\(\int F^{a+b (c+d x)^n} (c+d x)^{-1+2 n} \, dx\) [372]

Optimal result

Integrand size = 25, antiderivative size = 63 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+2 n} \, dx=-\frac {F^{a+b (c+d x)^n}}{b^2 d n \log ^2(F)}+\frac {F^{a+b (c+d x)^n} (c+d x)^n}{b d n \log (F)} \]

[Out]

-F^(a+b*(d*x+c)^n)/b^2/d/n/ln(F)^2+F^(a+b*(d*x+c)^n)*(d*x+c)^n/b/d/n/ln(F)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2244, 2240} \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+2 n} \, dx=\frac {(c+d x)^n F^{a+b (c+d x)^n}}{b d n \log (F)}-\frac {F^{a+b (c+d x)^n}}{b^2 d n \log ^2(F)} \]

[In]

Int[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 + 2*n),x]

[Out]

-(F^(a + b*(c + d*x)^n)/(b^2*d*n*Log[F]^2)) + (F^(a + b*(c + d*x)^n)*(c + d*x)^n)/(b*d*n*Log[F])

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2244

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^Simplify[m
- n]*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && IntegerQ[2*Simplify[(m + 1)/n]] && Lt
Q[0, Simplify[(m + 1)/n], 5] &&  !RationalQ[m] && SumSimplerQ[m, -n]

Rubi steps \begin{align*} \text {integral}& = \frac {F^{a+b (c+d x)^n} (c+d x)^n}{b d n \log (F)}-\frac {\int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx}{b \log (F)} \\ & = -\frac {F^{a+b (c+d x)^n}}{b^2 d n \log ^2(F)}+\frac {F^{a+b (c+d x)^n} (c+d x)^n}{b d n \log (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 = 32, normalized size of antiderivative = 0.51 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+2 n} \, dx=-\frac {F^a \Gamma \left (2,-b (c+d x)^n \log (F)\right )}{b^2 d n \log ^2(F)} \]

[In]

Integrate[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 + 2*n),x]

[Out]

-((F^a*Gamma[2, -(b*(c + d*x)^n*Log[F])])/(b^2*d*n*Log[F]^2))

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.65

method	result	size
risch	\(\frac {\left (b \left (d x +c \right )^{n} \ln \left (F \right )-1\right ) F^{a +b \left (d x +c \right )^{n}}}{b^{2} n d \ln \left (F \right )^{2}}\)	\(41\)
norman	\(\frac {{\mathrm e}^{n \ln \left (d x +c \right )} {\mathrm e}^{\left (a +b \,{\mathrm e}^{n \ln \left (d x +c \right )}\right ) \ln \left (F \right )}}{d b n \ln \left (F \right )}-\frac {{\mathrm e}^{\left (a +b \,{\mathrm e}^{n \ln \left (d x +c \right )}\right ) \ln \left (F \right )}}{b^{2} n d \ln \left (F \right )^{2}}\)	\(74\)

[In]

int(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+2*n),x,method=_RETURNVERBOSE)

[Out]

(b*(d*x+c)^n*ln(F)-1)/b^2/n/d/ln(F)^2*F^(a+b*(d*x+c)^n)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+2 n} \, dx=\frac {{\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) - 1\right )} e^{\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) + a \log \left (F\right )\right )}}{b^{2} d n \log \left (F\right )^{2}} \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+2*n),x, algorithm="fricas")

[Out]

((d*x + c)^n*b*log(F) - 1)*e^((d*x + c)^n*b*log(F) + a*log(F))/(b^2*d*n*log(F)^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (49) = 98\).

Time = 4.18 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.38 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+2 n} \, dx=\begin {cases} \frac {x}{c} & \text {for}\: F = 1 \wedge b = 0 \wedge d = 0 \wedge n = 0 \\F^{a} \left (\frac {c \left (c + d x\right )^{2 n - 1}}{2 d n} + \frac {x \left (c + d x\right )^{2 n - 1}}{2 n}\right ) & \text {for}\: b = 0 \\F^{a + b c^{n}} c^{2 n - 1} x & \text {for}\: d = 0 \\\frac {F^{a + b} \log {\left (\frac {c}{d} + x \right )}}{d} & \text {for}\: n = 0 \\\frac {c \left (c + d x\right )^{2 n - 1}}{2 d n} + \frac {x \left (c + d x\right )^{2 n - 1}}{2 n} & \text {for}\: F = 1 \\\frac {F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{n}}{b d n \log {\left (F \right )}} - \frac {F^{a + b \left (c + d x\right )^{n}}}{b^{2} d n \log {\left (F \right )}^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(F**(a+b*(d*x+c)**n)*(d*x+c)**(-1+2*n),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.71 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+2 n} \, dx=\frac {{\left ({\left (d x + c\right )}^{n} F^{a} b \log \left (F\right ) - F^{a}\right )} F^{{\left (d x + c\right )}^{n} b}}{b^{2} d n \log \left (F\right )^{2}} \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+2*n),x, algorithm="maxima")

[Out]

((d*x + c)^n*F^a*b*log(F) - F^a)*F^((d*x + c)^n*b)/(b^2*d*n*log(F)^2)

Giac [F]

\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+2 n} \, dx=\int { {\left (d x + c\right )}^{2 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+2*n),x, algorithm="giac")

[Out]

integrate((d*x + c)^(2*n - 1)*F^((d*x + c)^n*b + a), x)

Mupad [F(-1)]

Timed out. \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+2 n} \, dx=\int F^{a+b\,{\left (c+d\,x\right )}^n}\,{\left (c+d\,x\right )}^{2\,n-1} \,d x \]

[In]

int(F^(a + b*(c + d*x)^n)*(c + d*x)^(2*n - 1),x)

[Out]

int(F^(a + b*(c + d*x)^n)*(c + d*x)^(2*n - 1), x)