3.1.11 \(\int \frac {F^{c (a+b x)}}{(d+e x)^5} \, dx\) [11]

3.1.11.1 Optimal result

Integrand size = 17, antiderivative size = 161 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^5} \, dx=-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}-\frac {b c F^{c (a+b x)} \log (F)}{12 e^2 (d+e x)^3}-\frac {b^2 c^2 F^{c (a+b x)} \log ^2(F)}{24 e^3 (d+e x)^2}-\frac {b^3 c^3 F^{c (a+b x)} \log ^3(F)}{24 e^4 (d+e x)}+\frac {b^4 c^4 F^{c \left (a-\frac {b d}{e}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log ^4(F)}{24 e^5} \]

output

-1/4*F^(c*(b*x+a))/e/(e*x+d)^4-1/12*b*c*F^(c*(b*x+a))*ln(F)/e^2/(e*x+d)^3- 
1/24*b^2*c^2*F^(c*(b*x+a))*ln(F)^2/e^3/(e*x+d)^2-1/24*b^3*c^3*F^(c*(b*x+a) 
)*ln(F)^3/e^4/(e*x+d)+1/24*b^4*c^4*F^(c*(a-b*d/e))*Ei(b*c*(e*x+d)*ln(F)/e) 
*ln(F)^4/e^5
 

3.1.11.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.75 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^5} \, dx=\frac {F^{a c} \left (b^4 c^4 F^{-\frac {b c d}{e}} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log ^4(F)-\frac {e F^{b c x} \left (6 e^3+2 b c e^2 (d+e x) \log (F)+b^2 c^2 e (d+e x)^2 \log ^2(F)+b^3 c^3 (d+e x)^3 \log ^3(F)\right )}{(d+e x)^4}\right )}{24 e^5} \]

input

Integrate[F^(c*(a + b*x))/(d + e*x)^5,x]
 

output

(F^(a*c)*((b^4*c^4*ExpIntegralEi[(b*c*(d + e*x)*Log[F])/e]*Log[F]^4)/F^((b 
*c*d)/e) - (e*F^(b*c*x)*(6*e^3 + 2*b*c*e^2*(d + e*x)*Log[F] + b^2*c^2*e*(d 
 + e*x)^2*Log[F]^2 + b^3*c^3*(d + e*x)^3*Log[F]^3))/(d + e*x)^4))/(24*e^5)
 

3.1.11.3 Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2608, 2608, 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.

\(\displaystyle \int \frac {F^{c (a+b x)}}{(d+e x)^5} \, dx\)

\(\Big \downarrow \) 2608

\(\displaystyle \frac {b c \log (F) \int \frac {F^{c (a+b x)}}{(d+e x)^4}dx}{4 e}-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}\)

\(\Big \downarrow \) 2608

\(\displaystyle \frac {b c \log (F) \left (\frac {b c \log (F) \int \frac {F^{c (a+b x)}}{(d+e x)^3}dx}{3 e}-\frac {F^{c (a+b x)}}{3 e (d+e x)^3}\right )}{4 e}-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}\)

\(\Big \downarrow \) 2608

\(\displaystyle \frac {b c \log (F) \left (\frac {b c \log (F) \left (\frac {b c \log (F) \int \frac {F^{c (a+b x)}}{(d+e x)^2}dx}{2 e}-\frac {F^{c (a+b x)}}{2 e (d+e x)^2}\right )}{3 e}-\frac {F^{c (a+b x)}}{3 e (d+e x)^3}\right )}{4 e}-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}\)

\(\Big \downarrow \) 2608

\(\displaystyle \frac {b c \log (F) \left (\frac {b c \log (F) \left (\frac {b c \log (F) \left (\frac {b c \log (F) \int \frac {F^{c (a+b x)}}{d+e x}dx}{e}-\frac {F^{c (a+b x)}}{e (d+e x)}\right )}{2 e}-\frac {F^{c (a+b x)}}{2 e (d+e x)^2}\right )}{3 e}-\frac {F^{c (a+b x)}}{3 e (d+e x)^3}\right )}{4 e}-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}\)

\(\Big \downarrow \) 2609

\(\displaystyle \frac {b c \log (F) \left (\frac {b c \log (F) \left (\frac {b c \log (F) \left (\frac {b c \log (F) F^{c \left (a-\frac {b d}{e}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right )}{e^2}-\frac {F^{c (a+b x)}}{e (d+e x)}\right )}{2 e}-\frac {F^{c (a+b x)}}{2 e (d+e x)^2}\right )}{3 e}-\frac {F^{c (a+b x)}}{3 e (d+e x)^3}\right )}{4 e}-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}\)

input

Int[F^(c*(a + b*x))/(d + e*x)^5,x]
 

output

-1/4*F^(c*(a + b*x))/(e*(d + e*x)^4) + (b*c*Log[F]*(-1/3*F^(c*(a + b*x))/( 
e*(d + e*x)^3) + (b*c*Log[F]*(-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))*ExpInt 
egralEi[(b*c*(d + e*x)*Log[F])/e]*Log[F])/e^2))/(2*e)))/(3*e)))/(4*e)
 

3.1.11.3.1 Defintions of rubi rules used

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]
 

3.1.11.4 Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.51

method	result	size
risch	\(-\frac {c^{4} b^{4} \ln \left (F \right )^{4} F^{b c x} F^{c a}}{4 e^{5} \left (b c x \ln \left (F \right )+\frac {b c \ln \left (F \right ) d}{e}\right )^{4}}-\frac {c^{4} b^{4} \ln \left (F \right )^{4} F^{b c x} F^{c a}}{12 e^{5} \left (b c x \ln \left (F \right )+\frac {b c \ln \left (F \right ) d}{e}\right )^{3}}-\frac {c^{4} b^{4} \ln \left (F \right )^{4} F^{b c x} F^{c a}}{24 e^{5} \left (b c x \ln \left (F \right )+\frac {b c \ln \left (F \right ) d}{e}\right )^{2}}-\frac {c^{4} b^{4} \ln \left (F \right )^{4} F^{b c x} F^{c a}}{24 e^{5} \left (b c x \ln \left (F \right )+\frac {b c \ln \left (F \right ) d}{e}\right )}-\frac {c^{4} b^{4} \ln \left (F \right )^{4} F^{\frac {c \left (a e -b d \right )}{e}} \operatorname {Ei}_{1}\left (-b c x \ln \left (F \right )-c a \ln \left (F \right )-\frac {-\ln \left (F \right ) a c e +\ln \left (F \right ) b c d}{e}\right )}{24 e^{5}}\)	\(243\)

input

int(F^(c*(b*x+a))/(e*x+d)^5,x,method=_RETURNVERBOSE)
 

output

-1/4*c^4*b^4*ln(F)^4/e^5*F^(b*c*x)*F^(c*a)/(b*c*x*ln(F)+b*c*ln(F)/e*d)^4-1 
/12*c^4*b^4*ln(F)^4/e^5*F^(b*c*x)*F^(c*a)/(b*c*x*ln(F)+b*c*ln(F)/e*d)^3-1/ 
24*c^4*b^4*ln(F)^4/e^5*F^(b*c*x)*F^(c*a)/(b*c*x*ln(F)+b*c*ln(F)/e*d)^2-1/2 
4*c^4*b^4*ln(F)^4/e^5*F^(b*c*x)*F^(c*a)/(b*c*x*ln(F)+b*c*ln(F)/e*d)-1/24*c 
^4*b^4*ln(F)^4/e^5*F^(c*(a*e-b*d)/e)*Ei(1,-b*c*x*ln(F)-c*a*ln(F)-(-ln(F)*a 
*c*e+ln(F)*b*c*d)/e)
 

3.1.11.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.86 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^5} \, dx=\frac {\frac {{\left (b^{4} c^{4} e^{4} x^{4} + 4 \, b^{4} c^{4} d e^{3} x^{3} + 6 \, b^{4} c^{4} d^{2} e^{2} x^{2} + 4 \, b^{4} c^{4} d^{3} e x + b^{4} c^{4} d^{4}\right )} {\rm Ei}\left (\frac {{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right ) \log \left (F\right )^{4}}{F^{\frac {b c d - a c e}{e}}} - {\left (6 \, e^{4} + {\left (b^{3} c^{3} e^{4} x^{3} + 3 \, b^{3} c^{3} d e^{3} x^{2} + 3 \, b^{3} c^{3} d^{2} e^{2} x + b^{3} c^{3} d^{3} e\right )} \log \left (F\right )^{3} + {\left (b^{2} c^{2} e^{4} x^{2} + 2 \, b^{2} c^{2} d e^{3} x + b^{2} c^{2} d^{2} e^{2}\right )} \log \left (F\right )^{2} + 2 \, {\left (b c e^{4} x + b c d e^{3}\right )} \log \left (F\right )\right )} F^{b c x + a c}}{24 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} \]

input

integrate(F^(c*(b*x+a))/(e*x+d)^5,x, algorithm="fricas")
 

output

1/24*((b^4*c^4*e^4*x^4 + 4*b^4*c^4*d*e^3*x^3 + 6*b^4*c^4*d^2*e^2*x^2 + 4*b 
^4*c^4*d^3*e*x + b^4*c^4*d^4)*Ei((b*c*e*x + b*c*d)*log(F)/e)*log(F)^4/F^(( 
b*c*d - a*c*e)/e) - (6*e^4 + (b^3*c^3*e^4*x^3 + 3*b^3*c^3*d*e^3*x^2 + 3*b^ 
3*c^3*d^2*e^2*x + b^3*c^3*d^3*e)*log(F)^3 + (b^2*c^2*e^4*x^2 + 2*b^2*c^2*d 
*e^3*x + b^2*c^2*d^2*e^2)*log(F)^2 + 2*(b*c*e^4*x + b*c*d*e^3)*log(F))*F^( 
b*c*x + a*c))/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x + d^4*e 
^5)
 

3.1.11.6 Sympy [F]

\[ \int \frac {F^{c (a+b x)}}{(d+e x)^5} \, dx=\int \frac {F^{c \left (a + b x\right )}}{\left (d + e x\right )^{5}}\, dx \]

input

integrate(F**(c*(b*x+a))/(e*x+d)**5,x)
 

output

Integral(F**(c*(a + b*x))/(d + e*x)**5, x)
 

3.1.11.7 Maxima [F]

\[ \int \frac {F^{c (a+b x)}}{(d+e x)^5} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{5}} \,d x } \]

input

integrate(F^(c*(b*x+a))/(e*x+d)^5,x, algorithm="maxima")
 

output

integrate(F^((b*x + a)*c)/(e*x + d)^5, x)
 

3.1.11.8 Giac [F]

\[ \int \frac {F^{c (a+b x)}}{(d+e x)^5} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{5}} \,d x } \]

input

integrate(F^(c*(b*x+a))/(e*x+d)^5,x, algorithm="giac")
 

output

integrate(F^((b*x + a)*c)/(e*x + d)^5, x)
 

3.1.11.9 Mupad [F(-1)]

Timed out. \[ \int \frac {F^{c (a+b x)}}{(d+e x)^5} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^5} \,d x \]

input

int(F^(c*(a + b*x))/(d + e*x)^5,x)
 

output

int(F^(c*(a + b*x))/(d + e*x)^5, x)