3.2.26 \(\int f^{a+b x+c x^2} \cos ^2(d+e x) \, dx\) [126]

Optimal. Leaf size=231 \[ \frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {Erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-2 i d+\frac {(2 e+i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {Erfi}\left (\frac {2 i e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{2 i d-\frac {(2 i e+b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {Erfi}\left (\frac {2 i e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}} \]

[Out]

1/8*exp(-2*I*d+1/4*(2*e+I*b*ln(f))^2/c/ln(f))*f^a*erfi(1/2*(-2*I*e+b*ln(f)+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*P
i^(1/2)/c^(1/2)/ln(f)^(1/2)+1/8*exp(2*I*d-1/4*(2*I*e+b*ln(f))^2/c/ln(f))*f^a*erfi(1/2*(2*I*e+b*ln(f)+2*c*x*ln(
f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)+1/4*f^(a-1/4*b^2/c)*erfi(1/2*(2*c*x+b)*ln(f)^(1/2)/c^(1/
2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4561, 2266, 2235, 2325} \begin {gather*} \frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {Erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {\sqrt {\pi } f^a e^{\frac {(2 e+i b \log (f))^2}{4 c \log (f)}-2 i d} \text {Erfi}\left (\frac {-b \log (f)-2 c x \log (f)+2 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a e^{2 i d-\frac {(b \log (f)+2 i e)^2}{4 c \log (f)}} \text {Erfi}\left (\frac {b \log (f)+2 c x \log (f)+2 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/(4*Sqrt[c]*Sqrt[Log[f]]) - (E^((-2*I
)*d + (2*e + I*b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[((2*I)*e - b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt
[Log[f]])])/(8*Sqrt[c]*Sqrt[Log[f]]) + (E^((2*I)*d - ((2*I)*e + b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[((
2*I)*e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(8*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 4561

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 ^2(d+e x) \, dx &=\int \left (\frac {1}{2} f^{a+b x+c x^2}+\frac {1}{4} e^{-2 i d-2 i e x} f^{a+b x+c x^2}+\frac {1}{4} e^{2 i d+2 i e x} f^{a+b x+c x^2}\right ) \, dx\\ &=\frac {1}{4} \int e^{-2 i d-2 i e x} f^{a+b x+c x^2} \, dx+\frac {1}{4} \int e^{2 i d+2 i e x} f^{a+b x+c x^2} \, dx+\frac {1}{2} \int f^{a+b x+c x^2} \, dx\\ &=\frac {1}{4} \int \exp \left (-2 i d+a \log (f)+c x^2 \log (f)-x (2 i e-b \log (f))\right ) \, dx+\frac {1}{4} \int \exp \left (2 i d+a \log (f)+c x^2 \log (f)+x (2 i e+b \log (f))\right ) \, dx+\frac {1}{2} f^{a-\frac {b^2}{4 c}} \int f^{\frac {(b+2 c x)^2}{4 c}} \, dx\\ &=\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {1}{4} \left (\exp \left (-2 i d+\frac {(2 e+i b \log (f))^2}{4 c \log (f)}\right ) f^a\right ) \int \exp \left (\frac {(-2 i e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx+\frac {1}{4} \left (e^{2 i d-\frac {(2 i e+b \log (f))^2}{4 c \log (f)}} f^a\right ) \int \exp \left (\frac {(2 i e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx\\ &=\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {\exp \left (-2 i d+\frac {(2 e+i b \log (f))^2}{4 c \log (f)}\right ) f^a \sqrt {\pi } \text {erfi}\left (\frac {2 i e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{2 i d-\frac {(2 i e+b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {2 i e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}\\ \end {align*}

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Mathematica [A]
time = 0.60, size = 204, normalized size = 0.88 \begin {gather*} \frac {e^{-\frac {i b e}{c}} f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \left (2 e^{\frac {i b e}{c}} \text {Erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )+e^{\frac {e (e+2 i b \log (f))}{c \log (f)}} \text {Erfi}\left (\frac {-2 i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (2 d)-i \sin (2 d))+e^{\frac {e^2}{c \log (f)}} \text {Erfi}\left (\frac {2 i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (2 d)+i \sin (2 d))\right )}{8 \sqrt {c} \sqrt {\log (f)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(f^(a - b^2/(4*c))*Sqrt[Pi]*(2*E^((I*b*e)/c)*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])] + E^((e*(e + (2*I)*b
*Log[f]))/(c*Log[f]))*Erfi[((-2*I)*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cos[2*d] - I*Sin[2*d]) +
E^(e^2/(c*Log[f]))*Erfi[((2*I)*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cos[2*d] + I*Sin[2*d])))/(8
*Sqrt[c]*E^((I*b*e)/c)*Sqrt[Log[f]])

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Maple [A]
time = 0.28, size = 217, normalized size = 0.94

method	result	size
risch	\(-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}-4 i \ln \left (f \right ) b e +8 i d \ln \left (f \right ) c -4 e^{2}}{4 \ln \left (f \right ) c}} \erf \left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )-2 i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{8 \sqrt {-c \ln \left (f \right )}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}+4 i \ln \left (f \right ) b e -8 i d \ln \left (f \right ) c -4 e^{2}}{4 \ln \left (f \right ) c}} \erf \left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {2 i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{8 \sqrt {-c \ln \left (f \right )}}-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} \erf \left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2-4*I*ln(f)*b*e+8*I*d*ln(f)*c-4*e^2)/ln(f)/c)/(-c*ln(f))^(1/2)*erf(-(-c*
ln(f))^(1/2)*x+1/2*(b*ln(f)-2*I*e)/(-c*ln(f))^(1/2))-1/8*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2+4*I*ln(f)*b*e-8*I*
d*ln(f)*c-4*e^2)/ln(f)/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*(2*I*e+b*ln(f))/(-c*ln(f))^(1/2))-1/4*P
i^(1/2)*f^a*f^(-1/4*b^2/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2/(-c*ln(f))^(1/2)*b*ln(f))

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.30, size = 399, normalized size = 1.73 \begin {gather*} \frac {\sqrt {\pi } {\left (f^{a} {\left (\cos \left (\frac {2 \, c d - b e}{c}\right ) + i \, \sin \left (\frac {2 \, c d - b e}{c}\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}} - \frac {1}{2} \, {\left (b \log \left (f\right ) + 2 i \, e\right )} \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}}\right ) e^{\left (\frac {e^{2}}{c \log \left (f\right )}\right )} + f^{a} {\left (\cos \left (\frac {2 \, c d - b e}{c}\right ) - i \, \sin \left (\frac {2 \, c d - b e}{c}\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}} - \frac {1}{2} \, {\left (b \log \left (f\right ) - 2 i \, e\right )} \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}}\right ) e^{\left (\frac {e^{2}}{c \log \left (f\right )}\right )} + f^{a} {\left (\cos \left (\frac {2 \, c d - b e}{c}\right ) + i \, \sin \left (\frac {2 \, c d - b e}{c}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) + b \log \left (f\right ) + 2 i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {e^{2}}{c \log \left (f\right )}\right )} + f^{a} {\left (\cos \left (\frac {2 \, c d - b e}{c}\right ) - i \, \sin \left (\frac {2 \, c d - b e}{c}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) + b \log \left (f\right ) - 2 i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {e^{2}}{c \log \left (f\right )}\right )} + 2 \, f^{a} \operatorname {erf}\left (-\frac {1}{2} \, b \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}} \log \left (f\right ) + x \overline {\sqrt {-c \log \left (f\right )}}\right ) - 2 \, f^{a} \operatorname {erf}\left (\frac {2 \, c x \log \left (f\right ) + b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right )}}\right )\right )}}{16 \, \sqrt {-c \log \left (f\right )} f^{\frac {b^{2}}{4 \, c}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sqrt(pi)*(f^a*(cos((2*c*d - b*e)/c) + I*sin((2*c*d - b*e)/c))*erf(x*conjugate(sqrt(-c*log(f))) - 1/2*(b*l
og(f) + 2*I*e)*conjugate(1/sqrt(-c*log(f))))*e^(e^2/(c*log(f))) + f^a*(cos((2*c*d - b*e)/c) - I*sin((2*c*d - b
*e)/c))*erf(x*conjugate(sqrt(-c*log(f))) - 1/2*(b*log(f) - 2*I*e)*conjugate(1/sqrt(-c*log(f))))*e^(e^2/(c*log(
f))) + f^a*(cos((2*c*d - b*e)/c) + I*sin((2*c*d - b*e)/c))*erf(1/2*(2*c*x*log(f) + b*log(f) + 2*I*e)*sqrt(-c*l
og(f))/(c*log(f)))*e^(e^2/(c*log(f))) + f^a*(cos((2*c*d - b*e)/c) - I*sin((2*c*d - b*e)/c))*erf(1/2*(2*c*x*log
(f) + b*log(f) - 2*I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(e^2/(c*log(f))) + 2*f^a*erf(-1/2*b*conjugate(1/sqrt(-c*
log(f)))*log(f) + x*conjugate(sqrt(-c*log(f)))) - 2*f^a*erf(1/2*(2*c*x*log(f) + b*log(f))/sqrt(-c*log(f))))/(s
qrt(-c*log(f))*f^(1/4*b^2/c))

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Fricas [A]
time = 2.39, size = 226, normalized size = 0.98 \begin {gather*} -\frac {\sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) - 2 i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + 4 \, {\left (2 i \, c d - i \, b e\right )} \log \left (f\right ) - 4 \, e^{2}}{4 \, c \log \left (f\right )}\right )} + \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) + 2 i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + 4 \, {\left (-2 i \, c d + i \, b e\right )} \log \left (f\right ) - 4 \, e^{2}}{4 \, c \log \left (f\right )}\right )} + \frac {2 \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{8 \, c \log \left (f\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(sqrt(pi)*sqrt(-c*log(f))*erf(1/2*((2*c*x + b)*log(f) - 2*I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(-1/4*((b^2
- 4*a*c)*log(f)^2 + 4*(2*I*c*d - I*b*e)*log(f) - 4*e^2)/(c*log(f))) + sqrt(pi)*sqrt(-c*log(f))*erf(1/2*((2*c*x
+ b)*log(f) + 2*I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(-1/4*((b^2 - 4*a*c)*log(f)^2 + 4*(-2*I*c*d + I*b*e)*log(f
) - 4*e^2)/(c*log(f))) + 2*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*(2*c*x + b)*sqrt(-c*log(f))/c)/f^(1/4*(b^2 - 4*a*c
)/c))/(c*log(f))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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