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

Optimal. Leaf size=217 \[ \frac {1}{6} b^3 d^3 e^2 \log ^3(F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{6 x}+b^2 d^2 e f \log ^2(F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{3 x^3}-\frac {b d e^2 \log (F) F^{a+b c+b d x}}{6 x^2}-\frac {e f F^{a+b c+b d x}}{x^2}-\frac {b d e f \log (F) F^{a+b c+b d x}}{x}+b d f^2 \log (F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {f^2 F^{a+b c+b d x}}{x} \]

[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

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Rubi [A]  time = 0.46, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2199, 2177, 2178} \[ \frac {1}{6} b^3 d^3 e^2 \log ^3(F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{6 x}+b^2 d^2 e f \log ^2(F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{3 x^3}-\frac {b d e^2 \log (F) F^{a+b c+b d x}}{6 x^2}-\frac {e f F^{a+b c+b d x}}{x^2}-\frac {b d e f \log (F) F^{a+b c+b d x}}{x}+b d f^2 \log (F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {f^2 F^{a+b c+b d x}}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^2*F^(a + b*c + b*d*x))/(3*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 2177

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] &&  !$UseGamma ===
True

Rule 2178

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

Rule 2199

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] &&  !$UseGamma === True

Rubi steps

\begin {align*} \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx &=\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*}

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Mathematica [A]  time = 0.23, size = 116, normalized size = 0.53 \[ \frac {F^{a+b c} \left (b d x^3 \log (F) \left (b^2 d^2 e^2 \log ^2(F)+6 b d e f \log (F)+6 f^2\right ) \text {Ei}(b d x \log (F))-F^{b d x} \left (b^2 d^2 e^2 x^2 \log ^2(F)+b d e x \log (F) (e+6 f x)+2 \left (e^2+3 e f x+3 f^2 x^2\right )\right )\right )}{6 x^3} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.42, size = 137, normalized size = 0.63 \[ \frac {{\left (b^{3} d^{3} e^{2} x^{3} \log \relax (F)^{3} + 6 \, b^{2} d^{2} e f x^{3} \log \relax (F)^{2} + 6 \, b d f^{2} x^{3} \log \relax (F)\right )} F^{b c + a} {\rm Ei}\left (b d x \log \relax (F)\right ) - {\left (b^{2} d^{2} e^{2} x^{2} \log \relax (F)^{2} + 6 \, f^{2} x^{2} + 6 \, e f x + 2 \, e^{2} + {\left (6 \, b d e f x^{2} + b d e^{2} x\right )} \log \relax (F)\right )} F^{b d x + b c + a}}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{2} F^{{\left (d x + c\right )} b + a}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.08, size = 290, normalized size = 1.34 \[ -\frac {b^{3} d^{3} e^{2} F^{a} F^{b c} \Ei \left (1, -b d x \ln \relax (F )+b c \ln \relax (F )+a \ln \relax (F )-\left (b c +a \right ) \ln \relax (F )\right ) \ln \relax (F )^{3}}{6}-b^{2} d^{2} e f \,F^{a} F^{b c} \Ei \left (1, -b d x \ln \relax (F )+b c \ln \relax (F )+a \ln \relax (F )-\left (b c +a \right ) \ln \relax (F )\right ) \ln \relax (F )^{2}-\frac {b^{2} d^{2} e^{2} F^{b d x} F^{b c +a} \ln \relax (F )^{2}}{6 x}-b d \,f^{2} F^{a} F^{b c} \Ei \left (1, -b d x \ln \relax (F )+b c \ln \relax (F )+a \ln \relax (F )-\left (b c +a \right ) \ln \relax (F )\right ) \ln \relax (F )-\frac {b d e f \,F^{b d x} F^{b c +a} \ln \relax (F )}{x}-\frac {b d \,e^{2} F^{b d x} F^{b c +a} \ln \relax (F )}{6 x^{2}}-\frac {f^{2} F^{b d x} F^{b c +a}}{x}-\frac {e f \,F^{b d x} F^{b c +a}}{x^{2}}-\frac {e^{2} F^{b d x} F^{b c +a}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.05, size = 85, normalized size = 0.39 \[ F^{b c + a} b^{3} d^{3} e^{2} \Gamma \left (-3, -b d x \log \relax (F)\right ) \log \relax (F)^{3} - 2 \, F^{b c + a} b^{2} d^{2} e f \Gamma \left (-2, -b d x \log \relax (F)\right ) \log \relax (F)^{2} + F^{b c + a} b d f^{2} \Gamma \left (-1, -b d x \log \relax (F)\right ) \log \relax (F) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.66, size = 202, normalized size = 0.93 \[ -\frac {F^{b\,d\,x}\,F^{a+b\,c}\,f^2}{x}-F^{a+b\,c}\,b^3\,d^3\,e^2\,{\ln \relax (F)}^3\,\left (F^{b\,d\,x}\,\left (\frac {1}{6\,b\,d\,x\,\ln \relax (F)}+\frac {1}{6\,b^2\,d^2\,x^2\,{\ln \relax (F)}^2}+\frac {1}{3\,b^3\,d^3\,x^3\,{\ln \relax (F)}^3}\right )+\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \relax (F)\right )}{6}\right )-F^{a+b\,c}\,b\,d\,f^2\,\ln \relax (F)\,\mathrm {expint}\left (-b\,d\,x\,\ln \relax (F)\right )-2\,F^{a+b\,c}\,b^2\,d^2\,e\,f\,{\ln \relax (F)}^2\,\left (\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \relax (F)\right )}{2}+F^{b\,d\,x}\,\left (\frac {1}{2\,b\,d\,x\,\ln \relax (F)}+\frac {1}{2\,b^2\,d^2\,x^2\,{\ln \relax (F)}^2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + b \left (c + d x\right )} \left (e + f x\right )^{2}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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