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

Optimal. Leaf size=128 \[ \frac{b^3 c^3 \log ^3(F) F^{c \left (a-\frac{b d}{e}\right )} \text{Ei}\left (\frac{b c (d+e x) \log (F)}{e}\right )}{6 e^4}-\frac{b^2 c^2 \log ^2(F) F^{c (a+b x)}}{6 e^3 (d+e x)}-\frac{b c \log (F) F^{c (a+b x)}}{6 e^2 (d+e x)^2}-\frac{F^{c (a+b x)}}{3 e (d+e x)^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.100237, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2177, 2178} \[ \frac{b^3 c^3 \log ^3(F) F^{c \left (a-\frac{b d}{e}\right )} \text{Ei}\left (\frac{b c (d+e x) \log (F)}{e}\right )}{6 e^4}-\frac{b^2 c^2 \log ^2(F) F^{c (a+b x)}}{6 e^3 (d+e x)}-\frac{b c \log (F) F^{c (a+b x)}}{6 e^2 (d+e x)^2}-\frac{F^{c (a+b x)}}{3 e (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))/(d + e*x)^4,x]

[Out]

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

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

Rubi steps

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

Mathematica [A]  time = 0.201428, size = 99, normalized size = 0.77 \[ \frac{F^{a c} \left (b^3 c^3 \log ^3(F) F^{-\frac{b c d}{e}} \text{Ei}\left (\frac{b c (d+e x) \log (F)}{e}\right )-\frac{e F^{b c x} \left (b^2 c^2 \log ^2(F) (d+e x)^2+b c e \log (F) (d+e x)+2 e^2\right )}{(d+e x)^3}\right )}{6 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.049, size = 199, normalized size = 1.6 \begin{align*} -{\frac{{b}^{3}{c}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}{F}^{bcx}{F}^{ac}}{3\,{e}^{4}} \left ( bcx\ln \left ( F \right ) +{\frac{\ln \left ( F \right ) bcd}{e}} \right ) ^{-3}}-{\frac{{b}^{3}{c}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}{F}^{bcx}{F}^{ac}}{6\,{e}^{4}} \left ( bcx\ln \left ( F \right ) +{\frac{\ln \left ( F \right ) bcd}{e}} \right ) ^{-2}}-{\frac{{b}^{3}{c}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}{F}^{bcx}{F}^{ac}}{6\,{e}^{4}} \left ( bcx\ln \left ( F \right ) +{\frac{\ln \left ( F \right ) bcd}{e}} \right ) ^{-1}}-{\frac{{b}^{3}{c}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}}{6\,{e}^{4}}{F}^{{\frac{c \left ( ae-bd \right ) }{e}}}{\it Ei} \left ( 1,-bcx\ln \left ( F \right ) -ac\ln \left ( F \right ) -{\frac{-\ln \left ( F \right ) ace+\ln \left ( F \right ) bcd}{e}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b^3*c^3*ln(F)^3/e^4*F^(b*c*x)*F^(a*c)/(b*c*x*ln(F)+1/e*ln(F)*b*c*d)^3-1/6*b^3*c^3*ln(F)^3/e^4*F^(b*c*x)*F
^(a*c)/(b*c*x*ln(F)+1/e*ln(F)*b*c*d)^2-1/6*b^3*c^3*ln(F)^3/e^4*F^(b*c*x)*F^(a*c)/(b*c*x*ln(F)+1/e*ln(F)*b*c*d)
-1/6*b^3*c^3*ln(F)^3/e^4*F^(c*(a*e-b*d)/e)*Ei(1,-b*c*x*ln(F)-a*c*ln(F)-(-ln(F)*a*c*e+ln(F)*b*c*d)/e)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.52107, size = 425, normalized size = 3.32 \begin{align*} \frac{\frac{{\left (b^{3} c^{3} e^{3} x^{3} + 3 \, b^{3} c^{3} d e^{2} x^{2} + 3 \, b^{3} c^{3} d^{2} e x + b^{3} c^{3} d^{3}\right )}{\rm Ei}\left (\frac{{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right ) \log \left (F\right )^{3}}{F^{\frac{b c d - a c e}{e}}} -{\left (2 \, e^{3} +{\left (b^{2} c^{2} e^{3} x^{2} + 2 \, b^{2} c^{2} d e^{2} x + b^{2} c^{2} d^{2} e\right )} \log \left (F\right )^{2} +{\left (b c e^{3} x + b c d e^{2}\right )} \log \left (F\right )\right )} F^{b c x + a c}}{6 \,{\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((b^3*c^3*e^3*x^3 + 3*b^3*c^3*d*e^2*x^2 + 3*b^3*c^3*d^2*e*x + b^3*c^3*d^3)*Ei((b*c*e*x + b*c*d)*log(F)/e)*
log(F)^3/F^((b*c*d - a*c*e)/e) - (2*e^3 + (b^2*c^2*e^3*x^2 + 2*b^2*c^2*d*e^2*x + b^2*c^2*d^2*e)*log(F)^2 + (b*
c*e^3*x + b*c*d*e^2)*log(F))*F^(b*c*x + a*c))/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{c \left (a + b x\right )}}{\left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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