3.14 \(\int \frac {\cosh ^2(a+b x)}{(c+d x)^3} \, dx\)

Optimal. Leaf size=112 \[ \frac {b^2 \cosh \left (2 a-\frac {2 b c}{d}\right ) \text {Chi}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {b^2 \sinh \left (2 a-\frac {2 b c}{d}\right ) \text {Shi}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}-\frac {b \sinh (a+b x) \cosh (a+b x)}{d^2 (c+d x)}-\frac {\cosh ^2(a+b x)}{2 d (c+d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3314, 31, 3312, 3303, 3298, 3301} \[ \frac {b^2 \cosh \left (2 a-\frac {2 b c}{d}\right ) \text {Chi}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {b^2 \sinh \left (2 a-\frac {2 b c}{d}\right ) \text {Shi}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}-\frac {b \sinh (a+b x) \cosh (a+b x)}{d^2 (c+d x)}-\frac {\cosh ^2(a+b x)}{2 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]^2/(c + d*x)^3,x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, 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 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 3314

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(b*Si
n[e + f*x])^n)/(d*(m + 1)), x] + (Dist[(b^2*f^2*n*(n - 1))/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin
[e + f*x])^(n - 2), x], x] - Dist[(f^2*n^2)/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x
], x] - Simp[(b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*(b*Sin[e + f*x])^(n - 1))/(d^2*(m + 1)*(m + 2)), x]) /; Fre
eQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]

Rubi steps

\begin {align*} \int \frac {\cosh ^2(a+b x)}{(c+d x)^3} \, dx &=-\frac {\cosh ^2(a+b x)}{2 d (c+d x)^2}-\frac {b \cosh (a+b x) \sinh (a+b x)}{d^2 (c+d x)}-\frac {b^2 \int \frac {1}{c+d x} \, dx}{d^2}+\frac {\left (2 b^2\right ) \int \frac {\cosh ^2(a+b x)}{c+d x} \, dx}{d^2}\\ &=-\frac {\cosh ^2(a+b x)}{2 d (c+d x)^2}-\frac {b^2 \log (c+d x)}{d^3}-\frac {b \cosh (a+b x) \sinh (a+b x)}{d^2 (c+d x)}+\frac {\left (2 b^2\right ) \int \left (\frac {1}{2 (c+d x)}+\frac {\cosh (2 a+2 b x)}{2 (c+d x)}\right ) \, dx}{d^2}\\ &=-\frac {\cosh ^2(a+b x)}{2 d (c+d x)^2}-\frac {b \cosh (a+b x) \sinh (a+b x)}{d^2 (c+d x)}+\frac {b^2 \int \frac {\cosh (2 a+2 b x)}{c+d x} \, dx}{d^2}\\ &=-\frac {\cosh ^2(a+b x)}{2 d (c+d x)^2}-\frac {b \cosh (a+b x) \sinh (a+b x)}{d^2 (c+d x)}+\frac {\left (b^2 \cosh \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cosh \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d^2}+\frac {\left (b^2 \sinh \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sinh \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d^2}\\ &=-\frac {\cosh ^2(a+b x)}{2 d (c+d x)^2}+\frac {b^2 \cosh \left (2 a-\frac {2 b c}{d}\right ) \text {Chi}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}-\frac {b \cosh (a+b x) \sinh (a+b x)}{d^2 (c+d x)}+\frac {b^2 \sinh \left (2 a-\frac {2 b c}{d}\right ) \text {Shi}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}\\ \end {align*}

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Mathematica [A]  time = 0.88, size = 102, normalized size = 0.91 \[ \frac {2 b^2 \cosh \left (2 a-\frac {2 b c}{d}\right ) \text {Chi}\left (\frac {2 b (c+d x)}{d}\right )+2 b^2 \sinh \left (2 a-\frac {2 b c}{d}\right ) \text {Shi}\left (\frac {2 b (c+d x)}{d}\right )-\frac {d \left (b (c+d x) \sinh (2 (a+b x))+d \cosh ^2(a+b x)\right )}{(c+d x)^2}}{2 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]^2/(c + d*x)^3,x]

[Out]

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

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fricas [B]  time = 0.54, size = 278, normalized size = 2.48 \[ -\frac {d^{2} \cosh \left (b x + a\right )^{2} + d^{2} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b d^{2} x + b c d\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + d^{2} - 2 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} {\rm Ei}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} {\rm Ei}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cosh \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} {\rm Ei}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} {\rm Ei}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sinh \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2/(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/4*(d^2*cosh(b*x + a)^2 + d^2*sinh(b*x + a)^2 + 4*(b*d^2*x + b*c*d)*cosh(b*x + a)*sinh(b*x + a) + d^2 - 2*((
b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*Ei(2*(b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*Ei(-2*(b*d*
x + b*c)/d))*cosh(-2*(b*c - a*d)/d) - 2*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*Ei(2*(b*d*x + b*c)/d) - (b^2*d^
2*x^2 + 2*b^2*c*d*x + b^2*c^2)*Ei(-2*(b*d*x + b*c)/d))*sinh(-2*(b*c - a*d)/d))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

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giac [B]  time = 0.13, size = 330, normalized size = 2.95 \[ \frac {4 \, b^{2} d^{2} x^{2} {\rm Ei}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) e^{\left (2 \, a - \frac {2 \, b c}{d}\right )} + 4 \, b^{2} d^{2} x^{2} {\rm Ei}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )} + 8 \, b^{2} c d x {\rm Ei}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) e^{\left (2 \, a - \frac {2 \, b c}{d}\right )} + 8 \, b^{2} c d x {\rm Ei}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )} + 4 \, b^{2} c^{2} {\rm Ei}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) e^{\left (2 \, a - \frac {2 \, b c}{d}\right )} + 4 \, b^{2} c^{2} {\rm Ei}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )} - 2 \, b d^{2} x e^{\left (2 \, b x + 2 \, a\right )} + 2 \, b d^{2} x e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, b c d e^{\left (2 \, b x + 2 \, a\right )} + 2 \, b c d e^{\left (-2 \, b x - 2 \, a\right )} - d^{2} e^{\left (2 \, b x + 2 \, a\right )} - d^{2} e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, d^{2}}{8 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2/(d*x+c)^3,x, algorithm="giac")

[Out]

1/8*(4*b^2*d^2*x^2*Ei(2*(b*d*x + b*c)/d)*e^(2*a - 2*b*c/d) + 4*b^2*d^2*x^2*Ei(-2*(b*d*x + b*c)/d)*e^(-2*a + 2*
b*c/d) + 8*b^2*c*d*x*Ei(2*(b*d*x + b*c)/d)*e^(2*a - 2*b*c/d) + 8*b^2*c*d*x*Ei(-2*(b*d*x + b*c)/d)*e^(-2*a + 2*
b*c/d) + 4*b^2*c^2*Ei(2*(b*d*x + b*c)/d)*e^(2*a - 2*b*c/d) + 4*b^2*c^2*Ei(-2*(b*d*x + b*c)/d)*e^(-2*a + 2*b*c/
d) - 2*b*d^2*x*e^(2*b*x + 2*a) + 2*b*d^2*x*e^(-2*b*x - 2*a) - 2*b*c*d*e^(2*b*x + 2*a) + 2*b*c*d*e^(-2*b*x - 2*
a) - d^2*e^(2*b*x + 2*a) - d^2*e^(-2*b*x - 2*a) - 2*d^2)/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

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maple [B]  time = 0.25, size = 299, normalized size = 2.67 \[ -\frac {1}{4 d \left (d x +c \right )^{2}}+\frac {b^{3} {\mathrm e}^{-2 b x -2 a} x}{4 d \left (b^{2} d^{2} x^{2}+2 b^{2} c d x +c^{2} b^{2}\right )}+\frac {b^{3} {\mathrm e}^{-2 b x -2 a} c}{4 d^{2} \left (b^{2} d^{2} x^{2}+2 b^{2} c d x +c^{2} b^{2}\right )}-\frac {b^{2} {\mathrm e}^{-2 b x -2 a}}{8 d \left (b^{2} d^{2} x^{2}+2 b^{2} c d x +c^{2} b^{2}\right )}-\frac {b^{2} {\mathrm e}^{-\frac {2 \left (d a -c b \right )}{d}} \Ei \left (1, 2 b x +2 a -\frac {2 \left (d a -c b \right )}{d}\right )}{2 d^{3}}-\frac {b^{2} {\mathrm e}^{2 b x +2 a}}{8 d^{3} \left (\frac {b c}{d}+b x \right )^{2}}-\frac {b^{2} {\mathrm e}^{2 b x +2 a}}{4 d^{3} \left (\frac {b c}{d}+b x \right )}-\frac {b^{2} {\mathrm e}^{\frac {2 d a -2 c b}{d}} \Ei \left (1, -2 b x -2 a -\frac {2 \left (-d a +c b \right )}{d}\right )}{2 d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^2/(d*x+c)^3,x)

[Out]

-1/4/d/(d*x+c)^2+1/4*b^3*exp(-2*b*x-2*a)/d/(b^2*d^2*x^2+2*b^2*c*d*x+b^2*c^2)*x+1/4*b^3*exp(-2*b*x-2*a)/d^2/(b^
2*d^2*x^2+2*b^2*c*d*x+b^2*c^2)*c-1/8*b^2*exp(-2*b*x-2*a)/d/(b^2*d^2*x^2+2*b^2*c*d*x+b^2*c^2)-1/2*b^2/d^3*exp(-
2*(a*d-b*c)/d)*Ei(1,2*b*x+2*a-2*(a*d-b*c)/d)-1/8*b^2/d^3*exp(2*b*x+2*a)/(b*c/d+b*x)^2-1/4*b^2/d^3*exp(2*b*x+2*
a)/(b*c/d+b*x)-1/2*b^2/d^3*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 = 0.60, size = 99, normalized size = 0.88 \[ -\frac {1}{4 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} - \frac {e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )} E_{3}\left (\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, {\left (d x + c\right )}^{2} d} - \frac {e^{\left (2 \, a - \frac {2 \, b c}{d}\right )} E_{3}\left (-\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{4 \, {\left (d x + c\right )}^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2/(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/4/(d^3*x^2 + 2*c*d^2*x + c^2*d) - 1/4*e^(-2*a + 2*b*c/d)*exp_integral_e(3, 2*(d*x + c)*b/d)/((d*x + c)^2*d)
- 1/4*e^(2*a - 2*b*c/d)*exp_integral_e(3, -2*(d*x + c)*b/d)/((d*x + c)^2*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^2/(c + d*x)^3,x)

[Out]

int(cosh(a + b*x)^2/(c + d*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**2/(d*x+c)**3,x)

[Out]

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

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