3.111 \(\int f^{a+b x} \cos ^2(d+f x^2) \, dx\)

Optimal. Leaf size=157 \[ \left (-\frac{1}{16}-\frac{i}{16}\right ) \sqrt{\pi } f^{a-\frac{1}{2}} e^{\frac{i b^2 \log ^2(f)}{8 f}+2 i d} \text{Erf}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (b \log (f)+4 i f x)}{\sqrt{f}}\right )-\left (\frac{1}{16}+\frac{i}{16}\right ) \sqrt{\pi } f^{a-\frac{1}{2}} e^{-\frac{1}{8} i \left (\frac{b^2 \log ^2(f)}{f}+16 d\right )} \text{Erfi}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (-b \log (f)+4 i f x)}{\sqrt{f}}\right )+\frac{f^{a+b x}}{2 b \log (f)} \]

[Out]

(-1/16 - I/16)*E^((2*I)*d + ((I/8)*b^2*Log[f]^2)/f)*f^(-1/2 + a)*Sqrt[Pi]*Erf[((1/4 + I/4)*((4*I)*f*x + b*Log[
f]))/Sqrt[f]] - ((1/16 + I/16)*f^(-1/2 + a)*Sqrt[Pi]*Erfi[((1/4 + I/4)*((4*I)*f*x - b*Log[f]))/Sqrt[f]])/E^((I
/8)*(16*d + (b^2*Log[f]^2)/f)) + f^(a + b*x)/(2*b*Log[f])

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Rubi [A]  time = 0.173463, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4473, 2194, 2287, 2234, 2204, 2205} \[ \left (-\frac{1}{16}-\frac{i}{16}\right ) \sqrt{\pi } f^{a-\frac{1}{2}} e^{\frac{i b^2 \log ^2(f)}{8 f}+2 i d} \text{Erf}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (b \log (f)+4 i f x)}{\sqrt{f}}\right )-\left (\frac{1}{16}+\frac{i}{16}\right ) \sqrt{\pi } f^{a-\frac{1}{2}} e^{-\frac{1}{8} i \left (\frac{b^2 \log ^2(f)}{f}+16 d\right )} \text{Erfi}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (-b \log (f)+4 i f x)}{\sqrt{f}}\right )+\frac{f^{a+b x}}{2 b \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x)*Cos[d + f*x^2]^2,x]

[Out]

(-1/16 - I/16)*E^((2*I)*d + ((I/8)*b^2*Log[f]^2)/f)*f^(-1/2 + a)*Sqrt[Pi]*Erf[((1/4 + I/4)*((4*I)*f*x + b*Log[
f]))/Sqrt[f]] - ((1/16 + I/16)*f^(-1/2 + a)*Sqrt[Pi]*Erfi[((1/4 + I/4)*((4*I)*f*x - b*Log[f]))/Sqrt[f]])/E^((I
/8)*(16*d + (b^2*Log[f]^2)/f)) + f^(a + b*x)/(2*b*Log[f])

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]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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 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 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]

Rubi steps

\begin{align*} \int f^{a+b x} \cos ^2\left (d+f x^2\right ) \, dx &=\int \left (\frac{1}{2} f^{a+b x}+\frac{1}{4} e^{-2 i d-2 i f x^2} f^{a+b x}+\frac{1}{4} e^{2 i d+2 i f x^2} f^{a+b x}\right ) \, dx\\ &=\frac{1}{4} \int e^{-2 i d-2 i f x^2} f^{a+b x} \, dx+\frac{1}{4} \int e^{2 i d+2 i f x^2} f^{a+b x} \, dx+\frac{1}{2} \int f^{a+b x} \, dx\\ &=\frac{f^{a+b x}}{2 b \log (f)}+\frac{1}{4} \int e^{-2 i d-2 i f x^2+a \log (f)+b x \log (f)} \, dx+\frac{1}{4} \int e^{2 i d+2 i f x^2+a \log (f)+b x \log (f)} \, dx\\ &=\frac{f^{a+b x}}{2 b \log (f)}+\frac{1}{4} \left (e^{2 i d+\frac{i b^2 \log ^2(f)}{8 f}} f^a\right ) \int e^{-\frac{i (4 i f x+b \log (f))^2}{8 f}} \, dx+\frac{1}{4} \left (e^{-\frac{1}{8} i \left (16 d+\frac{b^2 \log ^2(f)}{f}\right )} f^a\right ) \int e^{\frac{i (-4 i f x+b \log (f))^2}{8 f}} \, dx\\ &=\left (-\frac{1}{16}-\frac{i}{16}\right ) e^{2 i d+\frac{i b^2 \log ^2(f)}{8 f}} f^{-\frac{1}{2}+a} \sqrt{\pi } \text{erf}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (4 i f x+b \log (f))}{\sqrt{f}}\right )-\left (\frac{1}{16}+\frac{i}{16}\right ) e^{-\frac{1}{8} i \left (16 d+\frac{b^2 \log ^2(f)}{f}\right )} f^{-\frac{1}{2}+a} \sqrt{\pi } \text{erfi}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (4 i f x-b \log (f))}{\sqrt{f}}\right )+\frac{f^{a+b x}}{2 b \log (f)}\\ \end{align*}

Mathematica [A]  time = 1.09174, size = 158, normalized size = 1.01 \[ \frac{1}{16} f^a \left (\frac{(1-i) \sqrt{\pi } e^{-\frac{i b^2 \log ^2(f)}{8 f}} (\cos (d)-i \sin (d))^2 \text{Erf}\left (\frac{(4+4 i) f x-(1-i) b \log (f)}{4 \sqrt{f}}\right )}{\sqrt{f}}+\frac{(1+i) \sqrt{\pi } e^{\frac{i b^2 \log ^2(f)}{8 f}} (\sin (2 d)-i \cos (2 d)) \text{Erfi}\left (\frac{(1-i) b \log (f)+(4+4 i) f x}{4 \sqrt{f}}\right )}{\sqrt{f}}+\frac{8 f^{b x}}{b \log (f)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^a*((8*f^(b*x))/(b*Log[f]) + ((1 - I)*Sqrt[Pi]*Erf[((4 + 4*I)*f*x - (1 - I)*b*Log[f])/(4*Sqrt[f])]*(Cos[d] -
I*Sin[d])^2)/(E^(((I/8)*b^2*Log[f]^2)/f)*Sqrt[f]) + ((1 + I)*E^(((I/8)*b^2*Log[f]^2)/f)*Sqrt[Pi]*Erfi[((4 + 4
*I)*f*x + (1 - I)*b*Log[f])/(4*Sqrt[f])]*((-I)*Cos[2*d] + Sin[2*d]))/Sqrt[f]))/16

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Maple [A]  time = 0.109, size = 139, normalized size = 0.9 \begin{align*} -{\frac{\sqrt{2}{f}^{a}\sqrt{\pi }}{16}{{\rm e}^{{\frac{-{\frac{i}{8}} \left ( \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+16\,df \right ) }{f}}}}{\it Erf} \left ( -\sqrt{2}\sqrt{if}x+{\frac{\ln \left ( f \right ) b\sqrt{2}}{4}{\frac{1}{\sqrt{if}}}} \right ){\frac{1}{\sqrt{if}}}}-{\frac{{f}^{a}\sqrt{\pi }}{8}{{\rm e}^{{\frac{{\frac{i}{8}} \left ( \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+16\,df \right ) }{f}}}}{\it Erf} \left ( -\sqrt{-2\,if}x+{\frac{b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{-2\,if}}}} \right ){\frac{1}{\sqrt{-2\,if}}}}+{\frac{{f}^{bx+a}}{2\,b\ln \left ( f \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*Pi^(1/2)*f^a*exp(-1/8*I*(ln(f)^2*b^2+16*d*f)/f)*2^(1/2)/(I*f)^(1/2)*erf(-2^(1/2)*(I*f)^(1/2)*x+1/4*ln(f)
*b*2^(1/2)/(I*f)^(1/2))-1/8*Pi^(1/2)*f^a*exp(1/8*I*(ln(f)^2*b^2+16*d*f)/f)/(-2*I*f)^(1/2)*erf(-(-2*I*f)^(1/2)*
x+1/2*ln(f)*b/(-2*I*f)^(1/2))+1/2*f^(b*x+a)/b/ln(f)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: IndexError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: IndexError

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Fricas [B]  time = 0.516288, size = 768, normalized size = 4.89 \begin{align*} \frac{2 \, \pi b \sqrt{\frac{f}{\pi }} e^{\left (\frac{-i \, b^{2} \log \left (f\right )^{2} + 8 \, a f \log \left (f\right ) - 16 i \, d f}{8 \, f}\right )} \operatorname{C}\left (\frac{{\left (4 \, f x + i \, b \log \left (f\right )\right )} \sqrt{\frac{f}{\pi }}}{2 \, f}\right ) \log \left (f\right ) - 2 \, \pi b \sqrt{\frac{f}{\pi }} e^{\left (\frac{i \, b^{2} \log \left (f\right )^{2} + 8 \, a f \log \left (f\right ) + 16 i \, d f}{8 \, f}\right )} \operatorname{C}\left (-\frac{{\left (4 \, f x - i \, b \log \left (f\right )\right )} \sqrt{\frac{f}{\pi }}}{2 \, f}\right ) \log \left (f\right ) - 2 i \, \pi b \sqrt{\frac{f}{\pi }} e^{\left (\frac{-i \, b^{2} \log \left (f\right )^{2} + 8 \, a f \log \left (f\right ) - 16 i \, d f}{8 \, f}\right )} \operatorname{S}\left (\frac{{\left (4 \, f x + i \, b \log \left (f\right )\right )} \sqrt{\frac{f}{\pi }}}{2 \, f}\right ) \log \left (f\right ) - 2 i \, \pi b \sqrt{\frac{f}{\pi }} e^{\left (\frac{i \, b^{2} \log \left (f\right )^{2} + 8 \, a f \log \left (f\right ) + 16 i \, d f}{8 \, f}\right )} \operatorname{S}\left (-\frac{{\left (4 \, f x - i \, b \log \left (f\right )\right )} \sqrt{\frac{f}{\pi }}}{2 \, f}\right ) \log \left (f\right ) + 8 \, f f^{b x + a}}{16 \, b f \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*pi*b*sqrt(f/pi)*e^(1/8*(-I*b^2*log(f)^2 + 8*a*f*log(f) - 16*I*d*f)/f)*fresnel_cos(1/2*(4*f*x + I*b*log
(f))*sqrt(f/pi)/f)*log(f) - 2*pi*b*sqrt(f/pi)*e^(1/8*(I*b^2*log(f)^2 + 8*a*f*log(f) + 16*I*d*f)/f)*fresnel_cos
(-1/2*(4*f*x - I*b*log(f))*sqrt(f/pi)/f)*log(f) - 2*I*pi*b*sqrt(f/pi)*e^(1/8*(-I*b^2*log(f)^2 + 8*a*f*log(f) -
16*I*d*f)/f)*fresnel_sin(1/2*(4*f*x + I*b*log(f))*sqrt(f/pi)/f)*log(f) - 2*I*pi*b*sqrt(f/pi)*e^(1/8*(I*b^2*lo
g(f)^2 + 8*a*f*log(f) + 16*I*d*f)/f)*fresnel_sin(-1/2*(4*f*x - I*b*log(f))*sqrt(f/pi)/f)*log(f) + 8*f*f^(b*x +
a))/(b*f*log(f))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.34673, size = 703, normalized size = 4.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*cos(-1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a*sgn(f) + 1/2*pi*a)*log(abs(f))/(4*b^2*log(abs(f))^2 + (pi*
b*sgn(f) - pi*b)^2) - (pi*b*sgn(f) - pi*b)*sin(-1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a*sgn(f) + 1/2*pi*a)/(
4*b^2*log(abs(f))^2 + (pi*b*sgn(f) - pi*b)^2))*e^(b*x*log(abs(f)) + a*log(abs(f))) - 1/2*I*(-2*I*e^(1/2*I*pi*b
*x*sgn(f) - 1/2*I*pi*b*x + 1/2*I*pi*a*sgn(f) - 1/2*I*pi*a)/(2*I*pi*b*sgn(f) - 2*I*pi*b + 4*b*log(abs(f))) + 2*
I*e^(-1/2*I*pi*b*x*sgn(f) + 1/2*I*pi*b*x - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a)/(-2*I*pi*b*sgn(f) + 2*I*pi*b + 4*b*
log(abs(f))))*e^(b*x*log(abs(f)) + a*log(abs(f))) - 1/8*sqrt(pi)*erf(-1/8*sqrt(f)*(8*x - (pi*b*sgn(f) - pi*b +
2*I*b*log(abs(f)))/f)*(-I*f/abs(f) + 1))*e^(1/16*I*pi^2*b^2*sgn(f)/f + 1/8*pi*b^2*log(abs(f))*sgn(f)/f - 1/16
*I*pi^2*b^2/f - 1/8*pi*b^2*log(abs(f))/f + 1/8*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a + a*log(
abs(f)) + 2*I*d)/(sqrt(f)*(-I*f/abs(f) + 1)) - 1/8*sqrt(pi)*erf(-1/8*sqrt(f)*(8*x + (pi*b*sgn(f) - pi*b + 2*I*
b*log(abs(f)))/f)*(I*f/abs(f) + 1))*e^(-1/16*I*pi^2*b^2*sgn(f)/f - 1/8*pi*b^2*log(abs(f))*sgn(f)/f + 1/16*I*pi
^2*b^2/f + 1/8*pi*b^2*log(abs(f))/f - 1/8*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a + a*log(abs(f
)) - 2*I*d)/(sqrt(f)*(I*f/abs(f) + 1))