3.12 \(\int \frac {\cosh ^2(a+b x)}{c+d x} \, dx\)

Optimal. Leaf size=78 \[ \frac {\cosh \left (2 a-\frac {2 b c}{d}\right ) \text {Chi}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}+\frac {\sinh \left (2 a-\frac {2 b c}{d}\right ) \text {Shi}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}+\frac {\log (c+d x)}{2 d} \]

[Out]

1/2*Chi(2*b*c/d+2*b*x)*cosh(2*a-2*b*c/d)/d+1/2*ln(d*x+c)/d+1/2*Shi(2*b*c/d+2*b*x)*sinh(2*a-2*b*c/d)/d

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Rubi [A]  time = 0.15, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3312, 3303, 3298, 3301} \[ \frac {\cosh \left (2 a-\frac {2 b c}{d}\right ) \text {Chi}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}+\frac {\sinh \left (2 a-\frac {2 b c}{d}\right ) \text {Shi}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}+\frac {\log (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]^2/(c + d*x),x]

[Out]

(Cosh[2*a - (2*b*c)/d]*CoshIntegral[(2*b*c)/d + 2*b*x])/(2*d) + Log[c + d*x]/(2*d) + (Sinh[2*a - (2*b*c)/d]*Si
nhIntegral[(2*b*c)/d + 2*b*x])/(2*d)

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])
)

Rubi steps

\begin {align*} \int \frac {\cosh ^2(a+b x)}{c+d x} \, dx &=\int \left (\frac {1}{2 (c+d x)}+\frac {\cosh (2 a+2 b x)}{2 (c+d x)}\right ) \, dx\\ &=\frac {\log (c+d x)}{2 d}+\frac {1}{2} \int \frac {\cosh (2 a+2 b x)}{c+d x} \, dx\\ &=\frac {\log (c+d x)}{2 d}+\frac {1}{2} \cosh \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\cosh \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx+\frac {1}{2} \sinh \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sinh \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx\\ &=\frac {\cosh \left (2 a-\frac {2 b c}{d}\right ) \text {Chi}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}+\frac {\log (c+d x)}{2 d}+\frac {\sinh \left (2 a-\frac {2 b c}{d}\right ) \text {Shi}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 64, normalized size = 0.82 \[ \frac {\cosh \left (2 a-\frac {2 b c}{d}\right ) \text {Chi}\left (\frac {2 b (c+d x)}{d}\right )+\sinh \left (2 a-\frac {2 b c}{d}\right ) \text {Shi}\left (\frac {2 b (c+d x)}{d}\right )+\log (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]^2/(c + d*x),x]

[Out]

(Cosh[2*a - (2*b*c)/d]*CoshIntegral[(2*b*(c + d*x))/d] + Log[c + d*x] + Sinh[2*a - (2*b*c)/d]*SinhIntegral[(2*
b*(c + d*x))/d])/(2*d)

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fricas [A]  time = 0.47, size = 104, normalized size = 1.33 \[ \frac {{\left ({\rm Ei}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\rm Ei}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cosh \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left ({\rm Ei}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\rm Ei}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sinh \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, \log \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2/(d*x+c),x, algorithm="fricas")

[Out]

1/4*((Ei(2*(b*d*x + b*c)/d) + Ei(-2*(b*d*x + b*c)/d))*cosh(-2*(b*c - a*d)/d) + (Ei(2*(b*d*x + b*c)/d) - Ei(-2*
(b*d*x + b*c)/d))*sinh(-2*(b*c - a*d)/d) + 2*log(d*x + c))/d

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giac [A]  time = 0.12, size = 68, normalized size = 0.87 \[ \frac {{\rm Ei}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) e^{\left (2 \, a - \frac {2 \, b c}{d}\right )} + {\rm Ei}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )} + 2 \, \log \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2/(d*x+c),x, algorithm="giac")

[Out]

1/4*(Ei(2*(b*d*x + b*c)/d)*e^(2*a - 2*b*c/d) + Ei(-2*(b*d*x + b*c)/d)*e^(-2*a + 2*b*c/d) + 2*log(d*x + c))/d

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maple [A]  time = 0.25, size = 97, normalized size = 1.24 \[ \frac {\ln \left (d x +c \right )}{2 d}-\frac {{\mathrm e}^{-\frac {2 \left (d a -c b \right )}{d}} \Ei \left (1, 2 b x +2 a -\frac {2 \left (d a -c b \right )}{d}\right )}{4 d}-\frac {{\mathrm e}^{\frac {2 d a -2 c b}{d}} \Ei \left (1, -2 b x -2 a -\frac {2 \left (-d a +c b \right )}{d}\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^2/(d*x+c),x)

[Out]

1/2*ln(d*x+c)/d-1/4/d*exp(-2*(a*d-b*c)/d)*Ei(1,2*b*x+2*a-2*(a*d-b*c)/d)-1/4/d*exp(2*(a*d-b*c)/d)*Ei(1,-2*b*x-2
*a-2*(-a*d+b*c)/d)

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maxima [A]  time = 0.53, size = 72, normalized size = 0.92 \[ -\frac {e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )} E_{1}\left (\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, d} - \frac {e^{\left (2 \, a - \frac {2 \, b c}{d}\right )} E_{1}\left (-\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, d} + \frac {\log \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2/(d*x+c),x, algorithm="maxima")

[Out]

-1/4*e^(-2*a + 2*b*c/d)*exp_integral_e(1, 2*(d*x + c)*b/d)/d - 1/4*e^(2*a - 2*b*c/d)*exp_integral_e(1, -2*(d*x
+ c)*b/d)/d + 1/2*log(d*x + c)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^2}{c+d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^2/(c + d*x),x)

[Out]

int(cosh(a + b*x)^2/(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**2/(d*x+c),x)

[Out]

Integral(cosh(a + b*x)**2/(c + d*x), x)

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