Integrand size = 22, antiderivative size = 136 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=-\frac {e^2 F^{a+b c+b d x}}{2 x^2}-\frac {2 e f F^{a+b c+b d x}}{x}+f^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {b d e^2 F^{a+b c+b d x} \log (F)}{2 x}+2 b d e f F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log (F)+\frac {1}{2} b^2 d^2 e^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log ^2(F) \] Output:
-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 = 0.39 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.56 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\frac {F^{a+b c} \left (-e F^{b d x} (e+4 f x+b d e x \log (F))+x^2 \operatorname {ExpIntegralEi}(b d x \log (F)) \left (2 f^2+4 b d e f \log (F)+b^2 d^2 e^2 \log ^2(F)\right )\right )}{2 x^2} \] Input:
Integrate[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^3,x]
Output:
(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 = 1.04 (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 |
\(\displaystyle \frac {1}{2} b^2 d^2 e^2 \log ^2(F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d e^2 \log (F) F^{a+b c+b d x}}{2 x}+2 b d e f \log (F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {2 e f F^{a+b c+b d x}}{x}+f^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))\) |
Input:
Int[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^3,x]
Output:
-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
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 = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.53
method | result | size |
risch | \(-\frac {\ln \left (F \right )^{2} F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) b^{2} d^{2} e^{2} x^{2}+4 \ln \left (F \right ) F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) b d e f \,x^{2}+F^{b d x} F^{b c +a} b d \,e^{2} x \ln \left (F \right )+2 F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) f^{2} x^{2}+4 F^{b d x} F^{b c +a} e f x +F^{b d x} F^{b c +a} e^{2}}{2 x^{2}}\) | \(208\) |
meijerg | \(F^{b c +a} f^{2} \left (\ln \left (x \right )+\ln \left (-b d \right )+\ln \left (\ln \left (F \right )\right )-\ln \left (-b d x \ln \left (F \right )\right )-\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )\right )-2 b d \ln \left (F \right ) F^{b c +a} f e \left (\frac {1}{b d x \ln \left (F \right )}+1-\ln \left (x \right )-\ln \left (-b d \right )-\ln \left (\ln \left (F \right )\right )-\frac {2+2 b d x \ln \left (F \right )}{2 b d x \ln \left (F \right )}+\frac {{\mathrm e}^{b d x \ln \left (F \right )}}{b d x \ln \left (F \right )}+\ln \left (-b d x \ln \left (F \right )\right )+\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )\right )+F^{b c +a} e^{2} \ln \left (F \right )^{2} b^{2} d^{2} \left (-\frac {1}{2 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {1}{b d x \ln \left (F \right )}-\frac {3}{4}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (-b d \right )}{2}+\frac {\ln \left (\ln \left (F \right )\right )}{2}+\frac {9 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+12 b d x \ln \left (F \right )+6}{12 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {\left (3 b d x \ln \left (F \right )+3\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{6 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {\ln \left (-b d x \ln \left (F \right )\right )}{2}-\frac {\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )}{2}\right )\) | \(314\) |
Input:
int(F^(a+b*(d*x+c))*(f*x+e)^2/x^3,x,method=_RETURNVERBOSE)
Output:
-1/2*(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+4*ln(F)*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*d*e*f*x^2+F^(b*d*x)*F^(b*c+a)*b*d*e^2*x*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))*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 = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\frac {{\left (b^{2} d^{2} e^{2} x^{2} \log \left (F\right )^{2} + 4 \, b d e f x^{2} \log \left (F\right ) + 2 \, f^{2} x^{2}\right )} F^{b c + a} {\rm Ei}\left (b d x \log \left (F\right )\right ) - {\left (b d e^{2} x \log \left (F\right ) + 4 \, e f x + e^{2}\right )} F^{b d x + b c + a}}{2 \, x^{2}} \] Input:
integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^3,x, algorithm="fricas")
Output:
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
\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\int \frac {F^{a + b \left (c + d x\right )} \left (e + f x\right )^{2}}{x^{3}}\, dx \] Input:
integrate(F**(a+b*(d*x+c))*(f*x+e)**2/x**3,x)
Output:
Integral(F**(a + b*(c + d*x))*(e + f*x)**2/x**3, x)
Time = 0.15 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.54 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=-F^{b c + a} b^{2} d^{2} e^{2} \Gamma \left (-2, -b d x \log \left (F\right )\right ) \log \left (F\right )^{2} + 2 \, F^{b c + a} b d e f \Gamma \left (-1, -b d x \log \left (F\right )\right ) \log \left (F\right ) + F^{b c + a} f^{2} {\rm Ei}\left (b d x \log \left (F\right )\right ) \] Input:
integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^3,x, algorithm="maxima")
Output:
-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) )
\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{2} F^{{\left (d x + c\right )} b + a}}{x^{3}} \,d x } \] Input:
integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^3,x, algorithm="giac")
Output:
integrate((f*x + e)^2*F^((d*x + c)*b + a)/x^3, x)
Time = 22.69 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.98 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=F^{a+b\,c}\,f^2\,\mathrm {ei}\left (b\,d\,x\,\ln \left (F\right )\right )-\frac {2\,F^{b\,d\,x}\,F^{a+b\,c}\,e\,f}{x}-F^{a+b\,c}\,b^2\,d^2\,e^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )}{2}+F^{b\,d\,x}\,\left (\frac {1}{2\,b\,d\,x\,\ln \left (F\right )}+\frac {1}{2\,b^2\,d^2\,x^2\,{\ln \left (F\right )}^2}\right )\right )-2\,F^{a+b\,c}\,b\,d\,e\,f\,\ln \left (F\right )\,\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right ) \] Input:
int((F^(a + b*(c + d*x))*(e + f*x)^2)/x^3,x)
Output:
F^(a + b*c)*f^2*ei(b*d*x*log(F)) - (2*F^(b*d*x)*F^(a + b*c)*e*f)/x - F^(a + b*c)*b^2*d^2*e^2*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))) - 2*F^(a + b*c)*b*d*e*f*log(F)*e xpint(-b*d*x*log(F))
Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.79 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\frac {f^{b c +a} \left (\mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) \mathrm {log}\left (f \right )^{2} b^{2} d^{2} e^{2} x^{2}+4 \mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) \mathrm {log}\left (f \right ) b d e f \,x^{2}+2 \mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) f^{2} x^{2}-f^{b d x} \mathrm {log}\left (f \right ) b d \,e^{2} x -f^{b d x} e^{2}-4 f^{b d x} e f x \right )}{2 x^{2}} \] Input:
int(F^(a+b*(d*x+c))*(f*x+e)^2/x^3,x)
Output:
(f**(a + b*c)*(ei(log(f)*b*d*x)*log(f)**2*b**2*d**2*e**2*x**2 + 4*ei(log(f )*b*d*x)*log(f)*b*d*e*f*x**2 + 2*ei(log(f)*b*d*x)*f**2*x**2 - f**(b*d*x)*l og(f)*b*d*e**2*x - f**(b*d*x)*e**2 - 4*f**(b*d*x)*e*f*x))/(2*x**2)