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

Optimal. Leaf size=69 \[ \text {Int}\left (\frac {\cosh \left (x^2+x+\frac {1}{4}\right )}{x},x\right )-\frac {1}{2} \sqrt {\pi } \text {erf}\left (\frac {1}{2} (-2 x-1)\right )+\frac {1}{2} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (2 x+1)\right )-\frac {\sinh \left (x^2+x+\frac {1}{4}\right )}{x} \]

[Out]

-sinh(1/4+x+x^2)/x+1/2*erf(1/2+x)*Pi^(1/2)+1/2*erfi(1/2+x)*Pi^(1/2)+Unintegrable(cosh(1/4+x+x^2)/x,x)

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Rubi [A]  time = 0.04, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\sinh \left (\frac {1}{4}+x+x^2\right )}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Sinh[1/4 + x + x^2]/x^2,x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 11.79, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (\frac {1}{4}+x+x^2\right )}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sinh[1/4 + x + x^2]/x^2,x]

[Out]

Integrate[Sinh[1/4 + x + x^2]/x^2, x]

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fricas [A]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sinh \left (x^{2} + x + \frac {1}{4}\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sinh(x^2 + x + 1/4)/x^2, x)

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (x^{2} + x + \frac {1}{4}\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(x^2 + x + 1/4)/x^2, x)

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maple [A]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (\frac {1}{4}+x +x^{2}\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (x^{2} + x + \frac {1}{4}\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(x^2 + x + 1/4)/x^2, x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {sinh}\left (x^2+x+\frac {1}{4}\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x + x^2 + 1/4)/x^2,x)

[Out]

int(sinh(x + x^2 + 1/4)/x^2, x)

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh {\left (x^{2} + x + \frac {1}{4} \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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