Integrand size = 50, antiderivative size = 71 \[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^4\right )^m \Gamma \left (1+4 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-4 m}}{b c \log (F)} \] Output:
F^(c*(a-b*d/e))*((e*x+d)^4)^m*GAMMA(1+4*m,-b*c*(e*x+d)*ln(F)/e)/b/c/ln(F)/ ((-b*c*(e*x+d)*ln(F)/e)^(4*m))
Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^4\right )^m \Gamma \left (1+4 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-4 m}}{b c \log (F)} \] Input:
Integrate[F^(c*(a + b*x))*(d^4 + 4*d^3*e*x + 6*d^2*e^2*x^2 + 4*d*e^3*x^3 + e^4*x^4)^m,x]
Output:
(F^(c*(a - (b*d)/e))*((d + e*x)^4)^m*Gamma[1 + 4*m, -((b*c*(d + e*x)*Log[F ])/e)])/(b*c*Log[F]*(-((b*c*(d + e*x)*Log[F])/e))^(4*m))
Time = 0.38 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2008, 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.
\(\displaystyle \int \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 2008 |
\(\displaystyle (d+e x)^{-4 m} \left ((d+e x)^4\right )^m \int F^{c (a+b x)} (d+e x)^{4 m}dx\) |
\(\Big \downarrow \) 2612 |
\(\displaystyle \frac {\left ((d+e x)^4\right )^m F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{-4 m} \Gamma \left (4 m+1,-\frac {b c (d+e x) \log (F)}{e}\right )}{b c \log (F)}\) |
Input:
Int[F^(c*(a + b*x))*(d^4 + 4*d^3*e*x + 6*d^2*e^2*x^2 + 4*d*e^3*x^3 + e^4*x ^4)^m,x]
Output:
(F^(c*(a - (b*d)/e))*((d + e*x)^4)^m*Gamma[1 + 4*m, -((b*c*(d + e*x)*Log[F ])/e)])/(b*c*Log[F]*(-((b*c*(d + e*x)*Log[F])/e))^(4*m))
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Simp[((a + b*x)^Exp on[Px, x])^p/(a + b*x)^(Expon[Px, x]*p) Int[u*(a + b*x)^(Expon[Px, x]*p), x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; !IntegerQ[p] && PolyQ[Px, x ] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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]
\[\int F^{c \left (b x +a \right )} \left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right )^{m}d x\]
Input:
int(F^(c*(b*x+a))*(e^4*x^4+4*d*e^3*x^3+6*d^2*e^2*x^2+4*d^3*e*x+d^4)^m,x)
Output:
int(F^(c*(b*x+a))*(e^4*x^4+4*d*e^3*x^3+6*d^2*e^2*x^2+4*d^3*e*x+d^4)^m,x)
\[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\int { {\left (e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \] Input:
integrate(F^((b*x+a)*c)*(e^4*x^4+4*d*e^3*x^3+6*d^2*e^2*x^2+4*d^3*e*x+d^4)^ m,x, algorithm="fricas")
Output:
integral((e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4)^m*F^(b* c*x + a*c), x)
\[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\int F^{c \left (a + b x\right )} \left (\left (d + e x\right )^{4}\right )^{m}\, dx \] Input:
integrate(F**((b*x+a)*c)*(e**4*x**4+4*d*e**3*x**3+6*d**2*e**2*x**2+4*d**3* e*x+d**4)**m,x)
Output:
Integral(F**(c*(a + b*x))*((d + e*x)**4)**m, x)
\[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\int { {\left (e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \] Input:
integrate(F^((b*x+a)*c)*(e^4*x^4+4*d*e^3*x^3+6*d^2*e^2*x^2+4*d^3*e*x+d^4)^ m,x, algorithm="maxima")
Output:
integrate((e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4)^m*F^(( b*x + a)*c), x)
\[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\int { {\left (e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \] Input:
integrate(F^((b*x+a)*c)*(e^4*x^4+4*d*e^3*x^3+6*d^2*e^2*x^2+4*d^3*e*x+d^4)^ m,x, algorithm="giac")
Output:
integrate((e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4)^m*F^(( b*x + a)*c), x)
Timed out. \[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4\right )}^m \,d x \] Input:
int(F^(c*(a + b*x))*(d^4 + e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e *x)^m,x)
Output:
int(F^(c*(a + b*x))*(d^4 + e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e *x)^m, x)
\[ \int F^{c (a+b x)} \left (d^4+4 d^3 e x+6 d^2 e^2 x^2+4 d e^3 x^3+e^4 x^4\right )^m \, dx=\frac {f^{a c} \left (f^{b c x} \left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right )^{m}-4 \left (\int \frac {f^{b c x} \left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right )^{m}}{e x +d}d x \right ) e m \right )}{\mathrm {log}\left (f \right ) b c} \] Input:
int(F^((b*x+a)*c)*(e^4*x^4+4*d*e^3*x^3+6*d^2*e^2*x^2+4*d^3*e*x+d^4)^m,x)
Output:
(f**(a*c)*(f**(b*c*x)*(d**4 + 4*d**3*e*x + 6*d**2*e**2*x**2 + 4*d*e**3*x** 3 + e**4*x**4)**m - 4*int((f**(b*c*x)*(d**4 + 4*d**3*e*x + 6*d**2*e**2*x** 2 + 4*d*e**3*x**3 + e**4*x**4)**m)/(d + e*x),x)*e*m))/(log(f)*b*c)