\(\int f^{a+b x} \sin (d+f x^2) \, dx\) [91]

Optimal result

Integrand size = 16, antiderivative size = 142 \[ \int f^{a+b x} \sin \left (d+f x^2\right ) \, dx=\frac {1}{4} (-1)^{3/4} e^{\frac {1}{4} i \left (4 d+\frac {b^2 \log ^2(f)}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt [4]{-1} (2 i f x+b \log (f))}{2 \sqrt {f}}\right )-\frac {1}{4} (-1)^{3/4} e^{-\frac {1}{4} i \left (4 d+\frac {b^2 \log ^2(f)}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt [4]{-1} (2 i f x-b \log (f))}{2 \sqrt {f}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.93 \[ \int f^{a+b x} \sin \left (d+f x^2\right ) \, dx=-\frac {1}{4} \sqrt [4]{-1} e^{-\frac {i b^2 \log ^2(f)}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \left (e^{\frac {i b^2 \log ^2(f)}{2 f}} \text {erfi}\left (\frac {\sqrt [4]{-1} (2 f x-i b \log (f))}{2 \sqrt {f}}\right ) (\cos (d)+i \sin (d))+\text {erfi}\left (\frac {(-1)^{3/4} (2 f x+i b \log (f))}{2 \sqrt {f}}\right ) (i \cos (d)+\sin (d))\right ) \] Input:

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

Output:

-1/4*((-1)^(1/4)*f^(-1/2 + a)*Sqrt[Pi]*(E^(((I/2)*b^2*Log[f]^2)/f)*Erfi[(( 
-1)^(1/4)*(2*f*x - I*b*Log[f]))/(2*Sqrt[f])]*(Cos[d] + I*Sin[d]) + Erfi[(( 
-1)^(3/4)*(2*f*x + I*b*Log[f]))/(2*Sqrt[f])]*(I*Cos[d] + Sin[d])))/E^(((I/ 
4)*b^2*Log[f]^2)/f)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.

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

\(\Big \downarrow \) 4975

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((-1)^(3/4)*E^((I/4)*(4*d + (b^2*Log[f]^2)/f))*f^(-1/2 + a)*Sqrt[Pi]*Erf[( 
(-1)^(1/4)*((2*I)*f*x + b*Log[f]))/(2*Sqrt[f])])/4 - ((-1)^(3/4)*f^(-1/2 + 
 a)*Sqrt[Pi]*Erfi[((-1)^(1/4)*((2*I)*f*x - b*Log[f]))/(2*Sqrt[f])])/(4*E^( 
(I/4)*(4*d + (b^2*Log[f]^2)/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 = 0.57 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {i \left (\ln \left (f \right )^{2} b^{2}+4 d f \right )}{4 f}} \operatorname {erf}\left (-\sqrt {-i f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-i f}}\right )}{4 \sqrt {-i f}}-\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {i \left (\ln \left (f \right )^{2} b^{2}+4 d f \right )}{4 f}} \operatorname {erf}\left (-\sqrt {i f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {i f}}\right )}{4 \sqrt {i f}}\)	\(116\)

Input:

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

Output:

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

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. 265 vs. \(2 (98) = 196\).

Time = 0.08 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.87 \[ \int f^{a+b x} \sin \left (d+f x^2\right ) \, dx=\frac {i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \left (f\right )^{2} + 4 \, a f \log \left (f\right ) - 4 i \, d f}{4 \, f}\right )} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \left (f\right )\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) + i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \left (f\right )^{2} + 4 \, a f \log \left (f\right ) + 4 i \, d f}{4 \, f}\right )} \operatorname {C}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \left (f\right )\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) + \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \left (f\right )^{2} + 4 \, a f \log \left (f\right ) - 4 i \, d f}{4 \, f}\right )} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \left (f\right )\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \left (f\right )^{2} + 4 \, a f \log \left (f\right ) + 4 i \, d f}{4 \, f}\right )} \operatorname {S}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \left (f\right )\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right )}{4 \, f} \] Input:

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

Output:

1/4*(I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*log(f)^2 + 4*a*f*log(f) - 4*I* 
d*f)/f)*fresnel_cos(1/2*sqrt(2)*(2*f*x + I*b*log(f))*sqrt(f/pi)/f) + I*sqr 
t(2)*pi*sqrt(f/pi)*e^(1/4*(I*b^2*log(f)^2 + 4*a*f*log(f) + 4*I*d*f)/f)*fre 
snel_cos(-1/2*sqrt(2)*(2*f*x - I*b*log(f))*sqrt(f/pi)/f) + sqrt(2)*pi*sqrt 
(f/pi)*e^(1/4*(-I*b^2*log(f)^2 + 4*a*f*log(f) - 4*I*d*f)/f)*fresnel_sin(1/ 
2*sqrt(2)*(2*f*x + I*b*log(f))*sqrt(f/pi)/f) - sqrt(2)*pi*sqrt(f/pi)*e^(1/ 
4*(I*b^2*log(f)^2 + 4*a*f*log(f) + 4*I*d*f)/f)*fresnel_sin(-1/2*sqrt(2)*(2 
*f*x - I*b*log(f))*sqrt(f/pi)/f))/f
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04 \[ \int f^{a+b x} \sin \left (d+f x^2\right ) \, dx=-\frac {\sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, d f}{4 \, f}\right ) + \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {2 i \, f x - b \log \left (f\right )}{2 \, \sqrt {i \, f}}\right ) + {\left (-\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, d f}{4 \, f}\right ) + \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {2 i \, f x + b \log \left (f\right )}{2 \, \sqrt {-i \, f}}\right )\right )}}{8 \, \sqrt {f}} \] Input:

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

Output:

-1/8*sqrt(2)*sqrt(pi)*((-(I + 1)*f^a*cos(1/4*(b^2*log(f)^2 + 4*d*f)/f) + ( 
I - 1)*f^a*sin(1/4*(b^2*log(f)^2 + 4*d*f)/f))*erf(1/2*(2*I*f*x - b*log(f)) 
/sqrt(I*f)) + (-(I - 1)*f^a*cos(1/4*(b^2*log(f)^2 + 4*d*f)/f) + (I + 1)*f^ 
a*sin(1/4*(b^2*log(f)^2 + 4*d*f)/f))*erf(1/2*(2*I*f*x + b*log(f))/sqrt(-I* 
f)))/sqrt(f)
 

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. 300 vs. \(2 (98) = 196\).

Time = 0.17 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.11 \[ \int f^{a+b x} \sin \left (d+f x^2\right ) \, dx=\frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{8} \, \sqrt {2} {\left (4 \, x - \frac {\pi b \mathrm {sgn}\left (f\right ) - \pi b + 2 i \, b \log \left ({\left | f \right |}\right )}{f}\right )} {\left (-\frac {i \, f}{{\left | f \right |}} + 1\right )} \sqrt {{\left | f \right |}}\right ) e^{\left (\frac {i \, \pi ^{2} b^{2} \mathrm {sgn}\left (f\right )}{8 \, f} + \frac {\pi b^{2} \log \left ({\left | f \right |}\right ) \mathrm {sgn}\left (f\right )}{4 \, f} - \frac {i \, \pi ^{2} b^{2}}{8 \, f} - \frac {\pi b^{2} \log \left ({\left | f \right |}\right )}{4 \, f} + \frac {i \, b^{2} \log \left ({\left | f \right |}\right )^{2}}{4 \, f} - \frac {1}{2} i \, \pi a \mathrm {sgn}\left (f\right ) + \frac {1}{2} i \, \pi a + a \log \left ({\left | f \right |}\right ) + i \, d\right )}}{4 \, {\left (-\frac {i \, f}{{\left | f \right |}} + 1\right )} \sqrt {{\left | f \right |}}} - \frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{8} \, \sqrt {2} {\left (4 \, x + \frac {\pi b \mathrm {sgn}\left (f\right ) - \pi b + 2 i \, b \log \left ({\left | f \right |}\right )}{f}\right )} {\left (\frac {i \, f}{{\left | f \right |}} + 1\right )} \sqrt {{\left | f \right |}}\right ) e^{\left (-\frac {i \, \pi ^{2} b^{2} \mathrm {sgn}\left (f\right )}{8 \, f} - \frac {\pi b^{2} \log \left ({\left | f \right |}\right ) \mathrm {sgn}\left (f\right )}{4 \, f} + \frac {i \, \pi ^{2} b^{2}}{8 \, f} + \frac {\pi b^{2} \log \left ({\left | f \right |}\right )}{4 \, f} - \frac {i \, b^{2} \log \left ({\left | f \right |}\right )^{2}}{4 \, f} - \frac {1}{2} i \, \pi a \mathrm {sgn}\left (f\right ) + \frac {1}{2} i \, \pi a + a \log \left ({\left | f \right |}\right ) - i \, d\right )}}{4 \, {\left (\frac {i \, f}{{\left | f \right |}} + 1\right )} \sqrt {{\left | f \right |}}} \] Input:

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

Output:

1/4*I*sqrt(2)*sqrt(pi)*erf(-1/8*sqrt(2)*(4*x - (pi*b*sgn(f) - pi*b + 2*I*b 
*log(abs(f)))/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(ab 
s(f))/f + 1/4*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a + a*l 
og(abs(f)) + I*d)/((-I*f/abs(f) + 1)*sqrt(abs(f))) - 1/4*I*sqrt(2)*sqrt(pi 
)*erf(-1/8*sqrt(2)*(4*x + (pi*b*sgn(f) - pi*b + 2*I*b*log(abs(f)))/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(ab 
s(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*l 
og(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a + a*log(abs(f)) - I*d)/((I 
*f/abs(f) + 1)*sqrt(abs(f)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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