3.304 \(\int e^{x^2} \cosh (a+b x+c x^2) \, dx\)

Optimal. Leaf size=115 \[ \frac {\sqrt {\pi } e^{a-\frac {b^2}{4 (c+1)}} \text {erfi}\left (\frac {b+2 (c+1) x}{2 \sqrt {c+1}}\right )}{4 \sqrt {c+1}}-\frac {\sqrt {\pi } e^{-a-\frac {b^2}{4 (1-c)}} \text {erfi}\left (\frac {b-2 (1-c) x}{2 \sqrt {1-c}}\right )}{4 \sqrt {1-c}} \]

[Out]

-1/4*exp(-a-1/4*b^2/(1-c))*erfi(1/2*(b-2*(1-c)*x)/(1-c)^(1/2))*Pi^(1/2)/(1-c)^(1/2)+1/4*exp(a-1/4*b^2/(1+c))*e
rfi(1/2*(b+2*(1+c)*x)/(1+c)^(1/2))*Pi^(1/2)/(1+c)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {5513, 2234, 2204} \[ \frac {\sqrt {\pi } e^{a-\frac {b^2}{4 (c+1)}} \text {Erfi}\left (\frac {b+2 (c+1) x}{2 \sqrt {c+1}}\right )}{4 \sqrt {c+1}}-\frac {\sqrt {\pi } e^{-a-\frac {b^2}{4 (1-c)}} \text {Erfi}\left (\frac {b-2 (1-c) x}{2 \sqrt {1-c}}\right )}{4 \sqrt {1-c}} \]

Antiderivative was successfully verified.

[In]

Int[E^x^2*Cosh[a + b*x + c*x^2],x]

[Out]

-(E^(-a - b^2/(4*(1 - c)))*Sqrt[Pi]*Erfi[(b - 2*(1 - c)*x)/(2*Sqrt[1 - c])])/(4*Sqrt[1 - c]) + (E^(a - b^2/(4*
(1 + c)))*Sqrt[Pi]*Erfi[(b + 2*(1 + c)*x)/(2*Sqrt[1 + c])])/(4*Sqrt[1 + c])

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 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 5513

Int[Cosh[v_]^(n_.)*(F_)^(u_), x_Symbol] :> Int[ExpandTrigToExp[F^u, Cosh[v]^n, x], x] /; FreeQ[F, x] && (Linea
rQ[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 0.37, size = 122, normalized size = 1.06 \[ \frac {\sqrt {\pi } e^{-\frac {b^2}{4 c+4}} \left (\sqrt {c-1} (c+1) e^{\frac {b^2 c}{2 \left (c^2-1\right )}} (\cosh (a)-\sinh (a)) \text {erf}\left (\frac {b+2 (c-1) x}{2 \sqrt {c-1}}\right )+(c-1) \sqrt {c+1} (\sinh (a)+\cosh (a)) \text {erfi}\left (\frac {b+2 (c+1) x}{2 \sqrt {c+1}}\right )\right )}{4 \left (c^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x^2*Cosh[a + b*x + c*x^2],x]

[Out]

(Sqrt[Pi]*(Sqrt[-1 + c]*(1 + c)*E^((b^2*c)/(2*(-1 + c^2)))*Erf[(b + 2*(-1 + c)*x)/(2*Sqrt[-1 + c])]*(Cosh[a] -
Sinh[a]) + (-1 + c)*Sqrt[1 + c]*Erfi[(b + 2*(1 + c)*x)/(2*Sqrt[1 + c])]*(Cosh[a] + Sinh[a])))/(4*(-1 + c^2)*E
^(b^2/(4 + 4*c)))

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fricas [A]  time = 0.55, size = 165, normalized size = 1.43 \[ \frac {\sqrt {\pi } {\left ({\left (c + 1\right )} \cosh \left (-\frac {b^{2} - 4 \, a c + 4 \, a}{4 \, {\left (c - 1\right )}}\right ) - {\left (c + 1\right )} \sinh \left (-\frac {b^{2} - 4 \, a c + 4 \, a}{4 \, {\left (c - 1\right )}}\right )\right )} \sqrt {c - 1} \operatorname {erf}\left (\frac {2 \, {\left (c - 1\right )} x + b}{2 \, \sqrt {c - 1}}\right ) - \sqrt {\pi } {\left ({\left (c - 1\right )} \cosh \left (-\frac {b^{2} - 4 \, a c - 4 \, a}{4 \, {\left (c + 1\right )}}\right ) + {\left (c - 1\right )} \sinh \left (-\frac {b^{2} - 4 \, a c - 4 \, a}{4 \, {\left (c + 1\right )}}\right )\right )} \sqrt {-c - 1} \operatorname {erf}\left (\frac {{\left (2 \, {\left (c + 1\right )} x + b\right )} \sqrt {-c - 1}}{2 \, {\left (c + 1\right )}}\right )}{4 \, {\left (c^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(pi)*((c + 1)*cosh(-1/4*(b^2 - 4*a*c + 4*a)/(c - 1)) - (c + 1)*sinh(-1/4*(b^2 - 4*a*c + 4*a)/(c - 1))
)*sqrt(c - 1)*erf(1/2*(2*(c - 1)*x + b)/sqrt(c - 1)) - sqrt(pi)*((c - 1)*cosh(-1/4*(b^2 - 4*a*c - 4*a)/(c + 1)
) + (c - 1)*sinh(-1/4*(b^2 - 4*a*c - 4*a)/(c + 1)))*sqrt(-c - 1)*erf(1/2*(2*(c + 1)*x + b)*sqrt(-c - 1)/(c + 1
)))/(c^2 - 1)

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giac [A]  time = 0.13, size = 101, normalized size = 0.88 \[ -\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c - 1} {\left (2 \, x + \frac {b}{c + 1}\right )}\right ) e^{\left (-\frac {b^{2} - 4 \, a c - 4 \, a}{4 \, {\left (c + 1\right )}}\right )}}{4 \, \sqrt {-c - 1}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c - 1} {\left (2 \, x + \frac {b}{c - 1}\right )}\right ) e^{\left (\frac {b^{2} - 4 \, a c + 4 \, a}{4 \, {\left (c - 1\right )}}\right )}}{4 \, \sqrt {c - 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(pi)*erf(-1/2*sqrt(-c - 1)*(2*x + b/(c + 1)))*e^(-1/4*(b^2 - 4*a*c - 4*a)/(c + 1))/sqrt(-c - 1) - 1/4
*sqrt(pi)*erf(-1/2*sqrt(c - 1)*(2*x + b/(c - 1)))*e^(1/4*(b^2 - 4*a*c + 4*a)/(c - 1))/sqrt(c - 1)

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maple [A]  time = 0.75, size = 105, normalized size = 0.91 \[ \frac {\sqrt {\pi }\, {\mathrm e}^{-\frac {4 a c -b^{2}-4 a}{4 \left (c -1\right )}} \erf \left (\sqrt {c -1}\, x +\frac {b}{2 \sqrt {c -1}}\right )}{4 \sqrt {c -1}}-\frac {\sqrt {\pi }\, {\mathrm e}^{\frac {4 a c -b^{2}+4 a}{4+4 c}} \erf \left (-\sqrt {-1-c}\, x +\frac {b}{2 \sqrt {-1-c}}\right )}{4 \sqrt {-1-c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*Pi^(1/2)*exp(-1/4*(4*a*c-b^2-4*a)/(c-1))/(c-1)^(1/2)*erf((c-1)^(1/2)*x+1/2*b/(c-1)^(1/2))-1/4*Pi^(1/2)*exp
(1/4*(4*a*c-b^2+4*a)/(1+c))/(-1-c)^(1/2)*erf(-(-1-c)^(1/2)*x+1/2*b/(-1-c)^(1/2))

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maxima [A]  time = 0.33, size = 89, normalized size = 0.77 \[ \frac {\sqrt {\pi } \operatorname {erf}\left (\sqrt {-c - 1} x - \frac {b}{2 \, \sqrt {-c - 1}}\right ) e^{\left (a - \frac {b^{2}}{4 \, {\left (c + 1\right )}}\right )}}{4 \, \sqrt {-c - 1}} + \frac {\sqrt {\pi } \operatorname {erf}\left (\sqrt {c - 1} x + \frac {b}{2 \, \sqrt {c - 1}}\right ) e^{\left (-a + \frac {b^{2}}{4 \, {\left (c - 1\right )}}\right )}}{4 \, \sqrt {c - 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(pi)*erf(sqrt(-c - 1)*x - 1/2*b/sqrt(-c - 1))*e^(a - 1/4*b^2/(c + 1))/sqrt(-c - 1) + 1/4*sqrt(pi)*erf(
sqrt(c - 1)*x + 1/2*b/sqrt(c - 1))*e^(-a + 1/4*b^2/(c - 1))/sqrt(c - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {e}}^{x^2}\,\mathrm {cosh}\left (c\,x^2+b\,x+a\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x^2)*cosh(a + b*x + c*x^2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x^{2}} \cosh {\left (a + b x + c x^{2} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x**2)*cosh(a + b*x + c*x**2), x)

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