\(\int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx\) [72]

Optimal result

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

[Out]

-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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2230, 2208, 2209} \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=\frac {1}{6} b^3 d^3 e^2 \log ^3(F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{6 x}+b^2 d^2 e f \log ^2(F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{3 x^3}-\frac {b d e^2 \log (F) F^{a+b c+b d x}}{6 x^2}-\frac {e f F^{a+b c+b d x}}{x^2}-\frac {b d e f \log (F) F^{a+b c+b d x}}{x}+b d f^2 \log (F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {f^2 F^{a+b c+b d x}}{x} \]

[In]

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

[Out]

-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)*ExpIntegral
Ei[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

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^2 F^{a+b c+b d x}}{x^4}+\frac {2 e f F^{a+b c+b d x}}{x^3}+\frac {f^2 F^{a+b c+b d x}}{x^2}\right ) \, dx \\ & = e^2 \int \frac {F^{a+b c+b d x}}{x^4} \, dx+(2 e f) \int \frac {F^{a+b c+b d x}}{x^3} \, dx+f^2 \int \frac {F^{a+b c+b d x}}{x^2} \, 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 {1}{3} \left (b d e^2 \log (F)\right ) \int \frac {F^{a+b c+b d x}}{x^3} \, dx+(b d e f \log (F)) \int \frac {F^{a+b c+b d x}}{x^2} \, dx+\left (b d f^2 \log (F)\right ) \int \frac {F^{a+b c+b d x}}{x} \, 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} \text {Ei}(b d x \log (F)) \log (F)+\frac {1}{6} \left (b^2 d^2 e^2 \log ^2(F)\right ) \int \frac {F^{a+b c+b d x}}{x^2} \, dx+\left (b^2 d^2 e f \log ^2(F)\right ) \int \frac {F^{a+b c+b d x}}{x} \, 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} \text {Ei}(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} \text {Ei}(b d x \log (F)) \log ^2(F)+\frac {1}{6} \left (b^3 d^3 e^2 \log ^3(F)\right ) \int \frac {F^{a+b c+b d x}}{x} \, 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} \text {Ei}(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} \text {Ei}(b d x \log (F)) \log ^2(F)+\frac {1}{6} b^3 d^3 e^2 F^{a+b c} \text {Ei}(b d x \log (F)) \log ^3(F) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.53 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=\frac {F^{a+b c} \left (b d x^3 \operatorname {ExpIntegralEi}(b d x \log (F)) \log (F) \left (6 f^2+6 b d e f \log (F)+b^2 d^2 e^2 \log ^2(F)\right )-F^{b d x} \left (2 \left (e^2+3 e f x+3 f^2 x^2\right )+b d e x (e+6 f x) \log (F)+b^2 d^2 e^2 x^2 \log ^2(F)\right )\right )}{6 x^3} \]

[In]

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

[Out]

(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)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.35

method	result	size
risch	\(-\frac {\ln \left (F \right )^{3} F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) b^{3} d^{3} e^{2} x^{3}+6 \ln \left (F \right )^{2} F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) b^{2} d^{2} e f \,x^{3}+\ln \left (F \right )^{2} F^{b d x} F^{c b +a} b^{2} d^{2} e^{2} x^{2}+6 \ln \left (F \right ) F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) b d \,f^{2} x^{3}+6 \ln \left (F \right ) F^{b d x} F^{c b +a} b d e f \,x^{2}+\ln \left (F \right ) F^{b d x} F^{c b +a} b d \,e^{2} x +6 F^{b d x} F^{c b +a} f^{2} x^{2}+6 F^{b d x} F^{c b +a} e f x +2 F^{b d x} F^{c b +a} e^{2}}{6 x^{3}}\)	\(294\)
meijerg	\(-b d \ln \left (F \right ) F^{c b +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 {Ei}_{1}\left (-b d x \ln \left (F \right )\right )\right )+2 \ln \left (F \right )^{2} b^{2} d^{2} F^{c b +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 {Ei}_{1}\left (-b d x \ln \left (F \right )\right )}{2}\right )-F^{c b +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 {Ei}_{1}\left (-b d x \ln \left (F \right )\right )}{6}\right )\)	\(478\)

[In]

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

[Out]

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

none

Time = 0.24 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.63 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=\frac {{\left (b^{3} d^{3} e^{2} x^{3} \log \left (F\right )^{3} + 6 \, b^{2} d^{2} e f x^{3} \log \left (F\right )^{2} + 6 \, b d f^{2} x^{3} \log \left (F\right )\right )} F^{b c + a} {\rm Ei}\left (b d x \log \left (F\right )\right ) - {\left (b^{2} d^{2} e^{2} x^{2} \log \left (F\right )^{2} + 6 \, f^{2} x^{2} + 6 \, e f x + 2 \, e^{2} + {\left (6 \, b d e f x^{2} + b d e^{2} x\right )} \log \left (F\right )\right )} F^{b d x + b c + a}}{6 \, x^{3}} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.39 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=F^{b c + a} b^{3} d^{3} e^{2} \Gamma \left (-3, -b d x \log \left (F\right )\right ) \log \left (F\right )^{3} - 2 \, F^{b c + a} b^{2} d^{2} e f \Gamma \left (-2, -b d x \log \left (F\right )\right ) \log \left (F\right )^{2} + F^{b c + a} b d f^{2} \Gamma \left (-1, -b d x \log \left (F\right )\right ) \log \left (F\right ) \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.93 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx=-\frac {F^{b\,d\,x}\,F^{a+b\,c}\,f^2}{x}-F^{a+b\,c}\,b^3\,d^3\,e^2\,{\ln \left (F\right )}^3\,\left (F^{b\,d\,x}\,\left (\frac {1}{6\,b\,d\,x\,\ln \left (F\right )}+\frac {1}{6\,b^2\,d^2\,x^2\,{\ln \left (F\right )}^2}+\frac {1}{3\,b^3\,d^3\,x^3\,{\ln \left (F\right )}^3}\right )+\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )}{6}\right )-F^{a+b\,c}\,b\,d\,f^2\,\ln \left (F\right )\,\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )-2\,F^{a+b\,c}\,b^2\,d^2\,e\,f\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )}{2}+F^{b\,d\,x}\,\left (\frac {1}{2\,b\,d\,x\,\ln \left (F\right )}+\frac {1}{2\,b^2\,d^2\,x^2\,{\ln \left (F\right )}^2}\right )\right ) \]

[In]

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

[Out]

- (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*log(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)))