3.1.8 \(\int \frac {F^{c (a+b x)}}{(d+e x)^2} \, dx\) [8]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2208, 2209} \begin {gather*} \frac {b c \log (F) F^{c \left (a-\frac {b d}{e}\right )} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right )}{e^2}-\frac {F^{c (a+b x)}}{e (d+e x)} \end {gather*}

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

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

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Mathematica [A]
time = 0.10, size = 55, normalized size = 0.96 \begin {gather*} \frac {F^{a c} \left (-\frac {e F^{b c x}}{d+e x}+b c F^{-\frac {b c d}{e}} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log (F)\right )}{e^2} \end {gather*}

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 = 0.08, size = 99, normalized size = 1.74

method	result	size
risch	\(-\frac {c b \ln \left (F \right ) F^{b c x} F^{c a}}{e^{2} \left (b c x \ln \left (F \right )+\frac {\ln \left (F \right ) b c d}{e}\right )}-\frac {c b \ln \left (F \right ) F^{\frac {c \left (a e -b d \right )}{e}} \expIntegral \left (1, -b c x \ln \left (F \right )-c a \ln \left (F \right )-\frac {-\ln \left (F \right ) a c e +\ln \left (F \right ) b c d}{e}\right )}{e^{2}}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))/(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.41, size = 77, normalized size = 1.35 \begin {gather*} -\frac {F^{b c x + a c} e - \frac {{\left (b c x e + b c d\right )} {\rm Ei}\left ({\left (b c x e + b c d\right )} e^{\left (-1\right )} \log \left (F\right )\right ) \log \left (F\right )}{F^{{\left (b c d - a c e\right )} e^{\left (-1\right )}}}}{x e^{3} + d e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {F^{c \left (a + b x\right )}}{\left (d + e x\right )^{2}}\, dx \end {gather*}

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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))/(d + e*x)^2,x)

[Out]

int(F^(c*(a + b*x))/(d + e*x)^2, x)

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