\(\int \frac {F^{c (a+b x)}}{d+e x+f x^2} \, dx\) [25]

Optimal result

Integrand size = 22, antiderivative size = 148 \[ \int \frac {F^{c (a+b x)}}{d+e x+f x^2} \, dx=\frac {F^{c \left (a-\frac {b \left (e-\sqrt {e^2-4 d f}\right )}{2 f}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e-\sqrt {e^2-4 d f}+2 f x\right ) \log (F)}{2 f}\right )}{\sqrt {e^2-4 d f}}-\frac {F^{c \left (a-\frac {b \left (e+\sqrt {e^2-4 d f}\right )}{2 f}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e+\sqrt {e^2-4 d f}+2 f x\right ) \log (F)}{2 f}\right )}{\sqrt {e^2-4 d f}} \] Output:

F^(c*(a-1/2*b*(e-(-4*d*f+e^2)^(1/2))/f))*Ei(1/2*b*c*(e-(-4*d*f+e^2)^(1/2)+ 
2*f*x)*ln(F)/f)/(-4*d*f+e^2)^(1/2)-F^(c*(a-1/2*b*(e+(-4*d*f+e^2)^(1/2))/f) 
)*Ei(1/2*b*c*(2*f*x+(-4*d*f+e^2)^(1/2)+e)*ln(F)/f)/(-4*d*f+e^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.86 \[ \int \frac {F^{c (a+b x)}}{d+e x+f x^2} \, dx=\frac {F^{a c-\frac {b c \left (e+\sqrt {e^2-4 d f}\right )}{2 f}} \left (F^{\frac {b c \sqrt {e^2-4 d f}}{f}} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e-\sqrt {e^2-4 d f}+2 f x\right ) \log (F)}{2 f}\right )-\operatorname {ExpIntegralEi}\left (\frac {b c \left (e+\sqrt {e^2-4 d f}+2 f x\right ) \log (F)}{2 f}\right )\right )}{\sqrt {e^2-4 d f}} \] Input:

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

Output:

(F^(a*c - (b*c*(e + Sqrt[e^2 - 4*d*f]))/(2*f))*(F^((b*c*Sqrt[e^2 - 4*d*f]) 
/f)*ExpIntegralEi[(b*c*(e - Sqrt[e^2 - 4*d*f] + 2*f*x)*Log[F])/(2*f)] - Ex 
pIntegralEi[(b*c*(e + Sqrt[e^2 - 4*d*f] + 2*f*x)*Log[F])/(2*f)]))/Sqrt[e^2 
 - 4*d*f]
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2698, 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.

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

\(\Big \downarrow \) 2698

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {F^{c \left (a-\frac {b \left (e-\sqrt {e^2-4 d f}\right )}{2 f}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e+2 f x-\sqrt {e^2-4 d f}\right ) \log (F)}{2 f}\right )}{\sqrt {e^2-4 d f}}-\frac {F^{c \left (a-\frac {b \left (\sqrt {e^2-4 d f}+e\right )}{2 f}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e+2 f x+\sqrt {e^2-4 d f}\right ) \log (F)}{2 f}\right )}{\sqrt {e^2-4 d f}}\)

Input:

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

Output:

(F^(c*(a - (b*(e - Sqrt[e^2 - 4*d*f]))/(2*f)))*ExpIntegralEi[(b*c*(e - Sqr 
t[e^2 - 4*d*f] + 2*f*x)*Log[F])/(2*f)])/Sqrt[e^2 - 4*d*f] - (F^(c*(a - (b* 
(e + Sqrt[e^2 - 4*d*f]))/(2*f)))*ExpIntegralEi[(b*c*(e + Sqrt[e^2 - 4*d*f] 
 + 2*f*x)*Log[F])/(2*f)])/Sqrt[e^2 - 4*d*f]
 

Defintions of rubi rules used

rule 2009

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

rule 2698

Int[(F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))/((a_.) + (b_.)*(x_) + (c_.)*(x_ 
)^2), x_Symbol] :> Int[ExpandIntegrand[F^(g*(d + e*x)^n), 1/(a + b*x + c*x^ 
2), x], x] /; FreeQ[{F, a, b, c, d, e, g, n}, x]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.51

method	result	size
risch	\(-\frac {b \,F^{\frac {\left (2 f a -b e +\sqrt {-b^{2} \left (4 d f -e^{2}\right )}\right ) c}{2 f}} \operatorname {expIntegral}_{1}\left (\frac {2 f a \ln \left (F \right ) c -\ln \left (F \right ) b c e +\sqrt {-4 b^{2} d f +b^{2} e^{2}}\, \ln \left (F \right ) c -2 f \left (b c x \ln \left (F \right )+a c \ln \left (F \right )\right )}{2 f}\right )}{\sqrt {-4 b^{2} d f +b^{2} e^{2}}}+\frac {b \,F^{-\frac {\left (-2 f a +b e +\sqrt {-b^{2} \left (4 d f -e^{2}\right )}\right ) c}{2 f}} \operatorname {expIntegral}_{1}\left (-\frac {-2 f a \ln \left (F \right ) c +\ln \left (F \right ) b c e +\sqrt {-4 b^{2} d f +b^{2} e^{2}}\, \ln \left (F \right ) c +2 f \left (b c x \ln \left (F \right )+a c \ln \left (F \right )\right )}{2 f}\right )}{\sqrt {-4 b^{2} d f +b^{2} e^{2}}}\)	\(223\)

Input:

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

Output:

-b/(-4*b^2*d*f+b^2*e^2)^(1/2)*F^(1/2/f*(2*f*a-b*e+(-b^2*(4*d*f-e^2))^(1/2) 
)*c)*Ei(1,1/2*(2*f*a*ln(F)*c-ln(F)*b*c*e+(-4*b^2*d*f+b^2*e^2)^(1/2)*ln(F)* 
c-2*f*(b*c*x*ln(F)+a*c*ln(F)))/f)+b/(-4*b^2*d*f+b^2*e^2)^(1/2)*F^(-1/2*(-2 
*f*a+b*e+(-b^2*(4*d*f-e^2))^(1/2))*c/f)*Ei(1,-1/2*(-2*f*a*ln(F)*c+ln(F)*b* 
c*e+(-4*b^2*d*f+b^2*e^2)^(1/2)*ln(F)*c+2*f*(b*c*x*ln(F)+a*c*ln(F)))/f)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (134) = 268\).

Time = 0.08 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.06 \[ \int \frac {F^{c (a+b x)}}{d+e x+f x^2} \, dx=-\frac {f \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}} {\rm Ei}\left (\frac {{\left (2 \, b c f x + b c e\right )} \log \left (F\right ) + f \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}}}{2 \, f}\right ) e^{\left (-\frac {{\left (b c e - 2 \, a c f\right )} \log \left (F\right ) + f \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}}}{2 \, f}\right )} - f \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}} {\rm Ei}\left (\frac {{\left (2 \, b c f x + b c e\right )} \log \left (F\right ) - f \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}}}{2 \, f}\right ) e^{\left (-\frac {{\left (b c e - 2 \, a c f\right )} \log \left (F\right ) - f \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}}}{2 \, f}\right )}}{{\left (b c e^{2} - 4 \, b c d f\right )} \log \left (F\right )} \] Input:

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

Output:

-(f*sqrt((b^2*c^2*e^2 - 4*b^2*c^2*d*f)*log(F)^2/f^2)*Ei(1/2*((2*b*c*f*x + 
b*c*e)*log(F) + f*sqrt((b^2*c^2*e^2 - 4*b^2*c^2*d*f)*log(F)^2/f^2))/f)*e^( 
-1/2*((b*c*e - 2*a*c*f)*log(F) + f*sqrt((b^2*c^2*e^2 - 4*b^2*c^2*d*f)*log( 
F)^2/f^2))/f) - f*sqrt((b^2*c^2*e^2 - 4*b^2*c^2*d*f)*log(F)^2/f^2)*Ei(1/2* 
((2*b*c*f*x + b*c*e)*log(F) - f*sqrt((b^2*c^2*e^2 - 4*b^2*c^2*d*f)*log(F)^ 
2/f^2))/f)*e^(-1/2*((b*c*e - 2*a*c*f)*log(F) - f*sqrt((b^2*c^2*e^2 - 4*b^2 
*c^2*d*f)*log(F)^2/f^2))/f))/((b*c*e^2 - 4*b*c*d*f)*log(F))
 

Sympy [F]

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

integrate(F**((b*x+a)*c)/(f*x**2+e*x+d),x)
 

Output:

Integral(F**(c*(a + b*x))/(d + e*x + f*x**2), x)
 

Maxima [F]

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

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

Output:

integrate(F^((b*x + a)*c)/(f*x^2 + e*x + d), x)
 

Giac [F]

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

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

Output:

integrate(F^((b*x + a)*c)/(f*x^2 + e*x + d), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

int(F^(c*(a + b*x))/(d + e*x + f*x^2),x)
 

Output:

int(F^(c*(a + b*x))/(d + e*x + f*x^2), x)
 

Reduce [F]

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

int(F^((b*x+a)*c)/(f*x^2+e*x+d),x)
 

Output:

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