\(\int \frac {F^{a+b (c+d x)^n}}{(c+d x)^2} \, dx\) [365]

Optimal result

Integrand size = 21, antiderivative size = 52 \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^2} \, dx=-\frac {F^a \Gamma \left (-\frac {1}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{\frac {1}{n}}}{d n (c+d x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 52, 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 = {2250} \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^2} \, dx=-\frac {F^a \left (-b \log (F) (c+d x)^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b (c+d x)^n \log (F)\right )}{d n (c+d x)} \]

[In]

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

[Out]

-((F^a*Gamma[-n^(-1), -(b*(c + d*x)^n*Log[F])]*(-(b*(c + d*x)^n*Log[F]))^n^(-1))/(d*n*(c + d*x)))

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {F^a \Gamma \left (-\frac {1}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{\frac {1}{n}}}{d n (c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^2} \, dx=-\frac {F^a \Gamma \left (-\frac {1}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{\frac {1}{n}}}{d n (c+d x)} \]

[In]

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

[Out]

-((F^a*Gamma[-n^(-1), -(b*(c + d*x)^n*Log[F])]*(-(b*(c + d*x)^n*Log[F]))^n^(-1))/(d*n*(c + d*x)))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.37 \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^2} \, dx=-\frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n}{2}}\,{\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n\right )}^{\frac {1}{2\,n}-\frac {1}{2}}\,{\mathrm {M}}_{\frac {1}{2\,n}+\frac {1}{2},-\frac {1}{2\,n}}\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n\right )}{d\,\left (c+d\,x\right )} \]

[In]

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

[Out]

-(F^a*exp((b*log(F)*(c + d*x)^n)/2)*(b*log(F)*(c + d*x)^n)^(1/(2*n) - 1/2)*whittakerM(1/(2*n) + 1/2, -1/(2*n),
b*log(F)*(c + d*x)^n))/(d*(c + d*x))