3.200 \(\int \cosh ^3(a+b x) \log (x) \, dx\)

Optimal. Leaf size=88 \[ -\frac {3 \sinh (a) \text {Chi}(b x)}{4 b}-\frac {\sinh (3 a) \text {Chi}(3 b x)}{12 b}-\frac {3 \cosh (a) \text {Shi}(b x)}{4 b}-\frac {\cosh (3 a) \text {Shi}(3 b x)}{12 b}+\frac {\log (x) \sinh ^3(a+b x)}{3 b}+\frac {\log (x) \sinh (a+b x)}{b} \]

[Out]

-3/4*cosh(a)*Shi(b*x)/b-1/12*cosh(3*a)*Shi(3*b*x)/b-3/4*Chi(b*x)*sinh(a)/b-1/12*Chi(3*b*x)*sinh(3*a)/b+ln(x)*s
inh(b*x+a)/b+1/3*ln(x)*sinh(b*x+a)^3/b

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Rubi [A]  time = 0.48, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {2633, 2554, 12, 6742, 3303, 3298, 3301, 3312} \[ -\frac {3 \sinh (a) \text {Chi}(b x)}{4 b}-\frac {\sinh (3 a) \text {Chi}(3 b x)}{12 b}-\frac {3 \cosh (a) \text {Shi}(b x)}{4 b}-\frac {\cosh (3 a) \text {Shi}(3 b x)}{12 b}+\frac {\log (x) \sinh ^3(a+b x)}{3 b}+\frac {\log (x) \sinh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]^3*Log[x],x]

[Out]

(-3*CoshIntegral[b*x]*Sinh[a])/(4*b) - (CoshIntegral[3*b*x]*Sinh[3*a])/(12*b) + (Log[x]*Sinh[a + b*x])/b + (Lo
g[x]*Sinh[a + b*x]^3)/(3*b) - (3*Cosh[a]*SinhIntegral[b*x])/(4*b) - (Cosh[3*a]*SinhIntegral[3*b*x])/(12*b)

Rule 12

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

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 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]

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \cosh ^3(a+b x) \log (x) \, dx &=\frac {\log (x) \sinh (a+b x)}{b}+\frac {\log (x) \sinh ^3(a+b x)}{3 b}-\int \frac {\sinh (a+b x) \left (3+\sinh ^2(a+b x)\right )}{3 b x} \, dx\\ &=\frac {\log (x) \sinh (a+b x)}{b}+\frac {\log (x) \sinh ^3(a+b x)}{3 b}-\frac {\int \frac {\sinh (a+b x) \left (3+\sinh ^2(a+b x)\right )}{x} \, dx}{3 b}\\ &=\frac {\log (x) \sinh (a+b x)}{b}+\frac {\log (x) \sinh ^3(a+b x)}{3 b}-\frac {\int \left (\frac {3 \sinh (a+b x)}{x}+\frac {\sinh ^3(a+b x)}{x}\right ) \, dx}{3 b}\\ &=\frac {\log (x) \sinh (a+b x)}{b}+\frac {\log (x) \sinh ^3(a+b x)}{3 b}-\frac {\int \frac {\sinh ^3(a+b x)}{x} \, dx}{3 b}-\frac {\int \frac {\sinh (a+b x)}{x} \, dx}{b}\\ &=\frac {\log (x) \sinh (a+b x)}{b}+\frac {\log (x) \sinh ^3(a+b x)}{3 b}-\frac {i \int \left (\frac {3 i \sinh (a+b x)}{4 x}-\frac {i \sinh (3 a+3 b x)}{4 x}\right ) \, dx}{3 b}-\frac {\cosh (a) \int \frac {\sinh (b x)}{x} \, dx}{b}-\frac {\sinh (a) \int \frac {\cosh (b x)}{x} \, dx}{b}\\ &=-\frac {\text {Chi}(b x) \sinh (a)}{b}+\frac {\log (x) \sinh (a+b x)}{b}+\frac {\log (x) \sinh ^3(a+b x)}{3 b}-\frac {\cosh (a) \text {Shi}(b x)}{b}-\frac {\int \frac {\sinh (3 a+3 b x)}{x} \, dx}{12 b}+\frac {\int \frac {\sinh (a+b x)}{x} \, dx}{4 b}\\ &=-\frac {\text {Chi}(b x) \sinh (a)}{b}+\frac {\log (x) \sinh (a+b x)}{b}+\frac {\log (x) \sinh ^3(a+b x)}{3 b}-\frac {\cosh (a) \text {Shi}(b x)}{b}+\frac {\cosh (a) \int \frac {\sinh (b x)}{x} \, dx}{4 b}-\frac {\cosh (3 a) \int \frac {\sinh (3 b x)}{x} \, dx}{12 b}+\frac {\sinh (a) \int \frac {\cosh (b x)}{x} \, dx}{4 b}-\frac {\sinh (3 a) \int \frac {\cosh (3 b x)}{x} \, dx}{12 b}\\ &=-\frac {3 \text {Chi}(b x) \sinh (a)}{4 b}-\frac {\text {Chi}(3 b x) \sinh (3 a)}{12 b}+\frac {\log (x) \sinh (a+b x)}{b}+\frac {\log (x) \sinh ^3(a+b x)}{3 b}-\frac {3 \cosh (a) \text {Shi}(b x)}{4 b}-\frac {\cosh (3 a) \text {Shi}(3 b x)}{12 b}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 66, normalized size = 0.75 \[ -\frac {9 \sinh (a) \text {Chi}(b x)+\sinh (3 a) \text {Chi}(3 b x)+9 \cosh (a) \text {Shi}(b x)+\cosh (3 a) \text {Shi}(3 b x)-9 \log (x) \sinh (a+b x)-\log (x) \sinh (3 (a+b x))}{12 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]^3*Log[x],x]

[Out]

-1/12*(9*CoshIntegral[b*x]*Sinh[a] + CoshIntegral[3*b*x]*Sinh[3*a] - 9*Log[x]*Sinh[a + b*x] - Log[x]*Sinh[3*(a
+ b*x)] + 9*Cosh[a]*SinhIntegral[b*x] + Cosh[3*a]*SinhIntegral[3*b*x])/b

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fricas [B]  time = 0.49, size = 587, normalized size = 6.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*log(x),x, algorithm="fricas")

[Out]

1/24*(6*cosh(b*x + a)*log(x)*sinh(b*x + a)^5 + log(x)*sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 + 3)*log(x)*sinh(
b*x + a)^4 - (Ei(3*b*x) + Ei(-3*b*x))*cosh(b*x + a)^3*sinh(3*a) - 9*(Ei(b*x) + Ei(-b*x))*cosh(b*x + a)^3*sinh(
a) - ((Ei(3*b*x) - Ei(-3*b*x))*cosh(3*a) + 9*(Ei(b*x) - Ei(-b*x))*cosh(a))*cosh(b*x + a)^3 - ((Ei(3*b*x) - Ei(
-3*b*x))*cosh(3*a) + 9*(Ei(b*x) - Ei(-b*x))*cosh(a) - 4*(5*cosh(b*x + a)^3 + 9*cosh(b*x + a))*log(x) + (Ei(3*b
*x) + Ei(-3*b*x))*sinh(3*a) + 9*(Ei(b*x) + Ei(-b*x))*sinh(a))*sinh(b*x + a)^3 - 3*((Ei(3*b*x) + Ei(-3*b*x))*co
sh(b*x + a)*sinh(3*a) + 9*(Ei(b*x) + Ei(-b*x))*cosh(b*x + a)*sinh(a) + ((Ei(3*b*x) - Ei(-3*b*x))*cosh(3*a) + 9
*(Ei(b*x) - Ei(-b*x))*cosh(a))*cosh(b*x + a) - (5*cosh(b*x + a)^4 + 18*cosh(b*x + a)^2 - 3)*log(x))*sinh(b*x +
a)^2 + (cosh(b*x + a)^6 + 9*cosh(b*x + a)^4 - 9*cosh(b*x + a)^2 - 1)*log(x) - 3*((Ei(3*b*x) + Ei(-3*b*x))*cos
h(b*x + a)^2*sinh(3*a) + 9*(Ei(b*x) + Ei(-b*x))*cosh(b*x + a)^2*sinh(a) + ((Ei(3*b*x) - Ei(-3*b*x))*cosh(3*a)
+ 9*(Ei(b*x) - Ei(-b*x))*cosh(a))*cosh(b*x + a)^2 - 2*(cosh(b*x + a)^5 + 6*cosh(b*x + a)^3 - 3*cosh(b*x + a))*
log(x))*sinh(b*x + a))/(b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a)^2*sinh(b*x + a) + 3*b*cosh(b*x + a)*sinh(b*x + a
)^2 + b*sinh(b*x + a)^3)

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giac [A]  time = 0.20, size = 104, normalized size = 1.18 \[ \frac {1}{24} \, {\left (\frac {e^{\left (3 \, b x + 3 \, a\right )}}{b} + \frac {9 \, e^{\left (b x + a\right )}}{b} - \frac {9 \, e^{\left (-b x - a\right )}}{b} - \frac {e^{\left (-3 \, b x - 3 \, a\right )}}{b}\right )} \log \relax (x) - \frac {{\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - 9 \, {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + 9 \, {\rm Ei}\left (b x\right ) e^{a}}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*log(x),x, algorithm="giac")

[Out]

1/24*(e^(3*b*x + 3*a)/b + 9*e^(b*x + a)/b - 9*e^(-b*x - a)/b - e^(-3*b*x - 3*a)/b)*log(x) - 1/24*(Ei(3*b*x)*e^
(3*a) - 9*Ei(-b*x)*e^(-a) - Ei(-3*b*x)*e^(-3*a) + 9*Ei(b*x)*e^a)/b

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maple [A]  time = 1.69, size = 116, normalized size = 1.32 \[ -\frac {3 \Ei \left (1, b x \right ) {\mathrm e}^{-a}}{8 b}+\frac {\Ei \left (1, -3 b x \right ) {\mathrm e}^{3 a}}{24 b}+\frac {3 \Ei \left (1, -b x \right ) {\mathrm e}^{a}}{8 b}-\frac {\Ei \left (1, 3 b x \right ) {\mathrm e}^{-3 a}}{24 b}+\left (-\frac {{\mathrm e}^{-3 b x -3 a}}{24 b}-\frac {3 \,{\mathrm e}^{-b x -a}}{8 b}+\frac {3 \,{\mathrm e}^{b x +a}}{8 b}+\frac {{\mathrm e}^{3 b x +3 a}}{24 b}\right ) \ln \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^3*ln(x),x)

[Out]

(1/24/b*exp(3*b*x+3*a)+3/8/b*exp(b*x+a)-3/8/b*exp(-b*x-a)-1/24/b*exp(-3*b*x-3*a))*ln(x)+1/24/b*exp(3*a)*Ei(1,-
3*b*x)-1/24/b*exp(-3*a)*Ei(1,3*b*x)-3/8/b*exp(-a)*Ei(1,b*x)+3/8/b*exp(a)*Ei(1,-b*x)

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maxima [A]  time = 0.77, size = 111, normalized size = 1.26 \[ \frac {1}{24} \, {\left (\frac {e^{\left (3 \, b x + 3 \, a\right )}}{b} + \frac {9 \, e^{\left (b x + a\right )}}{b} - \frac {9 \, e^{\left (-b x - a\right )}}{b} - \frac {e^{\left (-3 \, b x - 3 \, a\right )}}{b}\right )} \log \relax (x) - \frac {{\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )}}{24 \, b} + \frac {3 \, {\rm Ei}\left (-b x\right ) e^{\left (-a\right )}}{8 \, b} + \frac {{\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )}}{24 \, b} - \frac {3 \, {\rm Ei}\left (b x\right ) e^{a}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*log(x),x, algorithm="maxima")

[Out]

1/24*(e^(3*b*x + 3*a)/b + 9*e^(b*x + a)/b - 9*e^(-b*x - a)/b - e^(-3*b*x - 3*a)/b)*log(x) - 1/24*Ei(3*b*x)*e^(
3*a)/b + 3/8*Ei(-b*x)*e^(-a)/b + 1/24*Ei(-3*b*x)*e^(-3*a)/b - 3/8*Ei(b*x)*e^a/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^3*log(x),x)

[Out]

int(cosh(a + b*x)^3*log(x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\relax (x )} \cosh ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**3*ln(x),x)

[Out]

Integral(log(x)*cosh(a + b*x)**3, x)

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