\(\int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx\) [15]

Optimal result

Integrand size = 28, antiderivative size = 57 \[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=-\frac {F^{c (a+b x)}}{e (d+e x)}+\frac {b c F^{c \left (a-\frac {b d}{e}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log (F)}{e^2} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {27, 2208, 2209} \[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=\frac {b c \log (F) F^{c \left (a-\frac {b d}{e}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right )}{e^2}-\frac {F^{c (a+b x)}}{e (d+e x)} \]

[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 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}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=\frac {F^{a c} \left (-\frac {e F^{b c x}}{d+e x}+b c F^{-\frac {b c d}{e}} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log (F)\right )}{e^2} \]

[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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.74

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

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.35 \[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=-\frac {F^{b c x + a c} e - \frac {{\left (b c e x + b c d\right )} {\rm Ei}\left (\frac {{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right ) \log \left (F\right )}{F^{\frac {b c d - a c e}{e}}}}{e^{3} x + d e^{2}} \]

[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)

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{d^2+2\,d\,e\,x+e^2\,x^2} \,d x \]

[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)