### 3.15 $$\int \frac{\cosh (\frac{1}{4}+x+x^2)}{x^2} \, dx$$

Optimal. Leaf size=68 $\text{Unintegrable}\left (\frac{\sinh \left (x^2+x+\frac{1}{4}\right )}{x},x\right )+\frac{1}{2} \sqrt{\pi } \text{Erf}\left (\frac{1}{2} (-2 x-1)\right )+\frac{1}{2} \sqrt{\pi } \text{Erfi}\left (\frac{1}{2} (2 x+1)\right )-\frac{\cosh \left (x^2+x+\frac{1}{4}\right )}{x}$

[Out]

-(Cosh[1/4 + x + x^2]/x) + (Sqrt[Pi]*Erf[(-1 - 2*x)/2])/2 + (Sqrt[Pi]*Erfi[(1 + 2*x)/2])/2 + Unintegrable[Sinh
[1/4 + x + x^2]/x, x]

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Rubi [A]  time = 0.0401572, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0., Rules used = {} $\int \frac{\cosh \left (\frac{1}{4}+x+x^2\right )}{x^2} \, dx$

Veriﬁcation is Not applicable to the result.

[In]

Int[Cosh[1/4 + x + x^2]/x^2,x]

[Out]

-(Cosh[1/4 + x + x^2]/x) + (Sqrt[Pi]*Erf[(-1 - 2*x)/2])/2 + (Sqrt[Pi]*Erfi[(1 + 2*x)/2])/2 + Defer[Int][Sinh[1
/4 + x + x^2]/x, x]

Rubi steps

\begin{align*} \int \frac{\cosh \left (\frac{1}{4}+x+x^2\right )}{x^2} \, dx &=-\frac{\cosh \left (\frac{1}{4}+x+x^2\right )}{x}+2 \int \sinh \left (\frac{1}{4}+x+x^2\right ) \, dx+\int \frac{\sinh \left (\frac{1}{4}+x+x^2\right )}{x} \, dx\\ &=-\frac{\cosh \left (\frac{1}{4}+x+x^2\right )}{x}-\int e^{-\frac{1}{4}-x-x^2} \, dx+\int e^{\frac{1}{4}+x+x^2} \, dx+\int \frac{\sinh \left (\frac{1}{4}+x+x^2\right )}{x} \, dx\\ &=-\frac{\cosh \left (\frac{1}{4}+x+x^2\right )}{x}-\int e^{-\frac{1}{4} (-1-2 x)^2} \, dx+\int e^{\frac{1}{4} (1+2 x)^2} \, dx+\int \frac{\sinh \left (\frac{1}{4}+x+x^2\right )}{x} \, dx\\ &=-\frac{\cosh \left (\frac{1}{4}+x+x^2\right )}{x}+\frac{1}{2} \sqrt{\pi } \text{erf}\left (\frac{1}{2} (-1-2 x)\right )+\frac{1}{2} \sqrt{\pi } \text{erfi}\left (\frac{1}{2} (1+2 x)\right )+\int \frac{\sinh \left (\frac{1}{4}+x+x^2\right )}{x} \, dx\\ \end{align*}

Mathematica [A]  time = 9.45823, size = 0, normalized size = 0. $\int \frac{\cosh \left (\frac{1}{4}+x+x^2\right )}{x^2} \, dx$

Veriﬁcation is Not applicable to the result.

[In]

Integrate[Cosh[1/4 + x + x^2]/x^2,x]

[Out]

Integrate[Cosh[1/4 + x + x^2]/x^2, x]

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Maple [A]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}\cosh \left ({\frac{1}{4}}+x+{x}^{2} \right ) }\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(1/4+x+x^2)/x^2,x)

[Out]

int(cosh(1/4+x+x^2)/x^2,x)

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh \left (x^{2} + x + \frac{1}{4}\right )}{x^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(1/4+x+x^2)/x^2,x, algorithm="maxima")

[Out]

integrate(cosh(x^2 + x + 1/4)/x^2, x)

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\cosh \left (x^{2} + x + \frac{1}{4}\right )}{x^{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(1/4+x+x^2)/x^2,x, algorithm="fricas")

[Out]

integral(cosh(x^2 + x + 1/4)/x^2, x)

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Sympy [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh{\left (x^{2} + x + \frac{1}{4} \right )}}{x^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(1/4+x+x**2)/x**2,x)

[Out]

Integral(cosh(x**2 + x + 1/4)/x**2, x)

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh \left (x^{2} + x + \frac{1}{4}\right )}{x^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(1/4+x+x^2)/x^2,x, algorithm="giac")

[Out]

integrate(cosh(x^2 + x + 1/4)/x^2, x)