\(\int F^{a+b (c+d x)^n} (c+d x)^{-1+4 n} \, dx\) [370]

Optimal result

Integrand size = 25, antiderivative size = 137 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+4 n} \, dx=-\frac {6 F^{a+b (c+d x)^n}}{b^4 d n \log ^4(F)}+\frac {6 F^{a+b (c+d x)^n} (c+d x)^n}{b^3 d n \log ^3(F)}-\frac {3 F^{a+b (c+d x)^n} (c+d x)^{2 n}}{b^2 d n \log ^2(F)}+\frac {F^{a+b (c+d x)^n} (c+d x)^{3 n}}{b d n \log (F)} \]

[Out]

-6*F^(a+b*(d*x+c)^n)/b^4/d/n/ln(F)^4+6*F^(a+b*(d*x+c)^n)*(d*x+c)^n/b^3/d/n/ln(F)^3-3*F^(a+b*(d*x+c)^n)*(d*x+c)
^(2*n)/b^2/d/n/ln(F)^2+F^(a+b*(d*x+c)^n)*(d*x+c)^(3*n)/b/d/n/ln(F)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2244, 2240} \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+4 n} \, dx=-\frac {6 F^{a+b (c+d x)^n}}{b^4 d n \log ^4(F)}+\frac {6 (c+d x)^n F^{a+b (c+d x)^n}}{b^3 d n \log ^3(F)}-\frac {3 (c+d x)^{2 n} F^{a+b (c+d x)^n}}{b^2 d n \log ^2(F)}+\frac {(c+d x)^{3 n} F^{a+b (c+d x)^n}}{b d n \log (F)} \]

[In]

Int[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 + 4*n),x]

[Out]

(-6*F^(a + b*(c + d*x)^n))/(b^4*d*n*Log[F]^4) + (6*F^(a + b*(c + d*x)^n)*(c + d*x)^n)/(b^3*d*n*Log[F]^3) - (3*
F^(a + b*(c + d*x)^n)*(c + d*x)^(2*n))/(b^2*d*n*Log[F]^2) + (F^(a + b*(c + d*x)^n)*(c + d*x)^(3*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)^{3 n}}{b d n \log (F)}-\frac {3 \int F^{a+b (c+d x)^n} (c+d x)^{-1+3 n} \, dx}{b \log (F)} \\ & = -\frac {3 F^{a+b (c+d x)^n} (c+d x)^{2 n}}{b^2 d n \log ^2(F)}+\frac {F^{a+b (c+d x)^n} (c+d x)^{3 n}}{b d n \log (F)}+\frac {6 \int F^{a+b (c+d x)^n} (c+d x)^{-1+2 n} \, dx}{b^2 \log ^2(F)} \\ & = \frac {6 F^{a+b (c+d x)^n} (c+d x)^n}{b^3 d n \log ^3(F)}-\frac {3 F^{a+b (c+d x)^n} (c+d x)^{2 n}}{b^2 d n \log ^2(F)}+\frac {F^{a+b (c+d x)^n} (c+d x)^{3 n}}{b d n \log (F)}-\frac {6 \int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx}{b^3 \log ^3(F)} \\ & = -\frac {6 F^{a+b (c+d x)^n}}{b^4 d n \log ^4(F)}+\frac {6 F^{a+b (c+d x)^n} (c+d x)^n}{b^3 d n \log ^3(F)}-\frac {3 F^{a+b (c+d x)^n} (c+d x)^{2 n}}{b^2 d n \log ^2(F)}+\frac {F^{a+b (c+d x)^n} (c+d x)^{3 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 = 0.23 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+4 n} \, dx=-\frac {F^a \Gamma \left (4,-b (c+d x)^n \log (F)\right )}{b^4 d n \log ^4(F)} \]

[In]

Integrate[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 + 4*n),x]

[Out]

-((F^a*Gamma[4, -(b*(c + d*x)^n*Log[F])])/(b^4*d*n*Log[F]^4))

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.56

method	result	size
risch	\(\frac {\left (b^{3} \left (d x +c \right )^{3 n} \ln \left (F \right )^{3}-3 b^{2} \left (d x +c \right )^{2 n} \ln \left (F \right )^{2}+6 b \left (d x +c \right )^{n} \ln \left (F \right )-6\right ) F^{a +b \left (d x +c \right )^{n}}}{b^{4} \ln \left (F \right )^{4} n d}\)	\(77\)

[In]

int(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+4*n),x,method=_RETURNVERBOSE)

[Out]

(((d*x+c)^n)^3*b^3*ln(F)^3-3*((d*x+c)^n)^2*b^2*ln(F)^2+6*b*(d*x+c)^n*ln(F)-6)/b^4/ln(F)^4/n/d*F^(a+b*(d*x+c)^n
)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.58 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+4 n} \, dx=\frac {{\left ({\left (d x + c\right )}^{3 \, n} b^{3} \log \left (F\right )^{3} - 3 \, {\left (d x + c\right )}^{2 \, n} b^{2} \log \left (F\right )^{2} + 6 \, {\left (d x + c\right )}^{n} b \log \left (F\right ) - 6\right )} e^{\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) + a \log \left (F\right )\right )}}{b^{4} d n \log \left (F\right )^{4}} \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+4*n),x, algorithm="fricas")

[Out]

((d*x + c)^(3*n)*b^3*log(F)^3 - 3*(d*x + c)^(2*n)*b^2*log(F)^2 + 6*(d*x + c)^n*b*log(F) - 6)*e^((d*x + c)^n*b*
log(F) + a*log(F))/(b^4*d*n*log(F)^4)

Sympy [A] (verification not implemented)

Time = 16.03 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.60 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+4 n} \, dx=\begin {cases} \frac {x}{c} & \text {for}\: F = 1 \wedge b = 0 \wedge d = 0 \wedge n = 0 \\F^{a} \left (\frac {c \left (c + d x\right )^{4 n - 1}}{4 d n} + \frac {x \left (c + d x\right )^{4 n - 1}}{4 n}\right ) & \text {for}\: b = 0 \\F^{a + b c^{n}} c^{4 n - 1} x & \text {for}\: d = 0 \\\frac {F^{a + b} \log {\left (\frac {c}{d} + x \right )}}{d} & \text {for}\: n = 0 \\\frac {c \left (c + d x\right )^{4 n - 1}}{4 d n} + \frac {x \left (c + d x\right )^{4 n - 1}}{4 n} & \text {for}\: F = 1 \\\frac {F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{3 n}}{b d n \log {\left (F \right )}} - \frac {3 F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{2 n}}{b^{2} d n \log {\left (F \right )}^{2}} + \frac {6 F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{n}}{b^{3} d n \log {\left (F \right )}^{3}} - \frac {6 F^{a + b \left (c + d x\right )^{n}}}{b^{4} d n \log {\left (F \right )}^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate(F**(a+b*(d*x+c)**n)*(d*x+c)**(-1+4*n),x)

[Out]

Piecewise((x/c, Eq(F, 1) & Eq(b, 0) & Eq(d, 0) & Eq(n, 0)), (F**a*(c*(c + d*x)**(4*n - 1)/(4*d*n) + x*(c + d*x
)**(4*n - 1)/(4*n)), Eq(b, 0)), (F**(a + b*c**n)*c**(4*n - 1)*x, Eq(d, 0)), (F**(a + b)*log(c/d + x)/d, Eq(n,
0)), (c*(c + d*x)**(4*n - 1)/(4*d*n) + x*(c + d*x)**(4*n - 1)/(4*n), Eq(F, 1)), (F**(a + b*(c + d*x)**n)*(c +
d*x)**(3*n)/(b*d*n*log(F)) - 3*F**(a + b*(c + d*x)**n)*(c + d*x)**(2*n)/(b**2*d*n*log(F)**2) + 6*F**(a + b*(c
+ d*x)**n)*(c + d*x)**n/(b**3*d*n*log(F)**3) - 6*F**(a + b*(c + d*x)**n)/(b**4*d*n*log(F)**4), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.64 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+4 n} \, dx=\frac {{\left ({\left (d x + c\right )}^{3 \, n} F^{a} b^{3} \log \left (F\right )^{3} - 3 \, {\left (d x + c\right )}^{2 \, n} F^{a} b^{2} \log \left (F\right )^{2} + 6 \, {\left (d x + c\right )}^{n} F^{a} b \log \left (F\right ) - 6 \, F^{a}\right )} F^{{\left (d x + c\right )}^{n} b}}{b^{4} d n \log \left (F\right )^{4}} \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+4*n),x, algorithm="maxima")

[Out]

((d*x + c)^(3*n)*F^a*b^3*log(F)^3 - 3*(d*x + c)^(2*n)*F^a*b^2*log(F)^2 + 6*(d*x + c)^n*F^a*b*log(F) - 6*F^a)*F
^((d*x + c)^n*b)/(b^4*d*n*log(F)^4)

Giac [F]

\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+4 n} \, dx=\int { {\left (d x + c\right )}^{4 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+4*n),x, algorithm="giac")

[Out]

integrate((d*x + c)^(4*n - 1)*F^((d*x + c)^n*b + a), x)

Mupad [F(-1)]

Timed out. \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+4 n} \, dx=\int F^{a+b\,{\left (c+d\,x\right )}^n}\,{\left (c+d\,x\right )}^{4\,n-1} \,d x \]

[In]

int(F^(a + b*(c + d*x)^n)*(c + d*x)^(4*n - 1),x)

[Out]

int(F^(a + b*(c + d*x)^n)*(c + d*x)^(4*n - 1), x)