Integrand size = 22, antiderivative size = 85 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=-\frac {e^2 F^{a+b c+b d x}}{x}+2 e f F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))+\frac {f^2 F^{a+b c+b d x}}{b d \log (F)}+b d e^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log (F) \] Output:
-e^2*F^(b*d*x+b*c+a)/x+2*e*f*F^(b*c+a)*Ei(b*d*x*ln(F))+f^2*F^(b*d*x+b*c+a) /b/d/ln(F)+b*d*e^2*F^(b*c+a)*Ei(b*d*x*ln(F))*ln(F)
Time = 0.42 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=F^{a+b c} \left (F^{b d x} \left (-\frac {e^2}{x}+\frac {f^2}{b d \log (F)}\right )+e \operatorname {ExpIntegralEi}(b d x \log (F)) (2 f+b d e \log (F))\right ) \] Input:
Integrate[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^2,x]
Output:
F^(a + b*c)*(F^(b*d*x)*(-(e^2/x) + f^2/(b*d*Log[F])) + e*ExpIntegralEi[b*d *x*Log[F]]*(2*f + b*d*e*Log[F]))
Time = 0.83 (sec) , antiderivative size = 85, 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.
\(\displaystyle \int \frac {(e+f x)^2 F^{a+b (c+d x)}}{x^2} \, dx\) |
\(\Big \downarrow \) 2629 |
\(\displaystyle \int \left (\frac {e^2 F^{a+b (c+d x)}}{x^2}+\frac {2 e f F^{a+b (c+d x)}}{x}+f^2 F^{a+b (c+d x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle b d e^2 \log (F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{x}+2 e f F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))+\frac {f^2 F^{a+b (c+d x)}}{b d \log (F)}\) |
Input:
Int[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^2,x]
Output:
-((e^2*F^(a + b*c + b*d*x))/x) + 2*e*f*F^(a + b*c)*ExpIntegralEi[b*d*x*Log [F]] + (f^2*F^(a + b*(c + d*x)))/(b*d*Log[F]) + b*d*e^2*F^(a + b*c)*ExpInt egralEi[b*d*x*Log[F]]*Log[F]
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]
Time = 0.09 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.80
method | result | size |
risch | \(-\frac {\ln \left (F \right )^{2} F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) b^{2} d^{2} e^{2} x +2 \ln \left (F \right ) F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) b d e f x +\ln \left (F \right ) F^{b d x} F^{b c +a} b d \,e^{2}-F^{b d x} F^{b c +a} f^{2} x}{\ln \left (F \right ) b d x}\) | \(153\) |
meijerg | \(-\frac {F^{b c +a} f^{2} \left (1-{\mathrm e}^{b d x \ln \left (F \right )}\right )}{b d \ln \left (F \right )}+2 F^{b c +a} f e \left (\ln \left (x \right )+\ln \left (-b d \right )+\ln \left (\ln \left (F \right )\right )-\ln \left (-b d x \ln \left (F \right )\right )-\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )\right )-F^{b c +a} e^{2} b d \ln \left (F \right ) \left (\frac {1}{b d x \ln \left (F \right )}+1-\ln \left (x \right )-\ln \left (-b d \right )-\ln \left (\ln \left (F \right )\right )-\frac {2+2 b d x \ln \left (F \right )}{2 b d x \ln \left (F \right )}+\frac {{\mathrm e}^{b d x \ln \left (F \right )}}{b d x \ln \left (F \right )}+\ln \left (-b d x \ln \left (F \right )\right )+\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )\right )\) | \(188\) |
Input:
int(F^(a+b*(d*x+c))*(f*x+e)^2/x^2,x,method=_RETURNVERBOSE)
Output:
-1/ln(F)/b/d*(ln(F)^2*F^(b*c)*F^a*Ei(1,ln(F)*b*c+ln(F)*a-b*d*x*ln(F)-(b*c+ a)*ln(F))*b^2*d^2*e^2*x+2*ln(F)*F^(b*c)*F^a*Ei(1,ln(F)*b*c+ln(F)*a-b*d*x*l n(F)-(b*c+a)*ln(F))*b*d*e*f*x+ln(F)*F^(b*d*x)*F^(b*c+a)*b*d*e^2-F^(b*d*x)* F^(b*c+a)*f^2*x)/x
Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=\frac {{\left (b^{2} d^{2} e^{2} x \log \left (F\right )^{2} + 2 \, b d e f x \log \left (F\right )\right )} F^{b c + a} {\rm Ei}\left (b d x \log \left (F\right )\right ) - {\left (b d e^{2} \log \left (F\right ) - f^{2} x\right )} F^{b d x + b c + a}}{b d x \log \left (F\right )} \] Input:
integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^2,x, algorithm="fricas")
Output:
((b^2*d^2*e^2*x*log(F)^2 + 2*b*d*e*f*x*log(F))*F^(b*c + a)*Ei(b*d*x*log(F) ) - (b*d*e^2*log(F) - f^2*x)*F^(b*d*x + b*c + a))/(b*d*x*log(F))
\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=\int \frac {F^{a + b \left (c + d x\right )} \left (e + f x\right )^{2}}{x^{2}}\, dx \] Input:
integrate(F**(a+b*(d*x+c))*(f*x+e)**2/x**2,x)
Output:
Integral(F**(a + b*(c + d*x))*(e + f*x)**2/x**2, x)
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=F^{b c + a} b d e^{2} \Gamma \left (-1, -b d x \log \left (F\right )\right ) \log \left (F\right ) + 2 \, F^{b c + a} e f {\rm Ei}\left (b d x \log \left (F\right )\right ) + \frac {F^{b d x + b c + a} f^{2}}{b d \log \left (F\right )} \] Input:
integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^2,x, algorithm="maxima")
Output:
F^(b*c + a)*b*d*e^2*gamma(-1, -b*d*x*log(F))*log(F) + 2*F^(b*c + a)*e*f*Ei (b*d*x*log(F)) + F^(b*d*x + b*c + a)*f^2/(b*d*log(F))
\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=\int { \frac {{\left (f x + e\right )}^{2} F^{{\left (d x + c\right )} b + a}}{x^{2}} \,d x } \] Input:
integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^2,x, algorithm="giac")
Output:
integrate((f*x + e)^2*F^((d*x + c)*b + a)/x^2, x)
Time = 22.69 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=2\,F^{a+b\,c}\,e\,f\,\mathrm {ei}\left (b\,d\,x\,\ln \left (F\right )\right )-\frac {F^{b\,d\,x}\,F^{a+b\,c}\,e^2}{x}+\frac {F^{a+b\,c+b\,d\,x}\,f^2}{b\,d\,\ln \left (F\right )}-F^{a+b\,c}\,b\,d\,e^2\,\ln \left (F\right )\,\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right ) \] Input:
int((F^(a + b*(c + d*x))*(e + f*x)^2)/x^2,x)
Output:
2*F^(a + b*c)*e*f*ei(b*d*x*log(F)) - (F^(b*d*x)*F^(a + b*c)*e^2)/x + (F^(a + b*c + b*d*x)*f^2)/(b*d*log(F)) - F^(a + b*c)*b*d*e^2*log(F)*expint(-b*d *x*log(F))
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=\frac {f^{b c +a} \left (\mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) \mathrm {log}\left (f \right )^{2} b^{2} d^{2} e^{2} x +2 \mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) \mathrm {log}\left (f \right ) b d e f x -f^{b d x} \mathrm {log}\left (f \right ) b d \,e^{2}+f^{b d x} f^{2} x \right )}{\mathrm {log}\left (f \right ) b d x} \] Input:
int(F^(a+b*(d*x+c))*(f*x+e)^2/x^2,x)
Output:
(f**(a + b*c)*(ei(log(f)*b*d*x)*log(f)**2*b**2*d**2*e**2*x + 2*ei(log(f)*b *d*x)*log(f)*b*d*e*f*x - f**(b*d*x)*log(f)*b*d*e**2 + f**(b*d*x)*f**2*x))/ (log(f)*b*d*x)