3.11 \(\int \cosh ^2(a+b x^2) \, dx\)

Optimal. Leaf size=78 \[ \frac{\sqrt{\frac{\pi }{2}} e^{-2 a} \text{Erf}\left (\sqrt{2} \sqrt{b} x\right )}{8 \sqrt{b}}+\frac{\sqrt{\frac{\pi }{2}} e^{2 a} \text{Erfi}\left (\sqrt{2} \sqrt{b} x\right )}{8 \sqrt{b}}+\frac{x}{2} \]

[Out]

x/2 + (Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[b]*x])/(8*Sqrt[b]*E^(2*a)) + (E^(2*a)*Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[b]*x])/(
8*Sqrt[b])

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Rubi [A]  time = 0.0429458, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5301, 5299, 2204, 2205} \[ \frac{\sqrt{\frac{\pi }{2}} e^{-2 a} \text{Erf}\left (\sqrt{2} \sqrt{b} x\right )}{8 \sqrt{b}}+\frac{\sqrt{\frac{\pi }{2}} e^{2 a} \text{Erfi}\left (\sqrt{2} \sqrt{b} x\right )}{8 \sqrt{b}}+\frac{x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x^2]^2,x]

[Out]

x/2 + (Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[b]*x])/(8*Sqrt[b]*E^(2*a)) + (E^(2*a)*Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[b]*x])/(
8*Sqrt[b])

Rule 5301

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a + b*Cosh[c + d*x^
n])^p, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 1] && IGtQ[p, 1]

Rule 5299

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[1/2, Int[E^(c + d*x^n), x], x] + Dist[1/2, Int[E^(-c - d*
x^n), x], x] /; FreeQ[{c, d}, x] && IGtQ[n, 1]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rubi steps

\begin{align*} \int \cosh ^2\left (a+b x^2\right ) \, dx &=\int \left (\frac{1}{2}+\frac{1}{2} \cosh \left (2 a+2 b x^2\right )\right ) \, dx\\ &=\frac{x}{2}+\frac{1}{2} \int \cosh \left (2 a+2 b x^2\right ) \, dx\\ &=\frac{x}{2}+\frac{1}{4} \int e^{-2 a-2 b x^2} \, dx+\frac{1}{4} \int e^{2 a+2 b x^2} \, dx\\ &=\frac{x}{2}+\frac{e^{-2 a} \sqrt{\frac{\pi }{2}} \text{erf}\left (\sqrt{2} \sqrt{b} x\right )}{8 \sqrt{b}}+\frac{e^{2 a} \sqrt{\frac{\pi }{2}} \text{erfi}\left (\sqrt{2} \sqrt{b} x\right )}{8 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.072337, size = 86, normalized size = 1.1 \[ \frac{\sqrt{\pi } (\cosh (2 a)-\sinh (2 a)) \text{Erf}\left (\sqrt{2} \sqrt{b} x\right )+\sqrt{\pi } (\sinh (2 a)+\cosh (2 a)) \text{Erfi}\left (\sqrt{2} \sqrt{b} x\right )+4 \sqrt{2} \sqrt{b} x}{8 \sqrt{2} \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x^2]^2,x]

[Out]

(4*Sqrt[2]*Sqrt[b]*x + Sqrt[Pi]*Erf[Sqrt[2]*Sqrt[b]*x]*(Cosh[2*a] - Sinh[2*a]) + Sqrt[Pi]*Erfi[Sqrt[2]*Sqrt[b]
*x]*(Cosh[2*a] + Sinh[2*a]))/(8*Sqrt[2]*Sqrt[b])

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Maple [A]  time = 0.029, size = 51, normalized size = 0.7 \begin{align*}{\frac{x}{2}}+{\frac{{{\rm e}^{-2\,a}}\sqrt{\pi }\sqrt{2}}{16}{\it Erf} \left ( x\sqrt{2}\sqrt{b} \right ){\frac{1}{\sqrt{b}}}}+{\frac{{{\rm e}^{2\,a}}\sqrt{\pi }}{8}{\it Erf} \left ( \sqrt{-2\,b}x \right ){\frac{1}{\sqrt{-2\,b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x^2+a)^2,x)

[Out]

1/2*x+1/16*exp(-2*a)*Pi^(1/2)*2^(1/2)/b^(1/2)*erf(x*2^(1/2)*b^(1/2))+1/8*exp(2*a)*Pi^(1/2)/(-2*b)^(1/2)*erf((-
2*b)^(1/2)*x)

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Maxima [A]  time = 1.67488, size = 76, normalized size = 0.97 \begin{align*} \frac{\sqrt{2} \sqrt{\pi } \operatorname{erf}\left (\sqrt{2} \sqrt{-b} x\right ) e^{\left (2 \, a\right )}}{16 \, \sqrt{-b}} + \frac{\sqrt{2} \sqrt{\pi } \operatorname{erf}\left (\sqrt{2} \sqrt{b} x\right ) e^{\left (-2 \, a\right )}}{16 \, \sqrt{b}} + \frac{1}{2} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sqrt(2)*sqrt(pi)*erf(sqrt(2)*sqrt(-b)*x)*e^(2*a)/sqrt(-b) + 1/16*sqrt(2)*sqrt(pi)*erf(sqrt(2)*sqrt(b)*x)*
e^(-2*a)/sqrt(b) + 1/2*x

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Fricas [A]  time = 1.66845, size = 225, normalized size = 2.88 \begin{align*} -\frac{\sqrt{2} \sqrt{\pi } \sqrt{-b}{\left (\cosh \left (2 \, a\right ) + \sinh \left (2 \, a\right )\right )} \operatorname{erf}\left (\sqrt{2} \sqrt{-b} x\right ) - \sqrt{2} \sqrt{\pi } \sqrt{b}{\left (\cosh \left (2 \, a\right ) - \sinh \left (2 \, a\right )\right )} \operatorname{erf}\left (\sqrt{2} \sqrt{b} x\right ) - 8 \, b x}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(sqrt(2)*sqrt(pi)*sqrt(-b)*(cosh(2*a) + sinh(2*a))*erf(sqrt(2)*sqrt(-b)*x) - sqrt(2)*sqrt(pi)*sqrt(b)*(c
osh(2*a) - sinh(2*a))*erf(sqrt(2)*sqrt(b)*x) - 8*b*x)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x**2+a)**2,x)

[Out]

Integral(cosh(a + b*x**2)**2, x)

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Giac [A]  time = 1.36541, size = 78, normalized size = 1. \begin{align*} -\frac{\sqrt{2} \sqrt{\pi } \operatorname{erf}\left (-\sqrt{2} \sqrt{-b} x\right ) e^{\left (2 \, a\right )}}{16 \, \sqrt{-b}} - \frac{\sqrt{2} \sqrt{\pi } \operatorname{erf}\left (-\sqrt{2} \sqrt{b} x\right ) e^{\left (-2 \, a\right )}}{16 \, \sqrt{b}} + \frac{1}{2} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*sqrt(2)*sqrt(pi)*erf(-sqrt(2)*sqrt(-b)*x)*e^(2*a)/sqrt(-b) - 1/16*sqrt(2)*sqrt(pi)*erf(-sqrt(2)*sqrt(b)*
x)*e^(-2*a)/sqrt(b) + 1/2*x