\(\int \frac {F^{a+b (c+d x)^n}}{(c+d x)^4} \, dx\) [367]

Optimal result

Integrand size = 21, antiderivative size = 54 \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^4} \, dx=-\frac {F^a \Gamma \left (-\frac {3}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{3/n}}{d n (c+d x)^3} \]

[Out]

-F^a*GAMMA(-3/n,-b*(d*x+c)^n*ln(F))*(-b*(d*x+c)^n*ln(F))^(3/n)/d/n/(d*x+c)^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2250} \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^4} \, dx=-\frac {F^a \left (-b \log (F) (c+d x)^n\right )^{3/n} \Gamma \left (-\frac {3}{n},-b (c+d x)^n \log (F)\right )}{d n (c+d x)^3} \]

[In]

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

[Out]

-((F^a*Gamma[-3/n, -(b*(c + d*x)^n*Log[F])]*(-(b*(c + d*x)^n*Log[F]))^(3/n))/(d*n*(c + d*x)^3))

Rule 2250

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

Rubi steps \begin{align*} \text {integral}& = -\frac {F^a \Gamma \left (-\frac {3}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{3/n}}{d n (c+d x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^4} \, dx=-\frac {F^a \Gamma \left (-\frac {3}{n},-b (c+d x)^n \log (F)\right ) \left (-b (c+d x)^n \log (F)\right )^{3/n}}{d n (c+d x)^3} \]

[In]

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

[Out]

-((F^a*Gamma[-3/n, -(b*(c + d*x)^n*Log[F])]*(-(b*(c + d*x)^n*Log[F]))^(3/n))/(d*n*(c + d*x)^3))

Maple [F]

\[\int \frac {F^{a +b \left (d x +c \right )^{n}}}{\left (d x +c \right )^{4}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(F**(a + b*(c + d*x)**n)/(c + d*x)**4, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.31 \[ \int \frac {F^{a+b (c+d x)^n}}{(c+d x)^4} \, dx=-\frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n}{2}}\,{\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n\right )}^{\frac {3}{2\,n}-\frac {1}{2}}\,{\mathrm {M}}_{\frac {3}{2\,n}+\frac {1}{2},-\frac {3}{2\,n}}\left (b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^n\right )}{3\,d\,{\left (c+d\,x\right )}^3} \]

[In]

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

[Out]

-(F^a*exp((b*log(F)*(c + d*x)^n)/2)*(b*log(F)*(c + d*x)^n)^(3/(2*n) - 1/2)*whittakerM(3/(2*n) + 1/2, -3/(2*n),
b*log(F)*(c + d*x)^n))/(3*d*(c + d*x)^3)