3.131 \(\int \frac {(c e+d e x)^3}{a+b \cosh ^{-1}(c+d x)} \, dx\)

Optimal. Leaf size=145 \[ -\frac {e^3 \sinh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{4 b d}-\frac {e^3 \sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{8 b d}+\frac {e^3 \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{4 b d}+\frac {e^3 \cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{8 b d} \]

[Out]

1/4*e^3*cosh(2*a/b)*Shi(2*(a+b*arccosh(d*x+c))/b)/b/d+1/8*e^3*cosh(4*a/b)*Shi(4*(a+b*arccosh(d*x+c))/b)/b/d-1/
4*e^3*Chi(2*(a+b*arccosh(d*x+c))/b)*sinh(2*a/b)/b/d-1/8*e^3*Chi(4*(a+b*arccosh(d*x+c))/b)*sinh(4*a/b)/b/d

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Rubi [A]  time = 0.33, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5866, 12, 5670, 5448, 3303, 3298, 3301} \[ -\frac {e^3 \sinh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \cosh ^{-1}(c+d x)\right )}{4 b d}-\frac {e^3 \sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 a}{b}+4 \cosh ^{-1}(c+d x)\right )}{8 b d}+\frac {e^3 \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \cosh ^{-1}(c+d x)\right )}{4 b d}+\frac {e^3 \cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \cosh ^{-1}(c+d x)\right )}{8 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^3*CoshIntegral[(2*a)/b + 2*ArcCosh[c + d*x]]*Sinh[(2*a)/b])/(4*b*d) - (e^3*CoshIntegral[(4*a)/b + 4*ArcCos
h[c + d*x]]*Sinh[(4*a)/b])/(8*b*d) + (e^3*Cosh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcCosh[c + d*x]])/(4*b*d) +
(e^3*Cosh[(4*a)/b]*SinhIntegral[(4*a)/b + 4*ArcCosh[c + d*x]])/(8*b*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 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 5670

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

Rule 5866

Int[((a_.) + ArcCosh[(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*ArcCosh[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)^3}{a+b \cosh ^{-1}(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^3 x^3}{a+b \cosh ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \frac {x^3}{a+b \cosh ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 (a+b x)}+\frac {\sinh (4 x)}{8 (a+b x)}\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \frac {\sinh (4 x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{8 d}+\frac {e^3 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{4 d}\\ &=\frac {\left (e^3 \cosh \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{4 d}+\frac {\left (e^3 \cosh \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{8 d}-\frac {\left (e^3 \sinh \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{4 d}-\frac {\left (e^3 \sinh \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{8 d}\\ &=-\frac {e^3 \text {Chi}\left (\frac {2 a}{b}+2 \cosh ^{-1}(c+d x)\right ) \sinh \left (\frac {2 a}{b}\right )}{4 b d}-\frac {e^3 \text {Chi}\left (\frac {4 a}{b}+4 \cosh ^{-1}(c+d x)\right ) \sinh \left (\frac {4 a}{b}\right )}{8 b d}+\frac {e^3 \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \cosh ^{-1}(c+d x)\right )}{4 b d}+\frac {e^3 \cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \cosh ^{-1}(c+d x)\right )}{8 b d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 109, normalized size = 0.75 \[ \frac {e^3 \left (-2 \sinh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )\right )+2 \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )\right )+\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\cosh ^{-1}(c+d x)\right )\right )\right )}{8 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*(-2*CoshIntegral[2*(a/b + ArcCosh[c + d*x])]*Sinh[(2*a)/b] - CoshIntegral[4*(a/b + ArcCosh[c + d*x])]*Sin
h[(4*a)/b] + 2*Cosh[(2*a)/b]*SinhIntegral[2*(a/b + ArcCosh[c + d*x])] + Cosh[(4*a)/b]*SinhIntegral[4*(a/b + Ar
cCosh[c + d*x])]))/(8*b*d)

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fricas [F]  time = 1.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}{b \operatorname {arcosh}\left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{3}}{b \operatorname {arcosh}\left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arccosh(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.34, size = 134, normalized size = 0.92 \[ \frac {\frac {e^{3} {\mathrm e}^{\frac {4 a}{b}} \Ei \left (1, 4 \,\mathrm {arccosh}\left (d x +c \right )+\frac {4 a}{b}\right )}{16 b}+\frac {e^{3} {\mathrm e}^{\frac {2 a}{b}} \Ei \left (1, 2 \,\mathrm {arccosh}\left (d x +c \right )+\frac {2 a}{b}\right )}{8 b}-\frac {e^{3} {\mathrm e}^{-\frac {2 a}{b}} \Ei \left (1, -2 \,\mathrm {arccosh}\left (d x +c \right )-\frac {2 a}{b}\right )}{8 b}-\frac {e^{3} {\mathrm e}^{-\frac {4 a}{b}} \Ei \left (1, -4 \,\mathrm {arccosh}\left (d x +c \right )-\frac {4 a}{b}\right )}{16 b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/16*e^3/b*exp(4*a/b)*Ei(1,4*arccosh(d*x+c)+4*a/b)+1/8*e^3/b*exp(2*a/b)*Ei(1,2*arccosh(d*x+c)+2*a/b)-1/8*
e^3/b*exp(-2*a/b)*Ei(1,-2*arccosh(d*x+c)-2*a/b)-1/16*e^3/b*exp(-4*a/b)*Ei(1,-4*arccosh(d*x+c)-4*a/b))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{3}}{b \operatorname {arcosh}\left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^3}{a+b\,\mathrm {acosh}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e**3*(Integral(c**3/(a + b*acosh(c + d*x)), x) + Integral(d**3*x**3/(a + b*acosh(c + d*x)), x) + Integral(3*c*
d**2*x**2/(a + b*acosh(c + d*x)), x) + Integral(3*c**2*d*x/(a + b*acosh(c + d*x)), x))

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