\(\int F^{a+b (c+d x)^n} (c+d x)^{-1-3 n} \, dx\) [377]

Optimal result

Integrand size = 25, antiderivative size = 139 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-3 n} \, dx=-\frac {F^{a+b (c+d x)^n} (c+d x)^{-3 n}}{3 d n}-\frac {b F^{a+b (c+d x)^n} (c+d x)^{-2 n} \log (F)}{6 d n}-\frac {b^2 F^{a+b (c+d x)^n} (c+d x)^{-n} \log ^2(F)}{6 d n}+\frac {b^3 F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^n \log (F)\right ) \log ^3(F)}{6 d n} \]

[Out]

-1/3*F^(a+b*(d*x+c)^n)/d/n/((d*x+c)^(3*n))-1/6*b*F^(a+b*(d*x+c)^n)*ln(F)/d/n/((d*x+c)^(2*n))-1/6*b^2*F^(a+b*(d
*x+c)^n)*ln(F)^2/d/n/((d*x+c)^n)+1/6*b^3*F^a*Ei(b*(d*x+c)^n*ln(F))*ln(F)^3/d/n

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 139, 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 = {2246, 2241} \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-3 n} \, dx=\frac {b^3 F^a \log ^3(F) \operatorname {ExpIntegralEi}\left (b (c+d x)^n \log (F)\right )}{6 d n}-\frac {b^2 \log ^2(F) (c+d x)^{-n} F^{a+b (c+d x)^n}}{6 d n}-\frac {(c+d x)^{-3 n} F^{a+b (c+d x)^n}}{3 d n}-\frac {b \log (F) (c+d x)^{-2 n} F^{a+b (c+d x)^n}}{6 d n} \]

[In]

Int[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 - 3*n),x]

[Out]

-1/3*F^(a + b*(c + d*x)^n)/(d*n*(c + d*x)^(3*n)) - (b*F^(a + b*(c + d*x)^n)*Log[F])/(6*d*n*(c + d*x)^(2*n)) -
(b^2*F^(a + b*(c + d*x)^n)*Log[F]^2)/(6*d*n*(c + d*x)^n) + (b^3*F^a*ExpIntegralEi[b*(c + d*x)^n*Log[F]]*Log[F]
^3)/(6*d*n)

Rule 2241

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

Rule 2246

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))), x] - Dist[b*n*(Log[F]/(m + 1)), 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]] && LtQ[-4, Simpl
ify[(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}}{3 d n}+\frac {1}{3} (b \log (F)) \int F^{a+b (c+d x)^n} (c+d x)^{-1-2 n} \, dx \\ & = -\frac {F^{a+b (c+d x)^n} (c+d x)^{-3 n}}{3 d n}-\frac {b F^{a+b (c+d x)^n} (c+d x)^{-2 n} \log (F)}{6 d n}+\frac {1}{6} \left (b^2 \log ^2(F)\right ) \int F^{a+b (c+d x)^n} (c+d x)^{-1-n} \, dx \\ & = -\frac {F^{a+b (c+d x)^n} (c+d x)^{-3 n}}{3 d n}-\frac {b F^{a+b (c+d x)^n} (c+d x)^{-2 n} \log (F)}{6 d n}-\frac {b^2 F^{a+b (c+d x)^n} (c+d x)^{-n} \log ^2(F)}{6 d n}+\frac {1}{6} \left (b^3 \log ^3(F)\right ) \int \frac {F^{a+b (c+d x)^n}}{c+d x} \, dx \\ & = -\frac {F^{a+b (c+d x)^n} (c+d x)^{-3 n}}{3 d n}-\frac {b F^{a+b (c+d x)^n} (c+d x)^{-2 n} \log (F)}{6 d n}-\frac {b^2 F^{a+b (c+d x)^n} (c+d x)^{-n} \log ^2(F)}{6 d n}+\frac {b^3 F^a \text {Ei}\left (b (c+d x)^n \log (F)\right ) \log ^3(F)}{6 d n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.22 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-3 n} \, dx=\frac {b^3 F^a \Gamma \left (-3,-b (c+d x)^n \log (F)\right ) \log ^3(F)}{d n} \]

[In]

Integrate[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 - 3*n),x]

[Out]

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

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99

method	result	size
risch	\(-\frac {F^{b \left (d x +c \right )^{n}} F^{a} \left (d x +c \right )^{-3 n}}{3 n d}-\frac {\ln \left (F \right ) b \,F^{b \left (d x +c \right )^{n}} F^{a} \left (d x +c \right )^{-2 n}}{6 n d}-\frac {\ln \left (F \right )^{2} b^{2} F^{b \left (d x +c \right )^{n}} F^{a} \left (d x +c \right )^{-n}}{6 n d}-\frac {\ln \left (F \right )^{3} b^{3} F^{a} \operatorname {Ei}_{1}\left (-b \left (d x +c \right )^{n} \ln \left (F \right )\right )}{6 n d}\)	\(137\)

[In]

int(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-3*n),x,method=_RETURNVERBOSE)

[Out]

-1/3/n/d*F^(b*(d*x+c)^n)*F^a/((d*x+c)^n)^3-1/6/n/d*ln(F)*b*F^(b*(d*x+c)^n)*F^a/((d*x+c)^n)^2-1/6/n/d*ln(F)^2*b
^2*F^(b*(d*x+c)^n)*F^a/((d*x+c)^n)-1/6/n/d*ln(F)^3*b^3*F^a*Ei(1,-b*(d*x+c)^n*ln(F))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.73 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-3 n} \, dx=\frac {{\left (d x + c\right )}^{3 \, n} F^{a} b^{3} {\rm Ei}\left ({\left (d x + c\right )}^{n} b \log \left (F\right )\right ) \log \left (F\right )^{3} - {\left ({\left (d x + c\right )}^{2 \, n} b^{2} \log \left (F\right )^{2} + {\left (d x + c\right )}^{n} b \log \left (F\right ) + 2\right )} e^{\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) + a \log \left (F\right )\right )}}{6 \, {\left (d x + c\right )}^{3 \, n} d n} \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-3*n),x, algorithm="fricas")

[Out]

1/6*((d*x + c)^(3*n)*F^a*b^3*Ei((d*x + c)^n*b*log(F))*log(F)^3 - ((d*x + c)^(2*n)*b^2*log(F)^2 + (d*x + c)^n*b
*log(F) + 2)*e^((d*x + c)^n*b*log(F) + a*log(F)))/((d*x + c)^(3*n)*d*n)

Sympy [F]

\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-3 n} \, dx=\int F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{- 3 n - 1}\, dx \]

[In]

integrate(F**(a+b*(d*x+c)**n)*(d*x+c)**(-1-3*n),x)

[Out]

Integral(F**(a + b*(c + d*x)**n)*(c + d*x)**(-3*n - 1), x)

Maxima [F]

\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-3 n} \, dx=\int { {\left (d x + c\right )}^{-3 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-3*n),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(-3*n - 1)*F^((d*x + c)^n*b + a), x)

Giac [F]

\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-3 n} \, dx=\int { {\left (d x + c\right )}^{-3 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-3*n),x, algorithm="giac")

[Out]

integrate((d*x + c)^(-3*n - 1)*F^((d*x + c)^n*b + a), x)

Mupad [F(-1)]

Timed out. \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-3 n} \, dx=\int \frac {F^{a+b\,{\left (c+d\,x\right )}^n}}{{\left (c+d\,x\right )}^{3\,n+1}} \,d x \]

[In]

int(F^(a + b*(c + d*x)^n)/(c + d*x)^(3*n + 1),x)

[Out]

int(F^(a + b*(c + d*x)^n)/(c + d*x)^(3*n + 1), x)