\(\int \frac {F^{a+b (c+d x)^n}}{c+d x} \, dx\) [374]

Optimal result

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

[Out]

F^a*Ei(b*(d*x+c)^n*ln(F))/d/n

Rubi [A] (verified)

Time = 0.02 (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} \[ \int \frac {F^{a+b (c+d x)^n}}{c+d x} \, dx=\frac {F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^n \log (F)\right )}{d n} \]

[In]

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

[Out]

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

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 (b (c+d x)^n \log (F)\right )}{d n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^n}}{c+d x} \, dx=\frac {F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^n \log (F)\right )}{d n} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18

method	result	size
risch	\(-\frac {F^{a} \operatorname {Ei}_{1}\left (-b \left (d x +c \right )^{n} \ln \left (F \right )\right )}{d n}\)	\(26\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^n}}{c+d x} \, dx=\frac {F^{a} {\rm Ei}\left ({\left (d x + c\right )}^{n} b \log \left (F\right )\right )}{d n} \]

[In]

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

[Out]

F^a*Ei((d*x + c)^n*b*log(F))/(d*n)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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