\(\int F^{a+\frac {b}{(c+d x)^3}} \, dx\) [355]

Optimal result

Integrand size = 13, antiderivative size = 47 \[ \int F^{a+\frac {b}{(c+d x)^3}} \, dx=\frac {F^a (c+d x) \Gamma \left (-\frac {1}{3},-\frac {b \log (F)}{(c+d x)^3}\right ) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}}}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2239} \[ \int F^{a+\frac {b}{(c+d x)^3}} \, dx=\frac {F^a (c+d x) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}} \Gamma \left (-\frac {1}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d} \]

[In]

Int[F^(a + b/(c + d*x)^3),x]

[Out]

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

Rule 2239

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(-F^a)*(c + d*x)*(Gamma[1/n, (-b)*(c + d
*x)^n*Log[F]]/(d*n*((-b)*(c + d*x)^n*Log[F])^(1/n))), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rubi steps \begin{align*} \text {integral}& = \frac {F^a (c+d x) \Gamma \left (-\frac {1}{3},-\frac {b \log (F)}{(c+d x)^3}\right ) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}}}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int F^{a+\frac {b}{(c+d x)^3}} \, dx=\frac {F^a (c+d x) \Gamma \left (-\frac {1}{3},-\frac {b \log (F)}{(c+d x)^3}\right ) \sqrt [3]{-\frac {b \log (F)}{(c+d x)^3}}}{3 d} \]

[In]

Integrate[F^(a + b/(c + d*x)^3),x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (41) = 82\).

Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.74 \[ \int F^{a+\frac {b}{(c+d x)^3}} \, dx=-\frac {F^{a} d \left (-\frac {b \log \left (F\right )}{d^{3}}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -\frac {b \log \left (F\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) - {\left (d x + c\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(F**(a + b/(c + d*x)**3), x)

Maxima [F]

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

[In]

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

[Out]

3*F^a*b*d*integrate(F^(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))*x/(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 +
4*c^3*d*x + c^4), x)*log(F) + F^a*F^(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))*x

Giac [F]

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

[In]

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

[Out]

integrate(F^(a + b/(d*x + c)^3), x)

Mupad [B] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.51 \[ \int F^{a+\frac {b}{(c+d x)^3}} \, dx=\frac {F^a\,\left (c+d\,x\right )\,\left (F^{\frac {b}{{\left (c+d\,x\right )}^3}}-\Gamma \left (\frac {2}{3},-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{1/3}+\Gamma \left (\frac {2}{3}\right )\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{1/3}\right )}{d} \]

[In]

int(F^(a + b/(c + d*x)^3),x)

[Out]

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