\(\int F^{a+b (c+d x)^n} (c+d x)^{-1+5 n} \, dx\) [369]

Optimal result

Integrand size = 25, antiderivative size = 94 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+5 n} \, dx=\frac {F^{a+b (c+d x)^n} \left (24-24 b (c+d x)^n \log (F)+12 b^2 (c+d x)^{2 n} \log ^2(F)-4 b^3 (c+d x)^{3 n} \log ^3(F)+b^4 (c+d x)^{4 n} \log ^4(F)\right )}{b^5 d n \log ^5(F)} \]

[Out]

F^(a+b*(d*x+c)^n)*(24-24*b*(d*x+c)^n*ln(F)+12*b^2*(d*x+c)^(2*n)*ln(F)^2-4*b^3*(d*x+c)^(3*n)*ln(F)^3+b^4*(d*x+c
)^(4*n)*ln(F)^4)/b^5/d/n/ln(F)^5

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 94, 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 = {2249} \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+5 n} \, dx=\frac {F^{a+b (c+d x)^n} \left (b^4 \log ^4(F) (c+d x)^{4 n}-4 b^3 \log ^3(F) (c+d x)^{3 n}+12 b^2 \log ^2(F) (c+d x)^{2 n}-24 b \log (F) (c+d x)^n+24\right )}{b^5 d n \log ^5(F)} \]

[In]

Int[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 + 5*n),x]

[Out]

(F^(a + b*(c + d*x)^n)*(24 - 24*b*(c + d*x)^n*Log[F] + 12*b^2*(c + d*x)^(2*n)*Log[F]^2 - 4*b^3*(c + d*x)^(3*n)
*Log[F]^3 + b^4*(c + d*x)^(4*n)*Log[F]^4))/(b^5*d*n*Log[F]^5)

Rule 2249

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*Lo
g[F]]]], x] /; IGtQ[p, 0]] /; FreeQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \frac {F^{a+b (c+d x)^n} \left (24-24 b (c+d x)^n \log (F)+12 b^2 (c+d x)^{2 n} \log ^2(F)-4 b^3 (c+d x)^{3 n} \log ^3(F)+b^4 (c+d x)^{4 n} \log ^4(F)\right )}{b^5 d n \log ^5(F)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.33 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+5 n} \, dx=\frac {F^a \Gamma \left (5,-b (c+d x)^n \log (F)\right )}{b^5 d n \log ^5(F)} \]

[In]

Integrate[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 + 5*n),x]

[Out]

(F^a*Gamma[5, -(b*(c + d*x)^n*Log[F])])/(b^5*d*n*Log[F]^5)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01

method	result	size
risch	\(\frac {F^{a +b \left (d x +c \right )^{n}} \left (24-24 b \left (d x +c \right )^{n} \ln \left (F \right )+12 b^{2} \left (d x +c \right )^{2 n} \ln \left (F \right )^{2}-4 b^{3} \left (d x +c \right )^{3 n} \ln \left (F \right )^{3}+b^{4} \left (d x +c \right )^{4 n} \ln \left (F \right )^{4}\right )}{b^{5} d n \ln \left (F \right )^{5}}\)	\(95\)

[In]

int(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+5*n),x,method=_RETURNVERBOSE)

[Out]

(((d*x+c)^n)^4*b^4*ln(F)^4-4*((d*x+c)^n)^3*b^3*ln(F)^3+12*((d*x+c)^n)^2*b^2*ln(F)^2-24*b*(d*x+c)^n*ln(F)+24)/b
^5/ln(F)^5/n/d*F^(a+b*(d*x+c)^n)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.04 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+5 n} \, dx=\frac {{\left ({\left (d x + c\right )}^{4 \, n} b^{4} \log \left (F\right )^{4} - 4 \, {\left (d x + c\right )}^{3 \, n} b^{3} \log \left (F\right )^{3} + 12 \, {\left (d x + c\right )}^{2 \, n} b^{2} \log \left (F\right )^{2} - 24 \, {\left (d x + c\right )}^{n} b \log \left (F\right ) + 24\right )} e^{\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) + a \log \left (F\right )\right )}}{b^{5} d n \log \left (F\right )^{5}} \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+5*n),x, algorithm="fricas")

[Out]

((d*x + c)^(4*n)*b^4*log(F)^4 - 4*(d*x + c)^(3*n)*b^3*log(F)^3 + 12*(d*x + c)^(2*n)*b^2*log(F)^2 - 24*(d*x + c
)^n*b*log(F) + 24)*e^((d*x + c)^n*b*log(F) + a*log(F))/(b^5*d*n*log(F)^5)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (92) = 184\).

Time = 24.65 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.69 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+5 n} \, dx=\begin {cases} \frac {x}{c} & \text {for}\: F = 1 \wedge b = 0 \wedge d = 0 \wedge n = 0 \\F^{a} \left (\frac {c \left (c + d x\right )^{5 n - 1}}{5 d n} + \frac {x \left (c + d x\right )^{5 n - 1}}{5 n}\right ) & \text {for}\: b = 0 \\F^{a + b c^{n}} c^{5 n - 1} x & \text {for}\: d = 0 \\\frac {F^{a + b} \log {\left (\frac {c}{d} + x \right )}}{d} & \text {for}\: n = 0 \\\frac {c \left (c + d x\right )^{5 n - 1}}{5 d n} + \frac {x \left (c + d x\right )^{5 n - 1}}{5 n} & \text {for}\: F = 1 \\\frac {F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{4 n}}{b d n \log {\left (F \right )}} - \frac {4 F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{3 n}}{b^{2} d n \log {\left (F \right )}^{2}} + \frac {12 F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{2 n}}{b^{3} d n \log {\left (F \right )}^{3}} - \frac {24 F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{n}}{b^{4} d n \log {\left (F \right )}^{4}} + \frac {24 F^{a + b \left (c + d x\right )^{n}}}{b^{5} d n \log {\left (F \right )}^{5}} & \text {otherwise} \end {cases} \]

[In]

integrate(F**(a+b*(d*x+c)**n)*(d*x+c)**(-1+5*n),x)

[Out]

Piecewise((x/c, Eq(F, 1) & Eq(b, 0) & Eq(d, 0) & Eq(n, 0)), (F**a*(c*(c + d*x)**(5*n - 1)/(5*d*n) + x*(c + d*x
)**(5*n - 1)/(5*n)), Eq(b, 0)), (F**(a + b*c**n)*c**(5*n - 1)*x, Eq(d, 0)), (F**(a + b)*log(c/d + x)/d, Eq(n,
0)), (c*(c + d*x)**(5*n - 1)/(5*d*n) + x*(c + d*x)**(5*n - 1)/(5*n), Eq(F, 1)), (F**(a + b*(c + d*x)**n)*(c +
d*x)**(4*n)/(b*d*n*log(F)) - 4*F**(a + b*(c + d*x)**n)*(c + d*x)**(3*n)/(b**2*d*n*log(F)**2) + 12*F**(a + b*(c
+ d*x)**n)*(c + d*x)**(2*n)/(b**3*d*n*log(F)**3) - 24*F**(a + b*(c + d*x)**n)*(c + d*x)**n/(b**4*d*n*log(F)**
4) + 24*F**(a + b*(c + d*x)**n)/(b**5*d*n*log(F)**5), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.15 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+5 n} \, dx=\frac {{\left ({\left (d x + c\right )}^{4 \, n} F^{a} b^{4} \log \left (F\right )^{4} - 4 \, {\left (d x + c\right )}^{3 \, n} F^{a} b^{3} \log \left (F\right )^{3} + 12 \, {\left (d x + c\right )}^{2 \, n} F^{a} b^{2} \log \left (F\right )^{2} - 24 \, {\left (d x + c\right )}^{n} F^{a} b \log \left (F\right ) + 24 \, F^{a}\right )} F^{{\left (d x + c\right )}^{n} b}}{b^{5} d n \log \left (F\right )^{5}} \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+5*n),x, algorithm="maxima")

[Out]

((d*x + c)^(4*n)*F^a*b^4*log(F)^4 - 4*(d*x + c)^(3*n)*F^a*b^3*log(F)^3 + 12*(d*x + c)^(2*n)*F^a*b^2*log(F)^2 -
24*(d*x + c)^n*F^a*b*log(F) + 24*F^a)*F^((d*x + c)^n*b)/(b^5*d*n*log(F)^5)

Giac [F]

\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+5 n} \, dx=\int { {\left (d x + c\right )}^{5 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+5*n),x, algorithm="giac")

[Out]

integrate((d*x + c)^(5*n - 1)*F^((d*x + c)^n*b + a), x)

Mupad [F(-1)]

Timed out. \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+5 n} \, dx=\int F^{a+b\,{\left (c+d\,x\right )}^n}\,{\left (c+d\,x\right )}^{5\,n-1} \,d x \]

[In]

int(F^(a + b*(c + d*x)^n)*(c + d*x)^(5*n - 1),x)

[Out]

int(F^(a + b*(c + d*x)^n)*(c + d*x)^(5*n - 1), x)