\(\int \frac {F^{c (a+b x)}}{(d+e x)^4} \, dx\) [10]

Optimal result

Integrand size = 17, antiderivative size = 128 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^4} \, dx=-\frac {F^{c (a+b x)}}{3 e (d+e x)^3}-\frac {b c F^{c (a+b x)} \log (F)}{6 e^2 (d+e x)^2}-\frac {b^2 c^2 F^{c (a+b x)} \log ^2(F)}{6 e^3 (d+e x)}+\frac {b^3 c^3 F^{c \left (a-\frac {b d}{e}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log ^3(F)}{6 e^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2208, 2209} \[ \int \frac {F^{c (a+b x)}}{(d+e x)^4} \, dx=\frac {b^3 c^3 \log ^3(F) F^{c \left (a-\frac {b d}{e}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right )}{6 e^4}-\frac {b^2 c^2 \log ^2(F) F^{c (a+b x)}}{6 e^3 (d+e x)}-\frac {b c \log (F) F^{c (a+b x)}}{6 e^2 (d+e x)^2}-\frac {F^{c (a+b x)}}{3 e (d+e x)^3} \]

[In]

Int[F^(c*(a + b*x))/(d + e*x)^4,x]

[Out]

-1/3*F^(c*(a + b*x))/(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 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]

Rubi steps \begin{align*} \text {integral}& = -\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] (verified)

Time = 0.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.77 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^4} \, dx=\frac {F^{a c} \left (b^3 c^3 F^{-\frac {b c d}{e}} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log ^3(F)-\frac {e F^{b c x} \left (2 e^2+b c e (d+e x) \log (F)+b^2 c^2 (d+e x)^2 \log ^2(F)\right )}{(d+e x)^3}\right )}{6 e^4} \]

[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] (verified)

Time = 0.35 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.55

method	result	size
risch	\(-\frac {c^{3} b^{3} \ln \left (F \right )^{3} F^{b c x} F^{c a}}{3 e^{4} \left (b c x \ln \left (F \right )+\frac {b c \ln \left (F \right ) d}{e}\right )^{3}}-\frac {c^{3} b^{3} \ln \left (F \right )^{3} F^{b c x} F^{c a}}{6 e^{4} \left (b c x \ln \left (F \right )+\frac {b c \ln \left (F \right ) d}{e}\right )^{2}}-\frac {c^{3} b^{3} \ln \left (F \right )^{3} F^{b c x} F^{c a}}{6 e^{4} \left (b c x \ln \left (F \right )+\frac {b c \ln \left (F \right ) d}{e}\right )}-\frac {c^{3} b^{3} \ln \left (F \right )^{3} F^{\frac {c \left (a e -b d \right )}{e}} \operatorname {Ei}_{1}\left (-b c x \ln \left (F \right )-c a \ln \left (F \right )-\frac {-\ln \left (F \right ) a c e +\ln \left (F \right ) b c d}{e}\right )}{6 e^{4}}\)	\(199\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.63 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^4} \, dx=\frac {\frac {{\left (b^{3} c^{3} e^{3} x^{3} + 3 \, b^{3} c^{3} d e^{2} x^{2} + 3 \, b^{3} c^{3} d^{2} e x + b^{3} c^{3} d^{3}\right )} {\rm Ei}\left (\frac {{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right ) \log \left (F\right )^{3}}{F^{\frac {b c d - a c e}{e}}} - {\left (2 \, e^{3} + {\left (b^{2} c^{2} e^{3} x^{2} + 2 \, b^{2} c^{2} d e^{2} x + b^{2} c^{2} d^{2} e\right )} \log \left (F\right )^{2} + {\left (b c e^{3} x + b c d e^{2}\right )} \log \left (F\right )\right )} F^{b c x + a c}}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]

[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]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

int(F^(c*(a + b*x))/(d + e*x)^4,x)

[Out]

int(F^(c*(a + b*x))/(d + e*x)^4, x)