\(\int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx\) [111]

Optimal result

Integrand size = 22, antiderivative size = 217 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=-\frac {e^2 F^{a+b c+b d x}}{3 x^3}-\frac {e f F^{a+b c+b d x}}{x^2}-\frac {f^2 F^{a+b c+b d x}}{x}-\frac {b d e^2 F^{a+b c+b d x} \log (F)}{6 x^2}-\frac {b d e f F^{a+b c+b d x} \log (F)}{x}+b d f^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log (F)-\frac {b^2 d^2 e^2 F^{a+b c+b d x} \log ^2(F)}{6 x}+b^2 d^2 e f F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log ^2(F)+\frac {1}{6} b^3 d^3 e^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log ^3(F) \] Output:

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

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.53 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=\frac {F^{a+b c} \left (b d x^3 \operatorname {ExpIntegralEi}(b d x \log (F)) \log (F) \left (6 f^2+6 b d e f \log (F)+b^2 d^2 e^2 \log ^2(F)\right )-F^{b d x} \left (2 \left (e^2+3 e f x+3 f^2 x^2\right )+b d e x (e+6 f x) \log (F)+b^2 d^2 e^2 x^2 \log ^2(F)\right )\right )}{6 x^3} \] Input:

Integrate[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^4,x]
 

Output:

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

Rubi [A] (verified)

Time = 1.25 (sec) , antiderivative size = 217, 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 = {2629, 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.

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

\(\Big \downarrow \) 2629

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{6} b^3 d^3 e^2 \log ^3(F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{6 x}+b^2 d^2 e f \log ^2(F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{3 x^3}-\frac {b d e^2 \log (F) F^{a+b c+b d x}}{6 x^2}-\frac {e f F^{a+b c+b d x}}{x^2}-\frac {b d e f \log (F) F^{a+b c+b d x}}{x}+b d f^2 \log (F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {f^2 F^{a+b c+b d x}}{x}\)

Input:

Int[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^4,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2629

Int[(F_)^(v_)*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandInte 
grand[F^v, Px*(d + e*x)^m, x], x] /; FreeQ[{F, d, e, m}, x] && PolynomialQ[ 
Px, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.35

method	result	size
risch	\(-\frac {F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) \ln \left (F \right )^{3} b^{3} d^{3} e^{2} x^{3}+6 F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) \ln \left (F \right )^{2} b^{2} d^{2} e f \,x^{3}+F^{b d x} F^{b c +a} b^{2} d^{2} e^{2} x^{2} \ln \left (F \right )^{2}+6 F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) \ln \left (F \right ) b d \,f^{2} x^{3}+6 F^{b d x} F^{b c +a} b d e f \,x^{2} \ln \left (F \right )+F^{b d x} F^{b c +a} b d \,e^{2} x \ln \left (F \right )+6 F^{b d x} F^{b c +a} f^{2} x^{2}+6 F^{b d x} F^{b c +a} e f x +2 F^{b d x} F^{b c +a} e^{2}}{6 x^{3}}\)	\(294\)
meijerg	\(-b d \ln \left (F \right ) F^{b c +a} f^{2} \left (\frac {1}{b d x \ln \left (F \right )}+1-\ln \left (x \right )-\ln \left (-b d \right )-\ln \left (\ln \left (F \right )\right )-\frac {2+2 b d x \ln \left (F \right )}{2 b d x \ln \left (F \right )}+\frac {{\mathrm e}^{b d x \ln \left (F \right )}}{b d x \ln \left (F \right )}+\ln \left (-b d x \ln \left (F \right )\right )+\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )\right )+2 \ln \left (F \right )^{2} b^{2} d^{2} F^{b c +a} f e \left (-\frac {1}{2 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {1}{b d x \ln \left (F \right )}-\frac {3}{4}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (-b d \right )}{2}+\frac {\ln \left (\ln \left (F \right )\right )}{2}+\frac {9 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+12 b d x \ln \left (F \right )+6}{12 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {\left (3 b d x \ln \left (F \right )+3\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{6 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {\ln \left (-b d x \ln \left (F \right )\right )}{2}-\frac {\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )}{2}\right )-F^{b c +a} e^{2} \ln \left (F \right )^{3} b^{3} d^{3} \left (\frac {1}{3 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}}+\frac {1}{2 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}+\frac {1}{2 b d x \ln \left (F \right )}+\frac {11}{36}-\frac {\ln \left (x \right )}{6}-\frac {\ln \left (-b d \right )}{6}-\frac {\ln \left (\ln \left (F \right )\right )}{6}-\frac {22 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}+36 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+36 b d x \ln \left (F \right )+24}{72 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}}+\frac {\left (4 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+4 b d x \ln \left (F \right )+8\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{24 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}}+\frac {\ln \left (-b d x \ln \left (F \right )\right )}{6}+\frac {\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )}{6}\right )\)	\(478\)

Input:

int(F^(a+b*(d*x+c))*(f*x+e)^2/x^4,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.63 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=\frac {{\left (b^{3} d^{3} e^{2} x^{3} \log \left (F\right )^{3} + 6 \, b^{2} d^{2} e f x^{3} \log \left (F\right )^{2} + 6 \, b d f^{2} x^{3} \log \left (F\right )\right )} F^{b c + a} {\rm Ei}\left (b d x \log \left (F\right )\right ) - {\left (b^{2} d^{2} e^{2} x^{2} \log \left (F\right )^{2} + 6 \, f^{2} x^{2} + 6 \, e f x + 2 \, e^{2} + {\left (6 \, b d e f x^{2} + b d e^{2} x\right )} \log \left (F\right )\right )} F^{b d x + b c + a}}{6 \, x^{3}} \] Input:

integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^4,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=\int \frac {F^{a + b \left (c + d x\right )} \left (e + f x\right )^{2}}{x^{4}}\, dx \] Input:

integrate(F**(a+b*(d*x+c))*(f*x+e)**2/x**4,x)
 

Output:

Integral(F**(a + b*(c + d*x))*(e + f*x)**2/x**4, x)
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.39 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=F^{b c + a} b^{3} d^{3} e^{2} \Gamma \left (-3, -b d x \log \left (F\right )\right ) \log \left (F\right )^{3} - 2 \, F^{b c + a} b^{2} d^{2} e f \Gamma \left (-2, -b d x \log \left (F\right )\right ) \log \left (F\right )^{2} + F^{b c + a} b d f^{2} \Gamma \left (-1, -b d x \log \left (F\right )\right ) \log \left (F\right ) \] Input:

integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^4,x, algorithm="maxima")
 

Output:

F^(b*c + a)*b^3*d^3*e^2*gamma(-3, -b*d*x*log(F))*log(F)^3 - 2*F^(b*c + a)* 
b^2*d^2*e*f*gamma(-2, -b*d*x*log(F))*log(F)^2 + F^(b*c + a)*b*d*f^2*gamma( 
-1, -b*d*x*log(F))*log(F)
 

Giac [F]

\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=\int { \frac {{\left (f x + e\right )}^{2} F^{{\left (d x + c\right )} b + a}}{x^{4}} \,d x } \] Input:

integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^4,x, algorithm="giac")
 

Output:

integrate((f*x + e)^2*F^((d*x + c)*b + a)/x^4, x)
 

Mupad [B] (verification not implemented)

Time = 22.88 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.93 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=-\frac {F^{b\,d\,x}\,F^{a+b\,c}\,f^2}{x}-F^{a+b\,c}\,b^3\,d^3\,e^2\,{\ln \left (F\right )}^3\,\left (F^{b\,d\,x}\,\left (\frac {1}{6\,b\,d\,x\,\ln \left (F\right )}+\frac {1}{6\,b^2\,d^2\,x^2\,{\ln \left (F\right )}^2}+\frac {1}{3\,b^3\,d^3\,x^3\,{\ln \left (F\right )}^3}\right )+\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )}{6}\right )-F^{a+b\,c}\,b\,d\,f^2\,\ln \left (F\right )\,\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )-2\,F^{a+b\,c}\,b^2\,d^2\,e\,f\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )}{2}+F^{b\,d\,x}\,\left (\frac {1}{2\,b\,d\,x\,\ln \left (F\right )}+\frac {1}{2\,b^2\,d^2\,x^2\,{\ln \left (F\right )}^2}\right )\right ) \] Input:

int((F^(a + b*(c + d*x))*(e + f*x)^2)/x^4,x)
 

Output:

- (F^(b*d*x)*F^(a + b*c)*f^2)/x - F^(a + b*c)*b^3*d^3*e^2*log(F)^3*(F^(b*d 
*x)*(1/(6*b*d*x*log(F)) + 1/(6*b^2*d^2*x^2*log(F)^2) + 1/(3*b^3*d^3*x^3*lo 
g(F)^3)) + expint(-b*d*x*log(F))/6) - F^(a + b*c)*b*d*f^2*log(F)*expint(-b 
*d*x*log(F)) - 2*F^(a + b*c)*b^2*d^2*e*f*log(F)^2*(expint(-b*d*x*log(F))/2 
 + F^(b*d*x)*(1/(2*b*d*x*log(F)) + 1/(2*b^2*d^2*x^2*log(F)^2)))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.80 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=\frac {f^{b c +a} \left (\mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) \mathrm {log}\left (f \right )^{3} b^{3} d^{3} e^{2} x^{3}+6 \mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) \mathrm {log}\left (f \right )^{2} b^{2} d^{2} e f \,x^{3}+6 \mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) \mathrm {log}\left (f \right ) b d \,f^{2} x^{3}-f^{b d x} \mathrm {log}\left (f \right )^{2} b^{2} d^{2} e^{2} x^{2}-f^{b d x} \mathrm {log}\left (f \right ) b d \,e^{2} x -6 f^{b d x} \mathrm {log}\left (f \right ) b d e f \,x^{2}-2 f^{b d x} e^{2}-6 f^{b d x} e f x -6 f^{b d x} f^{2} x^{2}\right )}{6 x^{3}} \] Input:

int(F^(a+b*(d*x+c))*(f*x+e)^2/x^4,x)
 

Output:

(f**(a + b*c)*(ei(log(f)*b*d*x)*log(f)**3*b**3*d**3*e**2*x**3 + 6*ei(log(f 
)*b*d*x)*log(f)**2*b**2*d**2*e*f*x**3 + 6*ei(log(f)*b*d*x)*log(f)*b*d*f**2 
*x**3 - f**(b*d*x)*log(f)**2*b**2*d**2*e**2*x**2 - f**(b*d*x)*log(f)*b*d*e 
**2*x - 6*f**(b*d*x)*log(f)*b*d*e*f*x**2 - 2*f**(b*d*x)*e**2 - 6*f**(b*d*x 
)*e*f*x - 6*f**(b*d*x)*f**2*x**2))/(6*x**3)