\(\int f^{a+c x^2} \cos ^3(d+e x) \, dx\) [130]

Optimal result

Integrand size = 18, antiderivative size = 293 \[ \int f^{a+c x^2} \cos ^3(d+e x) \, dx=-\frac {3 e^{-i d+\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-3 i d+\frac {9 e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {3 i e-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {3 e^{i d+\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{3 i d+\frac {9 e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {3 i e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}} \] Output:

-3/16*exp(-I*d+1/4*e^2/c/ln(f))*f^a*Pi^(1/2)*erfi(1/2*(I*e-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+9/4*e^2/c/ln(f))*f^ 
a*Pi^(1/2)*erfi(1/2*(3*I*e-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^2/c/ln(f))*f^a*Pi^(1/2)*erfi(1/2*(I*e+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+9/4*e^2/c/ln(f) 
)*f^a*Pi^(1/2)*erfi(1/2*(3*I*e+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))/c^(1/2)/l 
n(f)^(1/2)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.74 \[ \int f^{a+c x^2} \cos ^3(d+e x) \, dx=\frac {e^{\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \left (3 \text {erfi}\left (\frac {-i e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (d)-i \sin (d))+3 \text {erfi}\left (\frac {i e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (d)+i \sin (d))+e^{\frac {2 e^2}{c \log (f)}} \left (\text {erfi}\left (\frac {-3 i e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (3 d)-i \sin (3 d))+\text {erfi}\left (\frac {3 i e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (3 d)+i \sin (3 d))\right )\right )}{16 \sqrt {c} \sqrt {\log (f)}} \] Input:

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

Output:

(E^(e^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*(3*Erfi[((-I)*e + 2*c*x*Log[f])/(2*Sqrt 
[c]*Sqrt[Log[f]])]*(Cos[d] - I*Sin[d]) + 3*Erfi[(I*e + 2*c*x*Log[f])/(2*Sq 
rt[c]*Sqrt[Log[f]])]*(Cos[d] + I*Sin[d]) + E^((2*e^2)/(c*Log[f]))*(Erfi[(( 
-3*I)*e + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cos[3*d] - I*Sin[3*d]) 
+ Erfi[((3*I)*e + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cos[3*d] + I*Si 
n[3*d]))))/(16*Sqrt[c]*Sqrt[Log[f]])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.

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

\(\Big \downarrow \) 4976

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 \sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)}-i d} \text {erfi}\left (\frac {-2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}-\frac {\sqrt {\pi } f^a e^{\frac {9 e^2}{4 c \log (f)}-3 i d} \text {erfi}\left (\frac {-2 c x \log (f)+3 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {3 \sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)}+i d} \text {erfi}\left (\frac {2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a e^{\frac {9 e^2}{4 c \log (f)}+3 i d} \text {erfi}\left (\frac {2 c x \log (f)+3 i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}\)

Input:

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

Output:

(-3*E^((-I)*d + e^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(I*e - 2*c*x*Log[f])/( 
2*Sqrt[c]*Sqrt[Log[f]])])/(16*Sqrt[c]*Sqrt[Log[f]]) - (E^((-3*I)*d + (9*e^ 
2)/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[((3*I)*e - 2*c*x*Log[f])/(2*Sqrt[c]*Sqr 
t[Log[f]])])/(16*Sqrt[c]*Sqrt[Log[f]]) + (3*E^(I*d + e^2/(4*c*Log[f]))*f^a 
*Sqrt[Pi]*Erfi[(I*e + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(16*Sqrt[c] 
*Sqrt[Log[f]]) + (E^((3*I)*d + (9*e^2)/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[((3 
*I)*e + 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 = 2.20 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {3 \left (4 i d \ln \left (f \right ) c -3 e^{2}\right )}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (\sqrt {-c \ln \left (f \right )}\, x +\frac {3 i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}+\frac {3 \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 i d \ln \left (f \right ) c -e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (\sqrt {-c \ln \left (f \right )}\, x +\frac {i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}-\frac {3 \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {4 i d \ln \left (f \right ) c +e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {3 i d \ln \left (f \right ) c +\frac {9 e^{2}}{4}}{c \ln \left (f \right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {3 i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}\)	\(242\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.96 \[ \int f^{a+c x^2} \cos ^3(d+e x) \, dx=-\frac {\sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) + 3 i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} + 12 i \, c d \log \left (f\right ) + 9 \, e^{2}}{4 \, c \log \left (f\right )}\right )} + 3 \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) + i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} + 4 i \, c d \log \left (f\right ) + e^{2}}{4 \, c \log \left (f\right )}\right )} + 3 \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) - i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} - 4 i \, c d \log \left (f\right ) + e^{2}}{4 \, c \log \left (f\right )}\right )} + \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) - 3 i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} - 12 i \, c d \log \left (f\right ) + 9 \, e^{2}}{4 \, c \log \left (f\right )}\right )}}{16 \, c \log \left (f\right )} \] Input:

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

Output:

-1/16*(sqrt(pi)*sqrt(-c*log(f))*erf(1/2*(2*c*x*log(f) + 3*I*e)*sqrt(-c*log 
(f))/(c*log(f)))*e^(1/4*(4*a*c*log(f)^2 + 12*I*c*d*log(f) + 9*e^2)/(c*log( 
f))) + 3*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*(2*c*x*log(f) + I*e)*sqrt(-c*log 
(f))/(c*log(f)))*e^(1/4*(4*a*c*log(f)^2 + 4*I*c*d*log(f) + e^2)/(c*log(f)) 
) + 3*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*(2*c*x*log(f) - I*e)*sqrt(-c*log(f) 
)/(c*log(f)))*e^(1/4*(4*a*c*log(f)^2 - 4*I*c*d*log(f) + e^2)/(c*log(f))) + 
 sqrt(pi)*sqrt(-c*log(f))*erf(1/2*(2*c*x*log(f) - 3*I*e)*sqrt(-c*log(f))/( 
c*log(f)))*e^(1/4*(4*a*c*log(f)^2 - 12*I*c*d*log(f) + 9*e^2)/(c*log(f))))/ 
(c*log(f))
 

Sympy [F]

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

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

Output:

Integral(f**(a + 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 = 0.07 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.39 \[ \int f^{a+c x^2} \cos ^3(d+e x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/32*sqrt(pi)*(f^a*(cos(3*d) - I*sin(3*d))*erf(x*conjugate(sqrt(-c*log(f) 
)) + 3/2*I*e*conjugate(1/sqrt(-c*log(f))))*e^(9/4*e^2/(c*log(f))) + f^a*(c 
os(3*d) + I*sin(3*d))*erf(x*conjugate(sqrt(-c*log(f))) - 3/2*I*e*conjugate 
(1/sqrt(-c*log(f))))*e^(9/4*e^2/(c*log(f))) - f^a*(cos(3*d) + I*sin(3*d))* 
erf(1/2*(2*c*x*log(f) + 3*I*e)/sqrt(-c*log(f)))*e^(9/4*e^2/(c*log(f))) - f 
^a*(cos(3*d) - I*sin(3*d))*erf(1/2*(2*c*x*log(f) - 3*I*e)/sqrt(-c*log(f))) 
*e^(9/4*e^2/(c*log(f))) + 3*f^a*(cos(d) - I*sin(d))*erf(x*conjugate(sqrt(- 
c*log(f))) + 1/2*I*e*conjugate(1/sqrt(-c*log(f))))*e^(1/4*e^2/(c*log(f))) 
+ 3*f^a*(cos(d) + I*sin(d))*erf(x*conjugate(sqrt(-c*log(f))) - 1/2*I*e*con 
jugate(1/sqrt(-c*log(f))))*e^(1/4*e^2/(c*log(f))) - 3*f^a*(cos(d) + I*sin( 
d))*erf(1/2*(2*c*x*log(f) + I*e)/sqrt(-c*log(f)))*e^(1/4*e^2/(c*log(f))) - 
 3*f^a*(cos(d) - I*sin(d))*erf(1/2*(2*c*x*log(f) - I*e)/sqrt(-c*log(f)))*e 
^(1/4*e^2/(c*log(f))))*sqrt(-c*log(f))/(c*log(f))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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