\(\int \frac {F^{c (a+b x)}}{d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3} \, dx\) [52]

Optimal result

Integrand size = 39, antiderivative size = 95 \[ \int \frac {F^{c (a+b x)}}{d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3} \, dx=-\frac {F^{c (a+b x)}}{2 e (d+e x)^2}-\frac {b c F^{c (a+b x)} \log (F)}{2 e^2 (d+e x)}+\frac {b^2 c^2 F^{c \left (a-\frac {b d}{e}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log ^2(F)}{2 e^3} \] Output:

-1/2*F^(c*(b*x+a))/e/(e*x+d)^2-1/2*b*c*F^(c*(b*x+a))*ln(F)/e^2/(e*x+d)+1/2 
*b^2*c^2*F^(c*(a-b*d/e))*Ei(b*c*(e*x+d)*ln(F)/e)*ln(F)^2/e^3
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int \frac {F^{c (a+b x)}}{d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3} \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left (b^2 c^2 (d+e x)^2 \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log ^2(F)-e F^{\frac {b c (d+e x)}{e}} (e+b c (d+e x) \log (F))\right )}{2 e^3 (d+e x)^2} \] Input:

Integrate[F^(c*(a + b*x))/(d^3 + 3*d^2*e*x + 3*d*e^2*x^2 + e^3*x^3),x]
 

Output:

(F^(c*(a - (b*d)/e))*(b^2*c^2*(d + e*x)^2*ExpIntegralEi[(b*c*(d + e*x)*Log 
[F])/e]*Log[F]^2 - e*F^((b*c*(d + e*x))/e)*(e + b*c*(d + e*x)*Log[F])))/(2 
*e^3*(d + e*x)^2)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2007, 2608, 2608, 2609}

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.

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

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {F^{c (a+b x)}}{(d+e x)^3}dx\)

\(\Big \downarrow \) 2608

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

\(\Big \downarrow \) 2608

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

\(\Big \downarrow \) 2609

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

Input:

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

Output:

-1/2*F^(c*(a + b*x))/(e*(d + e*x)^2) + (b*c*Log[F]*(-(F^(c*(a + b*x))/(e*( 
d + e*x))) + (b*c*F^(c*(a - (b*d)/e))*ExpIntegralEi[(b*c*(d + e*x)*Log[F]) 
/e]*Log[F])/e^2))/(2*e)
 

Defintions of rubi rules used

rule 2007

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, 
x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex 
pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol 
yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
 

rule 2608

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m 
_), x_Symbol] :> Simp[(c + d*x)^(m + 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))) 
, x] - Simp[f*g*n*(Log[F]/(d*(m + 1)))   Int[(c + d*x)^(m + 1)*(b*F^(g*(e + 
 f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && In 
tegerQ[2*m] &&  !TrueQ[$UseGamma]
 

rule 2609

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si 
mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F 
reeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.63

method	result	size
risch	\(-\frac {\ln \left (F \right )^{2} b^{2} c^{2} F^{b c x} F^{a c}}{2 e^{3} \left (b c x \ln \left (F \right )+\frac {b c \ln \left (F \right ) d}{e}\right )^{2}}-\frac {\ln \left (F \right )^{2} b^{2} c^{2} F^{b c x} F^{a c}}{2 e^{3} \left (b c x \ln \left (F \right )+\frac {b c \ln \left (F \right ) d}{e}\right )}-\frac {\ln \left (F \right )^{2} b^{2} c^{2} F^{\frac {c \left (e a -b d \right )}{e}} \operatorname {expIntegral}_{1}\left (-b c x \ln \left (F \right )-a c \ln \left (F \right )-\frac {-e a \ln \left (F \right ) c +b c d \ln \left (F \right )}{e}\right )}{2 e^{3}}\)	\(155\)

Input:

int(F^(c*(b*x+a))/(e^3*x^3+3*d*e^2*x^2+3*d^2*e*x+d^3),x,method=_RETURNVERB 
OSE)
 

Output:

-1/2*ln(F)^2*b^2*c^2/e^3*F^(b*c*x)*F^(a*c)/(b*c*x*ln(F)+b*c*ln(F)/e*d)^2-1 
/2*ln(F)^2*b^2*c^2/e^3*F^(b*c*x)*F^(a*c)/(b*c*x*ln(F)+b*c*ln(F)/e*d)-1/2*l 
n(F)^2*b^2*c^2/e^3*F^(c*(a*e-b*d)/e)*Ei(1,-b*c*x*ln(F)-a*c*ln(F)-(-e*a*ln( 
F)*c+b*c*d*ln(F))/e)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.41 \[ \int \frac {F^{c (a+b x)}}{d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3} \, dx=\frac {\frac {{\left (b^{2} c^{2} e^{2} x^{2} + 2 \, b^{2} c^{2} d e x + b^{2} c^{2} d^{2}\right )} {\rm Ei}\left (\frac {{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right ) \log \left (F\right )^{2}}{F^{\frac {b c d - a c e}{e}}} - {\left (e^{2} + {\left (b c e^{2} x + b c d e\right )} \log \left (F\right )\right )} F^{b c x + a c}}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \] Input:

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

Output:

1/2*((b^2*c^2*e^2*x^2 + 2*b^2*c^2*d*e*x + b^2*c^2*d^2)*Ei((b*c*e*x + b*c*d 
)*log(F)/e)*log(F)^2/F^((b*c*d - a*c*e)/e) - (e^2 + (b*c*e^2*x + b*c*d*e)* 
log(F))*F^(b*c*x + a*c))/(e^5*x^2 + 2*d*e^4*x + d^2*e^3)
 

Sympy [F]

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

integrate(F**((b*x+a)*c)/(e**3*x**3+3*d*e**2*x**2+3*d**2*e*x+d**3),x)
 

Output:

Integral(F**(c*(a + b*x))/(d + e*x)**3, x)
 

Maxima [F]

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

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

Output:

integrate(F^((b*x + a)*c)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)
 

Giac [F]

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

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

Output:

integrate(F^((b*x + a)*c)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)
 

Mupad [F(-1)]

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

int(F^(c*(a + b*x))/(d^3 + e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x),x)
 

Output:

int(F^(c*(a + b*x))/(d^3 + e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x), x)
 

Reduce [F]

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

int(F^((b*x+a)*c)/(e^3*x^3+3*d*e^2*x^2+3*d^2*e*x+d^3),x)
 

Output:

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