3.134 \(\int f^{a+b x+c x^2} \cos (a+b x+e x^2) \, dx\)

Optimal. Leaf size=209 \[ \frac {\sqrt {\pi } \exp \left (-\left ((-\log (f)+i) \left (a-\frac {b^2 (-\log (f)+i)}{-4 c \log (f)+4 i e}\right )\right )\right ) \text {erf}\left (\frac {b (-\log (f)+i)+2 x (-c \log (f)+i e)}{2 \sqrt {-c \log (f)+i e}}\right )}{4 \sqrt {-c \log (f)+i e}}+\frac {\sqrt {\pi } \exp \left ((\log (f)+i) \left (a-\frac {b^2 (\log (f)+i)}{4 c \log (f)+4 i e}\right )\right ) \text {erfi}\left (\frac {b (\log (f)+i)+2 x (c \log (f)+i e)}{2 \sqrt {c \log (f)+i e}}\right )}{4 \sqrt {c \log (f)+i e}} \]

[Out]

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

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Rubi [A]  time = 0.48, antiderivative size = 209, 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 = {4473, 2287, 2234, 2205, 2204} \[ \frac {\sqrt {\pi } \exp \left (-(-\log (f)+i) \left (a-\frac {b^2 (-\log (f)+i)}{-4 c \log (f)+4 i e}\right )\right ) \text {Erf}\left (\frac {b (-\log (f)+i)+2 x (-c \log (f)+i e)}{2 \sqrt {-c \log (f)+i e}}\right )}{4 \sqrt {-c \log (f)+i e}}+\frac {\sqrt {\pi } \exp \left ((\log (f)+i) \left (a-\frac {b^2 (\log (f)+i)}{4 c \log (f)+4 i e}\right )\right ) \text {Erfi}\left (\frac {b (\log (f)+i)+2 x (c \log (f)+i e)}{2 \sqrt {c \log (f)+i e}}\right )}{4 \sqrt {c \log (f)+i e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Pi]*Erf[(b*(I - Log[f]) + 2*x*(I*e - c*Log[f]))/(2*Sqrt[I*e - c*Log[f]])])/(4*E^((I - Log[f])*(a - (b^2*
(I - Log[f]))/((4*I)*e - 4*c*Log[f])))*Sqrt[I*e - c*Log[f]]) + (E^((I + Log[f])*(a - (b^2*(I + Log[f]))/((4*I)
*e + 4*c*Log[f])))*Sqrt[Pi]*Erfi[(b*(I + Log[f]) + 2*x*(I*e + c*Log[f]))/(2*Sqrt[I*e + c*Log[f]])])/(4*Sqrt[I*
e + 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 4473

Int[Cos[v_]^(n_.)*(F_)^(u_), x_Symbol] :> Int[ExpandTrigToExp[F^u, Cos[v]^n, x], x] /; FreeQ[F, x] && (LinearQ
[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} \cos \left (a+b x+e x^2\right ) \, dx &=\int \left (\frac {1}{2} e^{-i a-i b x-i e x^2} f^{a+b x+c x^2}+\frac {1}{2} e^{i a+i b x+i e x^2} f^{a+b x+c x^2}\right ) \, dx\\ &=\frac {1}{2} \int e^{-i a-i b x-i e x^2} f^{a+b x+c x^2} \, dx+\frac {1}{2} \int e^{i a+i b x+i e x^2} f^{a+b x+c x^2} \, dx\\ &=\frac {1}{2} \int \exp \left (-a (i-\log (f))-b x (i-\log (f))-x^2 (i e-c \log (f))\right ) \, dx+\frac {1}{2} \int \exp \left (a (i+\log (f))+b x (i+\log (f))+x^2 (i e+c \log (f))\right ) \, dx\\ &=\frac {1}{2} \exp \left (-(i-\log (f)) \left (a-\frac {b^2 (i-\log (f))}{4 i e-4 c \log (f)}\right )\right ) \int \exp \left (\frac {(-b (i-\log (f))+2 x (-i e+c \log (f)))^2}{4 (-i e+c \log (f))}\right ) \, dx+\frac {1}{2} \exp \left ((i+\log (f)) \left (a-\frac {b^2 (i+\log (f))}{4 i e+4 c \log (f)}\right )\right ) \int \exp \left (\frac {(b (i+\log (f))+2 x (i e+c \log (f)))^2}{4 (i e+c \log (f))}\right ) \, dx\\ &=\frac {\exp \left (-(i-\log (f)) \left (a-\frac {b^2 (i-\log (f))}{4 i e-4 c \log (f)}\right )\right ) \sqrt {\pi } \text {erf}\left (\frac {b (i-\log (f))+2 x (i e-c \log (f))}{2 \sqrt {i e-c \log (f)}}\right )}{4 \sqrt {i e-c \log (f)}}+\frac {\exp \left ((i+\log (f)) \left (a-\frac {b^2 (i+\log (f))}{4 i e+4 c \log (f)}\right )\right ) \sqrt {\pi } \text {erfi}\left (\frac {b (i+\log (f))+2 x (i e+c \log (f))}{2 \sqrt {i e+c \log (f)}}\right )}{4 \sqrt {i e+c \log (f)}}\\ \end {align*}

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Mathematica [A]  time = 1.81, size = 325, normalized size = 1.56 \[ -\frac {i \sqrt {\pi } e^{-\frac {b^2 c \log ^3(f)}{2 \left (c^2 \log ^2(f)+e^2\right )}} f^{a-\frac {b^2}{2 (e-i c \log (f))}} \left ((\cos (a)+i \sin (a)) (e+i c \log (f)) \sqrt {c \log (f)+i e} \exp \left (\frac {1}{4} b^2 \left (\frac {\log ^2(f)}{c \log (f)-i e}+\frac {1}{c \log (f)+i e}\right )\right ) \text {erfi}\left (\frac {\log (f) (b+2 c x)+i (b+2 e x)}{2 \sqrt {c \log (f)+i e}}\right )-(\cos (a)-i \sin (a)) (e-i c \log (f)) \sqrt {c \log (f)-i e} f^{\frac {i b^2 c \log (f)}{c^2 \log ^2(f)+e^2}} \exp \left (\frac {1}{4} b^2 \left (\frac {\log ^2(f)}{c \log (f)+i e}+\frac {1}{c \log (f)-i e}\right )\right ) \text {erfi}\left (\frac {\log (f) (b+2 c x)-i (b+2 e x)}{2 \sqrt {c \log (f)-i e}}\right )\right )}{4 \left (c^2 \log ^2(f)+e^2\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/4*I)*f^(a - b^2/(2*(e - I*c*Log[f])))*Sqrt[Pi]*(-(E^((b^2*(((-I)*e + c*Log[f])^(-1) + Log[f]^2/(I*e + c*L
og[f])))/4)*f^((I*b^2*c*Log[f])/(e^2 + c^2*Log[f]^2))*Erfi[((-I)*(b + 2*e*x) + (b + 2*c*x)*Log[f])/(2*Sqrt[(-I
)*e + c*Log[f]])]*(e - I*c*Log[f])*Sqrt[(-I)*e + c*Log[f]]*(Cos[a] - I*Sin[a])) + E^((b^2*(Log[f]^2/((-I)*e +
c*Log[f]) + (I*e + c*Log[f])^(-1)))/4)*Erfi[(I*(b + 2*e*x) + (b + 2*c*x)*Log[f])/(2*Sqrt[I*e + c*Log[f]])]*(e
+ I*c*Log[f])*Sqrt[I*e + c*Log[f]]*(Cos[a] + I*Sin[a])))/(E^((b^2*c*Log[f]^3)/(2*(e^2 + c^2*Log[f]^2)))*(e^2 +
c^2*Log[f]^2))

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fricas [B]  time = 2.13, size = 381, normalized size = 1.82 \[ -\frac {\sqrt {\pi } {\left (c \log \relax (f) - i \, e\right )} \sqrt {-c \log \relax (f) - i \, e} \operatorname {erf}\left (\frac {{\left (2 \, e^{2} x + {\left (2 \, c^{2} x + b c\right )} \log \relax (f)^{2} + b e + {\left (i \, b c - i \, b e\right )} \log \relax (f)\right )} \sqrt {-c \log \relax (f) - i \, e}}{2 \, {\left (c^{2} \log \relax (f)^{2} + e^{2}\right )}}\right ) e^{\left (-\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} \log \relax (f)^{3} + i \, b^{2} e - 4 i \, a e^{2} - {\left (-2 i \, b^{2} c + 4 i \, a c^{2} + i \, b^{2} e\right )} \log \relax (f)^{2} - {\left (b^{2} c - 2 \, b^{2} e + 4 \, a e^{2}\right )} \log \relax (f)}{4 \, {\left (c^{2} \log \relax (f)^{2} + e^{2}\right )}}\right )} + \sqrt {\pi } {\left (c \log \relax (f) + i \, e\right )} \sqrt {-c \log \relax (f) + i \, e} \operatorname {erf}\left (\frac {{\left (2 \, e^{2} x + {\left (2 \, c^{2} x + b c\right )} \log \relax (f)^{2} + b e + {\left (-i \, b c + i \, b e\right )} \log \relax (f)\right )} \sqrt {-c \log \relax (f) + i \, e}}{2 \, {\left (c^{2} \log \relax (f)^{2} + e^{2}\right )}}\right ) e^{\left (-\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} \log \relax (f)^{3} - i \, b^{2} e + 4 i \, a e^{2} - {\left (2 i \, b^{2} c - 4 i \, a c^{2} - i \, b^{2} e\right )} \log \relax (f)^{2} - {\left (b^{2} c - 2 \, b^{2} e + 4 \, a e^{2}\right )} \log \relax (f)}{4 \, {\left (c^{2} \log \relax (f)^{2} + e^{2}\right )}}\right )}}{4 \, {\left (c^{2} \log \relax (f)^{2} + e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(sqrt(pi)*(c*log(f) - I*e)*sqrt(-c*log(f) - I*e)*erf(1/2*(2*e^2*x + (2*c^2*x + b*c)*log(f)^2 + b*e + (I*b
*c - I*b*e)*log(f))*sqrt(-c*log(f) - I*e)/(c^2*log(f)^2 + e^2))*e^(-1/4*((b^2*c - 4*a*c^2)*log(f)^3 + I*b^2*e
- 4*I*a*e^2 - (-2*I*b^2*c + 4*I*a*c^2 + I*b^2*e)*log(f)^2 - (b^2*c - 2*b^2*e + 4*a*e^2)*log(f))/(c^2*log(f)^2
+ e^2)) + sqrt(pi)*(c*log(f) + I*e)*sqrt(-c*log(f) + I*e)*erf(1/2*(2*e^2*x + (2*c^2*x + b*c)*log(f)^2 + b*e +
(-I*b*c + I*b*e)*log(f))*sqrt(-c*log(f) + I*e)/(c^2*log(f)^2 + e^2))*e^(-1/4*((b^2*c - 4*a*c^2)*log(f)^3 - I*b
^2*e + 4*I*a*e^2 - (2*I*b^2*c - 4*I*a*c^2 - I*b^2*e)*log(f)^2 - (b^2*c - 2*b^2*e + 4*a*e^2)*log(f))/(c^2*log(f
)^2 + e^2)))/(c^2*log(f)^2 + e^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(c*x^2 + b*x + a)*cos(e*x^2 + b*x + a), x)

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maple [A]  time = 0.20, size = 215, normalized size = 1.03 \[ -\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 a e -b^{2}+4 i \ln \relax (f ) a c -2 i \ln \relax (f ) b^{2}+\ln \relax (f )^{2} b^{2}}{4 \left (-i e +c \ln \relax (f )\right )}} \erf \left (-\sqrt {i e -c \ln \relax (f )}\, x +\frac {-i b +b \ln \relax (f )}{2 \sqrt {i e -c \ln \relax (f )}}\right )}{4 \sqrt {i e -c \ln \relax (f )}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {-4 a e +b^{2}+4 i \ln \relax (f ) a c -2 i \ln \relax (f ) b^{2}-\ln \relax (f )^{2} b^{2}}{4 i e +4 c \ln \relax (f )}} \erf \left (-\sqrt {-i e -c \ln \relax (f )}\, x +\frac {i b +b \ln \relax (f )}{2 \sqrt {-i e -c \ln \relax (f )}}\right )}{4 \sqrt {-i e -c \ln \relax (f )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.39, size = 1018, normalized size = 4.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(sqrt(pi)*sqrt(2*c^2*log(f)^2 + 2*e^2)*((I*f^(1/4*b^2*c/(c^2*log(f)^2 + e^2))*f^a*cos(-1/4*(b^2*e - 4*a*e
^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)) + f^(1/4*b^2*c/(c^2*log(f)^2 + e^2))*f^a*sin(
-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)))*erf(1/2*(2*(c*log(f) - I*
e)*x + b*log(f) - I*b)*sqrt(-c*log(f) + I*e)/(c*log(f) - I*e)) + (-I*f^(1/4*b^2*c/(c^2*log(f)^2 + e^2))*f^a*co
s(-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)) + f^(1/4*b^2*c/(c^2*log(
f)^2 + e^2))*f^a*sin(-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)))*erf(
1/2*(2*(c*log(f) + I*e)*x + b*log(f) + I*b)*sqrt(-c*log(f) - I*e)/(c*log(f) + I*e)))*sqrt(c*log(f) + sqrt(c^2*
log(f)^2 + e^2)) - sqrt(pi)*sqrt(2*c^2*log(f)^2 + 2*e^2)*((f^(1/4*b^2*c/(c^2*log(f)^2 + e^2))*f^a*cos(-1/4*(b^
2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)) - I*f^(1/4*b^2*c/(c^2*log(f)^2 + e
^2))*f^a*sin(-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)))*erf(1/2*(2*(
c*log(f) - I*e)*x + b*log(f) - I*b)*sqrt(-c*log(f) + I*e)/(c*log(f) - I*e)) + (f^(1/4*b^2*c/(c^2*log(f)^2 + e^
2))*f^a*cos(-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)) + I*f^(1/4*b^2
*c/(c^2*log(f)^2 + e^2))*f^a*sin(-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 +
e^2)))*erf(1/2*(2*(c*log(f) + I*e)*x + b*log(f) + I*b)*sqrt(-c*log(f) - I*e)/(c*log(f) + I*e)))*sqrt(-c*log(f
) + sqrt(c^2*log(f)^2 + e^2)))/(c^2*e^(1/4*b^2*c*log(f)^3/(c^2*log(f)^2 + e^2) + 1/2*b^2*e*log(f)/(c^2*log(f)^
2 + e^2))*log(f)^2 + e^2*e^(1/4*b^2*c*log(f)^3/(c^2*log(f)^2 + e^2) + 1/2*b^2*e*log(f)/(c^2*log(f)^2 + e^2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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