\(\int \frac {F^{a+b (c+d x)^n}}{(c+d x)^3} \, dx\) [366]

Optimal result

Integrand size = 21, antiderivative size = 54 \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^3} \, dx=-\frac {F^a \Gamma \left (-\frac {2}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{2/n}}{d n (c+d x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, 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} \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^3} \, dx=-\frac {F^a \left (-b \log (F) (c+d x)^n\right )^{2/n} \Gamma \left (-\frac {2}{n},-b (c+d x)^n \log (F)\right )}{d n (c+d x)^2} \]

[In]

Int[F^(a + b*(c + d*x)^n)/(c + d*x)^3,x]

[Out]

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

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 {2}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{2/n}}{d n (c+d x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^3} \, dx=-\frac {F^a \Gamma \left (-\frac {2}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{2/n}}{d n (c+d x)^2} \]

[In]

Integrate[F^(a + b*(c + d*x)^n)/(c + d*x)^3,x]

[Out]

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

Maple [F]

\[\int \frac {F^{a +b \left (d x +c \right )^{n}}}{\left (d x +c \right )^{3}}d x\]

[In]

int(F^(a+b*(d*x+c)^n)/(d*x+c)^3,x)

[Out]

int(F^(a+b*(d*x+c)^n)/(d*x+c)^3,x)

Fricas [F]

\[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^3} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{n} b + a}}{{\left (d x + c\right )}^{3}} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^n)/(d*x+c)^3,x, algorithm="fricas")

[Out]

integral(F^((d*x + c)^n*b + a)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

Sympy [F]

\[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^3} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{n}}}{\left (c + d x\right )^{3}}\, dx \]

[In]

integrate(F**(a+b*(d*x+c)**n)/(d*x+c)**3,x)

[Out]

Integral(F**(a + b*(c + d*x)**n)/(c + d*x)**3, x)

Maxima [F]

\[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^3} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{n} b + a}}{{\left (d x + c\right )}^{3}} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^n)/(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^n*b + a)/(d*x + c)^3, x)

Giac [F]

\[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^3} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{n} b + a}}{{\left (d x + c\right )}^{3}} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^n)/(d*x+c)^3,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^n*b + a)/(d*x + c)^3, x)

Mupad [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.24 \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^3} \, dx=-\frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n}{2}}\,{\mathrm {M}}_{\frac {1}{n}+\frac {1}{2},-\frac {1}{n}}\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n\right )\,{\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n\right )}^{\frac {1}{n}-\frac {1}{2}}}{2\,d\,{\left (c+d\,x\right )}^2} \]

[In]

int(F^(a + b*(c + d*x)^n)/(c + d*x)^3,x)

[Out]

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