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

Optimal. Leaf size=213 \[ \frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 b d}-\frac{3 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b d}+\frac{e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b d}-\frac{e^4 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 b d}+\frac{3 e^4 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b d}-\frac{e^4 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b d} \]

[Out]

(e^4*Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c + d*x])/b])/(8*b*d) - (3*e^4*Cosh[(3*a)/b]*CoshIntegral[(3*(a + b
*ArcSinh[c + d*x]))/b])/(16*b*d) + (e^4*Cosh[(5*a)/b]*CoshIntegral[(5*(a + b*ArcSinh[c + d*x]))/b])/(16*b*d) -
(e^4*Sinh[a/b]*SinhIntegral[(a + b*ArcSinh[c + d*x])/b])/(8*b*d) + (3*e^4*Sinh[(3*a)/b]*SinhIntegral[(3*(a +
b*ArcSinh[c + d*x]))/b])/(16*b*d) - (e^4*Sinh[(5*a)/b]*SinhIntegral[(5*(a + b*ArcSinh[c + d*x]))/b])/(16*b*d)

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Rubi [A]  time = 0.468893, antiderivative size = 209, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {5865, 12, 5669, 5448, 3303, 3298, 3301} \[ \frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b d}-\frac{3 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{16 b d}+\frac{e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{16 b d}-\frac{e^4 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b d}+\frac{3 e^4 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{16 b d}-\frac{e^4 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{16 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c + d*x]])/(8*b*d) - (3*e^4*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*A
rcSinh[c + d*x]])/(16*b*d) + (e^4*Cosh[(5*a)/b]*CoshIntegral[(5*a)/b + 5*ArcSinh[c + d*x]])/(16*b*d) - (e^4*Si
nh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]])/(8*b*d) + (3*e^4*Sinh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[
c + d*x]])/(16*b*d) - (e^4*Sinh[(5*a)/b]*SinhIntegral[(5*a)/b + 5*ArcSinh[c + d*x]])/(16*b*d)

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
]

Rule 12

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

Rule 5669

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

Rule 5448

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.302412, size = 151, normalized size = 0.71 \[ \frac{e^4 \left (2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )}{16 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*(2*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c + d*x]] - 3*Cosh[(3*a)/b]*CoshIntegral[3*(a/b + ArcSinh[c + d*x
])] + Cosh[(5*a)/b]*CoshIntegral[5*(a/b + ArcSinh[c + d*x])] - 2*Sinh[a/b]*SinhIntegral[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])]))/(16*b*d)

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Maple [A]  time = 0.164, size = 194, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( -{\frac{{e}^{4}}{32\,b}{{\rm e}^{5\,{\frac{a}{b}}}}{\it Ei} \left ( 1,5\,{\it Arcsinh} \left ( dx+c \right ) +5\,{\frac{a}{b}} \right ) }+{\frac{3\,{e}^{4}}{32\,b}{{\rm e}^{3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,3\,{\it Arcsinh} \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) }-{\frac{{e}^{4}}{16\,b}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( dx+c \right ) +{\frac{a}{b}} \right ) }-{\frac{{e}^{4}}{16\,b}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\it Arcsinh} \left ( dx+c \right ) -{\frac{a}{b}} \right ) }+{\frac{3\,{e}^{4}}{32\,b}{{\rm e}^{-3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-3\,{\it Arcsinh} \left ( dx+c \right ) -3\,{\frac{a}{b}} \right ) }-{\frac{{e}^{4}}{32\,b}{{\rm e}^{-5\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-5\,{\it Arcsinh} \left ( dx+c \right ) -5\,{\frac{a}{b}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/32*e^4/b*exp(5*a/b)*Ei(1,5*arcsinh(d*x+c)+5*a/b)+3/32*e^4/b*exp(3*a/b)*Ei(1,3*arcsinh(d*x+c)+3*a/b)-1/
16*e^4/b*exp(a/b)*Ei(1,arcsinh(d*x+c)+a/b)-1/16*e^4/b*exp(-a/b)*Ei(1,-arcsinh(d*x+c)-a/b)+3/32*e^4/b*exp(-3*a/
b)*Ei(1,-3*arcsinh(d*x+c)-3*a/b)-1/32*e^4/b*exp(-5*a/b)*Ei(1,-5*arcsinh(d*x+c)-5*a/b))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{4}}{b \operatorname{arsinh}\left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 \operatorname{arsinh}\left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{4}}{b \operatorname{arsinh}\left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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