3.73 \(\int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx\)

Optimal. Leaf size=321 \[ \frac {1}{24} b^4 d^4 e^2 \log ^4(F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {b^3 d^3 e^2 \log ^3(F) F^{a+b c+b d x}}{24 x}+\frac {1}{3} b^3 d^3 e f \log ^3(F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{24 x^2}-\frac {b^2 d^2 e f \log ^2(F) F^{a+b c+b d x}}{3 x}+\frac {1}{2} b^2 d^2 f^2 \log ^2(F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{4 x^4}-\frac {b d e^2 \log (F) F^{a+b c+b d x}}{12 x^3}-\frac {2 e f F^{a+b c+b d x}}{3 x^3}-\frac {b d e f \log (F) F^{a+b c+b d x}}{3 x^2}-\frac {f^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d f^2 \log (F) F^{a+b c+b d x}}{2 x} \]

[Out]

-1/4*e^2*F^(b*d*x+b*c+a)/x^4-2/3*e*f*F^(b*d*x+b*c+a)/x^3-1/2*f^2*F^(b*d*x+b*c+a)/x^2-1/12*b*d*e^2*F^(b*d*x+b*c
+a)*ln(F)/x^3-1/3*b*d*e*f*F^(b*d*x+b*c+a)*ln(F)/x^2-1/2*b*d*f^2*F^(b*d*x+b*c+a)*ln(F)/x-1/24*b^2*d^2*e^2*F^(b*
d*x+b*c+a)*ln(F)^2/x^2-1/3*b^2*d^2*e*f*F^(b*d*x+b*c+a)*ln(F)^2/x+1/2*b^2*d^2*f^2*F^(b*c+a)*Ei(b*d*x*ln(F))*ln(
F)^2-1/24*b^3*d^3*e^2*F^(b*d*x+b*c+a)*ln(F)^3/x+1/3*b^3*d^3*e*f*F^(b*c+a)*Ei(b*d*x*ln(F))*ln(F)^3+1/24*b^4*d^4
*e^2*F^(b*c+a)*Ei(b*d*x*ln(F))*ln(F)^4

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Rubi [A]  time = 0.58, antiderivative size = 321, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2199, 2177, 2178} \[ \frac {1}{24} b^4 d^4 e^2 \log ^4(F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{24 x^2}-\frac {b^3 d^3 e^2 \log ^3(F) F^{a+b c+b d x}}{24 x}+\frac {1}{3} b^3 d^3 e f \log ^3(F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {b^2 d^2 e f \log ^2(F) F^{a+b c+b d x}}{3 x}+\frac {1}{2} b^2 d^2 f^2 \log ^2(F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{4 x^4}-\frac {b d e^2 \log (F) F^{a+b c+b d x}}{12 x^3}-\frac {2 e f F^{a+b c+b d x}}{3 x^3}-\frac {b d e f \log (F) F^{a+b c+b d x}}{3 x^2}-\frac {f^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d f^2 \log (F) F^{a+b c+b d x}}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^5,x]

[Out]

-(e^2*F^(a + b*c + b*d*x))/(4*x^4) - (2*e*f*F^(a + b*c + b*d*x))/(3*x^3) - (f^2*F^(a + b*c + b*d*x))/(2*x^2) -
(b*d*e^2*F^(a + b*c + b*d*x)*Log[F])/(12*x^3) - (b*d*e*f*F^(a + b*c + b*d*x)*Log[F])/(3*x^2) - (b*d*f^2*F^(a
+ b*c + b*d*x)*Log[F])/(2*x) - (b^2*d^2*e^2*F^(a + b*c + b*d*x)*Log[F]^2)/(24*x^2) - (b^2*d^2*e*f*F^(a + b*c +
b*d*x)*Log[F]^2)/(3*x) + (b^2*d^2*f^2*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^2)/2 - (b^3*d^3*e^2*F^(a
+ b*c + b*d*x)*Log[F]^3)/(24*x) + (b^3*d^3*e*f*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^3)/3 + (b^4*d^4
*e^2*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^4)/24

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^5} \, dx &=\int \left (\frac {e^2 F^{a+b c+b d x}}{x^5}+\frac {2 e f F^{a+b c+b d x}}{x^4}+\frac {f^2 F^{a+b c+b d x}}{x^3}\right ) \, dx\\ &=e^2 \int \frac {F^{a+b c+b d x}}{x^5} \, dx+(2 e f) \int \frac {F^{a+b c+b d x}}{x^4} \, dx+f^2 \int \frac {F^{a+b c+b d x}}{x^3} \, dx\\ &=-\frac {e^2 F^{a+b c+b d x}}{4 x^4}-\frac {2 e f F^{a+b c+b d x}}{3 x^3}-\frac {f^2 F^{a+b c+b d x}}{2 x^2}+\frac {1}{4} \left (b d e^2 \log (F)\right ) \int \frac {F^{a+b c+b d x}}{x^4} \, dx+\frac {1}{3} (2 b d e f \log (F)) \int \frac {F^{a+b c+b d x}}{x^3} \, dx+\frac {1}{2} \left (b d f^2 \log (F)\right ) \int \frac {F^{a+b c+b d x}}{x^2} \, dx\\ &=-\frac {e^2 F^{a+b c+b d x}}{4 x^4}-\frac {2 e f F^{a+b c+b d x}}{3 x^3}-\frac {f^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d e^2 F^{a+b c+b d x} \log (F)}{12 x^3}-\frac {b d e f F^{a+b c+b d x} \log (F)}{3 x^2}-\frac {b d f^2 F^{a+b c+b d x} \log (F)}{2 x}+\frac {1}{12} \left (b^2 d^2 e^2 \log ^2(F)\right ) \int \frac {F^{a+b c+b d x}}{x^3} \, dx+\frac {1}{3} \left (b^2 d^2 e f \log ^2(F)\right ) \int \frac {F^{a+b c+b d x}}{x^2} \, dx+\frac {1}{2} \left (b^2 d^2 f^2 \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}}{4 x^4}-\frac {2 e f F^{a+b c+b d x}}{3 x^3}-\frac {f^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d e^2 F^{a+b c+b d x} \log (F)}{12 x^3}-\frac {b d e f F^{a+b c+b d x} \log (F)}{3 x^2}-\frac {b d f^2 F^{a+b c+b d x} \log (F)}{2 x}-\frac {b^2 d^2 e^2 F^{a+b c+b d x} \log ^2(F)}{24 x^2}-\frac {b^2 d^2 e f F^{a+b c+b d x} \log ^2(F)}{3 x}+\frac {1}{2} b^2 d^2 f^2 F^{a+b c} \text {Ei}(b d x \log (F)) \log ^2(F)+\frac {1}{24} \left (b^3 d^3 e^2 \log ^3(F)\right ) \int \frac {F^{a+b c+b d x}}{x^2} \, dx+\frac {1}{3} \left (b^3 d^3 e f \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}}{4 x^4}-\frac {2 e f F^{a+b c+b d x}}{3 x^3}-\frac {f^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d e^2 F^{a+b c+b d x} \log (F)}{12 x^3}-\frac {b d e f F^{a+b c+b d x} \log (F)}{3 x^2}-\frac {b d f^2 F^{a+b c+b d x} \log (F)}{2 x}-\frac {b^2 d^2 e^2 F^{a+b c+b d x} \log ^2(F)}{24 x^2}-\frac {b^2 d^2 e f F^{a+b c+b d x} \log ^2(F)}{3 x}+\frac {1}{2} b^2 d^2 f^2 F^{a+b c} \text {Ei}(b d x \log (F)) \log ^2(F)-\frac {b^3 d^3 e^2 F^{a+b c+b d x} \log ^3(F)}{24 x}+\frac {1}{3} b^3 d^3 e f F^{a+b c} \text {Ei}(b d x \log (F)) \log ^3(F)+\frac {1}{24} \left (b^4 d^4 e^2 \log ^4(F)\right ) \int \frac {F^{a+b c+b d x}}{x} \, dx\\ &=-\frac {e^2 F^{a+b c+b d x}}{4 x^4}-\frac {2 e f F^{a+b c+b d x}}{3 x^3}-\frac {f^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d e^2 F^{a+b c+b d x} \log (F)}{12 x^3}-\frac {b d e f F^{a+b c+b d x} \log (F)}{3 x^2}-\frac {b d f^2 F^{a+b c+b d x} \log (F)}{2 x}-\frac {b^2 d^2 e^2 F^{a+b c+b d x} \log ^2(F)}{24 x^2}-\frac {b^2 d^2 e f F^{a+b c+b d x} \log ^2(F)}{3 x}+\frac {1}{2} b^2 d^2 f^2 F^{a+b c} \text {Ei}(b d x \log (F)) \log ^2(F)-\frac {b^3 d^3 e^2 F^{a+b c+b d x} \log ^3(F)}{24 x}+\frac {1}{3} b^3 d^3 e f F^{a+b c} \text {Ei}(b d x \log (F)) \log ^3(F)+\frac {1}{24} b^4 d^4 e^2 F^{a+b c} \text {Ei}(b d x \log (F)) \log ^4(F)\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 156, normalized size = 0.49 \[ \frac {F^{a+b c} \left (b^2 d^2 x^4 \log ^2(F) \left (b^2 d^2 e^2 \log ^2(F)+8 b d e f \log (F)+12 f^2\right ) \text {Ei}(b d x \log (F))-F^{b d x} \left (b^3 d^3 e^2 x^3 \log ^3(F)+b^2 d^2 e x^2 \log ^2(F) (e+8 f x)+2 b d x \log (F) \left (e^2+4 e f x+6 f^2 x^2\right )+2 \left (3 e^2+8 e f x+6 f^2 x^2\right )\right )\right )}{24 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^5,x]

[Out]

(F^(a + b*c)*(b^2*d^2*x^4*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^2*(12*f^2 + 8*b*d*e*f*Log[F] + b^2*d^2*e^2*Log[F]
^2) - F^(b*d*x)*(2*(3*e^2 + 8*e*f*x + 6*f^2*x^2) + 2*b*d*x*(e^2 + 4*e*f*x + 6*f^2*x^2)*Log[F] + b^2*d^2*e*x^2*
(e + 8*f*x)*Log[F]^2 + b^3*d^3*e^2*x^3*Log[F]^3)))/(24*x^4)

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fricas [A]  time = 0.43, size = 186, normalized size = 0.58 \[ \frac {{\left (b^{4} d^{4} e^{2} x^{4} \log \relax (F)^{4} + 8 \, b^{3} d^{3} e f x^{4} \log \relax (F)^{3} + 12 \, b^{2} d^{2} f^{2} x^{4} \log \relax (F)^{2}\right )} F^{b c + a} {\rm Ei}\left (b d x \log \relax (F)\right ) - {\left (b^{3} d^{3} e^{2} x^{3} \log \relax (F)^{3} + 12 \, f^{2} x^{2} + 16 \, e f x + {\left (8 \, b^{2} d^{2} e f x^{3} + b^{2} d^{2} e^{2} x^{2}\right )} \log \relax (F)^{2} + 6 \, e^{2} + 2 \, {\left (6 \, b d f^{2} x^{3} + 4 \, b d e f x^{2} + b d e^{2} x\right )} \log \relax (F)\right )} F^{b d x + b c + a}}{24 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^5,x, algorithm="fricas")

[Out]

1/24*((b^4*d^4*e^2*x^4*log(F)^4 + 8*b^3*d^3*e*f*x^4*log(F)^3 + 12*b^2*d^2*f^2*x^4*log(F)^2)*F^(b*c + a)*Ei(b*d
*x*log(F)) - (b^3*d^3*e^2*x^3*log(F)^3 + 12*f^2*x^2 + 16*e*f*x + (8*b^2*d^2*e*f*x^3 + b^2*d^2*e^2*x^2)*log(F)^
2 + 6*e^2 + 2*(6*b*d*f^2*x^3 + 4*b*d*e*f*x^2 + b*d*e^2*x)*log(F))*F^(b*d*x + b*c + a))/x^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{2} F^{{\left (d x + c\right )} b + a}}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^5,x, algorithm="giac")

[Out]

integrate((f*x + e)^2*F^((d*x + c)*b + a)/x^5, x)

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maple [A]  time = 0.09, size = 382, normalized size = 1.19 \[ -\frac {b^{4} d^{4} e^{2} F^{a} F^{b c} \Ei \left (1, -b d x \ln \relax (F )+b c \ln \relax (F )+a \ln \relax (F )-\left (b c +a \right ) \ln \relax (F )\right ) \ln \relax (F )^{4}}{24}-\frac {b^{3} d^{3} e f \,F^{a} F^{b c} \Ei \left (1, -b d x \ln \relax (F )+b c \ln \relax (F )+a \ln \relax (F )-\left (b c +a \right ) \ln \relax (F )\right ) \ln \relax (F )^{3}}{3}-\frac {b^{3} d^{3} e^{2} F^{b d x} F^{b c +a} \ln \relax (F )^{3}}{24 x}-\frac {b^{2} d^{2} f^{2} F^{a} F^{b c} \Ei \left (1, -b d x \ln \relax (F )+b c \ln \relax (F )+a \ln \relax (F )-\left (b c +a \right ) \ln \relax (F )\right ) \ln \relax (F )^{2}}{2}-\frac {b^{2} d^{2} e f \,F^{b d x} F^{b c +a} \ln \relax (F )^{2}}{3 x}-\frac {b^{2} d^{2} e^{2} F^{b d x} F^{b c +a} \ln \relax (F )^{2}}{24 x^{2}}-\frac {b d \,f^{2} F^{b d x} F^{b c +a} \ln \relax (F )}{2 x}-\frac {b d e f \,F^{b d x} F^{b c +a} \ln \relax (F )}{3 x^{2}}-\frac {b d \,e^{2} F^{b d x} F^{b c +a} \ln \relax (F )}{12 x^{3}}-\frac {f^{2} F^{b d x} F^{b c +a}}{2 x^{2}}-\frac {2 e f \,F^{b d x} F^{b c +a}}{3 x^{3}}-\frac {e^{2} F^{b d x} F^{b c +a}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b*(d*x+c))*(f*x+e)^2/x^5,x)

[Out]

-2/3*e*f*F^(b*d*x)*F^(b*c+a)/x^3-1/3*ln(F)*b*d*e*f*F^(b*d*x)*F^(b*c+a)/x^2-1/3*ln(F)^2*b^2*d^2*e*f*F^(b*d*x)*F
^(b*c+a)/x-1/3*ln(F)^3*b^3*d^3*e*f*F^(b*c)*F^a*Ei(1,-b*d*x*ln(F)+b*c*ln(F)+a*ln(F)-(b*c+a)*ln(F))-1/24*ln(F)^4
*b^4*d^4*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))-1/2*ln(F)*b*d*f^2*F^(b*d*x)*F^(b*c
+a)/x-1/2*ln(F)^2*b^2*d^2*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))-1/2*f^2*F^(b*d*x)
*F^(b*c+a)/x^2-1/4*e^2*F^(b*d*x)*F^(b*c+a)/x^4-1/12*ln(F)*b*d*e^2*F^(b*d*x)*F^(b*c+a)/x^3-1/24*ln(F)^2*b^2*d^2
*e^2*F^(b*d*x)*F^(b*c+a)/x^2-1/24*ln(F)^3*b^3*d^3*e^2*F^(b*d*x)*F^(b*c+a)/x

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maxima [A]  time = 1.01, size = 93, normalized size = 0.29 \[ -F^{b c + a} b^{4} d^{4} e^{2} \Gamma \left (-4, -b d x \log \relax (F)\right ) \log \relax (F)^{4} + 2 \, F^{b c + a} b^{3} d^{3} e f \Gamma \left (-3, -b d x \log \relax (F)\right ) \log \relax (F)^{3} - F^{b c + a} b^{2} d^{2} f^{2} \Gamma \left (-2, -b d x \log \relax (F)\right ) \log \relax (F)^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^5,x, algorithm="maxima")

[Out]

-F^(b*c + a)*b^4*d^4*e^2*gamma(-4, -b*d*x*log(F))*log(F)^4 + 2*F^(b*c + a)*b^3*d^3*e*f*gamma(-3, -b*d*x*log(F)
)*log(F)^3 - F^(b*c + a)*b^2*d^2*f^2*gamma(-2, -b*d*x*log(F))*log(F)^2

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mupad [B]  time = 3.68, size = 258, normalized size = 0.80 \[ -F^{a+b\,c}\,b^2\,d^2\,f^2\,{\ln \relax (F)}^2\,\left (\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \relax (F)\right )}{2}+F^{b\,d\,x}\,\left (\frac {1}{2\,b\,d\,x\,\ln \relax (F)}+\frac {1}{2\,b^2\,d^2\,x^2\,{\ln \relax (F)}^2}\right )\right )-F^{a+b\,c}\,b^4\,d^4\,e^2\,{\ln \relax (F)}^4\,\left (F^{b\,d\,x}\,\left (\frac {1}{24\,b\,d\,x\,\ln \relax (F)}+\frac {1}{24\,b^2\,d^2\,x^2\,{\ln \relax (F)}^2}+\frac {1}{12\,b^3\,d^3\,x^3\,{\ln \relax (F)}^3}+\frac {1}{4\,b^4\,d^4\,x^4\,{\ln \relax (F)}^4}\right )+\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \relax (F)\right )}{24}\right )-2\,F^{a+b\,c}\,b^3\,d^3\,e\,f\,{\ln \relax (F)}^3\,\left (F^{b\,d\,x}\,\left (\frac {1}{6\,b\,d\,x\,\ln \relax (F)}+\frac {1}{6\,b^2\,d^2\,x^2\,{\ln \relax (F)}^2}+\frac {1}{3\,b^3\,d^3\,x^3\,{\ln \relax (F)}^3}\right )+\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \relax (F)\right )}{6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((F^(a + b*(c + d*x))*(e + f*x)^2)/x^5,x)

[Out]

- F^(a + b*c)*b^2*d^2*f^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))) - F^(a + b*c)*b^4*d^4*e^2*log(F)^4*(F^(b*d*x)*(1/(24*b*d*x*log(F)) + 1/(24*b^2*d^2*x^2*log(F)^2)
+ 1/(12*b^3*d^3*x^3*log(F)^3) + 1/(4*b^4*d^4*x^4*log(F)^4)) + expint(-b*d*x*log(F))/24) - 2*F^(a + b*c)*b^3*d^
3*e*f*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)) + exp
int(-b*d*x*log(F))/6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + b \left (c + d x\right )} \left (e + f x\right )^{2}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c))*(f*x+e)**2/x**5,x)

[Out]

Integral(F**(a + b*(c + d*x))*(e + f*x)**2/x**5, x)

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