3.2.68 \(\int \frac {(c e+d e x)^4}{(a+b \sinh ^{-1}(c+d x))^3} \, dx\) [168]

Optimal. Leaf size=320 \[ -\frac {e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{16 b^3 d}-\frac {27 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}+\frac {25 e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {e^4 \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{16 b^3 d}+\frac {27 e^4 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {25 e^4 \sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d} \]

[Out]

-2*e^4*(d*x+c)^3/b^2/d/(a+b*arcsinh(d*x+c))-5/2*e^4*(d*x+c)^5/b^2/d/(a+b*arcsinh(d*x+c))+1/16*e^4*Chi((a+b*arc
sinh(d*x+c))/b)*cosh(a/b)/b^3/d-27/32*e^4*Chi(3*(a+b*arcsinh(d*x+c))/b)*cosh(3*a/b)/b^3/d+25/32*e^4*Chi(5*(a+b
*arcsinh(d*x+c))/b)*cosh(5*a/b)/b^3/d-1/16*e^4*Shi((a+b*arcsinh(d*x+c))/b)*sinh(a/b)/b^3/d+27/32*e^4*Shi(3*(a+
b*arcsinh(d*x+c))/b)*sinh(3*a/b)/b^3/d-25/32*e^4*Shi(5*(a+b*arcsinh(d*x+c))/b)*sinh(5*a/b)/b^3/d-1/2*e^4*(d*x+
c)^4*(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcsinh(d*x+c))^2

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Rubi [A]
time = 0.58, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {5859, 12, 5779, 5818, 5780, 5556, 3384, 3379, 3382} \begin {gather*} \frac {e^4 \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{16 b^3 d}-\frac {27 e^4 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}+\frac {25 e^4 \cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {e^4 \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{16 b^3 d}+\frac {27 e^4 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {25 e^4 \sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e^4 \sqrt {(c+d x)^2+1} (c+d x)^4}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4/(a + b*ArcSinh[c + d*x])^3,x]

[Out]

-1/2*(e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(b*d*(a + b*ArcSinh[c + d*x])^2) - (2*e^4*(c + d*x)^3)/(b^2*d*(a
+ b*ArcSinh[c + d*x])) - (5*e^4*(c + d*x)^5)/(2*b^2*d*(a + b*ArcSinh[c + d*x])) + (e^4*Cosh[a/b]*CoshIntegral[
(a + b*ArcSinh[c + d*x])/b])/(16*b^3*d) - (27*e^4*Cosh[(3*a)/b]*CoshIntegral[(3*(a + b*ArcSinh[c + d*x]))/b])/
(32*b^3*d) + (25*e^4*Cosh[(5*a)/b]*CoshIntegral[(5*(a + b*ArcSinh[c + d*x]))/b])/(32*b^3*d) - (e^4*Sinh[a/b]*S
inhIntegral[(a + b*ArcSinh[c + d*x])/b])/(16*b^3*d) + (27*e^4*Sinh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c +
d*x]))/b])/(32*b^3*d) - (25*e^4*Sinh[(5*a)/b]*SinhIntegral[(5*(a + b*ArcSinh[c + d*x]))/b])/(32*b^3*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 3379

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 3382

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 3384

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 5556

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 5779

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 5780

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

Rule 5818

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

Rule 5859

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

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Mathematica [A]
time = 0.92, size = 316, normalized size = 0.99 \begin {gather*} \frac {e^4 \left (-\frac {16 b^2 (c+d x)^4 \sqrt {1+(c+d x)^2}}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac {16 b \left (-4 (c+d x)^3-5 (c+d x)^5\right )}{a+b \sinh ^{-1}(c+d x)}+48 \left (-\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+\cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-\sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )+25 \left (2 \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-3 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )-2 \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+3 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )\right )}{32 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4/(a + b*ArcSinh[c + d*x])^3,x]

[Out]

(e^4*((-16*b^2*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x])^2 + (16*b*(-4*(c + d*x)^3 - 5*(c +
d*x)^5))/(a + b*ArcSinh[c + d*x]) + 48*(-(Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c + d*x]]) + Cosh[(3*a)/b]*Cosh
Integral[3*(a/b + ArcSinh[c + d*x])] + Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]] - Sinh[(3*a)/b]*SinhInte
gral[3*(a/b + ArcSinh[c + d*x])]) + 25*(2*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c + d*x]] - 3*Cosh[(3*a)/b]*Cos
hIntegral[3*(a/b + ArcSinh[c + d*x])] + Cosh[(5*a)/b]*CoshIntegral[5*(a/b + ArcSinh[c + d*x])] - 2*Sinh[a/b]*S
inhIntegral[a/b + ArcSinh[c + d*x]] + 3*Sinh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c + d*x])] - Sinh[(5*a)/b]
*SinhIntegral[5*(a/b + ArcSinh[c + d*x])])))/(32*b^3*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(895\) vs. \(2(302)=604\).
time = 6.74, size = 896, normalized size = 2.80

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(896\)
default	\(\text {Expression too large to display}\)	\(896\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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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((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*((5*a*d^11 + b*d^11)*x^11*e^4 + 11*(5*a*c*d^10 + b*c*d^10)*x^10*e^4 + (5*(55*c^2*d^9 + 3*d^9)*a + (55*c^2
*d^9 + 3*d^9)*b)*x^9*e^4 + 3*(5*(55*c^3*d^8 + 9*c*d^8)*a + (55*c^3*d^8 + 9*c*d^8)*b)*x^8*e^4 + 3*(5*(110*c^4*d
^7 + 36*c^2*d^7 + d^7)*a + (110*c^4*d^7 + 36*c^2*d^7 + d^7)*b)*x^7*e^4 + 21*(5*(22*c^5*d^6 + 12*c^3*d^6 + c*d^
6)*a + (22*c^5*d^6 + 12*c^3*d^6 + c*d^6)*b)*x^6*e^4 + (5*(462*c^6*d^5 + 378*c^4*d^5 + 63*c^2*d^5 + d^5)*a + (4
62*c^6*d^5 + 378*c^4*d^5 + 63*c^2*d^5 + d^5)*b)*x^5*e^4 + (5*(330*c^7*d^4 + 378*c^5*d^4 + 105*c^3*d^4 + 5*c*d^
4)*a + (330*c^7*d^4 + 378*c^5*d^4 + 105*c^3*d^4 + 5*c*d^4)*b)*x^4*e^4 + (5*(165*c^8*d^3 + 252*c^6*d^3 + 105*c^
4*d^3 + 10*c^2*d^3)*a + (165*c^8*d^3 + 252*c^6*d^3 + 105*c^4*d^3 + 10*c^2*d^3)*b)*x^3*e^4 + (5*(55*c^9*d^2 + 1
08*c^7*d^2 + 63*c^5*d^2 + 10*c^3*d^2)*a + (55*c^9*d^2 + 108*c^7*d^2 + 63*c^5*d^2 + 10*c^3*d^2)*b)*x^2*e^4 + (5
*(11*c^10*d + 27*c^8*d + 21*c^6*d + 5*c^4*d)*a + (11*c^10*d + 27*c^8*d + 21*c^6*d + 5*c^4*d)*b)*x*e^4 + ((5*a*
d^8 + b*d^8)*x^8*e^4 + 8*(5*a*c*d^7 + b*c*d^7)*x^7*e^4 + (4*(35*c^2*d^6 + 2*d^6)*a + (28*c^2*d^6 + d^6)*b)*x^6
*e^4 + 2*(4*(35*c^3*d^5 + 6*c*d^5)*a + (28*c^3*d^5 + 3*c*d^5)*b)*x^5*e^4 + ((350*c^4*d^4 + 120*c^2*d^4 + 3*d^4
)*a + 5*(14*c^4*d^4 + 3*c^2*d^4)*b)*x^4*e^4 + 4*((70*c^5*d^3 + 40*c^3*d^3 + 3*c*d^3)*a + (14*c^5*d^3 + 5*c^3*d
^3)*b)*x^3*e^4 + (2*(70*c^6*d^2 + 60*c^4*d^2 + 9*c^2*d^2)*a + (28*c^6*d^2 + 15*c^4*d^2)*b)*x^2*e^4 + 2*(2*(10*
c^7*d + 12*c^5*d + 3*c^3*d)*a + (4*c^7*d + 3*c^5*d)*b)*x*e^4 + ((5*c^8 + 8*c^6 + 3*c^4)*a + (c^8 + c^6)*b)*e^4
)*(d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2) + (3*(5*a*d^9 + b*d^9)*x^9*e^4 + 27*(5*a*c*d^8 + b*c*d^8)*x^8*e^4 + ((54
0*c^2*d^7 + 31*d^7)*a + (108*c^2*d^7 + 5*d^7)*b)*x^7*e^4 + 7*((180*c^3*d^6 + 31*c*d^6)*a + (36*c^3*d^6 + 5*c*d
^6)*b)*x^6*e^4 + ((1890*c^4*d^5 + 651*c^2*d^5 + 20*d^5)*a + (378*c^4*d^5 + 105*c^2*d^5 + 2*d^5)*b)*x^5*e^4 + (
5*(378*c^5*d^4 + 217*c^3*d^4 + 20*c*d^4)*a + (378*c^5*d^4 + 175*c^3*d^4 + 10*c*d^4)*b)*x^4*e^4 + ((1260*c^6*d^
3 + 1085*c^4*d^3 + 200*c^2*d^3 + 4*d^3)*a + (252*c^6*d^3 + 175*c^4*d^3 + 20*c^2*d^3)*b)*x^3*e^4 + ((540*c^7*d^
2 + 651*c^5*d^2 + 200*c^3*d^2 + 12*c*d^2)*a + (108*c^7*d^2 + 105*c^5*d^2 + 20*c^3*d^2)*b)*x^2*e^4 + ((135*c^8*
d + 217*c^6*d + 100*c^4*d + 12*c^2*d)*a + (27*c^8*d + 35*c^6*d + 10*c^4*d)*b)*x*e^4 + ((15*c^9 + 31*c^7 + 20*c
^5 + 4*c^3)*a + (3*c^9 + 5*c^7 + 2*c^5)*b)*e^4)*(d^2*x^2 + 2*c*d*x + c^2 + 1) + (5*(c^11 + 3*c^9 + 3*c^7 + c^5
)*a + (c^11 + 3*c^9 + 3*c^7 + c^5)*b)*e^4 + (5*b*d^11*x^11*e^4 + 55*b*c*d^10*x^10*e^4 + 5*(55*c^2*d^9 + 3*d^9)
*b*x^9*e^4 + 15*(55*c^3*d^8 + 9*c*d^8)*b*x^8*e^4 + 15*(110*c^4*d^7 + 36*c^2*d^7 + d^7)*b*x^7*e^4 + 105*(22*c^5
*d^6 + 12*c^3*d^6 + c*d^6)*b*x^6*e^4 + 5*(462*c^6*d^5 + 378*c^4*d^5 + 63*c^2*d^5 + d^5)*b*x^5*e^4 + 5*(330*c^7
*d^4 + 378*c^5*d^4 + 105*c^3*d^4 + 5*c*d^4)*b*x^4*e^4 + 5*(165*c^8*d^3 + 252*c^6*d^3 + 105*c^4*d^3 + 10*c^2*d^
3)*b*x^3*e^4 + 5*(55*c^9*d^2 + 108*c^7*d^2 + 63*c^5*d^2 + 10*c^3*d^2)*b*x^2*e^4 + 5*(11*c^10*d + 27*c^8*d + 21
*c^6*d + 5*c^4*d)*b*x*e^4 + 5*(c^11 + 3*c^9 + 3*c^7 + c^5)*b*e^4 + (5*b*d^8*x^8*e^4 + 40*b*c*d^7*x^7*e^4 + 4*(
35*c^2*d^6 + 2*d^6)*b*x^6*e^4 + 8*(35*c^3*d^5 + 6*c*d^5)*b*x^5*e^4 + (350*c^4*d^4 + 120*c^2*d^4 + 3*d^4)*b*x^4
*e^4 + 4*(70*c^5*d^3 + 40*c^3*d^3 + 3*c*d^3)*b*x^3*e^4 + 2*(70*c^6*d^2 + 60*c^4*d^2 + 9*c^2*d^2)*b*x^2*e^4 + 4
*(10*c^7*d + 12*c^5*d + 3*c^3*d)*b*x*e^4 + (5*c^8 + 8*c^6 + 3*c^4)*b*e^4)*(d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2)
+ (15*b*d^9*x^9*e^4 + 135*b*c*d^8*x^8*e^4 + (540*c^2*d^7 + 31*d^7)*b*x^7*e^4 + 7*(180*c^3*d^6 + 31*c*d^6)*b*x^
6*e^4 + (1890*c^4*d^5 + 651*c^2*d^5 + 20*d^5)*b*x^5*e^4 + 5*(378*c^5*d^4 + 217*c^3*d^4 + 20*c*d^4)*b*x^4*e^4 +
(1260*c^6*d^3 + 1085*c^4*d^3 + 200*c^2*d^3 + 4*d^3)*b*x^3*e^4 + (540*c^7*d^2 + 651*c^5*d^2 + 200*c^3*d^2 + 12
*c*d^2)*b*x^2*e^4 + (135*c^8*d + 217*c^6*d + 100*c^4*d + 12*c^2*d)*b*x*e^4 + (15*c^9 + 31*c^7 + 20*c^5 + 4*c^3
)*b*e^4)*(d^2*x^2 + 2*c*d*x + c^2 + 1) + (15*b*d^10*x^10*e^4 + 150*b*c*d^9*x^9*e^4 + (675*c^2*d^8 + 38*d^8)*b*
x^8*e^4 + 8*(225*c^3*d^7 + 38*c*d^7)*b*x^7*e^4 + 2*(1575*c^4*d^6 + 532*c^2*d^6 + 16*d^6)*b*x^6*e^4 + 4*(945*c^
5*d^5 + 532*c^3*d^5 + 48*c*d^5)*b*x^5*e^4 + (3150*c^6*d^4 + 2660*c^4*d^4 + 480*c^2*d^4 + 9*d^4)*b*x^4*e^4 + 4*
(450*c^7*d^3 + 532*c^5*d^3 + 160*c^3*d^3 + 9*c*d^3)*b*x^3*e^4 + (675*c^8*d^2 + 1064*c^6*d^2 + 480*c^4*d^2 + 54
*c^2*d^2)*b*x^2*e^4 + 2*(75*c^9*d + 152*c^7*d + 96*c^5*d + 18*c^3*d)*b*x*e^4 + (15*c^10 + 38*c^8 + 32*c^6 + 9*
c^4)*b*e^4)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + (3*(5*a*d^10
+ b*d^10)*x^10*e^4 + 30*(5*a*c*d^9 + b*c*d^9)*x^9*e^4 + ((675*c^2*d^8 + 38*d^8)*a + (135*c^2*d^8 + 7*d^8)*b)*
x^8*e^4 + 8*((225*c^3*d^7 + 38*c*d^7)*a + (45*c^3*d^7 + 7*c*d^7)*b)*x^7*e^4 + (2*(1575*c^4*d^6 + 532*c^2*d^6 +
16*d^6)*a + (630*c^4*d^6 + 196*c^2*d^6 + 5*d^6)*b)*x^6*e^4 + 2*(2*(945*c^5*d^5 + 532*c^3*d^5 + 48*c*d^5)*a +
(378*c^5*d^5 + 196*c^3*d^5 + 15*c*d^5)*b)*x^5*e^4 + ((3150*c^6*d^4 + 2660*c^4*d^4 + 480*c^2*d^4 + 9*d^4)*a + (
630*c^6*d^4 + 490*c^4*d^4 + 75*c^2*d^4 + d^4)*b...

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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((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4/(a+b*asinh(d*x+c))**3,x)

[Out]

e**4*(Integral(c**4/(a**3 + 3*a**2*b*asinh(c + d*x) + 3*a*b**2*asinh(c + d*x)**2 + b**3*asinh(c + d*x)**3), x)
+ Integral(d**4*x**4/(a**3 + 3*a**2*b*asinh(c + d*x) + 3*a*b**2*asinh(c + d*x)**2 + b**3*asinh(c + d*x)**3),
x) + Integral(4*c*d**3*x**3/(a**3 + 3*a**2*b*asinh(c + d*x) + 3*a*b**2*asinh(c + d*x)**2 + b**3*asinh(c + d*x)
**3), x) + Integral(6*c**2*d**2*x**2/(a**3 + 3*a**2*b*asinh(c + d*x) + 3*a*b**2*asinh(c + d*x)**2 + b**3*asinh
(c + d*x)**3), x) + Integral(4*c**3*d*x/(a**3 + 3*a**2*b*asinh(c + d*x) + 3*a*b**2*asinh(c + d*x)**2 + b**3*as
inh(c + d*x)**3), x))

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Giac [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((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^3,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,e+d\,e\,x\right )}^4}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^4/(a + b*asinh(c + d*x))^3,x)

[Out]

int((c*e + d*e*x)^4/(a + b*asinh(c + d*x))^3, x)

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