\(\int f^{a+b x+c x^2} \cos ^3(d+e x) \, dx\) [139]

Optimal result

Integrand size = 21, antiderivative size = 346 \[ \int f^{a+b x+c x^2} \cos ^3(d+e x) \, dx=-\frac {3 e^{-i d+\frac {(e+i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-3 i d+\frac {(3 e+i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {3 i e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {3 e^{i d+\frac {(e-i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{3 i d-\frac {(3 i e+b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {3 i e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}} \] Output:

-3/16*exp(-I*d+1/4*(e+I*b*ln(f))^2/c/ln(f))*f^a*Pi^(1/2)*erfi(1/2*(I*e-b*l 
n(f)-2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))/c^(1/2)/ln(f)^(1/2)-1/16*exp(-3*I*d 
+1/4*(3*e+I*b*ln(f))^2/c/ln(f))*f^a*Pi^(1/2)*erfi(1/2*(3*I*e-b*ln(f)-2*c*x 
*ln(f))/c^(1/2)/ln(f)^(1/2))/c^(1/2)/ln(f)^(1/2)+3/16*exp(I*d+1/4*(e-I*b*l 
n(f))^2/c/ln(f))*f^a*Pi^(1/2)*erfi(1/2*(I*e+b*ln(f)+2*c*x*ln(f))/c^(1/2)/l 
n(f)^(1/2))/c^(1/2)/ln(f)^(1/2)+1/16*exp(3*I*d-1/4*(3*I*e+b*ln(f))^2/c/ln( 
f))*f^a*Pi^(1/2)*erfi(1/2*(3*I*e+b*ln(f)+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2)) 
/c^(1/2)/ln(f)^(1/2)
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.12 \[ \int f^{a+b x+c x^2} \cos ^3(d+e x) \, dx=\frac {e^{\frac {e (e-6 i b \log (f))}{4 c \log (f)}} f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \left (e^{\frac {e (2 e+3 i b \log (f))}{c \log (f)}} \cos (3 d) \text {erfi}\left (\frac {-3 i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )+e^{\frac {2 e^2}{c \log (f)}} \cos (3 d) \text {erfi}\left (\frac {3 i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )+3 e^{\frac {2 i b e}{c}} \text {erfi}\left (\frac {-i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (d)-i \sin (d))+3 e^{\frac {i b e}{c}} \text {erfi}\left (\frac {i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (d)+i \sin (d))-i e^{\frac {e (2 e+3 i b \log (f))}{c \log (f)}} \text {erfi}\left (\frac {-3 i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) \sin (3 d)+i e^{\frac {2 e^2}{c \log (f)}} \text {erfi}\left (\frac {3 i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) \sin (3 d)\right )}{16 \sqrt {c} \sqrt {\log (f)}} \] Input:

Integrate[f^(a + b*x + c*x^2)*Cos[d + e*x]^3,x]
 

Output:

(E^((e*(e - (6*I)*b*Log[f]))/(4*c*Log[f]))*f^(a - b^2/(4*c))*Sqrt[Pi]*(E^( 
(e*(2*e + (3*I)*b*Log[f]))/(c*Log[f]))*Cos[3*d]*Erfi[((-3*I)*e + (b + 2*c* 
x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])] + E^((2*e^2)/(c*Log[f]))*Cos[3*d]*Erf 
i[((3*I)*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])] + 3*E^(((2*I)*b 
*e)/c)*Erfi[((-I)*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cos[d 
] - I*Sin[d]) + 3*E^((I*b*e)/c)*Erfi[(I*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c] 
*Sqrt[Log[f]])]*(Cos[d] + I*Sin[d]) - I*E^((e*(2*e + (3*I)*b*Log[f]))/(c*L 
og[f]))*Erfi[((-3*I)*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*Sin 
[3*d] + I*E^((2*e^2)/(c*Log[f]))*Erfi[((3*I)*e + (b + 2*c*x)*Log[f])/(2*Sq 
rt[c]*Sqrt[Log[f]])]*Sin[3*d]))/(16*Sqrt[c]*Sqrt[Log[f]])
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 346, 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.

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

\(\Big \downarrow \) 4976

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

\(\Big \downarrow \) 2009

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

Input:

Int[f^(a + b*x + c*x^2)*Cos[d + e*x]^3,x]
 

Output:

(-3*E^((-I)*d + (e + I*b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(I*e - 
b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(16*Sqrt[c]*Sqrt[Log[f 
]]) - (E^((-3*I)*d + (3*e + I*b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[ 
((3*I)*e - b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(16*Sqrt[c] 
*Sqrt[Log[f]]) + (3*E^(I*d + (e - I*b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi] 
*Erfi[(I*e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(16*Sqrt[ 
c]*Sqrt[Log[f]]) + (E^((3*I)*d - ((3*I)*e + b*Log[f])^2/(4*c*Log[f]))*f^a* 
Sqrt[Pi]*Erfi[((3*I)*e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]]) 
])/(16*Sqrt[c]*Sqrt[Log[f]])
 

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]
 

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 340, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{\frac {\frac {3 i \ln \left (f \right ) b e}{2}-3 i d \ln \left (f \right ) c +\frac {9 e^{2}}{4}}{c \ln \left (f \right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )-3 i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}-\frac {3 \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{\frac {2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c +e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )-i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}-\frac {3 \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{-\frac {2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c -e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{-\frac {3 \left (2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c -3 e^{2}\right )}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {3 i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}\)	\(340\)

Input:

int(f^(c*x^2+b*x+a)*cos(e*x+d)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/16*Pi^(1/2)*f^a*f^(-1/4*b^2/c)*exp(3/4*(2*I*ln(f)*b*e-4*I*d*ln(f)*c+3*e 
^2)/ln(f)/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*(b*ln(f)-3*I*e)/ 
(-c*ln(f))^(1/2))-3/16*Pi^(1/2)*f^a*f^(-1/4*b^2/c)*exp(1/4*(2*I*ln(f)*b*e- 
4*I*d*ln(f)*c+e^2)/ln(f)/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*( 
b*ln(f)-I*e)/(-c*ln(f))^(1/2))-3/16*Pi^(1/2)*f^a*f^(-1/4*b^2/c)*exp(-1/4*( 
2*I*ln(f)*b*e-4*I*d*ln(f)*c-e^2)/ln(f)/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f)) 
^(1/2)*x+1/2*(I*e+b*ln(f))/(-c*ln(f))^(1/2))-1/16*Pi^(1/2)*f^a*f^(-1/4*b^2 
/c)*exp(-3/4*(2*I*ln(f)*b*e-4*I*d*ln(f)*c-3*e^2)/ln(f)/c)/(-c*ln(f))^(1/2) 
*erf(-(-c*ln(f))^(1/2)*x+1/2*(3*I*e+b*ln(f))/(-c*ln(f))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 344, normalized size of antiderivative = 0.99 \[ \int f^{a+b x+c x^2} \cos ^3(d+e x) \, dx=-\frac {3 \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) - i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - e^{2} + 2 \, {\left (2 i \, c d - i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )} + 3 \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) + i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - e^{2} + 2 \, {\left (-2 i \, c d + i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )} + \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) - 3 i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 9 \, e^{2} + 6 \, {\left (2 i \, c d - i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )} + \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) + 3 i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 9 \, e^{2} + 6 \, {\left (-2 i \, c d + i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )}}{16 \, c \log \left (f\right )} \] Input:

integrate(f^(c*x^2+b*x+a)*cos(e*x+d)^3,x, algorithm="fricas")
 

Output:

-1/16*(3*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*((2*c*x + b)*log(f) - I*e)*sqrt( 
-c*log(f))/(c*log(f)))*e^(-1/4*((b^2 - 4*a*c)*log(f)^2 - e^2 + 2*(2*I*c*d 
- I*b*e)*log(f))/(c*log(f))) + 3*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*((2*c*x 
+ b)*log(f) + I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(-1/4*((b^2 - 4*a*c)*log( 
f)^2 - e^2 + 2*(-2*I*c*d + I*b*e)*log(f))/(c*log(f))) + sqrt(pi)*sqrt(-c*l 
og(f))*erf(1/2*((2*c*x + b)*log(f) - 3*I*e)*sqrt(-c*log(f))/(c*log(f)))*e^ 
(-1/4*((b^2 - 4*a*c)*log(f)^2 - 9*e^2 + 6*(2*I*c*d - I*b*e)*log(f))/(c*log 
(f))) + sqrt(pi)*sqrt(-c*log(f))*erf(1/2*((2*c*x + b)*log(f) + 3*I*e)*sqrt 
(-c*log(f))/(c*log(f)))*e^(-1/4*((b^2 - 4*a*c)*log(f)^2 - 9*e^2 + 6*(-2*I* 
c*d + I*b*e)*log(f))/(c*log(f))))/(c*log(f))
 

Sympy [F]

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

integrate(f**(c*x**2+b*x+a)*cos(e*x+d)**3,x)
 

Output:

Integral(f**(a + b*x + c*x**2)*cos(d + e*x)**3, x)
 

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 680, normalized size of antiderivative = 1.97 \[ \int f^{a+b x+c x^2} \cos ^3(d+e x) \, dx=\text {Too large to display} \] Input:

integrate(f^(c*x^2+b*x+a)*cos(e*x+d)^3,x, algorithm="maxima")
 

Output:

-1/32*sqrt(pi)*(f^a*(cos(-3/2*(2*c*d - b*e)/c) - I*sin(-3/2*(2*c*d - b*e)/ 
c))*erf(x*conjugate(sqrt(-c*log(f))) - 1/2*(b*log(f) + 3*I*e)*conjugate(1/ 
sqrt(-c*log(f))))*e^(9/4*e^2/(c*log(f))) + f^a*(cos(-3/2*(2*c*d - b*e)/c) 
+ I*sin(-3/2*(2*c*d - b*e)/c))*erf(x*conjugate(sqrt(-c*log(f))) - 1/2*(b*l 
og(f) - 3*I*e)*conjugate(1/sqrt(-c*log(f))))*e^(9/4*e^2/(c*log(f))) + f^a* 
(cos(-3/2*(2*c*d - b*e)/c) - I*sin(-3/2*(2*c*d - b*e)/c))*erf(1/2*(2*c*x*l 
og(f) + b*log(f) + 3*I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(9/4*e^2/(c*log(f) 
)) + f^a*(cos(-3/2*(2*c*d - b*e)/c) + I*sin(-3/2*(2*c*d - b*e)/c))*erf(1/2 
*(2*c*x*log(f) + b*log(f) - 3*I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(9/4*e^2/ 
(c*log(f))) + 3*f^a*(cos(-1/2*(2*c*d - b*e)/c) - I*sin(-1/2*(2*c*d - b*e)/ 
c))*erf(x*conjugate(sqrt(-c*log(f))) - 1/2*(b*log(f) + I*e)*conjugate(1/sq 
rt(-c*log(f))))*e^(1/4*e^2/(c*log(f))) + 3*f^a*(cos(-1/2*(2*c*d - b*e)/c) 
+ I*sin(-1/2*(2*c*d - b*e)/c))*erf(x*conjugate(sqrt(-c*log(f))) - 1/2*(b*l 
og(f) - I*e)*conjugate(1/sqrt(-c*log(f))))*e^(1/4*e^2/(c*log(f))) + 3*f^a* 
(cos(-1/2*(2*c*d - b*e)/c) - I*sin(-1/2*(2*c*d - b*e)/c))*erf(1/2*(2*c*x*l 
og(f) + b*log(f) + I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(1/4*e^2/(c*log(f))) 
 + 3*f^a*(cos(-1/2*(2*c*d - b*e)/c) + I*sin(-1/2*(2*c*d - b*e)/c))*erf(1/2 
*(2*c*x*log(f) + b*log(f) - I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(1/4*e^2/(c 
*log(f))))*sqrt(-c*log(f))/(c*f^(1/4*b^2/c)*log(f))
 

Giac [F]

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

integrate(f^(c*x^2+b*x+a)*cos(e*x+d)^3,x, algorithm="giac")
 

Output:

integrate(f^(c*x^2 + b*x + a)*cos(e*x + d)^3, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

f**a*int(f**(b*x + c*x**2)*cos(d + e*x)**3,x)