinh \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 b^3 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {2 e^4 (c+d x)^3}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {5 e^4 (c+d x)^5}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^2 \sqrt {1+(c+d x)^2}}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {25 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e^4 \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right ) \sinh \left (\frac {a}{b}\right )}{48 b^4 d}+\frac {27 e^4 \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac {3 a}{b}\right )}{32 b^4 d}-\frac {125 e^4 \text {Chi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac {5 a}{b}\right )}{96 b^4 d}+\frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{48 b^4 d}-\frac {27 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{32 b^4 d}+\frac {125 e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{96 b^4 d}\\ \end {align*}

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Mathematica [A]  time = 1.90, size = 410, normalized size = 1.00 \[ -\frac {e^4 \left (\frac {32 b^3 \sqrt {(c+d x)^2+1} (c+d x)^4}{\left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {16 b^2 \left (-5 (c+d x)^5-4 (c+d x)^3\right )}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+384 \left (\sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+544 \left (-3 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+\sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )+125 \left (10 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-5 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )-10 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+5 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )+\frac {16 b \sqrt {(c+d x)^2+1} \left (25 (c+d x)^4+12 (c+d x)^2\right )}{a+b \sinh ^{-1}(c+d x)}\right )}{96 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4/(a + b*ArcSinh[c + d*x])^4,x]

[Out]

-1/96*(e^4*((32*b^3*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x])^3 - (16*b^2*(-4*(c + d*x)^3 -
5*(c + d*x)^5))/(a + b*ArcSinh[c + d*x])^2 + (16*b*Sqrt[1 + (c + d*x)^2]*(12*(c + d*x)^2 + 25*(c + d*x)^4))/(a
+ b*ArcSinh[c + d*x]) + 384*(CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b] - Cosh[a/b]*SinhIntegral[a/b + Ar
cSinh[c + d*x]]) + 544*(-3*CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b] + CoshIntegral[3*(a/b + ArcSinh[c +
d*x])]*Sinh[(3*a)/b] + 3*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]] - Cosh[(3*a)/b]*SinhIntegral[3*(a/b +
ArcSinh[c + d*x])]) + 125*(10*CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b] - 5*CoshIntegral[3*(a/b + ArcSinh
[c + d*x])]*Sinh[(3*a)/b] + CoshIntegral[5*(a/b + ArcSinh[c + d*x])]*Sinh[(5*a)/b] - 10*Cosh[a/b]*SinhIntegral
[a/b + ArcSinh[c + d*x]] + 5*Cosh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c + d*x])] - Cosh[(5*a)/b]*SinhIntegr
al[5*(a/b + ArcSinh[c + d*x])])))/(b^4*d)

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}{b^{4} \operatorname {arsinh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname {arsinh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname {arsinh}\left (d x + c\right ) + a^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^4,x, algorithm="fricas")

[Out]

integral((d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4)/(b^4*arcsinh(d*x + c)^4
+ 4*a*b^3*arcsinh(d*x + c)^3 + 6*a^2*b^2*arcsinh(d*x + c)^2 + 4*a^3*b*arcsinh(d*x + c) + a^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{4}}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^4/(b*arcsinh(d*x + c) + a)^4, x)

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maple [B]  time = 0.43, size = 1244, normalized size = 3.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^4,x)

[Out]

1/d*(1/192*(16*(d*x+c)^5-16*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+20*(d*x+c)^3-12*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+5*d*x+
5*c-(1+(d*x+c)^2)^(1/2))*e^4*(25*b^2*arcsinh(d*x+c)^2+50*a*b*arcsinh(d*x+c)-5*arcsinh(d*x+c)*b^2+25*a^2-5*a*b+
2*b^2)/b^3/(b^3*arcsinh(d*x+c)^3+3*a*b^2*arcsinh(d*x+c)^2+3*a^2*b*arcsinh(d*x+c)+a^3)+125/192*e^4/b^4*exp(5*a/
b)*Ei(1,5*arcsinh(d*x+c)+5*a/b)-1/64*(-4*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+4*(d*x+c)^3-(1+(d*x+c)^2)^(1/2)+3*d*x+3
*c)*e^4*(9*b^2*arcsinh(d*x+c)^2+18*a*b*arcsinh(d*x+c)-3*arcsinh(d*x+c)*b^2+9*a^2-3*a*b+2*b^2)/b^3/(b^3*arcsinh
(d*x+c)^3+3*a*b^2*arcsinh(d*x+c)^2+3*a^2*b*arcsinh(d*x+c)+a^3)-27/64*e^4/b^4*exp(3*a/b)*Ei(1,3*arcsinh(d*x+c)+
3*a/b)+1/96*(-(1+(d*x+c)^2)^(1/2)+d*x+c)*e^4*(b^2*arcsinh(d*x+c)^2+2*a*b*arcsinh(d*x+c)-arcsinh(d*x+c)*b^2+a^2
-a*b+2*b^2)/b^3/(b^3*arcsinh(d*x+c)^3+3*a*b^2*arcsinh(d*x+c)^2+3*a^2*b*arcsinh(d*x+c)+a^3)+1/96*e^4/b^4*exp(a/
b)*Ei(1,arcsinh(d*x+c)+a/b)-1/48/b*e^4*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^3-1/96/b^2*e^4*(d*x+c+
(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^2-1/96/b^3*e^4*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))-1/96
/b^4*e^4*exp(-a/b)*Ei(1,-arcsinh(d*x+c)-a/b)+1/32/b*e^4*(4*(d*x+c)^3+3*d*x+3*c+4*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)
+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^3+3/64/b^2*e^4*(4*(d*x+c)^3+3*d*x+3*c+4*(d*x+c)^2*(1+(d*x+c)^2)^(1/
2)+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^2+9/64/b^3*e^4*(4*(d*x+c)^3+3*d*x+3*c+4*(d*x+c)^2*(1+(d*x+c)^2)^(
1/2)+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))+27/64/b^4*e^4*exp(-3*a/b)*Ei(1,-3*arcsinh(d*x+c)-3*a/b)-1/96/b*
e^4*(16*(d*x+c)^5+20*(d*x+c)^3+16*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+5*d*x+5*c+12*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+(1+
(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^3-5/192/b^2*e^4*(16*(d*x+c)^5+20*(d*x+c)^3+16*(d*x+c)^4*(1+(d*x+c)^2)^(
1/2)+5*d*x+5*c+12*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^2-25/192/b^3*e^4*(16
*(d*x+c)^5+20*(d*x+c)^3+16*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+5*d*x+5*c+12*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+(1+(d*x+c)
^2)^(1/2))/(a+b*arcsinh(d*x+c))-125/192/b^4*e^4*exp(-5*a/b)*Ei(1,-5*arcsinh(d*x+c)-5*a/b))

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^4}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^4/(a + b*asinh(c + d*x))^4,x)

[Out]

int((c*e + d*e*x)^4/(a + b*asinh(c + d*x))^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{4} \left (\int \frac {c^{4}}{a^{4} + 4 a^{3} b \operatorname {asinh}{\left (c + d x \right )} + 6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )} + 4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )} + b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}\, dx + \int \frac {d^{4} x^{4}}{a^{4} + 4 a^{3} b \operatorname {asinh}{\left (c + d x \right )} + 6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )} + 4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )} + b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}\, dx + \int \frac {4 c d^{3} x^{3}}{a^{4} + 4 a^{3} b \operatorname {asinh}{\left (c + d x \right )} + 6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )} + 4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )} + b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}\, dx + \int \frac {6 c^{2} d^{2} x^{2}}{a^{4} + 4 a^{3} b \operatorname {asinh}{\left (c + d x \right )} + 6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )} + 4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )} + b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}\, dx + \int \frac {4 c^{3} d x}{a^{4} + 4 a^{3} b \operatorname {asinh}{\left (c + d x \right )} + 6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )} + 4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )} + b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4/(a+b*asinh(d*x+c))**4,x)

[Out]

e**4*(Integral(c**4/(a**4 + 4*a**3*b*asinh(c + d*x) + 6*a**2*b**2*asinh(c + d*x)**2 + 4*a*b**3*asinh(c + d*x)*
*3 + b**4*asinh(c + d*x)**4), x) + Integral(d**4*x**4/(a**4 + 4*a**3*b*asinh(c + d*x) + 6*a**2*b**2*asinh(c +
d*x)**2 + 4*a*b**3*asinh(c + d*x)**3 + b**4*asinh(c + d*x)**4), x) + Integral(4*c*d**3*x**3/(a**4 + 4*a**3*b*a
sinh(c + d*x) + 6*a**2*b**2*asinh(c + d*x)**2 + 4*a*b**3*asinh(c + d*x)**3 + b**4*asinh(c + d*x)**4), x) + Int
egral(6*c**2*d**2*x**2/(a**4 + 4*a**3*b*asinh(c + d*x) + 6*a**2*b**2*asinh(c + d*x)**2 + 4*a*b**3*asinh(c + d*
x)**3 + b**4*asinh(c + d*x)**4), x) + Integral(4*c**3*d*x/(a**4 + 4*a**3*b*asinh(c + d*x) + 6*a**2*b**2*asinh(
c + d*x)**2 + 4*a*b**3*asinh(c + d*x)**3 + b**4*asinh(c + d*x)**4), x))

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