Integrand size = 25, antiderivative size = 114 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+6 n} \, dx=-\frac {F^{a+b (c+d x)^n} \left (120-120 b (c+d x)^n \log (F)+60 b^2 (c+d x)^{2 n} \log ^2(F)-20 b^3 (c+d x)^{3 n} \log ^3(F)+5 b^4 (c+d x)^{4 n} \log ^4(F)-b^5 (c+d x)^{5 n} \log ^5(F)\right )}{b^6 d n \log ^6(F)} \] Output:
-F^(a+b*(d*x+c)^n)*(120-120*b*(d*x+c)^n*ln(F)+60*b^2*(d*x+c)^(2*n)*ln(F)^2 -20*b^3*(d*x+c)^(3*n)*ln(F)^3+5*b^4*(d*x+c)^(4*n)*ln(F)^4-b^5*(d*x+c)^(5*n )*ln(F)^5)/b^6/d/n/ln(F)^6
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.28 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+6 n} \, dx=-\frac {F^a \Gamma \left (6,-b (c+d x)^n \log (F)\right )}{b^6 d n \log ^6(F)} \] Input:
Integrate[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 + 6*n),x]
Output:
-((F^a*Gamma[6, -(b*(c + d*x)^n*Log[F])])/(b^6*d*n*Log[F]^6))
Time = 0.43 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2647}
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.
\(\displaystyle \int (c+d x)^{6 n-1} F^{a+b (c+d x)^n} \, dx\) |
\(\Big \downarrow \) 2647 |
\(\displaystyle -\frac {F^{a+b (c+d x)^n} \left (-b^5 \log ^5(F) (c+d x)^{5 n}+5 b^4 \log ^4(F) (c+d x)^{4 n}-20 b^3 \log ^3(F) (c+d x)^{3 n}+60 b^2 \log ^2(F) (c+d x)^{2 n}-120 b \log (F) (c+d x)^n+120\right )}{b^6 d n \log ^6(F)}\) |
Input:
Int[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 + 6*n),x]
Output:
-((F^(a + b*(c + d*x)^n)*(120 - 120*b*(c + d*x)^n*Log[F] + 60*b^2*(c + d*x )^(2*n)*Log[F]^2 - 20*b^3*(c + d*x)^(3*n)*Log[F]^3 + 5*b^4*(c + d*x)^(4*n) *Log[F]^4 - b^5*(c + d*x)^(5*n)*Log[F]^5))/(b^6*d*n*Log[F]^6))
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> With[{p = Simplify[(m + 1)/n]}, Simp[(-F^a)*((f/d)^m/(d*n* ((-b)*Log[F])^p))*Simplify[FunctionExpand[Gamma[p, (-b)*(c + d*x)^n*Log[F]] ]], x] /; IGtQ[p, 0]] /; FreeQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0] && !TrueQ[$UseGamma]
Time = 0.33 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.99
method | result | size |
risch | \(\frac {\left (b^{5} \left (d x +c \right )^{5 n} \ln \left (F \right )^{5}-5 b^{4} \left (d x +c \right )^{4 n} \ln \left (F \right )^{4}+20 b^{3} \left (d x +c \right )^{3 n} \ln \left (F \right )^{3}-60 b^{2} \left (d x +c \right )^{2 n} \ln \left (F \right )^{2}+120 b \left (d x +c \right )^{n} \ln \left (F \right )-120\right ) F^{a +b \left (d x +c \right )^{n}}}{b^{6} \ln \left (F \right )^{6} n d}\) | \(113\) |
Input:
int(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+6*n),x,method=_RETURNVERBOSE)
Output:
(((d*x+c)^n)^5*b^5*ln(F)^5-5*((d*x+c)^n)^4*b^4*ln(F)^4+20*((d*x+c)^n)^3*b^ 3*ln(F)^3-60*((d*x+c)^n)^2*b^2*ln(F)^2+120*b*(d*x+c)^n*ln(F)-120)/b^6/ln(F )^6/n/d*F^(a+b*(d*x+c)^n)
Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+6 n} \, dx=\frac {{\left ({\left (d x + c\right )}^{5 \, n} b^{5} \log \left (F\right )^{5} - 5 \, {\left (d x + c\right )}^{4 \, n} b^{4} \log \left (F\right )^{4} + 20 \, {\left (d x + c\right )}^{3 \, n} b^{3} \log \left (F\right )^{3} - 60 \, {\left (d x + c\right )}^{2 \, n} b^{2} \log \left (F\right )^{2} + 120 \, {\left (d x + c\right )}^{n} b \log \left (F\right ) - 120\right )} e^{\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) + a \log \left (F\right )\right )}}{b^{6} d n \log \left (F\right )^{6}} \] Input:
integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+6*n),x, algorithm="fricas")
Output:
((d*x + c)^(5*n)*b^5*log(F)^5 - 5*(d*x + c)^(4*n)*b^4*log(F)^4 + 20*(d*x + c)^(3*n)*b^3*log(F)^3 - 60*(d*x + c)^(2*n)*b^2*log(F)^2 + 120*(d*x + c)^n *b*log(F) - 120)*e^((d*x + c)^n*b*log(F) + a*log(F))/(b^6*d*n*log(F)^6)
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (112) = 224\).
Time = 41.76 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.52 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+6 n} \, dx=\begin {cases} \frac {x}{c} & \text {for}\: F = 1 \wedge b = 0 \wedge d = 0 \wedge n = 0 \\F^{a} \left (\frac {c \left (c + d x\right )^{6 n - 1}}{6 d n} + \frac {x \left (c + d x\right )^{6 n - 1}}{6 n}\right ) & \text {for}\: b = 0 \\F^{a + b c^{n}} c^{6 n - 1} x & \text {for}\: d = 0 \\\frac {F^{a + b} \log {\left (\frac {c}{d} + x \right )}}{d} & \text {for}\: n = 0 \\\frac {c \left (c + d x\right )^{6 n - 1}}{6 d n} + \frac {x \left (c + d x\right )^{6 n - 1}}{6 n} & \text {for}\: F = 1 \\\frac {F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{5 n}}{b d n \log {\left (F \right )}} - \frac {5 F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{4 n}}{b^{2} d n \log {\left (F \right )}^{2}} + \frac {20 F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{3 n}}{b^{3} d n \log {\left (F \right )}^{3}} - \frac {60 F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{2 n}}{b^{4} d n \log {\left (F \right )}^{4}} + \frac {120 F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{n}}{b^{5} d n \log {\left (F \right )}^{5}} - \frac {120 F^{a + b \left (c + d x\right )^{n}}}{b^{6} d n \log {\left (F \right )}^{6}} & \text {otherwise} \end {cases} \] Input:
integrate(F**(a+b*(d*x+c)**n)*(d*x+c)**(-1+6*n),x)
Output:
Piecewise((x/c, Eq(F, 1) & Eq(b, 0) & Eq(d, 0) & Eq(n, 0)), (F**a*(c*(c + d*x)**(6*n - 1)/(6*d*n) + x*(c + d*x)**(6*n - 1)/(6*n)), Eq(b, 0)), (F**(a + b*c**n)*c**(6*n - 1)*x, Eq(d, 0)), (F**(a + b)*log(c/d + x)/d, Eq(n, 0) ), (c*(c + d*x)**(6*n - 1)/(6*d*n) + x*(c + d*x)**(6*n - 1)/(6*n), Eq(F, 1 )), (F**(a + b*(c + d*x)**n)*(c + d*x)**(5*n)/(b*d*n*log(F)) - 5*F**(a + b *(c + d*x)**n)*(c + d*x)**(4*n)/(b**2*d*n*log(F)**2) + 20*F**(a + b*(c + d *x)**n)*(c + d*x)**(3*n)/(b**3*d*n*log(F)**3) - 60*F**(a + b*(c + d*x)**n) *(c + d*x)**(2*n)/(b**4*d*n*log(F)**4) + 120*F**(a + b*(c + d*x)**n)*(c + d*x)**n/(b**5*d*n*log(F)**5) - 120*F**(a + b*(c + d*x)**n)/(b**6*d*n*log(F )**6), True))
Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.13 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+6 n} \, dx=\frac {{\left ({\left (d x + c\right )}^{5 \, n} F^{a} b^{5} \log \left (F\right )^{5} - 5 \, {\left (d x + c\right )}^{4 \, n} F^{a} b^{4} \log \left (F\right )^{4} + 20 \, {\left (d x + c\right )}^{3 \, n} F^{a} b^{3} \log \left (F\right )^{3} - 60 \, {\left (d x + c\right )}^{2 \, n} F^{a} b^{2} \log \left (F\right )^{2} + 120 \, {\left (d x + c\right )}^{n} F^{a} b \log \left (F\right ) - 120 \, F^{a}\right )} F^{{\left (d x + c\right )}^{n} b}}{b^{6} d n \log \left (F\right )^{6}} \] Input:
integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+6*n),x, algorithm="maxima")
Output:
((d*x + c)^(5*n)*F^a*b^5*log(F)^5 - 5*(d*x + c)^(4*n)*F^a*b^4*log(F)^4 + 2 0*(d*x + c)^(3*n)*F^a*b^3*log(F)^3 - 60*(d*x + c)^(2*n)*F^a*b^2*log(F)^2 + 120*(d*x + c)^n*F^a*b*log(F) - 120*F^a)*F^((d*x + c)^n*b)/(b^6*d*n*log(F) ^6)
\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+6 n} \, dx=\int { {\left (d x + c\right )}^{6 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \] Input:
integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+6*n),x, algorithm="giac")
Output:
integrate((d*x + c)^(6*n - 1)*F^((d*x + c)^n*b + a), x)
Timed out. \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+6 n} \, dx=\int F^{a+b\,{\left (c+d\,x\right )}^n}\,{\left (c+d\,x\right )}^{6\,n-1} \,d x \] Input:
int(F^(a + b*(c + d*x)^n)*(c + d*x)^(6*n - 1),x)
Output:
int(F^(a + b*(c + d*x)^n)*(c + d*x)^(6*n - 1), x)
Time = 0.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.98 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+6 n} \, dx=\frac {f^{\left (d x +c \right )^{n} b +a} \left (\left (d x +c \right )^{5 n} \mathrm {log}\left (f \right )^{5} b^{5}-5 \left (d x +c \right )^{4 n} \mathrm {log}\left (f \right )^{4} b^{4}+20 \left (d x +c \right )^{3 n} \mathrm {log}\left (f \right )^{3} b^{3}-60 \left (d x +c \right )^{2 n} \mathrm {log}\left (f \right )^{2} b^{2}+120 \left (d x +c \right )^{n} \mathrm {log}\left (f \right ) b -120\right )}{\mathrm {log}\left (f \right )^{6} b^{6} d n} \] Input:
int(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+6*n),x)
Output:
(f**((c + d*x)**n*b + a)*((c + d*x)**(5*n)*log(f)**5*b**5 - 5*(c + d*x)**( 4*n)*log(f)**4*b**4 + 20*(c + d*x)**(3*n)*log(f)**3*b**3 - 60*(c + d*x)**( 2*n)*log(f)**2*b**2 + 120*(c + d*x)**n*log(f)*b - 120))/(log(f)**6*b**6*d* n)