3.1.55 \(\int \frac {(a+b x^2)^2 \cosh (c+d x)}{x^4} \, dx\) [55]

Optimal. Leaf size=133 \[ -\frac {a^2 \cosh (c+d x)}{3 x^3}-\frac {2 a b \cosh (c+d x)}{x}-\frac {a^2 d^2 \cosh (c+d x)}{6 x}+2 a b d \text {Chi}(d x) \sinh (c)+\frac {1}{6} a^2 d^3 \text {Chi}(d x) \sinh (c)+\frac {b^2 \sinh (c+d x)}{d}-\frac {a^2 d \sinh (c+d x)}{6 x^2}+2 a b d \cosh (c) \text {Shi}(d x)+\frac {1}{6} a^2 d^3 \cosh (c) \text {Shi}(d x) \]

[Out]

-1/3*a^2*cosh(d*x+c)/x^3-2*a*b*cosh(d*x+c)/x-1/6*a^2*d^2*cosh(d*x+c)/x+2*a*b*d*cosh(c)*Shi(d*x)+1/6*a^2*d^3*co
sh(c)*Shi(d*x)+2*a*b*d*Chi(d*x)*sinh(c)+1/6*a^2*d^3*Chi(d*x)*sinh(c)+b^2*sinh(d*x+c)/d-1/6*a^2*d*sinh(d*x+c)/x
^2

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Rubi [A]
time = 0.18, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {5395, 2717, 3378, 3384, 3379, 3382} \begin {gather*} \frac {1}{6} a^2 d^3 \sinh (c) \text {Chi}(d x)+\frac {1}{6} a^2 d^3 \cosh (c) \text {Shi}(d x)-\frac {a^2 d^2 \cosh (c+d x)}{6 x}-\frac {a^2 \cosh (c+d x)}{3 x^3}-\frac {a^2 d \sinh (c+d x)}{6 x^2}+2 a b d \sinh (c) \text {Chi}(d x)+2 a b d \cosh (c) \text {Shi}(d x)-\frac {2 a b \cosh (c+d x)}{x}+\frac {b^2 \sinh (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*Cosh[c + d*x])/x^4,x]

[Out]

-1/3*(a^2*Cosh[c + d*x])/x^3 - (2*a*b*Cosh[c + d*x])/x - (a^2*d^2*Cosh[c + d*x])/(6*x) + 2*a*b*d*CoshIntegral[
d*x]*Sinh[c] + (a^2*d^3*CoshIntegral[d*x]*Sinh[c])/6 + (b^2*Sinh[c + d*x])/d - (a^2*d*Sinh[c + d*x])/(6*x^2) +
2*a*b*d*Cosh[c]*SinhIntegral[d*x] + (a^2*d^3*Cosh[c]*SinhIntegral[d*x])/6

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3378

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 3379

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 3382

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 3384

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 5395

Int[Cosh[(c_.) + (d_.)*(x_)]*((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[Cosh[c + d*x], (e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \cosh (c+d x)}{x^4} \, dx &=\int \left (b^2 \cosh (c+d x)+\frac {a^2 \cosh (c+d x)}{x^4}+\frac {2 a b \cosh (c+d x)}{x^2}\right ) \, dx\\ &=a^2 \int \frac {\cosh (c+d x)}{x^4} \, dx+(2 a b) \int \frac {\cosh (c+d x)}{x^2} \, dx+b^2 \int \cosh (c+d x) \, dx\\ &=-\frac {a^2 \cosh (c+d x)}{3 x^3}-\frac {2 a b \cosh (c+d x)}{x}+\frac {b^2 \sinh (c+d x)}{d}+\frac {1}{3} \left (a^2 d\right ) \int \frac {\sinh (c+d x)}{x^3} \, dx+(2 a b d) \int \frac {\sinh (c+d x)}{x} \, dx\\ &=-\frac {a^2 \cosh (c+d x)}{3 x^3}-\frac {2 a b \cosh (c+d x)}{x}+\frac {b^2 \sinh (c+d x)}{d}-\frac {a^2 d \sinh (c+d x)}{6 x^2}+\frac {1}{6} \left (a^2 d^2\right ) \int \frac {\cosh (c+d x)}{x^2} \, dx+(2 a b d \cosh (c)) \int \frac {\sinh (d x)}{x} \, dx+(2 a b d \sinh (c)) \int \frac {\cosh (d x)}{x} \, dx\\ &=-\frac {a^2 \cosh (c+d x)}{3 x^3}-\frac {2 a b \cosh (c+d x)}{x}-\frac {a^2 d^2 \cosh (c+d x)}{6 x}+2 a b d \text {Chi}(d x) \sinh (c)+\frac {b^2 \sinh (c+d x)}{d}-\frac {a^2 d \sinh (c+d x)}{6 x^2}+2 a b d \cosh (c) \text {Shi}(d x)+\frac {1}{6} \left (a^2 d^3\right ) \int \frac {\sinh (c+d x)}{x} \, dx\\ &=-\frac {a^2 \cosh (c+d x)}{3 x^3}-\frac {2 a b \cosh (c+d x)}{x}-\frac {a^2 d^2 \cosh (c+d x)}{6 x}+2 a b d \text {Chi}(d x) \sinh (c)+\frac {b^2 \sinh (c+d x)}{d}-\frac {a^2 d \sinh (c+d x)}{6 x^2}+2 a b d \cosh (c) \text {Shi}(d x)+\frac {1}{6} \left (a^2 d^3 \cosh (c)\right ) \int \frac {\sinh (d x)}{x} \, dx+\frac {1}{6} \left (a^2 d^3 \sinh (c)\right ) \int \frac {\cosh (d x)}{x} \, dx\\ &=-\frac {a^2 \cosh (c+d x)}{3 x^3}-\frac {2 a b \cosh (c+d x)}{x}-\frac {a^2 d^2 \cosh (c+d x)}{6 x}+2 a b d \text {Chi}(d x) \sinh (c)+\frac {1}{6} a^2 d^3 \text {Chi}(d x) \sinh (c)+\frac {b^2 \sinh (c+d x)}{d}-\frac {a^2 d \sinh (c+d x)}{6 x^2}+2 a b d \cosh (c) \text {Shi}(d x)+\frac {1}{6} a^2 d^3 \cosh (c) \text {Shi}(d x)\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 114, normalized size = 0.86 \begin {gather*} \frac {1}{6} \left (-\frac {2 a^2 \cosh (c+d x)}{x^3}-\frac {12 a b \cosh (c+d x)}{x}-\frac {a^2 d^2 \cosh (c+d x)}{x}+a d \left (12 b+a d^2\right ) \text {Chi}(d x) \sinh (c)+\frac {6 b^2 \sinh (c+d x)}{d}-\frac {a^2 d \sinh (c+d x)}{x^2}+a d \left (12 b+a d^2\right ) \cosh (c) \text {Shi}(d x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*Cosh[c + d*x])/x^4,x]

[Out]

((-2*a^2*Cosh[c + d*x])/x^3 - (12*a*b*Cosh[c + d*x])/x - (a^2*d^2*Cosh[c + d*x])/x + a*d*(12*b + a*d^2)*CoshIn
tegral[d*x]*Sinh[c] + (6*b^2*Sinh[c + d*x])/d - (a^2*d*Sinh[c + d*x])/x^2 + a*d*(12*b + a*d^2)*Cosh[c]*SinhInt
egral[d*x])/6

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Maple [A]
time = 0.83, size = 222, normalized size = 1.67

method	result	size
risch	\(-\frac {b^{2} {\mathrm e}^{-d x -c}}{2 d}+\frac {d^{3} a^{2} {\mathrm e}^{-c} \expIntegral \left (1, d x \right )}{12}-\frac {d^{2} a^{2} {\mathrm e}^{-d x -c}}{12 x}+\frac {d \,a^{2} {\mathrm e}^{-d x -c}}{12 x^{2}}-\frac {a^{2} {\mathrm e}^{-d x -c}}{6 x^{3}}-\frac {a b \,{\mathrm e}^{-d x -c}}{x}+d a b \,{\mathrm e}^{-c} \expIntegral \left (1, d x \right )-\frac {d^{3} a^{2} {\mathrm e}^{c} \expIntegral \left (1, -d x \right )}{12}-\frac {a^{2} {\mathrm e}^{d x +c}}{6 x^{3}}+\frac {{\mathrm e}^{d x +c} b^{2}}{2 d}-\frac {a b \,{\mathrm e}^{d x +c}}{x}-d a b \,{\mathrm e}^{c} \expIntegral \left (1, -d x \right )-\frac {d \,a^{2} {\mathrm e}^{d x +c}}{12 x^{2}}-\frac {d^{2} a^{2} {\mathrm e}^{d x +c}}{12 x}\)	\(222\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*cosh(d*x+c)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/2/d*b^2*exp(-d*x-c)+1/12*d^3*a^2*exp(-c)*Ei(1,d*x)-1/12*d^2*a^2*exp(-d*x-c)/x+1/12*d*a^2*exp(-d*x-c)/x^2-1/
6*a^2*exp(-d*x-c)/x^3-a*b*exp(-d*x-c)/x+d*a*b*exp(-c)*Ei(1,d*x)-1/12*d^3*a^2*exp(c)*Ei(1,-d*x)-1/6*a^2/x^3*exp
(d*x+c)+1/2/d*exp(d*x+c)*b^2-a*b/x*exp(d*x+c)-d*a*b*exp(c)*Ei(1,-d*x)-1/12*d*a^2/x^2*exp(d*x+c)-1/12*d^2*a^2/x
*exp(d*x+c)

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Maxima [A]
time = 0.33, size = 135, normalized size = 1.02 \begin {gather*} \frac {1}{6} \, {\left (a^{2} d^{2} e^{\left (-c\right )} \Gamma \left (-2, d x\right ) - a^{2} d^{2} e^{c} \Gamma \left (-2, -d x\right ) - 6 \, a b {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 6 \, a b {\rm Ei}\left (d x\right ) e^{c} - \frac {3 \, {\left (d x e^{c} - e^{c}\right )} b^{2} e^{\left (d x\right )}}{d^{2}} - \frac {3 \, {\left (d x + 1\right )} b^{2} e^{\left (-d x - c\right )}}{d^{2}}\right )} d + \frac {1}{3} \, {\left (3 \, b^{2} x - \frac {6 \, a b x^{2} + a^{2}}{x^{3}}\right )} \cosh \left (d x + c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*cosh(d*x+c)/x^4,x, algorithm="maxima")

[Out]

1/6*(a^2*d^2*e^(-c)*gamma(-2, d*x) - a^2*d^2*e^c*gamma(-2, -d*x) - 6*a*b*Ei(-d*x)*e^(-c) + 6*a*b*Ei(d*x)*e^c -
3*(d*x*e^c - e^c)*b^2*e^(d*x)/d^2 - 3*(d*x + 1)*b^2*e^(-d*x - c)/d^2)*d + 1/3*(3*b^2*x - (6*a*b*x^2 + a^2)/x^
3)*cosh(d*x + c)

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Fricas [A]
time = 0.37, size = 171, normalized size = 1.29 \begin {gather*} -\frac {2 \, {\left (2 \, a^{2} d + {\left (a^{2} d^{3} + 12 \, a b d\right )} x^{2}\right )} \cosh \left (d x + c\right ) - {\left ({\left (a^{2} d^{4} + 12 \, a b d^{2}\right )} x^{3} {\rm Ei}\left (d x\right ) - {\left (a^{2} d^{4} + 12 \, a b d^{2}\right )} x^{3} {\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) + 2 \, {\left (a^{2} d^{2} x - 6 \, b^{2} x^{3}\right )} \sinh \left (d x + c\right ) - {\left ({\left (a^{2} d^{4} + 12 \, a b d^{2}\right )} x^{3} {\rm Ei}\left (d x\right ) + {\left (a^{2} d^{4} + 12 \, a b d^{2}\right )} x^{3} {\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{12 \, d x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*cosh(d*x+c)/x^4,x, algorithm="fricas")

[Out]

-1/12*(2*(2*a^2*d + (a^2*d^3 + 12*a*b*d)*x^2)*cosh(d*x + c) - ((a^2*d^4 + 12*a*b*d^2)*x^3*Ei(d*x) - (a^2*d^4 +
12*a*b*d^2)*x^3*Ei(-d*x))*cosh(c) + 2*(a^2*d^2*x - 6*b^2*x^3)*sinh(d*x + c) - ((a^2*d^4 + 12*a*b*d^2)*x^3*Ei(
d*x) + (a^2*d^4 + 12*a*b*d^2)*x^3*Ei(-d*x))*sinh(c))/(d*x^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right )^{2} \cosh {\left (c + d x \right )}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*cosh(d*x+c)/x**4,x)

[Out]

Integral((a + b*x**2)**2*cosh(c + d*x)/x**4, x)

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Giac [A]
time = 0.42, size = 236, normalized size = 1.77 \begin {gather*} -\frac {a^{2} d^{4} x^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a^{2} d^{4} x^{3} {\rm Ei}\left (d x\right ) e^{c} + 12 \, a b d^{2} x^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 12 \, a b d^{2} x^{3} {\rm Ei}\left (d x\right ) e^{c} + a^{2} d^{3} x^{2} e^{\left (d x + c\right )} + a^{2} d^{3} x^{2} e^{\left (-d x - c\right )} + a^{2} d^{2} x e^{\left (d x + c\right )} + 12 \, a b d x^{2} e^{\left (d x + c\right )} - 6 \, b^{2} x^{3} e^{\left (d x + c\right )} - a^{2} d^{2} x e^{\left (-d x - c\right )} + 12 \, a b d x^{2} e^{\left (-d x - c\right )} + 6 \, b^{2} x^{3} e^{\left (-d x - c\right )} + 2 \, a^{2} d e^{\left (d x + c\right )} + 2 \, a^{2} d e^{\left (-d x - c\right )}}{12 \, d x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*cosh(d*x+c)/x^4,x, algorithm="giac")

[Out]

-1/12*(a^2*d^4*x^3*Ei(-d*x)*e^(-c) - a^2*d^4*x^3*Ei(d*x)*e^c + 12*a*b*d^2*x^3*Ei(-d*x)*e^(-c) - 12*a*b*d^2*x^3
*Ei(d*x)*e^c + a^2*d^3*x^2*e^(d*x + c) + a^2*d^3*x^2*e^(-d*x - c) + a^2*d^2*x*e^(d*x + c) + 12*a*b*d*x^2*e^(d*
x + c) - 6*b^2*x^3*e^(d*x + c) - a^2*d^2*x*e^(-d*x - c) + 12*a*b*d*x^2*e^(-d*x - c) + 6*b^2*x^3*e^(-d*x - c) +
2*a^2*d*e^(d*x + c) + 2*a^2*d*e^(-d*x - c))/(d*x^3)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (b\,x^2+a\right )}^2}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(c + d*x)*(a + b*x^2)^2)/x^4,x)

[Out]

int((cosh(c + d*x)*(a + b*x^2)^2)/x^4, x)

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