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

Optimal. Leaf size=137 \[ \frac{6 (c+d x)^n F^{a+b (c+d x)^n}}{b^3 d n \log ^3(F)}-\frac{3 (c+d x)^{2 n} F^{a+b (c+d x)^n}}{b^2 d n \log ^2(F)}-\frac{6 F^{a+b (c+d x)^n}}{b^4 d n \log ^4(F)}+\frac{(c+d x)^{3 n} F^{a+b (c+d x)^n}}{b d n \log (F)} \]

[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])

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Rubi [A]  time = 0.167854, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2213, 2209} \[ \frac{6 (c+d x)^n F^{a+b (c+d x)^n}}{b^3 d n \log ^3(F)}-\frac{3 (c+d x)^{2 n} F^{a+b (c+d x)^n}}{b^2 d n \log ^2(F)}-\frac{6 F^{a+b (c+d x)^n}}{b^4 d n \log ^4(F)}+\frac{(c+d x)^{3 n} F^{a+b (c+d x)^n}}{b d n \log (F)} \]

Antiderivative was successfully verified.

[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 2213

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]

Rule 2209

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]

Rubi steps

\begin{align*} \int F^{a+b (c+d x)^n} (c+d x)^{-1+4 n} \, dx &=\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]  time = 0.0061688, size = 32, normalized size = 0.23 \[ -\frac{F^a \text{Gamma}\left (4,-b \log (F) (c+d x)^n\right )}{b^4 d n \log ^4(F)} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.019, size = 77, normalized size = 0.6 \begin{align*}{\frac{ \left ( \left ( \left ( dx+c \right ) ^{n} \right ) ^{3}{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}-3\, \left ( \left ( dx+c \right ) ^{n} \right ) ^{2}{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}+6\,b \left ( dx+c \right ) ^{n}\ln \left ( F \right ) -6 \right ){F}^{a+b \left ( dx+c \right ) ^{n}}}{{b}^{4} \left ( \ln \left ( F \right ) \right ) ^{4}nd}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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
)

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Maxima [A]  time = 1.04289, size = 117, normalized size = 0.85 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.60181, size = 201, normalized size = 1.47 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{4 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)