\(\int f^{a+c x^2} \sin ^2(d+e x) \, dx\) [98]

Optimal result

Integrand size = 18, antiderivative size = 171 \[ \int f^{a+c x^2} \sin ^2(d+e x) \, dx=\frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{-2 i d+\frac {e^2}{c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e-c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{2 i d+\frac {e^2}{c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}} \] Output:

1/4*f^a*Pi^(1/2)*erfi(c^(1/2)*x*ln(f)^(1/2))/c^(1/2)/ln(f)^(1/2)+1/8*exp(- 
2*I*d+e^2/c/ln(f))*f^a*Pi^(1/2)*erfi((I*e-c*x*ln(f))/c^(1/2)/ln(f)^(1/2))/ 
c^(1/2)/ln(f)^(1/2)-1/8*exp(2*I*d+e^2/c/ln(f))*f^a*Pi^(1/2)*erfi((I*e+c*x* 
ln(f))/c^(1/2)/ln(f)^(1/2))/c^(1/2)/ln(f)^(1/2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.77 \[ \int f^{a+c x^2} \sin ^2(d+e x) \, dx=\frac {f^a \sqrt {\pi } \left (2 \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )-e^{\frac {e^2}{c \log (f)}} \left (\text {erfi}\left (\frac {-i e+c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right ) (\cos (2 d)-i \sin (2 d))+\text {erfi}\left (\frac {i e+c x \log (f)}{\sqrt {c} \sqrt {\log (f)}}\right ) (\cos (2 d)+i \sin (2 d))\right )\right )}{8 \sqrt {c} \sqrt {\log (f)}} \] Input:

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

Output:

(f^a*Sqrt[Pi]*(2*Erfi[Sqrt[c]*x*Sqrt[Log[f]]] - E^(e^2/(c*Log[f]))*(Erfi[( 
(-I)*e + c*x*Log[f])/(Sqrt[c]*Sqrt[Log[f]])]*(Cos[2*d] - I*Sin[2*d]) + Erf 
i[(I*e + c*x*Log[f])/(Sqrt[c]*Sqrt[Log[f]])]*(Cos[2*d] + I*Sin[2*d]))))/(8 
*Sqrt[c]*Sqrt[Log[f]])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 171, 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 = {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.

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

\(\Big \downarrow \) 4975

\(\displaystyle \int \left (-\frac {1}{4} e^{-2 i d-2 i e x} f^{a+c x^2}-\frac {1}{4} e^{2 i d+2 i e x} f^{a+c x^2}+\frac {1}{2} f^{a+c x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\pi } f^a e^{\frac {e^2}{c \log (f)}-2 i d} \text {erfi}\left (\frac {-c x \log (f)+i e}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}-\frac {\sqrt {\pi } f^a e^{\frac {e^2}{c \log (f)}+2 i d} \text {erfi}\left (\frac {c x \log (f)+i e}{\sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}}\)

Input:

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

Output:

(f^a*Sqrt[Pi]*Erfi[Sqrt[c]*x*Sqrt[Log[f]]])/(4*Sqrt[c]*Sqrt[Log[f]]) + (E^ 
((-2*I)*d + e^2/(c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(I*e - c*x*Log[f])/(Sqrt[c]* 
Sqrt[Log[f]])])/(8*Sqrt[c]*Sqrt[Log[f]]) - (E^((2*I)*d + e^2/(c*Log[f]))*f 
^a*Sqrt[Pi]*Erfi[(I*e + c*x*Log[f])/(Sqrt[c]*Sqrt[Log[f]])])/(8*Sqrt[c]*Sq 
rt[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 = 0.90 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.85

method	result	size
risch	\(-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {2 i d \ln \left (f \right ) c -e^{2}}{\ln \left (f \right ) c}} \operatorname {erf}\left (\sqrt {-c \ln \left (f \right )}\, x +\frac {i e}{\sqrt {-c \ln \left (f \right )}}\right )}{8 \sqrt {-c \ln \left (f \right )}}+\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {2 i d \ln \left (f \right ) c +e^{2}}{\ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {i e}{\sqrt {-c \ln \left (f \right )}}\right )}{8 \sqrt {-c \ln \left (f \right )}}+\frac {f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-c \ln \left (f \right )}\, x \right )}{4 \sqrt {-c \ln \left (f \right )}}\)	\(145\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94 \[ \int f^{a+c x^2} \sin ^2(d+e x) \, dx=-\frac {2 \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x\right ) - \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (c x \log \left (f\right ) + i \, e\right )} \sqrt {-c \log \left (f\right )}}{c \log \left (f\right )}\right ) e^{\left (\frac {a c \log \left (f\right )^{2} + 2 i \, c d \log \left (f\right ) + e^{2}}{c \log \left (f\right )}\right )} - \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (c x \log \left (f\right ) - i \, e\right )} \sqrt {-c \log \left (f\right )}}{c \log \left (f\right )}\right ) e^{\left (\frac {a c \log \left (f\right )^{2} - 2 i \, c d \log \left (f\right ) + e^{2}}{c \log \left (f\right )}\right )}}{8 \, c \log \left (f\right )} \] Input:

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

Output:

-1/8*(2*sqrt(pi)*sqrt(-c*log(f))*f^a*erf(sqrt(-c*log(f))*x) - sqrt(pi)*sqr 
t(-c*log(f))*erf((c*x*log(f) + I*e)*sqrt(-c*log(f))/(c*log(f)))*e^((a*c*lo 
g(f)^2 + 2*I*c*d*log(f) + e^2)/(c*log(f))) - sqrt(pi)*sqrt(-c*log(f))*erf( 
(c*x*log(f) - I*e)*sqrt(-c*log(f))/(c*log(f)))*e^((a*c*log(f)^2 - 2*I*c*d* 
log(f) + e^2)/(c*log(f))))/(c*log(f))
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

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

Time = 0.05 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.38 \[ \int f^{a+c x^2} \sin ^2(d+e x) \, dx=-\frac {\sqrt {\pi } {\left (f^{a} {\left (\cos \left (2 \, d\right ) - i \, \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}} + i \, e \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}}\right ) e^{\left (\frac {e^{2}}{c \log \left (f\right )}\right )} + f^{a} {\left (\cos \left (2 \, d\right ) + i \, \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}} - i \, e \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}}\right ) e^{\left (\frac {e^{2}}{c \log \left (f\right )}\right )} - f^{a} {\left (\cos \left (2 \, d\right ) + i \, \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (\frac {c x \log \left (f\right ) + i \, e}{\sqrt {-c \log \left (f\right )}}\right ) e^{\left (\frac {e^{2}}{c \log \left (f\right )}\right )} - f^{a} {\left (\cos \left (2 \, d\right ) - i \, \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (\frac {c x \log \left (f\right ) - i \, e}{\sqrt {-c \log \left (f\right )}}\right ) e^{\left (\frac {e^{2}}{c \log \left (f\right )}\right )} - 2 \, f^{a} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}}\right ) - 2 \, f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x\right )\right )}}{16 \, \sqrt {-c \log \left (f\right )}} \] Input:

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

Output:

-1/16*sqrt(pi)*(f^a*(cos(2*d) - I*sin(2*d))*erf(x*conjugate(sqrt(-c*log(f) 
)) + I*e*conjugate(1/sqrt(-c*log(f))))*e^(e^2/(c*log(f))) + f^a*(cos(2*d) 
+ I*sin(2*d))*erf(x*conjugate(sqrt(-c*log(f))) - I*e*conjugate(1/sqrt(-c*l 
og(f))))*e^(e^2/(c*log(f))) - f^a*(cos(2*d) + I*sin(2*d))*erf((c*x*log(f) 
+ I*e)/sqrt(-c*log(f)))*e^(e^2/(c*log(f))) - f^a*(cos(2*d) - I*sin(2*d))*e 
rf((c*x*log(f) - I*e)/sqrt(-c*log(f)))*e^(e^2/(c*log(f))) - 2*f^a*erf(x*co 
njugate(sqrt(-c*log(f)))) - 2*f^a*erf(sqrt(-c*log(f))*x))/sqrt(-c*log(f))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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