Integrand size = 21, antiderivative size = 98 \[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=-\frac {e F^{c \left (a-\frac {b d}{e}\right ) n-c n (a+b x)} \left (F^{c (a+b x)}\right )^n \sqrt [3]{d+e x} \Gamma \left (\frac {7}{3},-\frac {b c n (d+e x) \log (F)}{e}\right )}{b^2 c^2 n^2 \log ^2(F) \sqrt [3]{-\frac {b c n (d+e x) \log (F)}{e}}} \] Output:
-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 )
Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80 \[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=-\frac {F^{-\frac {b c n (d+e x)}{e}} \left (F^{c (a+b x)}\right )^n (d+e x)^{7/3} \Gamma \left (\frac {7}{3},-\frac {b c n (d+e x) \log (F)}{e}\right )}{e \left (-\frac {b c n (d+e x) \log (F)}{e}\right )^{7/3}} \] Input:
Integrate[(F^(c*(a + b*x)))^n*(d + e*x)^(4/3),x]
Output:
-(((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)))
Time = 0.49 (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 = {2613, 2612}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (d+e x)^{4/3} \left (F^{c (a+b x)}\right )^n \, dx\) |
\(\Big \downarrow \) 2613 |
\(\displaystyle F^{-c n (a+b x)} \left (F^{c (a+b x)}\right )^n \int F^{c n (a+b x)} (d+e x)^{4/3}dx\) |
\(\Big \downarrow \) 2612 |
\(\displaystyle -\frac {e \sqrt [3]{d+e x} \left (F^{c (a+b x)}\right )^n F^{c n \left (a-\frac {b d}{e}\right )-c n (a+b x)} \Gamma \left (\frac {7}{3},-\frac {b c n (d+e x) \log (F)}{e}\right )}{b^2 c^2 n^2 \log ^2(F) \sqrt [3]{-\frac {b c n \log (F) (d+e x)}{e}}}\) |
Input:
Int[(F^(c*(a + b*x)))^n*(d + e*x)^(4/3),x]
Output:
-((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)*Log[F])/e)])/(b^2*c^2*n^2*Log[F]^2*(-(( b*c*n*(d + e*x)*Log[F])/e))^(1/3)))
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]
Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_)*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[(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]
\[\int \left (F^{c \left (b x +a \right )}\right )^{n} \left (e x +d \right )^{\frac {4}{3}}d x\]
Input:
int((F^(c*(b*x+a)))^n*(e*x+d)^(4/3),x)
Output:
int((F^(c*(b*x+a)))^n*(e*x+d)^(4/3),x)
Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.36 \[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=\frac {\frac {4 \, \left (-\frac {b c n \log \left (F\right )}{e}\right )^{\frac {2}{3}} e^{2} \Gamma \left (\frac {1}{3}, -\frac {{\left (b c e n x + b c d n\right )} \log \left (F\right )}{e}\right )}{F^{\frac {{\left (b c d - a c e\right )} n}{e}}} - 3 \, {\left (4 \, b c e n \log \left (F\right ) - 3 \, {\left (b^{2} c^{2} e n^{2} x + b^{2} c^{2} d n^{2}\right )} \log \left (F\right )^{2}\right )} {\left (e x + d\right )}^{\frac {1}{3}} F^{b c n x + a c n}}{9 \, b^{3} c^{3} n^{3} \log \left (F\right )^{3}} \] Input:
integrate((F^((b*x+a)*c))^n*(e*x+d)^(4/3),x, algorithm="fricas")
Output:
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)
Timed out. \[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=\text {Timed out} \] Input:
integrate((F**((b*x+a)*c))**n*(e*x+d)**(4/3),x)
Output:
Timed out
\[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=\int { {\left (e x + d\right )}^{\frac {4}{3}} {\left (F^{{\left (b x + a\right )} c}\right )}^{n} \,d x } \] Input:
integrate((F^((b*x+a)*c))^n*(e*x+d)^(4/3),x, algorithm="maxima")
Output:
integrate((e*x + d)^(4/3)*F^((b*x + a)*c*n), x)
\[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=\int { {\left (e x + d\right )}^{\frac {4}{3}} {\left (F^{{\left (b x + a\right )} c}\right )}^{n} \,d x } \] Input:
integrate((F^((b*x+a)*c))^n*(e*x+d)^(4/3),x, algorithm="giac")
Output:
integrate((e*x + d)^(4/3)*(F^((b*x + a)*c))^n, x)
Timed out. \[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=\int {\left (F^{c\,\left (a+b\,x\right )}\right )}^n\,{\left (d+e\,x\right )}^{4/3} \,d x \] Input:
int((F^(c*(a + b*x)))^n*(d + e*x)^(4/3),x)
Output:
int((F^(c*(a + b*x)))^n*(d + e*x)^(4/3), x)
\[ \int \left (F^{c (a+b x)}\right )^n (d+e x)^{4/3} \, dx=f^{a c n} \left (\left (\int f^{b c n x} \left (e x +d \right )^{\frac {1}{3}} x d x \right ) e +\left (\int f^{b c n x} \left (e x +d \right )^{\frac {1}{3}}d x \right ) d \right ) \] Input:
int((F^((b*x+a)*c))^n*(e*x+d)^(4/3),x)
Output:
f**(a*c*n)*(int(f**(b*c*n*x)*(d + e*x)**(1/3)*x,x)*e + int(f**(b*c*n*x)*(d + e*x)**(1/3),x)*d)