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

Optimal. Leaf size=57 \[ \frac{b c \log (F) F^{c \left (a-\frac{b d}{e}\right )} \text{ExpIntegralEi}\left (\frac{b c \log (F) (d+e x)}{e}\right )}{e^2}-\frac{F^{c (a+b x)}}{e (d+e x)} \]

[Out]

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

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Rubi [A]  time = 0.0775856, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118 \[ \frac{b c \log (F) F^{c \left (a-\frac{b d}{e}\right )} \text{ExpIntegralEi}\left (\frac{b c \log (F) (d+e x)}{e}\right )}{e^2}-\frac{F^{c (a+b x)}}{e (d+e x)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 9.91969, size = 51, normalized size = 0.89 \[ - \frac{F^{c \left (a + b x\right )}}{e \left (d + e x\right )} + \frac{F^{\frac{c \left (a e - b d\right )}{e}} b c \log{\left (F \right )} \operatorname{Ei}{\left (\frac{b c \left (d + e x\right ) \log{\left (F \right )}}{e} \right )}}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(F**(c*(b*x+a))/(e*x+d)**2,x)

[Out]

-F**(c*(a + b*x))/(e*(d + e*x)) + F**(c*(a*e - b*d)/e)*b*c*log(F)*Ei(b*c*(d + e*
x)*log(F)/e)/e**2

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Mathematica [A]  time = 0.118037, size = 55, normalized size = 0.96 \[ \frac{F^{a c} \left (b c \log (F) F^{-\frac{b c d}{e}} \text{ExpIntegralEi}\left (\frac{b c \log (F) (d+e x)}{e}\right )-\frac{e F^{b c x}}{d+e x}\right )}{e^2} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [A]  time = 0.034, size = 97, normalized size = 1.7 \[ -{\frac{{F}^{c \left ( bx+a \right ) }cb\ln \left ( F \right ) }{{e}^{2}} \left ( bcx\ln \left ( F \right ) +{\frac{\ln \left ( F \right ) bcd}{e}} \right ) ^{-1}}-{\frac{cb\ln \left ( F \right ) }{{e}^{2}}{F}^{{\frac{c \left ( ea-bd \right ) }{e}}}{\it Ei} \left ( 1,-bcx\ln \left ( F \right ) -\ln \left ( F \right ) ac-{\frac{-eac\ln \left ( F \right ) +\ln \left ( F \right ) bcd}{e}} \right ) } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(F^(c*(b*x+a))/(e*x+d)^2,x)

[Out]

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

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Maxima [A]  time = 0.786934, size = 59, normalized size = 1.04 \[ -\frac{F^{a c} exp_integral_e\left (2, -\frac{{\left (e x + d\right )} b c \log \left (F\right )}{e}\right )}{{\left (e x + d\right )} F^{\frac{b c d}{e}} e} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-F^(a*c)*exp_integral_e(2, -(e*x + d)*b*c*log(F)/e)/((e*x + d)*F^(b*c*d/e)*e)

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Fricas [A]  time = 0.245465, size = 104, normalized size = 1.82 \[ -\frac{F^{b c x + a c} e - \frac{{\left (b c e x + b c d\right )}{\rm Ei}\left (\frac{{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right ) \log \left (F\right )}{F^{\frac{b c d - a c e}{e}}}}{e^{3} x + d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-(F^(b*c*x + a*c)*e - (b*c*e*x + b*c*d)*Ei((b*c*e*x + b*c*d)*log(F)/e)*log(F)/F^
((b*c*d - a*c*e)/e))/(e^3*x + d*e^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(F**(c*(b*x+a))/(e*x+d)**2,x)

[Out]

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

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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