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

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {27, 2177, 2178} \[ \frac {b c \log (F) F^{c \left (a-\frac {b d}{e}\right )} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right )}{e^2}-\frac {F^{c (a+b x)}}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))/(d^2 + 2*d*e*x + e^2*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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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^2+2 d e x+e^2 x^2} \, dx &=\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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 55, normalized size = 0.96 \[ \frac {F^{a c} \left (b c \log (F) F^{-\frac {b c d}{e}} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right )-\frac {e F^{b c x}}{d+e x}\right )}{e^2} \]

Antiderivative was successfully verified.

[In]

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

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 77, normalized size = 1.35 \[ -\frac {F^{b c x + a c} e - \frac {{\left (b c e x + b c d\right )} {\rm Ei}\left (\frac {{\left (b c e x + b c d\right )} \log \relax (F)}{e}\right ) \log \relax (F)}{F^{\frac {b c d - a c e}{e}}}}{e^{3} x + d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.07, size = 99, normalized size = 1.74 \[ -\frac {b c \,F^{a c} F^{b c x} \ln \relax (F )}{\left (b c x \ln \relax (F )+\frac {b c d \ln \relax (F )}{e}\right ) e^{2}}-\frac {b c \,F^{\frac {\left (a e -b d \right ) c}{e}} \Ei \left (1, -b c x \ln \relax (F )-a c \ln \relax (F )-\frac {-a c e \ln \relax (F )+b c d \ln \relax (F )}{e}\right ) \ln \relax (F )}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________