\(\int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx\) [356]

Optimal result

Integrand size = 21, antiderivative size = 49 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\frac {F^a \Gamma \left (\frac {1}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}}} \]

[Out]

1/3*F^a*GAMMA(1/3,-b*ln(F)/(d*x+c)^3)/d/(d*x+c)/(-b*ln(F)/(d*x+c)^3)^(1/3)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, 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} \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\frac {F^a \Gamma \left (\frac {1}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}}} \]

[In]

Int[F^(a + b/(c + d*x)^3)/(c + d*x)^2,x]

[Out]

(F^a*Gamma[1/3, -((b*Log[F])/(c + d*x)^3)])/(3*d*(c + d*x)*(-((b*Log[F])/(c + d*x)^3))^(1/3))

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}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\frac {F^a \Gamma \left (\frac {1}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d (c+d x) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}}} \]

[In]

Integrate[F^(a + b/(c + d*x)^3)/(c + d*x)^2,x]

[Out]

(F^a*Gamma[1/3, -((b*Log[F])/(c + d*x)^3)])/(3*d*(c + d*x)*(-((b*Log[F])/(c + d*x)^3))^(1/3))

Maple [F]

\[\int \frac {F^{a +\frac {b}{\left (d x +c \right )^{3}}}}{\left (d x +c \right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.20 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=-\frac {F^{a} d \left (-\frac {b \log \left (F\right )}{d^{3}}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -\frac {b \log \left (F\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )}{3 \, b \log \left (F\right )} \]

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^2,x, algorithm="fricas")

[Out]

-1/3*F^a*d*(-b*log(F)/d^3)^(2/3)*gamma(1/3, -b*log(F)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))/(b*log(F))

Sympy [F]

\[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\int \frac {F^{a + \frac {b}{\left (c + d x\right )^{3}}}}{\left (c + d x\right )^{2}}\, dx \]

[In]

integrate(F**(a+b/(d*x+c)**3)/(d*x+c)**2,x)

[Out]

Integral(F**(a + b/(c + d*x)**3)/(c + d*x)**2, x)

Maxima [F]

\[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{3}}}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^3)/(d*x+c)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.18 \[ \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{(c+d x)^2} \, dx=\frac {F^a\,\left (3\,\Gamma \left (\frac {2}{3}\right )\,\Gamma \left (\frac {1}{3},-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )-2\,\pi \,\sqrt {3}\right )}{9\,d\,\Gamma \left (\frac {2}{3}\right )\,\left (c+d\,x\right )\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{1/3}} \]

[In]

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

[Out]

(F^a*(3*gamma(2/3)*igamma(1/3, -(b*log(F))/(c + d*x)^3) - 2*3^(1/2)*pi))/(9*d*gamma(2/3)*(c + d*x)*(-(b*log(F)
)/(c + d*x)^3)^(1/3))