Integrand size = 22, antiderivative size = 136 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=-\frac {e^2 F^{a+b c+b d x}}{2 x^2}-\frac {2 e f F^{a+b c+b d x}}{x}+f^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {b d e^2 F^{a+b c+b d x} \log (F)}{2 x}+2 b d e f F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log (F)+\frac {1}{2} b^2 d^2 e^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log ^2(F) \]
-1/2*e^2*F^(b*d*x+b*c+a)/x^2-2*e*f*F^(b*d*x+b*c+a)/x+f^2*F^(b*c+a)*Ei(b*d* x*ln(F))-1/2*b*d*e^2*F^(b*d*x+b*c+a)*ln(F)/x+2*b*d*e*f*F^(b*c+a)*Ei(b*d*x* ln(F))*ln(F)+1/2*b^2*d^2*e^2*F^(b*c+a)*Ei(b*d*x*ln(F))*ln(F)^2
Time = 0.33 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.56 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\frac {F^{a+b c} \left (-e F^{b d x} (e+4 f x+b d e x \log (F))+x^2 \operatorname {ExpIntegralEi}(b d x \log (F)) \left (2 f^2+4 b d e f \log (F)+b^2 d^2 e^2 \log ^2(F)\right )\right )}{2 x^2} \]
(F^(a + b*c)*(-(e*F^(b*d*x)*(e + 4*f*x + b*d*e*x*Log[F])) + x^2*ExpIntegra lEi[b*d*x*Log[F]]*(2*f^2 + 4*b*d*e*f*Log[F] + b^2*d^2*e^2*Log[F]^2)))/(2*x ^2)
Time = 0.64 (sec) , antiderivative size = 136, 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.
\(\displaystyle \int \frac {(e+f x)^2 F^{a+b (c+d x)}}{x^3} \, dx\) |
\(\Big \downarrow \) 2629 |
\(\displaystyle \int \left (\frac {e^2 F^{a+b (c+d x)}}{x^3}+\frac {2 e f F^{a+b (c+d x)}}{x^2}+\frac {f^2 F^{a+b (c+d x)}}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} b^2 d^2 e^2 \log ^2(F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d e^2 \log (F) F^{a+b c+b d x}}{2 x}+2 b d e f \log (F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {2 e f F^{a+b c+b d x}}{x}+f^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))\) |
-1/2*(e^2*F^(a + b*c + b*d*x))/x^2 - (2*e*f*F^(a + b*c + b*d*x))/x + f^2*F ^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]] - (b*d*e^2*F^(a + b*c + b*d*x)*Log[ F])/(2*x) + 2*b*d*e*f*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F] + (b^ 2*d^2*e^2*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^2)/2
3.1.71.3.1 Defintions of rubi rules used
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 = 0.23 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.53
method | result | size |
risch | \(-\frac {\ln \left (F \right )^{2} F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) b^{2} d^{2} e^{2} x^{2}+4 \ln \left (F \right ) F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) b d e f \,x^{2}+\ln \left (F \right ) F^{b d x} F^{c b +a} b d \,e^{2} x +2 F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) f^{2} x^{2}+4 F^{b d x} F^{c b +a} e f x +F^{b d x} F^{c b +a} e^{2}}{2 x^{2}}\) | \(208\) |
meijerg | \(F^{c b +a} f^{2} \left (\ln \left (x \right )+\ln \left (-b d \right )+\ln \left (\ln \left (F \right )\right )-\ln \left (-b d x \ln \left (F \right )\right )-\operatorname {Ei}_{1}\left (-b d x \ln \left (F \right )\right )\right )-2 b d \ln \left (F \right ) F^{c b +a} f e \left (\frac {1}{b d x \ln \left (F \right )}+1-\ln \left (x \right )-\ln \left (-b d \right )-\ln \left (\ln \left (F \right )\right )-\frac {2+2 b d x \ln \left (F \right )}{2 b d x \ln \left (F \right )}+\frac {{\mathrm e}^{b d x \ln \left (F \right )}}{b d x \ln \left (F \right )}+\ln \left (-b d x \ln \left (F \right )\right )+\operatorname {Ei}_{1}\left (-b d x \ln \left (F \right )\right )\right )+F^{c b +a} e^{2} \ln \left (F \right )^{2} b^{2} d^{2} \left (-\frac {1}{2 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {1}{b d x \ln \left (F \right )}-\frac {3}{4}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (-b d \right )}{2}+\frac {\ln \left (\ln \left (F \right )\right )}{2}+\frac {9 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+12 b d x \ln \left (F \right )+6}{12 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {\left (3 b d x \ln \left (F \right )+3\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{6 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {\ln \left (-b d x \ln \left (F \right )\right )}{2}-\frac {\operatorname {Ei}_{1}\left (-b d x \ln \left (F \right )\right )}{2}\right )\) | \(314\) |
-1/2*(ln(F)^2*F^(c*b)*F^a*Ei(1,c*b*ln(F)+a*ln(F)-b*d*x*ln(F)-(b*c+a)*ln(F) )*b^2*d^2*e^2*x^2+4*ln(F)*F^(c*b)*F^a*Ei(1,c*b*ln(F)+a*ln(F)-b*d*x*ln(F)-( b*c+a)*ln(F))*b*d*e*f*x^2+ln(F)*F^(b*d*x)*F^(b*c+a)*b*d*e^2*x+2*F^(c*b)*F^ a*Ei(1,c*b*ln(F)+a*ln(F)-b*d*x*ln(F)-(b*c+a)*ln(F))*f^2*x^2+4*F^(b*d*x)*F^ (b*c+a)*e*f*x+F^(b*d*x)*F^(b*c+a)*e^2)/x^2
Time = 0.23 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\frac {{\left (b^{2} d^{2} e^{2} x^{2} \log \left (F\right )^{2} + 4 \, b d e f x^{2} \log \left (F\right ) + 2 \, f^{2} x^{2}\right )} F^{b c + a} {\rm Ei}\left (b d x \log \left (F\right )\right ) - {\left (b d e^{2} x \log \left (F\right ) + 4 \, e f x + e^{2}\right )} F^{b d x + b c + a}}{2 \, x^{2}} \]
1/2*((b^2*d^2*e^2*x^2*log(F)^2 + 4*b*d*e*f*x^2*log(F) + 2*f^2*x^2)*F^(b*c + a)*Ei(b*d*x*log(F)) - (b*d*e^2*x*log(F) + 4*e*f*x + e^2)*F^(b*d*x + b*c + a))/x^2
\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\int \frac {F^{a + b \left (c + d x\right )} \left (e + f x\right )^{2}}{x^{3}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.54 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=-F^{b c + a} b^{2} d^{2} e^{2} \Gamma \left (-2, -b d x \log \left (F\right )\right ) \log \left (F\right )^{2} + 2 \, F^{b c + a} b d e f \Gamma \left (-1, -b d x \log \left (F\right )\right ) \log \left (F\right ) + F^{b c + a} f^{2} {\rm Ei}\left (b d x \log \left (F\right )\right ) \]
-F^(b*c + a)*b^2*d^2*e^2*gamma(-2, -b*d*x*log(F))*log(F)^2 + 2*F^(b*c + a) *b*d*e*f*gamma(-1, -b*d*x*log(F))*log(F) + F^(b*c + a)*f^2*Ei(b*d*x*log(F) )
\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{2} F^{{\left (d x + c\right )} b + a}}{x^{3}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.98 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=F^{a+b\,c}\,f^2\,\mathrm {ei}\left (b\,d\,x\,\ln \left (F\right )\right )-\frac {2\,F^{b\,d\,x}\,F^{a+b\,c}\,e\,f}{x}-F^{a+b\,c}\,b^2\,d^2\,e^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )}{2}+F^{b\,d\,x}\,\left (\frac {1}{2\,b\,d\,x\,\ln \left (F\right )}+\frac {1}{2\,b^2\,d^2\,x^2\,{\ln \left (F\right )}^2}\right )\right )-2\,F^{a+b\,c}\,b\,d\,e\,f\,\ln \left (F\right )\,\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right ) \]