3.2.15 \(\int f^{a+b x} \cos ^3(d+e x+f x^2) \, dx\) [115]

3.2.15.1 Optimal result
3.2.15.2 Mathematica [A] (verified)
3.2.15.3 Rubi [A] (verified)
3.2.15.4 Maple [A] (verified)
3.2.15.5 Fricas [B] (verification not implemented)
3.2.15.6 Sympy [F]
3.2.15.7 Maxima [A] (verification not implemented)
3.2.15.8 Giac [B] (verification not implemented)
3.2.15.9 Mupad [F(-1)]

3.2.15.1 Optimal result

Integrand size = 21, antiderivative size = 340 \[ \int f^{a+b x} \cos ^3\left (d+e x+f x^2\right ) \, dx=-\frac {3}{16} \sqrt [4]{-1} e^{\frac {1}{4} i \left (4 d+\frac {(i e+b \log (f))^2}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt [4]{-1} (i e+2 i f x+b \log (f))}{2 \sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) e^{3 i d+\frac {i (3 i e+b \log (f))^2}{12 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{6}} \text {erf}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (3 i e+6 i f x+b \log (f))}{\sqrt {6} \sqrt {f}}\right )-\frac {3}{16} \sqrt [4]{-1} e^{-i d+\frac {i (e+i b \log (f))^2}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt [4]{-1} (i e+2 i f x-b \log (f))}{2 \sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) e^{-3 i d+\frac {i (3 e+i b \log (f))^2}{12 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{6}} \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (3 i e+6 i f x-b \log (f))}{\sqrt {6} \sqrt {f}}\right ) \]

output
(-1/96-1/96*I)*exp(3*I*d+1/12*I*(3*I*e+b*ln(f))^2/f)*f^(-1/2+a)*erf((1/12+ 
1/12*I)*(3*I*e+6*I*f*x+b*ln(f))*6^(1/2)/f^(1/2))*6^(1/2)*Pi^(1/2)-(1/96+1/ 
96*I)*exp(-3*I*d+1/12*I*(3*e+I*b*ln(f))^2/f)*f^(-1/2+a)*erfi((1/12+1/12*I) 
*(3*I*e+6*I*f*x-b*ln(f))*6^(1/2)/f^(1/2))*6^(1/2)*Pi^(1/2)-3/16*(-1)^(1/4) 
*exp(1/4*I*(4*d+(I*e+b*ln(f))^2/f))*f^(-1/2+a)*erf(1/2*(-1)^(1/4)*(I*e+2*I 
*f*x+b*ln(f))/f^(1/2))*Pi^(1/2)-3/16*(-1)^(1/4)*exp(-I*d+1/4*I*(e+I*b*ln(f 
))^2/f)*f^(-1/2+a)*erfi(1/2*(-1)^(1/4)*(I*e+2*I*f*x-b*ln(f))/f^(1/2))*Pi^( 
1/2)
 
3.2.15.2 Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 322, normalized size of antiderivative = 0.95 \[ \int f^{a+b x} \cos ^3\left (d+e x+f x^2\right ) \, dx=\frac {1}{48} \sqrt [4]{-1} e^{-\frac {i \left (3 e^2+b^2 \log ^2(f)\right )}{4 f}} f^{a-\frac {b e+f}{2 f}} \sqrt {\pi } \left (9 e^{\frac {i \left (e^2+b^2 \log ^2(f)\right )}{2 f}} \text {erfi}\left (\frac {\sqrt [4]{-1} (e+2 f x-i b \log (f))}{2 \sqrt {f}}\right ) (-i \cos (d)+\sin (d))+e^{\frac {i e^2}{f}} \left (-9 \text {erfi}\left (\frac {(-1)^{3/4} (e+2 f x+i b \log (f))}{2 \sqrt {f}}\right ) (\cos (d)-i \sin (d))-\sqrt {3} e^{\frac {i \left (3 e^2+b^2 \log ^2(f)\right )}{6 f}} \text {erfi}\left (\frac {(-1)^{3/4} (3 e+6 f x+i b \log (f))}{2 \sqrt {3} \sqrt {f}}\right ) (\cos (3 d)-i \sin (3 d))\right )+\sqrt {3} e^{\frac {i b^2 \log ^2(f)}{3 f}} \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (3 e+6 f x-i b \log (f))}{\sqrt {6} \sqrt {f}}\right ) (-i \cos (3 d)+\sin (3 d))\right ) \]

input
Integrate[f^(a + b*x)*Cos[d + e*x + f*x^2]^3,x]
 
output
((-1)^(1/4)*f^(a - (b*e + f)/(2*f))*Sqrt[Pi]*(9*E^(((I/2)*(e^2 + b^2*Log[f 
]^2))/f)*Erfi[((-1)^(1/4)*(e + 2*f*x - I*b*Log[f]))/(2*Sqrt[f])]*((-I)*Cos 
[d] + Sin[d]) + E^((I*e^2)/f)*(-9*Erfi[((-1)^(3/4)*(e + 2*f*x + I*b*Log[f] 
))/(2*Sqrt[f])]*(Cos[d] - I*Sin[d]) - Sqrt[3]*E^(((I/6)*(3*e^2 + b^2*Log[f 
]^2))/f)*Erfi[((-1)^(3/4)*(3*e + 6*f*x + I*b*Log[f]))/(2*Sqrt[3]*Sqrt[f])] 
*(Cos[3*d] - I*Sin[3*d])) + Sqrt[3]*E^(((I/3)*b^2*Log[f]^2)/f)*Erfi[((1/2 
+ I/2)*(3*e + 6*f*x - I*b*Log[f]))/(Sqrt[6]*Sqrt[f])]*((-I)*Cos[3*d] + Sin 
[3*d])))/(48*E^(((I/4)*(3*e^2 + b^2*Log[f]^2))/f))
 
3.2.15.3 Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4976, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int f^{a+b x} \cos ^3\left (d+e x+f x^2\right ) \, dx\)

\(\Big \downarrow \) 4976

\(\displaystyle \int \left (\frac {3}{8} f^{a+b x} \exp \left (-3 i \left (d+e x+f x^2\right )+2 i d+2 i e x+2 i f x^2\right )+\frac {3}{8} f^{a+b x} \exp \left (-3 i \left (d+e x+f x^2\right )+4 i d+4 i e x+4 i f x^2\right )+\frac {1}{8} f^{a+b x} \exp \left (-3 i \left (d+e x+f x^2\right )+6 i d+6 i e x+6 i f x^2\right )+\frac {1}{8} f^{a+b x} e^{-3 i \left (d+e x+f x^2\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3}{16} \sqrt [4]{-1} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {1}{4} i \left (4 d+\frac {(b \log (f)+i e)^2}{f}\right )} \text {erf}\left (\frac {\sqrt [4]{-1} (b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{\frac {i (b \log (f)+3 i e)^2}{12 f}+3 i d} \text {erf}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b \log (f)+3 i e+6 i f x)}{\sqrt {6} \sqrt {f}}\right )-\frac {3}{16} \sqrt [4]{-1} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {i (e+i b \log (f))^2}{4 f}-i d} \text {erfi}\left (\frac {\sqrt [4]{-1} (-b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{\frac {i (3 e+i b \log (f))^2}{12 f}-3 i d} \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (-b \log (f)+3 i e+6 i f x)}{\sqrt {6} \sqrt {f}}\right )\)

input
Int[f^(a + b*x)*Cos[d + e*x + f*x^2]^3,x]
 
output
(-3*(-1)^(1/4)*E^((I/4)*(4*d + (I*e + b*Log[f])^2/f))*f^(-1/2 + a)*Sqrt[Pi 
]*Erf[((-1)^(1/4)*(I*e + (2*I)*f*x + b*Log[f]))/(2*Sqrt[f])])/16 - (1/16 + 
 I/16)*E^((3*I)*d + ((I/12)*((3*I)*e + b*Log[f])^2)/f)*f^(-1/2 + a)*Sqrt[P 
i/6]*Erf[((1/2 + I/2)*((3*I)*e + (6*I)*f*x + b*Log[f]))/(Sqrt[6]*Sqrt[f])] 
 - (3*(-1)^(1/4)*E^((-I)*d + ((I/4)*(e + I*b*Log[f])^2)/f)*f^(-1/2 + a)*Sq 
rt[Pi]*Erfi[((-1)^(1/4)*(I*e + (2*I)*f*x - b*Log[f]))/(2*Sqrt[f])])/16 - ( 
1/16 + I/16)*E^((-3*I)*d + ((I/12)*(3*e + I*b*Log[f])^2)/f)*f^(-1/2 + a)*S 
qrt[Pi/6]*Erfi[((1/2 + I/2)*((3*I)*e + (6*I)*f*x - b*Log[f]))/(Sqrt[6]*Sqr 
t[f])]
 

3.2.15.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4976
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]
 
3.2.15.4 Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 315, normalized size of antiderivative = 0.93

method result size
risch \(-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{-\frac {i \left (\ln \left (f \right )^{2} b^{2}+36 d f -9 e^{2}\right )}{12 f}} \sqrt {3}\, \operatorname {erf}\left (-\sqrt {3}\, \sqrt {i f}\, x +\frac {\left (b \ln \left (f \right )-3 i e \right ) \sqrt {3}}{6 \sqrt {i f}}\right )}{48 \sqrt {i f}}-\frac {3 \sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{-\frac {i \left (\ln \left (f \right )^{2} b^{2}+4 d f -e^{2}\right )}{4 f}} \operatorname {erf}\left (-\sqrt {i f}\, x +\frac {b \ln \left (f \right )-i e}{2 \sqrt {i f}}\right )}{16 \sqrt {i f}}-\frac {3 \sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{\frac {i \left (\ln \left (f \right )^{2} b^{2}+4 d f -e^{2}\right )}{4 f}} \operatorname {erf}\left (-\sqrt {-i f}\, x +\frac {i e +b \ln \left (f \right )}{2 \sqrt {-i f}}\right )}{16 \sqrt {-i f}}-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{\frac {i \left (\ln \left (f \right )^{2} b^{2}+36 d f -9 e^{2}\right )}{12 f}} \operatorname {erf}\left (-\sqrt {-3 i f}\, x +\frac {3 i e +b \ln \left (f \right )}{2 \sqrt {-3 i f}}\right )}{16 \sqrt {-3 i f}}\) \(315\)

input
int(f^(b*x+a)*cos(f*x^2+e*x+d)^3,x,method=_RETURNVERBOSE)
 
output
-1/48*Pi^(1/2)*f^a*f^(-1/2/f*b*e)*exp(-1/12*I*(ln(f)^2*b^2+36*d*f-9*e^2)/f 
)*3^(1/2)/(I*f)^(1/2)*erf(-3^(1/2)*(I*f)^(1/2)*x+1/6*(b*ln(f)-3*I*e)*3^(1/ 
2)/(I*f)^(1/2))-3/16*Pi^(1/2)*f^a*f^(-1/2/f*b*e)*exp(-1/4*I*(ln(f)^2*b^2+4 
*d*f-e^2)/f)/(I*f)^(1/2)*erf(-(I*f)^(1/2)*x+1/2*(b*ln(f)-I*e)/(I*f)^(1/2)) 
-3/16*Pi^(1/2)*f^a*f^(-1/2/f*b*e)*exp(1/4*I*(ln(f)^2*b^2+4*d*f-e^2)/f)/(-I 
*f)^(1/2)*erf(-(-I*f)^(1/2)*x+1/2*(I*e+b*ln(f))/(-I*f)^(1/2))-1/16*Pi^(1/2 
)*f^a*f^(-1/2/f*b*e)*exp(1/12*I*(ln(f)^2*b^2+36*d*f-9*e^2)/f)/(-3*I*f)^(1/ 
2)*erf(-(-3*I*f)^(1/2)*x+1/2*(3*I*e+b*ln(f))/(-3*I*f)^(1/2))
 
3.2.15.5 Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 629 vs. \(2 (220) = 440\).

Time = 0.27 (sec) , antiderivative size = 629, normalized size of antiderivative = 1.85 \[ \int f^{a+b x} \cos ^3\left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \left (f\right )^{2} + 9 i \, e^{2} - 36 i \, d f - 6 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{12 \, f}\right )} \operatorname {C}\left (\frac {\sqrt {6} {\left (6 \, f x + i \, b \log \left (f\right ) + 3 \, e\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) - \sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \left (f\right )^{2} - 9 i \, e^{2} + 36 i \, d f - 6 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{12 \, f}\right )} \operatorname {C}\left (-\frac {\sqrt {6} {\left (6 \, f x - i \, b \log \left (f\right ) + 3 \, e\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) + 9 \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \left (f\right )^{2} + i \, e^{2} - 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right )} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \left (f\right ) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - 9 \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \left (f\right )^{2} - i \, e^{2} + 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right )} \operatorname {C}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \left (f\right ) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - i \, \sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \left (f\right )^{2} + 9 i \, e^{2} - 36 i \, d f - 6 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{12 \, f}\right )} \operatorname {S}\left (\frac {\sqrt {6} {\left (6 \, f x + i \, b \log \left (f\right ) + 3 \, e\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) - i \, \sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \left (f\right )^{2} - 9 i \, e^{2} + 36 i \, d f - 6 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{12 \, f}\right )} \operatorname {S}\left (-\frac {\sqrt {6} {\left (6 \, f x - i \, b \log \left (f\right ) + 3 \, e\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) - 9 i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \left (f\right )^{2} + i \, e^{2} - 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right )} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \left (f\right ) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - 9 i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \left (f\right )^{2} - i \, e^{2} + 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right )} \operatorname {S}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \left (f\right ) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right )}{48 \, f} \]

input
integrate(f^(b*x+a)*cos(f*x^2+e*x+d)^3,x, algorithm="fricas")
 
output
1/48*(sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(-I*b^2*log(f)^2 + 9*I*e^2 - 36*I*d*f 
- 6*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(1/6*sqrt(6)*(6*f*x + I*b*log(f) + 
 3*e)*sqrt(f/pi)/f) - sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(I*b^2*log(f)^2 - 9*I* 
e^2 + 36*I*d*f - 6*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(-1/6*sqrt(6)*(6*f* 
x - I*b*log(f) + 3*e)*sqrt(f/pi)/f) + 9*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b 
^2*log(f)^2 + I*e^2 - 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(1/2 
*sqrt(2)*(2*f*x + I*b*log(f) + e)*sqrt(f/pi)/f) - 9*sqrt(2)*pi*sqrt(f/pi)* 
e^(1/4*(I*b^2*log(f)^2 - I*e^2 + 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fres 
nel_cos(-1/2*sqrt(2)*(2*f*x - I*b*log(f) + e)*sqrt(f/pi)/f) - I*sqrt(6)*pi 
*sqrt(f/pi)*e^(1/12*(-I*b^2*log(f)^2 + 9*I*e^2 - 36*I*d*f - 6*(b*e - 2*a*f 
)*log(f))/f)*fresnel_sin(1/6*sqrt(6)*(6*f*x + I*b*log(f) + 3*e)*sqrt(f/pi) 
/f) - I*sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(I*b^2*log(f)^2 - 9*I*e^2 + 36*I*d*f 
 - 6*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(-1/6*sqrt(6)*(6*f*x - I*b*log(f) 
 + 3*e)*sqrt(f/pi)/f) - 9*I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*log(f)^2 
+ I*e^2 - 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(1/2*sqrt(2)*(2* 
f*x + I*b*log(f) + e)*sqrt(f/pi)/f) - 9*I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(I* 
b^2*log(f)^2 - I*e^2 + 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(-1 
/2*sqrt(2)*(2*f*x - I*b*log(f) + e)*sqrt(f/pi)/f))/f
 
3.2.15.6 Sympy [F]

\[ \int f^{a+b x} \cos ^3\left (d+e x+f x^2\right ) \, dx=\int f^{a + b x} \cos ^{3}{\left (d + e x + f x^{2} \right )}\, dx \]

input
integrate(f**(b*x+a)*cos(f*x**2+e*x+d)**3,x)
 
output
Integral(f**(a + b*x)*cos(d + e*x + f*x**2)**3, x)
 
3.2.15.7 Maxima [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.11 \[ \int f^{a+b x} \cos ^3\left (d+e x+f x^2\right ) \, dx=-\frac {9^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right ) - \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (6 i \, f x - b \log \left (f\right ) + 3 i \, e\right )} \sqrt {3 i \, f}}{6 \, f}\right ) + {\left (\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right ) + \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (6 i \, f x + b \log \left (f\right ) + 3 i \, e\right )} \sqrt {-3 i \, f}}{6 \, f}\right )\right )} f^{\frac {3}{2}} - 9 \, \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right ) + \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, f x - b \log \left (f\right ) + i \, e\right )} \sqrt {i \, f}}{2 \, f}\right ) + {\left (-\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right ) - \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, f x + b \log \left (f\right ) + i \, e\right )} \sqrt {-i \, f}}{2 \, f}\right )\right )} f^{\frac {3}{2}}}{96 \, f^{2} f^{\frac {b e}{2 \, f}}} \]

input
integrate(f^(b*x+a)*cos(f*x^2+e*x+d)^3,x, algorithm="maxima")
 
output
-1/96*(9^(1/4)*sqrt(2)*sqrt(pi)*((-(I - 1)*f^a*cos(1/12*(b^2*log(f)^2 - 9* 
e^2 + 36*d*f)/f) - (I + 1)*f^a*sin(1/12*(b^2*log(f)^2 - 9*e^2 + 36*d*f)/f) 
)*erf(1/6*I*(6*I*f*x - b*log(f) + 3*I*e)*sqrt(3*I*f)/f) + ((I + 1)*f^a*cos 
(1/12*(b^2*log(f)^2 - 9*e^2 + 36*d*f)/f) + (I - 1)*f^a*sin(1/12*(b^2*log(f 
)^2 - 9*e^2 + 36*d*f)/f))*erf(1/6*I*(6*I*f*x + b*log(f) + 3*I*e)*sqrt(-3*I 
*f)/f))*f^(3/2) - 9*sqrt(2)*sqrt(pi)*(((I - 1)*f^a*cos(1/4*(b^2*log(f)^2 - 
 e^2 + 4*d*f)/f) + (I + 1)*f^a*sin(1/4*(b^2*log(f)^2 - e^2 + 4*d*f)/f))*er 
f(1/2*I*(2*I*f*x - b*log(f) + I*e)*sqrt(I*f)/f) + (-(I + 1)*f^a*cos(1/4*(b 
^2*log(f)^2 - e^2 + 4*d*f)/f) - (I - 1)*f^a*sin(1/4*(b^2*log(f)^2 - e^2 + 
4*d*f)/f))*erf(1/2*I*(2*I*f*x + b*log(f) + I*e)*sqrt(-I*f)/f))*f^(3/2))/(f 
^2*f^(1/2*b*e/f))
 
3.2.15.8 Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 751 vs. \(2 (220) = 440\).

Time = 0.49 (sec) , antiderivative size = 751, normalized size of antiderivative = 2.21 \[ \int f^{a+b x} \cos ^3\left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \]

input
integrate(f^(b*x+a)*cos(f*x^2+e*x+d)^3,x, algorithm="giac")
 
output
3/16*I*sqrt(2)*sqrt(pi)*erf(-1/8*I*sqrt(2)*(4*x - (pi*b*sgn(f) - pi*b + 2* 
I*b*log(abs(f)) - 2*e)/f)*(I*f/abs(f) + 1)*sqrt(abs(f)))*e^(1/8*I*pi^2*b^2 
*sgn(f)/f + 1/4*pi*b^2*log(abs(f))*sgn(f)/f - 1/8*I*pi^2*b^2/f - 1/4*pi*b^ 
2*log(abs(f))/f + 1/4*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/4*I*pi 
*b*e*sgn(f)/f + 1/2*I*pi*a - 1/4*I*pi*b*e/f + a*log(abs(f)) - 1/2*b*e*log( 
abs(f))/f + I*d - 1/4*I*e^2/f)/((I*f/abs(f) + 1)*sqrt(abs(f))) + 1/48*I*sq 
rt(6)*sqrt(pi)*erf(-1/24*I*sqrt(6)*sqrt(f)*(12*x - (pi*b*sgn(f) - pi*b + 2 
*I*b*log(abs(f)) - 6*e)/f)*(I*f/abs(f) + 1))*e^(1/24*I*pi^2*b^2*sgn(f)/f + 
 1/12*pi*b^2*log(abs(f))*sgn(f)/f - 1/24*I*pi^2*b^2/f - 1/12*pi*b^2*log(ab 
s(f))/f + 1/12*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/4*I*pi*b*e*sg 
n(f)/f + 1/2*I*pi*a - 1/4*I*pi*b*e/f + a*log(abs(f)) - 1/2*b*e*log(abs(f)) 
/f + 3*I*d - 3/4*I*e^2/f)/(sqrt(f)*(I*f/abs(f) + 1)) - 1/48*I*sqrt(6)*sqrt 
(pi)*erf(1/24*I*sqrt(6)*sqrt(f)*(12*x + (pi*b*sgn(f) - pi*b + 2*I*b*log(ab 
s(f)) + 6*e)/f)*(-I*f/abs(f) + 1))*e^(-1/24*I*pi^2*b^2*sgn(f)/f - 1/12*pi* 
b^2*log(abs(f))*sgn(f)/f + 1/24*I*pi^2*b^2/f + 1/12*pi*b^2*log(abs(f))/f - 
 1/12*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/4*I*pi*b*e*sgn(f)/f + 
1/2*I*pi*a - 1/4*I*pi*b*e/f + a*log(abs(f)) - 1/2*b*e*log(abs(f))/f - 3*I* 
d + 3/4*I*e^2/f)/(sqrt(f)*(-I*f/abs(f) + 1)) - 3/16*I*sqrt(2)*sqrt(pi)*erf 
(1/8*I*sqrt(2)*(4*x + (pi*b*sgn(f) - pi*b + 2*I*b*log(abs(f)) + 2*e)/f)*(- 
I*f/abs(f) + 1)*sqrt(abs(f)))*e^(-1/8*I*pi^2*b^2*sgn(f)/f - 1/4*pi*b^2*...
 
3.2.15.9 Mupad [F(-1)]

Timed out. \[ \int f^{a+b x} \cos ^3\left (d+e x+f x^2\right ) \, dx=\int f^{a+b\,x}\,{\cos \left (f\,x^2+e\,x+d\right )}^3 \,d x \]

input
int(f^(a + b*x)*cos(d + e*x + f*x^2)^3,x)
 
output
int(f^(a + b*x)*cos(d + e*x + f*x^2)^3, x)