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

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

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5513, 2287, 2234, 2205, 2204} \[ \frac {1}{4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {(e-b \log (f))^2}{4 f}-d} \text {Erf}\left (\frac {-b \log (f)+e+2 f x}{2 \sqrt {f}}\right )+\frac {1}{4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{d-\frac {(b \log (f)+e)^2}{4 f}} \text {Erfi}\left (\frac {b \log (f)+e+2 f x}{2 \sqrt {f}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x)*Cosh[d + e*x + f*x^2],x]

[Out]

(E^(-d + (e - b*Log[f])^2/(4*f))*f^(-1/2 + a)*Sqrt[Pi]*Erf[(e + 2*f*x - b*Log[f])/(2*Sqrt[f])])/4 + (E^(d - (e
+ b*Log[f])^2/(4*f))*f^(-1/2 + a)*Sqrt[Pi]*Erfi[(e + 2*f*x + b*Log[f])/(2*Sqrt[f])])/4

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]

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

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Mathematica [A]  time = 0.28, size = 123, normalized size = 1.07 \[ \frac {1}{4} \sqrt {\pi } f^{a-\frac {b e+f}{2 f}} e^{-\frac {b^2 \log ^2(f)+e^2}{4 f}} \left ((\cosh (d)-\sinh (d)) e^{\frac {b^2 \log ^2(f)+e^2}{2 f}} \text {erf}\left (\frac {-b \log (f)+e+2 f x}{2 \sqrt {f}}\right )+(\sinh (d)+\cosh (d)) \text {erfi}\left (\frac {b \log (f)+e+2 f x}{2 \sqrt {f}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.64, size = 251, normalized size = 2.18 \[ -\frac {\sqrt {\pi } \sqrt {-f} \cosh \left (\frac {b^{2} \log \relax (f)^{2} + e^{2} - 4 \, d f + 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right ) \operatorname {erf}\left (\frac {{\left (2 \, f x + b \log \relax (f) + e\right )} \sqrt {-f}}{2 \, f}\right ) + \sqrt {\pi } \sqrt {f} \cosh \left (\frac {b^{2} \log \relax (f)^{2} + e^{2} - 4 \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right ) \operatorname {erf}\left (-\frac {2 \, f x - b \log \relax (f) + e}{2 \, \sqrt {f}}\right ) - \sqrt {\pi } \sqrt {-f} \operatorname {erf}\left (\frac {{\left (2 \, f x + b \log \relax (f) + e\right )} \sqrt {-f}}{2 \, f}\right ) \sinh \left (\frac {b^{2} \log \relax (f)^{2} + e^{2} - 4 \, d f + 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right ) + \sqrt {\pi } \sqrt {f} \operatorname {erf}\left (-\frac {2 \, f x - b \log \relax (f) + e}{2 \, \sqrt {f}}\right ) \sinh \left (\frac {b^{2} \log \relax (f)^{2} + e^{2} - 4 \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 134, normalized size = 1.17 \[ -\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {f} {\left (2 \, x - \frac {b \log \relax (f) - e}{f}\right )}\right ) e^{\left (\frac {b^{2} \log \relax (f)^{2} + 4 \, a f \log \relax (f) - 2 \, b e \log \relax (f) - 4 \, d f + e^{2}}{4 \, f}\right )}}{4 \, \sqrt {f}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-f} {\left (2 \, x + \frac {b \log \relax (f) + e}{f}\right )}\right ) e^{\left (-\frac {b^{2} \log \relax (f)^{2} - 4 \, a f \log \relax (f) + 2 \, b e \log \relax (f) - 4 \, d f + e^{2}}{4 \, f}\right )}}{4 \, \sqrt {-f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 126, normalized size = 1.10 \[ -\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {\ln \relax (f )^{2} b^{2}-2 \ln \relax (f ) b e -4 d f +e^{2}}{4 f}} \erf \left (-\sqrt {f}\, x +\frac {b \ln \relax (f )-e}{2 \sqrt {f}}\right )}{4 \sqrt {f}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \relax (f )^{2} b^{2}+2 \ln \relax (f ) b e -4 d f +e^{2}}{4 f}} \erf \left (-\sqrt {-f}\, x +\frac {e +b \ln \relax (f )}{2 \sqrt {-f}}\right )}{4 \sqrt {-f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 102, normalized size = 0.89 \[ \frac {1}{4} \, \sqrt {\pi } f^{a - \frac {1}{2}} \operatorname {erf}\left (\sqrt {f} x - \frac {b \log \relax (f) - e}{2 \, \sqrt {f}}\right ) e^{\left (-d + \frac {{\left (b \log \relax (f) - e\right )}^{2}}{4 \, f}\right )} + \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-f} x - \frac {b \log \relax (f) + e}{2 \, \sqrt {-f}}\right ) e^{\left (d - \frac {{\left (b \log \relax (f) + e\right )}^{2}}{4 \, f}\right )}}{4 \, \sqrt {-f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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