\(\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{c+d x} \, dx\) [320]

Optimal result

Integrand size = 21, antiderivative size = 22 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{c+d x} \, dx=-\frac {F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right )}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2241} \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{c+d x} \, dx=-\frac {F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right )}{2 d} \]

[In]

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

[Out]

-1/2*(F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)^2])/d

Rule 2241

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[F^a*(ExpIntegralEi[
b*(c + d*x)^n*Log[F]]/(f*n)), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {F^a \text {Ei}\left (\frac {b \log (F)}{(c+d x)^2}\right )}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{c+d x} \, dx=-\frac {F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right )}{2 d} \]

[In]

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

[Out]

-1/2*(F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)^2])/d

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05

method	result	size
risch	\(\frac {F^{a} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (F \right )}{\left (d x +c \right )^{2}}\right )}{2 d}\)	\(23\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{c+d x} \, dx=-\frac {F^{a} {\rm Ei}\left (\frac {b \log \left (F\right )}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{c+d x} \, dx=-\frac {F^a\,\mathrm {ei}\left (\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}{2\,d} \]

[In]

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

[Out]

-(F^a*ei((b*log(F))/(c + d*x)^2))/(2*d)