3.4.99 \(\int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^3} \, dx\) [399]

3.4.99.1 Optimal result

Integrand size = 21, antiderivative size = 267 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^3} \, dx=\frac {d^2 F^{a+\frac {b}{c+d x}}}{2 f (d e-c f)^2}-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}-\frac {b d^2 F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^3}+\frac {b d F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^2 (e+f x)}-\frac {b d^2 F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log (F)}{(d e-c f)^3}+\frac {b^2 d^2 f F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log ^2(F)}{2 (d e-c f)^4} \]

output

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

3.4.99.2 Mathematica [F]

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

input

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

output

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

3.4.99.3 Rubi [A] (verified)

Time = 1.94 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2653, 7293, 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.

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

\(\Big \downarrow \) 2653

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.4.99.3.1 Defintions of rubi rules used

rule 2009

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

rule 2653

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))*((e_.) + (f_.)*(x_))^(m_), x_ 
Symbol] :> Simp[(e + f*x)^(m + 1)*(F^(a + b/(c + d*x))/(f*(m + 1))), x] + S 
imp[b*d*(Log[F]/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*(F^(a + b/(c + d*x))/( 
c + d*x)^2), x], x] /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] 
 && ILtQ[m, -1]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.4.99.4 Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 506, normalized size of antiderivative = 1.90

method	result	size
risch	\(-\frac {b \,d^{2} \ln \left (F \right ) F^{a} F^{\frac {b}{d x +c}}}{\left (c f -d e \right )^{3} \left (\frac {b \ln \left (F \right )}{d x +c}+a \ln \left (F \right )-\frac {\ln \left (F \right ) a c f}{c f -d e}+\frac {\ln \left (F \right ) a d e}{c f -d e}-\frac {\ln \left (F \right ) b f}{c f -d e}\right )}-\frac {b \,d^{2} \ln \left (F \right ) F^{\frac {a c f -a d e +b f}{c f -d e}} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (F \right )}{d x +c}-a \ln \left (F \right )-\frac {-\ln \left (F \right ) a c f +a e d \ln \left (F \right )-\ln \left (F \right ) b f}{c f -d e}\right )}{\left (c f -d e \right )^{3}}-\frac {b^{2} d^{2} \ln \left (F \right )^{2} f \,F^{a} F^{\frac {b}{d x +c}}}{2 \left (c f -d e \right )^{4} \left (\frac {b \ln \left (F \right )}{d x +c}+a \ln \left (F \right )-\frac {\ln \left (F \right ) a c f}{c f -d e}+\frac {\ln \left (F \right ) a d e}{c f -d e}-\frac {\ln \left (F \right ) b f}{c f -d e}\right )^{2}}-\frac {b^{2} d^{2} \ln \left (F \right )^{2} f \,F^{a} F^{\frac {b}{d x +c}}}{2 \left (c f -d e \right )^{4} \left (\frac {b \ln \left (F \right )}{d x +c}+a \ln \left (F \right )-\frac {\ln \left (F \right ) a c f}{c f -d e}+\frac {\ln \left (F \right ) a d e}{c f -d e}-\frac {\ln \left (F \right ) b f}{c f -d e}\right )}-\frac {b^{2} d^{2} \ln \left (F \right )^{2} f \,F^{\frac {a c f -a d e +b f}{c f -d e}} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (F \right )}{d x +c}-a \ln \left (F \right )-\frac {-\ln \left (F \right ) a c f +a e d \ln \left (F \right )-\ln \left (F \right ) b f}{c f -d e}\right )}{2 \left (c f -d e \right )^{4}}\)	\(506\)

input

int(F^(a+b/(d*x+c))/(f*x+e)^3,x,method=_RETURNVERBOSE)
 

output

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

3.4.99.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (257) = 514\).

Time = 0.33 (sec) , antiderivative size = 555, normalized size of antiderivative = 2.08 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^3} \, dx=\frac {{\left ({\left (b^{2} d^{2} f^{3} x^{2} + 2 \, b^{2} d^{2} e f^{2} x + b^{2} d^{2} e^{2} f\right )} \log \left (F\right )^{2} - 2 \, {\left (b d^{3} e^{3} - b c d^{2} e^{2} f + {\left (b d^{3} e f^{2} - b c d^{2} f^{3}\right )} x^{2} + 2 \, {\left (b d^{3} e^{2} f - b c d^{2} e f^{2}\right )} x\right )} \log \left (F\right )\right )} F^{\frac {a d e - {\left (a c + b\right )} f}{d e - c f}} {\rm Ei}\left (\frac {{\left (b d f x + b d e\right )} \log \left (F\right )}{c d e - c^{2} f + {\left (d^{2} e - c d f\right )} x}\right ) + {\left (2 \, c d^{3} e^{3} - 5 \, c^{2} d^{2} e^{2} f + 4 \, c^{3} d e f^{2} - c^{4} f^{3} + {\left (d^{4} e^{2} f - 2 \, c d^{3} e f^{2} + c^{2} d^{2} f^{3}\right )} x^{2} + 2 \, {\left (d^{4} e^{3} - 2 \, c d^{3} e^{2} f + c^{2} d^{2} e f^{2}\right )} x - {\left (b c d^{2} e^{2} f - b c^{2} d e f^{2} + {\left (b d^{3} e f^{2} - b c d^{2} f^{3}\right )} x^{2} + {\left (b d^{3} e^{2} f - b c^{2} d f^{3}\right )} x\right )} \log \left (F\right )\right )} F^{\frac {a d x + a c + b}{d x + c}}}{2 \, {\left (d^{4} e^{6} - 4 \, c d^{3} e^{5} f + 6 \, c^{2} d^{2} e^{4} f^{2} - 4 \, c^{3} d e^{3} f^{3} + c^{4} e^{2} f^{4} + {\left (d^{4} e^{4} f^{2} - 4 \, c d^{3} e^{3} f^{3} + 6 \, c^{2} d^{2} e^{2} f^{4} - 4 \, c^{3} d e f^{5} + c^{4} f^{6}\right )} x^{2} + 2 \, {\left (d^{4} e^{5} f - 4 \, c d^{3} e^{4} f^{2} + 6 \, c^{2} d^{2} e^{3} f^{3} - 4 \, c^{3} d e^{2} f^{4} + c^{4} e f^{5}\right )} x\right )}} \]

input

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

output

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

3.4.99.6 Sympy [F]

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

input

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

output

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

3.4.99.7 Maxima [F]

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

input

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

output

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

3.4.99.8 Giac [F]

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

input

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

output

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

3.4.99.9 Mupad [F(-1)]

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

input

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

output

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