\(\int F^{c (a+b x)} (d+e x)^{4/3} \, dx\) [49]

Optimal result

Integrand size = 19, antiderivative size = 71 \[ \int F^{c (a+b x)} (d+e x)^{4/3} \, dx=-\frac {e F^{c \left (a-\frac {b d}{e}\right )} \sqrt [3]{d+e x} \Gamma \left (\frac {7}{3},-\frac {b c (d+e x) \log (F)}{e}\right )}{b^2 c^2 \log ^2(F) \sqrt [3]{-\frac {b c (d+e x) \log (F)}{e}}} \]

[Out]

-e*F^(c*(a-b*d/e))*(e*x+d)^(1/3)*GAMMA(7/3,-b*c*(e*x+d)*ln(F)/e)/b^2/c^2/ln(F)^2/(-b*c*(e*x+d)*ln(F)/e)^(1/3)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2212} \[ \int F^{c (a+b x)} (d+e x)^{4/3} \, dx=-\frac {e \sqrt [3]{d+e x} F^{c \left (a-\frac {b d}{e}\right )} \Gamma \left (\frac {7}{3},-\frac {b c (d+e x) \log (F)}{e}\right )}{b^2 c^2 \log ^2(F) \sqrt [3]{-\frac {b c \log (F) (d+e x)}{e}}} \]

[In]

Int[F^(c*(a + b*x))*(d + e*x)^(4/3),x]

[Out]

-((e*F^(c*(a - (b*d)/e))*(d + e*x)^(1/3)*Gamma[7/3, -((b*c*(d + e*x)*Log[F])/e)])/(b^2*c^2*Log[F]^2*(-((b*c*(d
+ e*x)*Log[F])/e))^(1/3)))

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\frac {e F^{c \left (a-\frac {b d}{e}\right )} \sqrt [3]{d+e x} \Gamma \left (\frac {7}{3},-\frac {b c (d+e x) \log (F)}{e}\right )}{b^2 c^2 \log ^2(F) \sqrt [3]{-\frac {b c (d+e x) \log (F)}{e}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.89 \[ \int F^{c (a+b x)} (d+e x)^{4/3} \, dx=-\frac {F^{c \left (a-\frac {b d}{e}\right )} (d+e x)^{7/3} \Gamma \left (\frac {7}{3},-\frac {b c (d+e x) \log (F)}{e}\right )}{e \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{7/3}} \]

[In]

Integrate[F^(c*(a + b*x))*(d + e*x)^(4/3),x]

[Out]

-((F^(c*(a - (b*d)/e))*(d + e*x)^(7/3)*Gamma[7/3, -((b*c*(d + e*x)*Log[F])/e)])/(e*(-((b*c*(d + e*x)*Log[F])/e
))^(7/3)))

Maple [F]

\[\int F^{c \left (b x +a \right )} \left (e x +d \right )^{\frac {4}{3}}d x\]

[In]

int(F^(c*(b*x+a))*(e*x+d)^(4/3),x)

[Out]

int(F^(c*(b*x+a))*(e*x+d)^(4/3),x)

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.65 \[ \int F^{c (a+b x)} (d+e x)^{4/3} \, dx=\frac {\frac {4 \, \left (-\frac {b c \log \left (F\right )}{e}\right )^{\frac {2}{3}} e^{2} \Gamma \left (\frac {1}{3}, -\frac {{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right )}{F^{\frac {b c d - a c e}{e}}} - 3 \, {\left (4 \, b c e \log \left (F\right ) - 3 \, {\left (b^{2} c^{2} e x + b^{2} c^{2} d\right )} \log \left (F\right )^{2}\right )} {\left (e x + d\right )}^{\frac {1}{3}} F^{b c x + a c}}{9 \, b^{3} c^{3} \log \left (F\right )^{3}} \]

[In]

integrate(F^(c*(b*x+a))*(e*x+d)^(4/3),x, algorithm="fricas")

[Out]

1/9*(4*(-b*c*log(F)/e)^(2/3)*e^2*gamma(1/3, -(b*c*e*x + b*c*d)*log(F)/e)/F^((b*c*d - a*c*e)/e) - 3*(4*b*c*e*lo
g(F) - 3*(b^2*c^2*e*x + b^2*c^2*d)*log(F)^2)*(e*x + d)^(1/3)*F^(b*c*x + a*c))/(b^3*c^3*log(F)^3)

Sympy [F]

\[ \int F^{c (a+b x)} (d+e x)^{4/3} \, dx=\int F^{c \left (a + b x\right )} \left (d + e x\right )^{\frac {4}{3}}\, dx \]

[In]

integrate(F**(c*(b*x+a))*(e*x+d)**(4/3),x)

[Out]

Integral(F**(c*(a + b*x))*(d + e*x)**(4/3), x)

Maxima [F]

\[ \int F^{c (a+b x)} (d+e x)^{4/3} \, dx=\int { {\left (e x + d\right )}^{\frac {4}{3}} F^{{\left (b x + a\right )} c} \,d x } \]

[In]

integrate(F^(c*(b*x+a))*(e*x+d)^(4/3),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int F^{c (a+b x)} (d+e x)^{4/3} \, dx=\int { {\left (e x + d\right )}^{\frac {4}{3}} F^{{\left (b x + a\right )} c} \,d x } \]

[In]

integrate(F^(c*(b*x+a))*(e*x+d)^(4/3),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} (d+e x)^{4/3} \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (d+e\,x\right )}^{4/3} \,d x \]

[In]

int(F^(c*(a + b*x))*(d + e*x)^(4/3),x)

[Out]

int(F^(c*(a + b*x))*(d + e*x)^(4/3), x)