3.16 \(\int \frac{F^{c (a+b x)}}{d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3} \, dx\)

Optimal. Leaf size=95 \[ \frac{b^2 c^2 \log ^2(F) F^{c \left (a-\frac{b d}{e}\right )} \text{Ei}\left (\frac{b c (d+e x) \log (F)}{e}\right )}{2 e^3}-\frac{b c \log (F) F^{c (a+b x)}}{2 e^2 (d+e x)}-\frac{F^{c (a+b x)}}{2 e (d+e x)^2} \]

[Out]

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

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Rubi [A]  time = 0.097557, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2187, 2177, 2178} \[ \frac{b^2 c^2 \log ^2(F) F^{c \left (a-\frac{b d}{e}\right )} \text{Ei}\left (\frac{b c (d+e x) \log (F)}{e}\right )}{2 e^3}-\frac{b c \log (F) F^{c (a+b x)}}{2 e^2 (d+e x)}-\frac{F^{c (a+b x)}}{2 e (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))/(d^3 + 3*d^2*e*x + 3*d*e^2*x^2 + e^3*x^3),x]

[Out]

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

Rule 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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

Mathematica [A]  time = 0.0505127, size = 88, normalized size = 0.93 \[ \frac{F^{c \left (a-\frac{b d}{e}\right )} \left (b^2 c^2 \log ^2(F) (d+e x)^2 \text{Ei}\left (\frac{b c (d+e x) \log (F)}{e}\right )-e F^{\frac{b c (d+e x)}{e}} (b c \log (F) (d+e x)+e)\right )}{2 e^3 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))/(d^3 + 3*d^2*e*x + 3*d*e^2*x^2 + e^3*x^3),x]

[Out]

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

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Maple [A]  time = 0.059, size = 155, normalized size = 1.6 \begin{align*} -{\frac{{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{F}^{bcx}{F}^{ac}}{2\,{e}^{3}} \left ( bcx\ln \left ( F \right ) +{\frac{\ln \left ( F \right ) bcd}{e}} \right ) ^{-2}}-{\frac{{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{F}^{bcx}{F}^{ac}}{2\,{e}^{3}} \left ( bcx\ln \left ( F \right ) +{\frac{\ln \left ( F \right ) bcd}{e}} \right ) ^{-1}}-{\frac{{b}^{2}{c}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}}{2\,{e}^{3}}{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^3*x^3+3*d*e^2*x^2+3*d^2*e*x+d^3),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e^3*x^3+3*d*e^2*x^2+3*d^2*e*x+d^3),x, algorithm="maxima")

[Out]

integrate(F^((b*x + a)*c)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)

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Fricas [A]  time = 1.49918, size = 278, normalized size = 2.93 \begin{align*} \frac{\frac{{\left (b^{2} c^{2} e^{2} x^{2} + 2 \, b^{2} c^{2} d e x + b^{2} c^{2} d^{2}\right )}{\rm Ei}\left (\frac{{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right ) \log \left (F\right )^{2}}{F^{\frac{b c d - a c e}{e}}} -{\left (e^{2} +{\left (b c e^{2} x + b c d e\right )} \log \left (F\right )\right )} F^{b c x + a c}}{2 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e^3*x^3+3*d*e^2*x^2+3*d^2*e*x+d^3),x, algorithm="fricas")

[Out]

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

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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 )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))/(e**3*x**3+3*d*e**2*x**2+3*d**2*e*x+d**3),x)

[Out]

Integral(F**(c*(a + b*x))/(d + e*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e^3*x^3+3*d*e^2*x^2+3*d^2*e*x+d^3),x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)