\(\int (F^{c (a+b x)})^n (d+e x)^{4/3} \, dx\) [50]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=-\frac {e F^{c \left (a-\frac {b d}{e}\right ) n-c n (a+b x)} \left (F^{c (a+b x)}\right )^n \sqrt [3]{d+e x} \Gamma \left (\frac {7}{3},-\frac {b c n (d+e x) \log (F)}{e}\right )}{b^2 c^2 n^2 \log ^2(F) \sqrt [3]{-\frac {b c n (d+e x) \log (F)}{e}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2213, 2212} \[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=-\frac {e \sqrt [3]{d+e x} \left (F^{c (a+b x)}\right )^n F^{c n \left (a-\frac {b d}{e}\right )-c n (a+b x)} \Gamma \left (\frac {7}{3},-\frac {b c n (d+e x) \log (F)}{e}\right )}{b^2 c^2 n^2 \log ^2(F) \sqrt [3]{-\frac {b c n \log (F) (d+e x)}{e}}} \]

[In]

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

[Out]

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

Rule 2213

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[(b*F^(g*(e +
f*x)))^n/F^(g*n*(e + f*x)), Int[(c + d*x)^m*F^(g*n*(e + f*x)), x], x] /; FreeQ[{F, b, c, d, e, f, g, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \left (F^{-c n (a+b x)} \left (F^{c (a+b x)}\right )^n\right ) \int F^{c n (a+b x)} (d+e x)^{4/3} \, dx \\ & = -\frac {e F^{c \left (a-\frac {b d}{e}\right ) n-c n (a+b x)} \left (F^{c (a+b x)}\right )^n \sqrt [3]{d+e x} \Gamma \left (\frac {7}{3},-\frac {b c n (d+e x) \log (F)}{e}\right )}{b^2 c^2 n^2 \log ^2(F) \sqrt [3]{-\frac {b c n (d+e x) \log (F)}{e}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80 \[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=-\frac {F^{-\frac {b c n (d+e x)}{e}} \left (F^{c (a+b x)}\right )^n (d+e x)^{7/3} \Gamma \left (\frac {7}{3},-\frac {b c n (d+e x) \log (F)}{e}\right )}{e \left (-\frac {b c n (d+e x) \log (F)}{e}\right )^{7/3}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.36 \[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=\frac {\frac {4 \, \left (-\frac {b c n \log \left (F\right )}{e}\right )^{\frac {2}{3}} e^{2} \Gamma \left (\frac {1}{3}, -\frac {{\left (b c e n x + b c d n\right )} \log \left (F\right )}{e}\right )}{F^{\frac {{\left (b c d - a c e\right )} n}{e}}} - 3 \, {\left (4 \, b c e n \log \left (F\right ) - 3 \, {\left (b^{2} c^{2} e n^{2} x + b^{2} c^{2} d n^{2}\right )} \log \left (F\right )^{2}\right )} {\left (e x + d\right )}^{\frac {1}{3}} F^{b c n x + a c n}}{9 \, b^{3} c^{3} n^{3} \log \left (F\right )^{3}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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