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

Optimal. Leaf size=171 \[ -\frac {\sqrt {\pi } f^a e^{\frac {e^2}{c \log (f)}-2 i d} \text {erfi}\left (\frac {-c x \log (f)+i e}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a e^{\frac {e^2}{c \log (f)}+2 i d} \text {erfi}\left (\frac {c x \log (f)+i e}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[Out]

1/8*exp(-2*I*d+e^2/c/ln(f))*f^a*erfi((-I*e+c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)+1/8*ex
p(2*I*d+e^2/c/ln(f))*f^a*erfi((I*e+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*erfi(x
*c^(1/2)*ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4473, 2204, 2287, 2234} \[ -\frac {\sqrt {\pi } f^a e^{\frac {e^2}{c \log (f)}-2 i d} \text {Erfi}\left (\frac {-c x \log (f)+i e}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a e^{\frac {e^2}{c \log (f)}+2 i d} \text {Erfi}\left (\frac {c x \log (f)+i e}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a \text {Erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + c*x^2)*Cos[d + e*x]^2,x]

[Out]

(f^a*Sqrt[Pi]*Erfi[Sqrt[c]*x*Sqrt[Log[f]]])/(4*Sqrt[c]*Sqrt[Log[f]]) - (E^((-2*I)*d + e^2/(c*Log[f]))*f^a*Sqrt
[Pi]*Erfi[(I*e - c*x*Log[f])/(Sqrt[c]*Sqrt[Log[f]])])/(8*Sqrt[c]*Sqrt[Log[f]]) + (E^((2*I)*d + e^2/(c*Log[f]))
*f^a*Sqrt[Pi]*Erfi[(I*e + c*x*Log[f])/(Sqrt[c]*Sqrt[Log[f]])])/(8*Sqrt[c]*Sqrt[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 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+c x^2} \cos ^2(d+e x) \, dx &=\int \left (\frac {1}{2} f^{a+c x^2}+\frac {1}{4} e^{-2 i d-2 i e x} f^{a+c x^2}+\frac {1}{4} e^{2 i d+2 i e x} f^{a+c x^2}\right ) \, dx\\ &=\frac {1}{4} \int e^{-2 i d-2 i e x} f^{a+c x^2} \, dx+\frac {1}{4} \int e^{2 i d+2 i e x} f^{a+c x^2} \, dx+\frac {1}{2} \int f^{a+c x^2} \, dx\\ &=\frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {1}{4} \int e^{-2 i d-2 i e x+a \log (f)+c x^2 \log (f)} \, dx+\frac {1}{4} \int e^{2 i d+2 i e x+a \log (f)+c x^2 \log (f)} \, dx\\ &=\frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {1}{4} \left (e^{-2 i d+\frac {e^2}{c \log (f)}} f^a\right ) \int e^{\frac {(-2 i e+2 c x \log (f))^2}{4 c \log (f)}} \, dx+\frac {1}{4} \left (e^{2 i d+\frac {e^2}{c \log (f)}} f^a\right ) \int e^{\frac {(2 i e+2 c x \log (f))^2}{4 c \log (f)}} \, dx\\ &=\frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-2 i d+\frac {e^2}{c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e-c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{2 i d+\frac {e^2}{c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 131, normalized size = 0.77 \[ \frac {\sqrt {\pi } f^a \left (2 \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )+e^{\frac {e^2}{c \log (f)}} \left ((\cos (2 d)-i \sin (2 d)) \text {erfi}\left (\frac {c x \log (f)-i e}{\sqrt {c} \sqrt {\log (f)}}\right )+(\cos (2 d)+i \sin (2 d)) \text {erfi}\left (\frac {c x \log (f)+i e}{\sqrt {c} \sqrt {\log (f)}}\right )\right )\right )}{8 \sqrt {c} \sqrt {\log (f)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 159, normalized size = 0.93 \[ -\frac {2 \, \sqrt {\pi } \sqrt {-c \log \relax (f)} f^{a} \operatorname {erf}\left (\sqrt {-c \log \relax (f)} x\right ) + \sqrt {\pi } \sqrt {-c \log \relax (f)} \operatorname {erf}\left (\frac {{\left (c x \log \relax (f) + i \, e\right )} \sqrt {-c \log \relax (f)}}{c \log \relax (f)}\right ) e^{\left (\frac {a c \log \relax (f)^{2} + 2 i \, c d \log \relax (f) + e^{2}}{c \log \relax (f)}\right )} + \sqrt {\pi } \sqrt {-c \log \relax (f)} \operatorname {erf}\left (\frac {{\left (c x \log \relax (f) - i \, e\right )} \sqrt {-c \log \relax (f)}}{c \log \relax (f)}\right ) e^{\left (\frac {a c \log \relax (f)^{2} - 2 i \, c d \log \relax (f) + e^{2}}{c \log \relax (f)}\right )}}{8 \, c \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.24, size = 145, normalized size = 0.85 \[ \frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {2 i d \ln \relax (f ) c -e^{2}}{\ln \relax (f ) c}} \erf \left (\sqrt {-c \ln \relax (f )}\, x +\frac {i e}{\sqrt {-c \ln \relax (f )}}\right )}{8 \sqrt {-c \ln \relax (f )}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {2 i d \ln \relax (f ) c +e^{2}}{\ln \relax (f ) c}} \erf \left (-\sqrt {-c \ln \relax (f )}\, x +\frac {i e}{\sqrt {-c \ln \relax (f )}}\right )}{8 \sqrt {-c \ln \relax (f )}}+\frac {f^{a} \sqrt {\pi }\, \erf \left (\sqrt {-c \ln \relax (f )}\, x \right )}{4 \sqrt {-c \ln \relax (f )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.36, size = 236, normalized size = 1.38 \[ \frac {\sqrt {\pi } {\left (f^{a} {\left (\cos \left (2 \, d\right ) - i \, \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \relax (f)}} + i \, e \overline {\frac {1}{\sqrt {-c \log \relax (f)}}}\right ) e^{\left (\frac {e^{2}}{c \log \relax (f)}\right )} + f^{a} {\left (\cos \left (2 \, d\right ) + i \, \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \relax (f)}} - i \, e \overline {\frac {1}{\sqrt {-c \log \relax (f)}}}\right ) e^{\left (\frac {e^{2}}{c \log \relax (f)}\right )} - f^{a} {\left (\cos \left (2 \, d\right ) + i \, \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (\frac {c x \log \relax (f) + i \, e}{\sqrt {-c \log \relax (f)}}\right ) e^{\left (\frac {e^{2}}{c \log \relax (f)}\right )} - f^{a} {\left (\cos \left (2 \, d\right ) - i \, \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (\frac {c x \log \relax (f) - i \, e}{\sqrt {-c \log \relax (f)}}\right ) e^{\left (\frac {e^{2}}{c \log \relax (f)}\right )} + 2 \, f^{a} \operatorname {erf}\left (x \overline {\sqrt {-c \log \relax (f)}}\right ) + 2 \, f^{a} \operatorname {erf}\left (\sqrt {-c \log \relax (f)} x\right )\right )}}{16 \, \sqrt {-c \log \relax (f)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + c*x^2)*cos(d + e*x)^2,x)

[Out]

int(f^(a + c*x^2)*cos(d + e*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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