3.311 \(\int f^{a+c x^2} \cosh (d+e x) \, dx\)

Optimal. Leaf size=133 \[ \frac {\sqrt {\pi } f^a e^{d-\frac {e^2}{4 c \log (f)}} \text {erfi}\left (\frac {2 c x \log (f)+e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {\sqrt {\pi } f^a e^{-\frac {e^2}{4 c \log (f)}-d} \text {erfi}\left (\frac {e-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[Out]

1/4*exp(-d-1/4*e^2/c/ln(f))*f^a*erfi(1/2*(-e+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)+1/
4*exp(d-1/4*e^2/c/ln(f))*f^a*erfi(1/2*(e+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5513, 2287, 2234, 2204} \[ \frac {\sqrt {\pi } f^a e^{d-\frac {e^2}{4 c \log (f)}} \text {Erfi}\left (\frac {2 c x \log (f)+e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {\sqrt {\pi } f^a e^{-\frac {e^2}{4 c \log (f)}-d} \text {Erfi}\left (\frac {e-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + c*x^2)*Cosh[d + e*x],x]

[Out]

-(E^(-d - e^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(e - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(4*Sqrt[c]*Sqrt[Lo
g[f]]) + (E^(d - e^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(e + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(4*Sqrt[c]*
Sqrt[Log[f]])

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 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, 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 f^{a+c x^2} \cosh (d+e x) \, dx &=\int \left (\frac {1}{2} e^{-d-e x} f^{a+c x^2}+\frac {1}{2} e^{d+e x} f^{a+c x^2}\right ) \, dx\\ &=\frac {1}{2} \int e^{-d-e x} f^{a+c x^2} \, dx+\frac {1}{2} \int e^{d+e x} f^{a+c x^2} \, dx\\ &=\frac {1}{2} \int e^{-d-e x+a \log (f)+c x^2 \log (f)} \, dx+\frac {1}{2} \int e^{d+e x+a \log (f)+c x^2 \log (f)} \, dx\\ &=\frac {1}{2} \left (e^{-d-\frac {e^2}{4 c \log (f)}} f^a\right ) \int e^{\frac {(-e+2 c x \log (f))^2}{4 c \log (f)}} \, dx+\frac {1}{2} \left (e^{d-\frac {e^2}{4 c \log (f)}} f^a\right ) \int e^{\frac {(e+2 c x \log (f))^2}{4 c \log (f)}} \, dx\\ &=-\frac {e^{-d-\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{d-\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 104, normalized size = 0.78 \[ \frac {\sqrt {\pi } f^a e^{-\frac {e^2}{4 c \log (f)}} \left ((\cosh (d)-\sinh (d)) \text {erfi}\left (\frac {2 c x \log (f)-e}{2 \sqrt {c} \sqrt {\log (f)}}\right )+(\sinh (d)+\cosh (d)) \text {erfi}\left (\frac {2 c x \log (f)+e}{2 \sqrt {c} \sqrt {\log (f)}}\right )\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + c*x^2)*Cosh[d + e*x],x]

[Out]

(f^a*Sqrt[Pi]*(Erfi[(-e + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cosh[d] - Sinh[d]) + Erfi[(e + 2*c*x*Log[f]
)/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cosh[d] + Sinh[d])))/(4*Sqrt[c]*E^(e^2/(4*c*Log[f]))*Sqrt[Log[f]])

________________________________________________________________________________________

fricas [B]  time = 0.62, size = 216, normalized size = 1.62 \[ -\frac {\sqrt {-c \log \relax (f)} {\left (\sqrt {\pi } \cosh \left (\frac {4 \, a c \log \relax (f)^{2} + 4 \, c d \log \relax (f) - e^{2}}{4 \, c \log \relax (f)}\right ) + \sqrt {\pi } \sinh \left (\frac {4 \, a c \log \relax (f)^{2} + 4 \, c d \log \relax (f) - e^{2}}{4 \, c \log \relax (f)}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \relax (f) + e\right )} \sqrt {-c \log \relax (f)}}{2 \, c \log \relax (f)}\right ) + \sqrt {-c \log \relax (f)} {\left (\sqrt {\pi } \cosh \left (\frac {4 \, a c \log \relax (f)^{2} - 4 \, c d \log \relax (f) - e^{2}}{4 \, c \log \relax (f)}\right ) + \sqrt {\pi } \sinh \left (\frac {4 \, a c \log \relax (f)^{2} - 4 \, c d \log \relax (f) - e^{2}}{4 \, c \log \relax (f)}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \relax (f) - e\right )} \sqrt {-c \log \relax (f)}}{2 \, c \log \relax (f)}\right )}{4 \, c \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(c*x^2+a)*cosh(e*x+d),x, algorithm="fricas")

[Out]

-1/4*(sqrt(-c*log(f))*(sqrt(pi)*cosh(1/4*(4*a*c*log(f)^2 + 4*c*d*log(f) - e^2)/(c*log(f))) + sqrt(pi)*sinh(1/4
*(4*a*c*log(f)^2 + 4*c*d*log(f) - e^2)/(c*log(f))))*erf(1/2*(2*c*x*log(f) + e)*sqrt(-c*log(f))/(c*log(f))) + s
qrt(-c*log(f))*(sqrt(pi)*cosh(1/4*(4*a*c*log(f)^2 - 4*c*d*log(f) - e^2)/(c*log(f))) + sqrt(pi)*sinh(1/4*(4*a*c
*log(f)^2 - 4*c*d*log(f) - e^2)/(c*log(f))))*erf(1/2*(2*c*x*log(f) - e)*sqrt(-c*log(f))/(c*log(f))))/(c*log(f)
)

________________________________________________________________________________________

giac [A]  time = 0.15, size = 132, normalized size = 0.99 \[ -\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \relax (f)} {\left (2 \, x + \frac {e}{c \log \relax (f)}\right )}\right ) e^{\left (\frac {4 \, a c \log \relax (f)^{2} + 4 \, c d \log \relax (f) - e^{2}}{4 \, c \log \relax (f)}\right )}}{4 \, \sqrt {-c \log \relax (f)}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \relax (f)} {\left (2 \, x - \frac {e}{c \log \relax (f)}\right )}\right ) e^{\left (\frac {4 \, a c \log \relax (f)^{2} - 4 \, c d \log \relax (f) - e^{2}}{4 \, c \log \relax (f)}\right )}}{4 \, \sqrt {-c \log \relax (f)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(c*x^2+a)*cosh(e*x+d),x, algorithm="giac")

[Out]

-1/4*sqrt(pi)*erf(-1/2*sqrt(-c*log(f))*(2*x + e/(c*log(f))))*e^(1/4*(4*a*c*log(f)^2 + 4*c*d*log(f) - e^2)/(c*l
og(f)))/sqrt(-c*log(f)) - 1/4*sqrt(pi)*erf(-1/2*sqrt(-c*log(f))*(2*x - e/(c*log(f))))*e^(1/4*(4*a*c*log(f)^2 -
 4*c*d*log(f) - e^2)/(c*log(f)))/sqrt(-c*log(f))

________________________________________________________________________________________

maple [A]  time = 0.16, size = 117, normalized size = 0.88 \[ \frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 d \ln \relax (f ) c +e^{2}}{4 \ln \relax (f ) c}} \erf \left (\sqrt {-c \ln \relax (f )}\, x +\frac {e}{2 \sqrt {-c \ln \relax (f )}}\right )}{4 \sqrt {-c \ln \relax (f )}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {4 d \ln \relax (f ) c -e^{2}}{4 \ln \relax (f ) c}} \erf \left (-\sqrt {-c \ln \relax (f )}\, x +\frac {e}{2 \sqrt {-c \ln \relax (f )}}\right )}{4 \sqrt {-c \ln \relax (f )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c*x^2+a)*cosh(e*x+d),x)

[Out]

1/4*Pi^(1/2)*f^a*exp(-1/4*(4*d*ln(f)*c+e^2)/ln(f)/c)/(-c*ln(f))^(1/2)*erf((-c*ln(f))^(1/2)*x+1/2*e/(-c*ln(f))^
(1/2))-1/4*Pi^(1/2)*f^a*exp(1/4*(4*d*ln(f)*c-e^2)/ln(f)/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*e/(-c*
ln(f))^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 105, normalized size = 0.79 \[ \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \relax (f)} x - \frac {e}{2 \, \sqrt {-c \log \relax (f)}}\right ) e^{\left (d - \frac {e^{2}}{4 \, c \log \relax (f)}\right )}}{4 \, \sqrt {-c \log \relax (f)}} + \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \relax (f)} x + \frac {e}{2 \, \sqrt {-c \log \relax (f)}}\right ) e^{\left (-d - \frac {e^{2}}{4 \, c \log \relax (f)}\right )}}{4 \, \sqrt {-c \log \relax (f)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(c*x^2+a)*cosh(e*x+d),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + c*x^2)*cosh(d + e*x),x)

[Out]

int(f^(a + c*x^2)*cosh(d + e*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(c*x**2+a)*cosh(e*x+d),x)

[Out]

Integral(f**(a + c*x**2)*cosh(d + e*x), x)

________________________________________________________________________________________