\(\int f^{a+c x^2} \cosh (d+e x) \, dx\) [311]

Optimal result

Integrand size = 16, antiderivative size = 133 \[ \int f^{a+c x^2} \cosh (d+e x) \, dx=-\frac {e^{-d-\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{d-\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[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] (verified)

Time = 0.14 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5624, 2325, 2266, 2235} \[ \int f^{a+c x^2} \cosh (d+e x) \, dx=\frac {\sqrt {\pi } f^a e^{d-\frac {e^2}{4 c \log (f)}} \text {erfi}\left (\frac {2 c x \log (f)+e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {\sqrt {\pi } f^a e^{-\frac {e^2}{4 c \log (f)}-d} \text {erfi}\left (\frac {e-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[In]

Int[f^(a + c*x^2)*Cosh[d + e*x],x]

[Out]

-1/4*(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]])])/(Sqrt[c]*Sqrt[
Log[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 2235

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 2266

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 2325

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 5624

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*} \text {integral}& = \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] (verified)

Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.78 \[ \int f^{a+c x^2} \cosh (d+e x) \, dx=\frac {e^{-\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \left (\text {erfi}\left (\frac {-e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cosh (d)-\sinh (d))+\text {erfi}\left (\frac {e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cosh (d)+\sinh (d))\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[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]])

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {\operatorname {erf}\left (\sqrt {-c \ln \left (f \right )}\, x +\frac {e}{2 \sqrt {-c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 d \ln \left (f \right ) c +e^{2}}{4 \ln \left (f \right ) c}}}{4 \sqrt {-c \ln \left (f \right )}}-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {e}{2 \sqrt {-c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {4 d \ln \left (f \right ) c -e^{2}}{4 \ln \left (f \right ) c}}}{4 \sqrt {-c \ln \left (f \right )}}\)	\(117\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (101) = 202\).

Time = 0.28 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.62 \[ \int f^{a+c x^2} \cosh (d+e x) \, dx=-\frac {\sqrt {-c \log \left (f\right )} {\left (\sqrt {\pi } \cosh \left (\frac {4 \, a c \log \left (f\right )^{2} + 4 \, c d \log \left (f\right ) - e^{2}}{4 \, c \log \left (f\right )}\right ) + \sqrt {\pi } \sinh \left (\frac {4 \, a c \log \left (f\right )^{2} + 4 \, c d \log \left (f\right ) - e^{2}}{4 \, c \log \left (f\right )}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) + e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) + \sqrt {-c \log \left (f\right )} {\left (\sqrt {\pi } \cosh \left (\frac {4 \, a c \log \left (f\right )^{2} - 4 \, c d \log \left (f\right ) - e^{2}}{4 \, c \log \left (f\right )}\right ) + \sqrt {\pi } \sinh \left (\frac {4 \, a c \log \left (f\right )^{2} - 4 \, c d \log \left (f\right ) - e^{2}}{4 \, c \log \left (f\right )}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) - e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right )}{4 \, c \log \left (f\right )} \]

[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)
)

Sympy [F]

\[ \int f^{a+c x^2} \cosh (d+e x) \, dx=\int f^{a + c x^{2}} \cosh {\left (d + e x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.79 \[ \int f^{a+c x^2} \cosh (d+e x) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {e}{2 \, \sqrt {-c \log \left (f\right )}}\right ) e^{\left (d - \frac {e^{2}}{4 \, c \log \left (f\right )}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} + \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x + \frac {e}{2 \, \sqrt {-c \log \left (f\right )}}\right ) e^{\left (-d - \frac {e^{2}}{4 \, c \log \left (f\right )}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99 \[ \int f^{a+c x^2} \cosh (d+e x) \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {e}{c \log \left (f\right )}\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} + 4 \, c d \log \left (f\right ) - e^{2}}{4 \, c \log \left (f\right )}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x - \frac {e}{c \log \left (f\right )}\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} - 4 \, c d \log \left (f\right ) - e^{2}}{4 \, c \log \left (f\right )}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} \]

[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))

Mupad [F(-1)]

Timed out. \[ \int f^{a+c x^2} \cosh (d+e x) \, dx=\int f^{c\,x^2+a}\,\mathrm {cosh}\left (d+e\,x\right ) \,d x \]

[In]

int(f^(a + c*x^2)*cosh(d + e*x),x)

[Out]

int(f^(a + c*x^2)*cosh(d + e*x), x)