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

Optimal. Leaf size=161 \[ \frac {\sqrt {\pi } f^a e^{\frac {(e-b \log (f))^2}{4 (f-c \log (f))}-d} \text {erf}\left (\frac {-b \log (f)+2 x (f-c \log (f))+e}{2 \sqrt {f-c \log (f)}}\right )}{4 \sqrt {f-c \log (f)}}+\frac {\sqrt {\pi } f^a e^{d-\frac {(b \log (f)+e)^2}{4 (c \log (f)+f)}} \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+f)+e}{2 \sqrt {c \log (f)+f}}\right )}{4 \sqrt {c \log (f)+f}} \]

[Out]

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

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Rubi [A]  time = 0.43, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5513, 2287, 2234, 2205, 2204} \[ \frac {\sqrt {\pi } f^a e^{\frac {(e-b \log (f))^2}{4 (f-c \log (f))}-d} \text {Erf}\left (\frac {-b \log (f)+2 x (f-c \log (f))+e}{2 \sqrt {f-c \log (f)}}\right )}{4 \sqrt {f-c \log (f)}}+\frac {\sqrt {\pi } f^a e^{d-\frac {(b \log (f)+e)^2}{4 (c \log (f)+f)}} \text {Erfi}\left (\frac {b \log (f)+2 x (c \log (f)+f)+e}{2 \sqrt {c \log (f)+f}}\right )}{4 \sqrt {c \log (f)+f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(-d + (e - b*Log[f])^2/(4*(f - c*Log[f])))*f^a*Sqrt[Pi]*Erf[(e - b*Log[f] + 2*x*(f - c*Log[f]))/(2*Sqrt[f -
c*Log[f]])])/(4*Sqrt[f - c*Log[f]]) + (E^(d - (e + b*Log[f])^2/(4*(f + c*Log[f])))*f^a*Sqrt[Pi]*Erfi[(e + b*L
og[f] + 2*x*(f + c*Log[f]))/(2*Sqrt[f + c*Log[f]])])/(4*Sqrt[f + c*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 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+c x^2} \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+c x^2}+\frac {1}{2} e^{d+e x+f x^2} f^{a+b x+c x^2}\right ) \, dx\\ &=\frac {1}{2} \int e^{-d-e x-f x^2} f^{a+b x+c x^2} \, dx+\frac {1}{2} \int e^{d+e x+f x^2} f^{a+b x+c x^2} \, dx\\ &=\frac {1}{2} \int \exp \left (-d+a \log (f)-x (e-b \log (f))-x^2 (f-c \log (f))\right ) \, dx+\frac {1}{2} \int \exp \left (d+a \log (f)+x (e+b \log (f))+x^2 (f+c \log (f))\right ) \, dx\\ &=\frac {1}{2} \left (e^{-d+\frac {(e-b \log (f))^2}{4 (f-c \log (f))}} f^a\right ) \int \exp \left (\frac {(-e+b \log (f)+2 x (-f+c \log (f)))^2}{4 (-f+c \log (f))}\right ) \, dx+\frac {1}{2} \left (e^{d-\frac {(e+b \log (f))^2}{4 (f+c \log (f))}} f^a\right ) \int \exp \left (\frac {(e+b \log (f)+2 x (f+c \log (f)))^2}{4 (f+c \log (f))}\right ) \, dx\\ &=\frac {e^{-d+\frac {(e-b \log (f))^2}{4 (f-c \log (f))}} f^a \sqrt {\pi } \text {erf}\left (\frac {e-b \log (f)+2 x (f-c \log (f))}{2 \sqrt {f-c \log (f)}}\right )}{4 \sqrt {f-c \log (f)}}+\frac {e^{d-\frac {(e+b \log (f))^2}{4 (f+c \log (f))}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e+b \log (f)+2 x (f+c \log (f))}{2 \sqrt {f+c \log (f)}}\right )}{4 \sqrt {f+c \log (f)}}\\ \end {align*}

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Mathematica [A]  time = 1.51, size = 251, normalized size = 1.56 \[ \frac {\sqrt {\pi } e^{-\frac {b^2 \log ^2(f)+e^2}{4 (c \log (f)+f)}} f^{a+\frac {b e f}{c^2 \log ^2(f)-f^2}} \left (\sqrt {f-c \log (f)} (c \log (f)+f) (\cosh (d)-\sinh (d)) f^{\frac {b e}{2 (c \log (f)+f)}} \exp \left (\frac {f \left (b^2 \log ^2(f)+e^2\right )}{2 \left (f^2-c^2 \log ^2(f)\right )}\right ) \text {erf}\left (\frac {-\log (f) (b+2 c x)+e+2 f x}{2 \sqrt {f-c \log (f)}}\right )+(f-c \log (f)) \sqrt {c \log (f)+f} (\sinh (d)+\cosh (d)) f^{\frac {b e}{2 f-2 c \log (f)}} \text {erfi}\left (\frac {\log (f) (b+2 c x)+e+2 f x}{2 \sqrt {c \log (f)+f}}\right )\right )}{4 \left (f^2-c^2 \log ^2(f)\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 0.54, size = 362, normalized size = 2.25 \[ -\frac {{\left (\sqrt {\pi } {\left (c \log \relax (f) + f\right )} \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \relax (f)^{2} + e^{2} - 4 \, d f + 2 \, {\left (2 \, c d - b e + 2 \, a f\right )} \log \relax (f)}{4 \, {\left (c \log \relax (f) - f\right )}}\right ) + \sqrt {\pi } {\left (c \log \relax (f) + f\right )} \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \relax (f)^{2} + e^{2} - 4 \, d f + 2 \, {\left (2 \, c d - b e + 2 \, a f\right )} \log \relax (f)}{4 \, {\left (c \log \relax (f) - f\right )}}\right )\right )} \sqrt {-c \log \relax (f) + f} \operatorname {erf}\left (-\frac {{\left (2 \, f x - {\left (2 \, c x + b\right )} \log \relax (f) + e\right )} \sqrt {-c \log \relax (f) + f}}{2 \, {\left (c \log \relax (f) - f\right )}}\right ) + {\left (\sqrt {\pi } {\left (c \log \relax (f) - f\right )} \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \relax (f)^{2} + e^{2} - 4 \, d f - 2 \, {\left (2 \, c d - b e + 2 \, a f\right )} \log \relax (f)}{4 \, {\left (c \log \relax (f) + f\right )}}\right ) + \sqrt {\pi } {\left (c \log \relax (f) - f\right )} \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \relax (f)^{2} + e^{2} - 4 \, d f - 2 \, {\left (2 \, c d - b e + 2 \, a f\right )} \log \relax (f)}{4 \, {\left (c \log \relax (f) + f\right )}}\right )\right )} \sqrt {-c \log \relax (f) - f} \operatorname {erf}\left (\frac {{\left (2 \, f x + {\left (2 \, c x + b\right )} \log \relax (f) + e\right )} \sqrt {-c \log \relax (f) - f}}{2 \, {\left (c \log \relax (f) + f\right )}}\right )}{4 \, {\left (c^{2} \log \relax (f)^{2} - f^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 209, normalized size = 1.30 \[ -\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \relax (f) - f} {\left (2 \, x + \frac {b \log \relax (f) + e}{c \log \relax (f) + f}\right )}\right ) e^{\left (-\frac {b^{2} \log \relax (f)^{2} - 4 \, a c \log \relax (f)^{2} - 4 \, c d \log \relax (f) - 4 \, a f \log \relax (f) + 2 \, b e \log \relax (f) - 4 \, d f + e^{2}}{4 \, {\left (c \log \relax (f) + f\right )}}\right )}}{4 \, \sqrt {-c \log \relax (f) - f}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \relax (f) + f} {\left (2 \, x + \frac {b \log \relax (f) - e}{c \log \relax (f) - f}\right )}\right ) e^{\left (-\frac {b^{2} \log \relax (f)^{2} - 4 \, a c \log \relax (f)^{2} + 4 \, c d \log \relax (f) + 4 \, a f \log \relax (f) - 2 \, b e \log \relax (f) - 4 \, d f + e^{2}}{4 \, {\left (c \log \relax (f) - f\right )}}\right )}}{4 \, \sqrt {-c \log \relax (f) + f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 186, normalized size = 1.16 \[ -\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \relax (f )^{2} b^{2}-2 \ln \relax (f ) b e +4 d \ln \relax (f ) c -4 d f +e^{2}}{4 \left (-f +c \ln \relax (f )\right )}} \erf \left (-x \sqrt {f -c \ln \relax (f )}+\frac {b \ln \relax (f )-e}{2 \sqrt {f -c \ln \relax (f )}}\right )}{4 \sqrt {f -c \ln \relax (f )}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \relax (f )^{2} b^{2}+2 \ln \relax (f ) b e -4 d \ln \relax (f ) c -4 d f +e^{2}}{4 \left (f +c \ln \relax (f )\right )}} \erf \left (-\sqrt {-c \ln \relax (f )-f}\, x +\frac {e +b \ln \relax (f )}{2 \sqrt {-c \ln \relax (f )-f}}\right )}{4 \sqrt {-c \ln \relax (f )-f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c*x^2+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*ln(f)*c-4*d*f+e^2)/(-f+c*ln(f)))/(f-c*ln(f))^(1/2)*erf
(-x*(f-c*ln(f))^(1/2)+1/2*(b*ln(f)-e)/(f-c*ln(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*ln(f)*c-4*d*f+e^2)/(f+c*ln(f)))/(-c*ln(f)-f)^(1/2)*erf(-(-c*ln(f)-f)^(1/2)*x+1/2*(e+b*ln(f))/(-c*ln(f)-f)^(1
/2))

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maxima [A]  time = 0.34, size = 151, normalized size = 0.94 \[ \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \relax (f) - f} x - \frac {b \log \relax (f) + e}{2 \, \sqrt {-c \log \relax (f) - f}}\right ) e^{\left (-\frac {{\left (b \log \relax (f) + e\right )}^{2}}{4 \, {\left (c \log \relax (f) + f\right )}} + d\right )}}{4 \, \sqrt {-c \log \relax (f) - f}} + \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \relax (f) + f} x - \frac {b \log \relax (f) - e}{2 \, \sqrt {-c \log \relax (f) + f}}\right ) e^{\left (-\frac {{\left (b \log \relax (f) - e\right )}^{2}}{4 \, {\left (c \log \relax (f) - f\right )}} - d\right )}}{4 \, \sqrt {-c \log \relax (f) + f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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