3.4.100 \(\int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^4} \, dx\) [400]

3.4.100.1 Optimal result

Integrand size = 21, antiderivative size = 460 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^4} \, dx=\frac {d^3 F^{a+\frac {b}{c+d x}}}{3 f (d e-c f)^3}-\frac {F^{a+\frac {b}{c+d x}}}{3 f (e+f x)^3}-\frac {5 b d^3 F^{a+\frac {b}{c+d x}} \log (F)}{6 (d e-c f)^4}+\frac {b d F^{a+\frac {b}{c+d x}} \log (F)}{6 (d e-c f)^2 (e+f x)^2}+\frac {2 b d^2 F^{a+\frac {b}{c+d x}} \log (F)}{3 (d e-c f)^3 (e+f x)}-\frac {b d^3 F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log (F)}{(d e-c f)^4}+\frac {b^2 d^3 f F^{a+\frac {b}{c+d x}} \log ^2(F)}{6 (d e-c f)^5}-\frac {b^2 d^2 f F^{a+\frac {b}{c+d x}} \log ^2(F)}{6 (d e-c f)^4 (e+f x)}+\frac {b^2 d^3 f F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log ^2(F)}{(d e-c f)^5}-\frac {b^3 d^3 f^2 F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log ^3(F)}{6 (d e-c f)^6} \]

output

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

3.4.100.2 Mathematica [F]

\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^4} \, dx=\int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^4} \, dx \]

input

Integrate[F^(a + b/(c + d*x))/(e + f*x)^4,x]
 

output

Integrate[F^(a + b/(c + d*x))/(e + f*x)^4, x]
 

3.4.100.3 Rubi [A] (verified)

Time = 3.66 (sec) , antiderivative size = 454, normalized size of antiderivative = 0.99, 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.

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

\(\Big \downarrow \) 2653

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {b d \log (F) \left (\frac {b^2 d^2 f^3 \log ^2(F) F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{2 (d e-c f)^6}-\frac {3 b d^2 f^2 \log (F) F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{(d e-c f)^5}+\frac {3 d^2 f F^{a-\frac {b f}{d e-c f}} \operatorname {ExpIntegralEi}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{(d e-c f)^4}-\frac {b d^2 f^2 \log (F) F^{a+\frac {b}{c+d x}}}{2 (d e-c f)^5}+\frac {5 d^2 f F^{a+\frac {b}{c+d x}}}{2 (d e-c f)^4}-\frac {d^2 F^{a+\frac {b}{c+d x}}}{b \log (F) (d e-c f)^3}+\frac {b d f^2 \log (F) F^{a+\frac {b}{c+d x}}}{2 (e+f x) (d e-c f)^4}-\frac {2 d f F^{a+\frac {b}{c+d x}}}{(e+f x) (d e-c f)^3}-\frac {f F^{a+\frac {b}{c+d x}}}{2 (e+f x)^2 (d e-c f)^2}\right )}{3 f}-\frac {F^{a+\frac {b}{c+d x}}}{3 f (e+f x)^3}\)

input

Int[F^(a + b/(c + d*x))/(e + f*x)^4,x]
 

output

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

3.4.100.3.1 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] 
]
 

3.4.100.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(921\) vs. \(2(444)=888\).

Time = 0.82 (sec) , antiderivative size = 922, normalized size of antiderivative = 2.00

method	result	size
risch	\(\frac {b^{2} d^{3} \ln \left (F \right )^{2} f \,F^{a} F^{\frac {b}{d x +c}}}{\left (c f -d e \right )^{5} \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^{3} \ln \left (F \right )^{2} f \,F^{a} F^{\frac {b}{d x +c}}}{\left (c f -d e \right )^{5} \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^{3} \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 )}{\left (c f -d e \right )^{5}}+\frac {b^{3} d^{3} \ln \left (F \right )^{3} f^{2} F^{a} F^{\frac {b}{d x +c}}}{3 \left (c f -d e \right )^{6} \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 )^{3}}+\frac {b^{3} d^{3} \ln \left (F \right )^{3} f^{2} F^{a} F^{\frac {b}{d x +c}}}{6 \left (c f -d e \right )^{6} \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^{3} d^{3} \ln \left (F \right )^{3} f^{2} F^{a} F^{\frac {b}{d x +c}}}{6 \left (c f -d e \right )^{6} \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^{3} d^{3} \ln \left (F \right )^{3} f^{2} 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 )}{6 \left (c f -d e \right )^{6}}+\frac {b \,d^{3} \ln \left (F \right ) F^{a} F^{\frac {b}{d x +c}}}{\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 \,d^{3} \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 )^{4}}\)	\(922\)

input

int(F^(a+b/(d*x+c))/(f*x+e)^4,x,method=_RETURNVERBOSE)
 

output

b^2*d^3*ln(F)^2*f/(c*f-d*e)^5*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+b^ 
2*d^3*ln(F)^2*f/(c*f-d*e)^5*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^2*d^ 
3*ln(F)^2*f/(c*f-d*e)^5*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))+1/3*b^3*d^3*ln 
(F)^3*f^2/(c*f-d*e)^6*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)^3+1/6*b^3*d^ 
3*ln(F)^3*f^2/(c*f-d*e)^6*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+1/6*b^ 
3*d^3*ln(F)^3*f^2/(c*f-d*e)^6*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)+1/6* 
b^3*d^3*ln(F)^3*f^2/(c*f-d*e)^6*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))+b*d^3* 
ln(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)*l 
n(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^3*ln(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))
 

3.4.100.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1376 vs. \(2 (444) = 888\).

Time = 0.31 (sec) , antiderivative size = 1376, normalized size of antiderivative = 2.99 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^4} \, dx=\text {Too large to display} \]

input

integrate(F^(a+b/(d*x+c))/(f*x+e)^4,x, algorithm="fricas")
 

output

-1/6*(((b^3*d^3*f^5*x^3 + 3*b^3*d^3*e*f^4*x^2 + 3*b^3*d^3*e^2*f^3*x + b^3* 
d^3*e^3*f^2)*log(F)^3 - 6*(b^2*d^4*e^4*f - b^2*c*d^3*e^3*f^2 + (b^2*d^4*e* 
f^4 - b^2*c*d^3*f^5)*x^3 + 3*(b^2*d^4*e^2*f^3 - b^2*c*d^3*e*f^4)*x^2 + 3*( 
b^2*d^4*e^3*f^2 - b^2*c*d^3*e^2*f^3)*x)*log(F)^2 + 6*(b*d^5*e^5 - 2*b*c*d^ 
4*e^4*f + b*c^2*d^3*e^3*f^2 + (b*d^5*e^2*f^3 - 2*b*c*d^4*e*f^4 + b*c^2*d^3 
*f^5)*x^3 + 3*(b*d^5*e^3*f^2 - 2*b*c*d^4*e^2*f^3 + b*c^2*d^3*e*f^4)*x^2 + 
3*(b*d^5*e^4*f - 2*b*c*d^4*e^3*f^2 + b*c^2*d^3*e^2*f^3)*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)) - (6*c*d^5*e^5 - 24*c^2*d^4*e^4*f + 38*c^3*d^3*e^3*f 
^2 - 30*c^4*d^2*e^2*f^3 + 12*c^5*d*e*f^4 - 2*c^6*f^5 + 2*(d^6*e^3*f^2 - 3* 
c*d^5*e^2*f^3 + 3*c^2*d^4*e*f^4 - c^3*d^3*f^5)*x^3 + 6*(d^6*e^4*f - 3*c*d^ 
5*e^3*f^2 + 3*c^2*d^4*e^2*f^3 - c^3*d^3*e*f^4)*x^2 + (b^2*c*d^3*e^3*f^2 - 
b^2*c^2*d^2*e^2*f^3 + (b^2*d^4*e*f^4 - b^2*c*d^3*f^5)*x^3 + (2*b^2*d^4*e^2 
*f^3 - b^2*c*d^3*e*f^4 - b^2*c^2*d^2*f^5)*x^2 + (b^2*d^4*e^3*f^2 + b^2*c*d 
^3*e^2*f^3 - 2*b^2*c^2*d^2*e*f^4)*x)*log(F)^2 + 6*(d^6*e^5 - 3*c*d^5*e^4*f 
 + 3*c^2*d^4*e^3*f^2 - c^3*d^3*e^2*f^3)*x - (6*b*c*d^4*e^4*f - 13*b*c^2*d^ 
3*e^3*f^2 + 8*b*c^3*d^2*e^2*f^3 - b*c^4*d*e*f^4 + 5*(b*d^5*e^2*f^3 - 2*b*c 
*d^4*e*f^4 + b*c^2*d^3*f^5)*x^3 + (11*b*d^5*e^3*f^2 - 18*b*c*d^4*e^2*f^3 + 
 3*b*c^2*d^3*e*f^4 + 4*b*c^3*d^2*f^5)*x^2 + (6*b*d^5*e^4*f - 2*b*c*d^4*e^3 
*f^2 - 15*b*c^2*d^3*e^2*f^3 + 12*b*c^3*d^2*e*f^4 - b*c^4*d*f^5)*x)*log(...
 

3.4.100.6 Sympy [F(-1)]

Timed out. \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^4} \, dx=\text {Timed out} \]

input

integrate(F**(a+b/(d*x+c))/(f*x+e)**4,x)
 

output

Timed out
 

3.4.100.7 Maxima [F]

\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^4} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{{\left (f x + e\right )}^{4}} \,d x } \]

input

integrate(F^(a+b/(d*x+c))/(f*x+e)^4,x, algorithm="maxima")
 

output

integrate(F^(a + b/(d*x + c))/(f*x + e)^4, x)
 

3.4.100.8 Giac [F]

\[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^4} \, dx=\int { \frac {F^{a + \frac {b}{d x + c}}}{{\left (f x + e\right )}^{4}} \,d x } \]

input

integrate(F^(a+b/(d*x+c))/(f*x+e)^4,x, algorithm="giac")
 

output

integrate(F^(a + b/(d*x + c))/(f*x + e)^4, x)
 

3.4.100.9 Mupad [F(-1)]

Timed out. \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^4} \, dx=\int \frac {F^{a+\frac {b}{c+d\,x}}}{{\left (e+f\,x\right )}^4} \,d x \]

input

int(F^(a + b/(c + d*x))/(e + f*x)^4,x)
 

output

int(F^(a + b/(c + d*x))/(e + f*x)^4, x)