3.161 \(\int \frac {a+b \cosh (e+f x)}{(c+d x)^3} \, dx\)

Optimal. Leaf size=123 \[ -\frac {a}{2 d (c+d x)^2}+\frac {b f^2 \text {Chi}\left (x f+\frac {c f}{d}\right ) \cosh \left (e-\frac {c f}{d}\right )}{2 d^3}+\frac {b f^2 \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (x f+\frac {c f}{d}\right )}{2 d^3}-\frac {b f \sinh (e+f x)}{2 d^2 (c+d x)}-\frac {b \cosh (e+f x)}{2 d (c+d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3317, 3297, 3303, 3298, 3301} \[ -\frac {a}{2 d (c+d x)^2}+\frac {b f^2 \text {Chi}\left (x f+\frac {c f}{d}\right ) \cosh \left (e-\frac {c f}{d}\right )}{2 d^3}+\frac {b f^2 \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (x f+\frac {c f}{d}\right )}{2 d^3}-\frac {b f \sinh (e+f x)}{2 d^2 (c+d x)}-\frac {b \cosh (e+f x)}{2 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

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 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

Rubi steps

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

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Mathematica [A]  time = 0.63, size = 95, normalized size = 0.77 \[ \frac {-\frac {d (a d+b f (c+d x) \sinh (e+f x)+b d \cosh (e+f x))}{(c+d x)^2}+b f^2 \text {Chi}\left (f \left (\frac {c}{d}+x\right )\right ) \cosh \left (e-\frac {c f}{d}\right )+b f^2 \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (f \left (\frac {c}{d}+x\right )\right )}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.54, size = 274, normalized size = 2.23 \[ -\frac {2 \, b d^{2} \cosh \left (f x + e\right ) + 2 \, a d^{2} - {\left ({\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) + {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} {\rm Ei}\left (-\frac {d f x + c f}{d}\right )\right )} \cosh \left (-\frac {d e - c f}{d}\right ) + 2 \, {\left (b d^{2} f x + b c d f\right )} \sinh \left (f x + e\right ) + {\left ({\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) - {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} {\rm Ei}\left (-\frac {d f x + c f}{d}\right )\right )} \sinh \left (-\frac {d e - c f}{d}\right )}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.13, size = 328, normalized size = 2.67 \[ \frac {b d^{2} f^{2} x^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (\frac {c f}{d} - e\right )} + b d^{2} f^{2} x^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (-\frac {c f}{d} + e\right )} + 2 \, b c d f^{2} x {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (\frac {c f}{d} - e\right )} + 2 \, b c d f^{2} x {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (-\frac {c f}{d} + e\right )} + b c^{2} f^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (\frac {c f}{d} - e\right )} + b c^{2} f^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (-\frac {c f}{d} + e\right )} - b d^{2} f x e^{\left (f x + e\right )} + b d^{2} f x e^{\left (-f x - e\right )} - b c d f e^{\left (f x + e\right )} + b c d f e^{\left (-f x - e\right )} - b d^{2} e^{\left (f x + e\right )} - b d^{2} e^{\left (-f x - e\right )} - 2 \, a d^{2}}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 296, normalized size = 2.41 \[ -\frac {a}{2 d \left (d x +c \right )^{2}}+\frac {f^{3} b \,{\mathrm e}^{-f x -e} x}{4 d \left (d^{2} f^{2} x^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}+\frac {f^{3} b \,{\mathrm e}^{-f x -e} c}{4 d^{2} \left (d^{2} f^{2} x^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}-\frac {f^{2} b \,{\mathrm e}^{-f x -e}}{4 d \left (d^{2} f^{2} x^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}-\frac {f^{2} b \,{\mathrm e}^{\frac {c f -d e}{d}} \Ei \left (1, f x +e +\frac {c f -d e}{d}\right )}{4 d^{3}}-\frac {f^{2} b \,{\mathrm e}^{f x +e}}{4 d^{3} \left (\frac {c f}{d}+f x \right )^{2}}-\frac {f^{2} b \,{\mathrm e}^{f x +e}}{4 d^{3} \left (\frac {c f}{d}+f x \right )}-\frac {f^{2} b \,{\mathrm e}^{-\frac {c f -d e}{d}} \Ei \left (1, -f x -e -\frac {c f -d e}{d}\right )}{4 d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 98, normalized size = 0.80 \[ -\frac {1}{2} \, b {\left (\frac {e^{\left (-e + \frac {c f}{d}\right )} E_{3}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{{\left (d x + c\right )}^{2} d} + \frac {e^{\left (e - \frac {c f}{d}\right )} E_{3}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{{\left (d x + c\right )}^{2} d}\right )} - \frac {a}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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