\(\int f^{a+b x+c x^2} \sin ^3(d+e x) \, dx\) [108]

Optimal result

Integrand size = 21, antiderivative size = 354 \[ \int f^{a+b x+c x^2} \sin ^3(d+e x) \, dx=-\frac {3 i e^{-i d+\frac {(e+i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {i e^{-3 i d+\frac {(3 e+i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {3 i e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}-\frac {3 i e^{i d+\frac {(e-i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}}+\frac {i e^{3 i d-\frac {(3 i e+b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {3 i e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{16 \sqrt {c} \sqrt {\log (f)}} \] Output:

-3/16*I*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 
*ln(f)-2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))/c^(1/2)/ln(f)^(1/2)+1/16*I*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*I*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*ln(f)+2*c*x*ln(f))/c^( 
1/2)/ln(f)^(1/2))/c^(1/2)/ln(f)^(1/2)+1/16*I*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 = 391, normalized size of antiderivative = 1.10 \[ \int f^{a+b x+c x^2} \sin ^3(d+e x) \, dx=\frac {e^{\frac {e (e-6 i b \log (f))}{4 c \log (f)}} f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \left (-i e^{\frac {e (2 e+3 i b \log (f))}{c \log (f)}} \cos (3 d) \text {erfi}\left (\frac {-3 i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )+i e^{\frac {2 e^2}{c \log (f)}} \cos (3 d) \text {erfi}\left (\frac {3 i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )+3 i e^{\frac {i b e}{c}} \text {erfi}\left (\frac {-i e-(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (d)+i \sin (d))+3 e^{\frac {2 i b e}{c}} \text {erfi}\left (\frac {-i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (i \cos (d)+\sin (d))-e^{\frac {e (2 e+3 i b \log (f))}{c \log (f)}} \text {erfi}\left (\frac {-3 i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) \sin (3 d)-e^{\frac {2 e^2}{c \log (f)}} \text {erfi}\left (\frac {3 i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) \sin (3 d)\right )}{16 \sqrt {c} \sqrt {\log (f)}} \] Input:

Integrate[f^(a + b*x + c*x^2)*Sin[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]*((-I 
)*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]])] + I*E^((2*e^2)/(c*Log[f]))*Cos[3 
*d]*Erfi[((3*I)*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])] + (3*I)* 
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]) + 3*E^(((2*I)*b*e)/c)*Erfi[((-I)*e + (b + 2*c*x)*Log[ 
f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(I*Cos[d] + Sin[d]) - E^((e*(2*e + (3*I)*b*L 
og[f]))/(c*Log[f]))*Erfi[((-3*I)*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[L 
og[f]])]*Sin[3*d] - E^((2*e^2)/(c*Log[f]))*Erfi[((3*I)*e + (b + 2*c*x)*Log 
[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*Sin[3*d]))/(16*Sqrt[c]*Sqrt[Log[f]])
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 354, 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 = {4975, 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.

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

\(\Big \downarrow \) 4975

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(((-3*I)/16)*E^((-I)*d + (e + I*b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erf 
i[(I*e - b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(Sqrt[c]*Sqrt 
[Log[f]]) + ((I/16)*E^((-3*I)*d + (3*e + I*b*Log[f])^2/(4*c*Log[f]))*f^a*S 
qrt[Pi]*Erfi[((3*I)*e - b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])] 
)/(Sqrt[c]*Sqrt[Log[f]]) - (((3*I)/16)*E^(I*d + (e - I*b*Log[f])^2/(4*c*Lo 
g[f]))*f^a*Sqrt[Pi]*Erfi[(I*e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[L 
og[f]])])/(Sqrt[c]*Sqrt[Log[f]]) + ((I/16)*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]])])/(Sqrt[c]*Sqrt[Log[f]])
 

Defintions of rubi rules used

rule 2009

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

rule 4975

Int[(F_)^(u_)*Sin[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sin[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 = 2.52 (sec) , antiderivative size = 344, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {i \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{-\frac {3 \left (2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c -3 e^{2}\right )}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {3 i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}+\frac {i \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{\frac {\frac {3 i \ln \left (f \right ) b e}{2}-3 i d \ln \left (f \right ) c +\frac {9 e^{2}}{4}}{c \ln \left (f \right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )-3 i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}-\frac {3 i \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{\frac {2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c +e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )-i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}+\frac {3 i \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{-\frac {2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c -e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 \sqrt {-c \ln \left (f \right )}}\)	\(344\)

Input:

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

Output:

-1/16*I*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))+1/16*I*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*I*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*I*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))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 346, normalized size of antiderivative = 0.98 \[ \int f^{a+b x+c x^2} \sin ^3(d+e x) \, dx=\frac {-3 i \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) - i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - e^{2} + 2 \, {\left (2 i \, c d - i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )} + 3 i \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) + i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - e^{2} + 2 \, {\left (-2 i \, c d + i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )} + i \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) - 3 i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 9 \, e^{2} + 6 \, {\left (2 i \, c d - i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )} - i \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) + 3 i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 9 \, e^{2} + 6 \, {\left (-2 i \, c d + i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )}}{16 \, c \log \left (f\right )} \] Input:

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

Output:

1/16*(-3*I*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*((2*c*x + b)*log(f) - I*e)*sqr 
t(-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*I*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))) + I*sqrt(pi)*sqr 
t(-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))) - I*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]

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

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

Output:

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

Maxima [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 680, normalized size of antiderivative = 1.92 \[ \int f^{a+b x+c x^2} \sin ^3(d+e x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/32*sqrt(pi)*(f^a*(I*cos(-3/2*(2*c*d - b*e)/c) + 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*(-I*cos(-3/2*(2*c*d - b*e)/ 
c) + 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 
*(I*cos(-3/2*(2*c*d - b*e)/c) + 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 
))) + f^a*(-I*cos(-3/2*(2*c*d - b*e)/c) + 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*(I*cos(-1/2*(2*c*d - b*e)/c) + 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/ 
sqrt(-c*log(f))))*e^(1/4*e^2/(c*log(f))) - 3*f^a*(-I*cos(-1/2*(2*c*d - b*e 
)/c) + 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/sqrt(-c*log(f))))*e^(1/4*e^2/(c*log(f))) - 3*f 
^a*(I*cos(-1/2*(2*c*d - b*e)/c) + 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 
))) - 3*f^a*(-I*cos(-1/2*(2*c*d - b*e)/c) + 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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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