\(\int \frac {F^{c (a+b x)} \log ^n(d x) (e+e n+e (-2+b c x \log (F)) \log (d x))}{x^3} \, dx\) [89]

Optimal result

Integrand size = 38, antiderivative size = 22 \[ \int \frac {F^{c (a+b x)} \log ^n(d x) (e+e n+e (-2+b c x \log (F)) \log (d x))}{x^3} \, dx=\frac {e F^{c (a+b x)} \log ^{1+n}(d x)}{x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2233} \[ \int \frac {F^{c (a+b x)} \log ^n(d x) (e+e n+e (-2+b c x \log (F)) \log (d x))}{x^3} \, dx=\frac {e \log ^{n+1}(d x) F^{c (a+b x)}}{x^2} \]

[In]

Int[(F^(c*(a + b*x))*Log[d*x]^n*(e + e*n + e*(-2 + b*c*x*Log[F])*Log[d*x]))/x^3,x]

[Out]

(e*F^(c*(a + b*x))*Log[d*x]^(1 + n))/x^2

Rule 2233

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

Rubi steps \begin{align*} \text {integral}& = \frac {e F^{c (a+b x)} \log ^{1+n}(d x)}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {F^{c (a+b x)} \log ^n(d x) (e+e n+e (-2+b c x \log (F)) \log (d x))}{x^3} \, dx=\frac {e F^{a c+b c x} \log ^{1+n}(d x)}{x^2} \]

[In]

Integrate[(F^(c*(a + b*x))*Log[d*x]^n*(e + e*n + e*(-2 + b*c*x*Log[F])*Log[d*x]))/x^3,x]

[Out]

(e*F^(a*c + b*c*x)*Log[d*x]^(1 + n))/x^2

Maple [A] (verified)

Time = 15.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14

method	result	size
parallelrisch	\(\frac {\ln \left (d x \right ) \ln \left (d x \right )^{n} F^{c \left (b x +a \right )} e}{x^{2}}\)	\(25\)
risch	\(\frac {F^{c \left (b x +a \right )} e \left (-i \pi \,\operatorname {csgn}\left (i d \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i d x \right )+i \pi \,\operatorname {csgn}\left (i d \right ) \operatorname {csgn}\left (i d x \right )^{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i d x \right )^{2}-i \pi \operatorname {csgn}\left (i d x \right )^{3}+2 \ln \left (d \right )+2 \ln \left (x \right )\right ) \left (\ln \left (d \right )+\ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i d x \right ) \left (-\operatorname {csgn}\left (i d x \right )+\operatorname {csgn}\left (i d \right )\right ) \left (-\operatorname {csgn}\left (i d x \right )+\operatorname {csgn}\left (i x \right )\right )}{2}\right )^{n}}{2 x^{2}}\)	\(136\)

[In]

int(F^(c*(b*x+a))*ln(d*x)^n*(e+e*n+e*(-2+b*c*x*ln(F))*ln(d*x))/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {F^{c (a+b x)} \log ^n(d x) (e+e n+e (-2+b c x \log (F)) \log (d x))}{x^3} \, dx=\frac {F^{b c x + a c} e \log \left (d x\right )^{n} \log \left (d x\right )}{x^{2}} \]

[In]

integrate(F^(c*(b*x+a))*log(d*x)^n*(e+e*n+e*(-2+b*c*x*log(F))*log(d*x))/x^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {F^{c (a+b x)} \log ^n(d x) (e+e n+e (-2+b c x \log (F)) \log (d x))}{x^3} \, dx=e \left (\int \frac {F^{a c + b c x} \log {\left (d x \right )}^{n}}{x^{3}}\, dx + \int \frac {F^{a c + b c x} n \log {\left (d x \right )}^{n}}{x^{3}}\, dx + \int \left (- \frac {2 F^{a c + b c x} \log {\left (d x \right )} \log {\left (d x \right )}^{n}}{x^{3}}\right )\, dx + \int \frac {F^{a c + b c x} b c \log {\left (F \right )} \log {\left (d x \right )} \log {\left (d x \right )}^{n}}{x^{2}}\, dx\right ) \]

[In]

integrate(F**(c*(b*x+a))*ln(d*x)**n*(e+e*n+e*(-2+b*c*x*ln(F))*ln(d*x))/x**3,x)

[Out]

e*(Integral(F**(a*c + b*c*x)*log(d*x)**n/x**3, x) + Integral(F**(a*c + b*c*x)*n*log(d*x)**n/x**3, x) + Integra
l(-2*F**(a*c + b*c*x)*log(d*x)*log(d*x)**n/x**3, x) + Integral(F**(a*c + b*c*x)*b*c*log(F)*log(d*x)*log(d*x)**
n/x**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {F^{c (a+b x)} \log ^n(d x) (e+e n+e (-2+b c x \log (F)) \log (d x))}{x^3} \, dx=\frac {{\left (F^{a c} e \log \left (d\right ) + F^{a c} e \log \left (x\right )\right )} e^{\left (b c x \log \left (F\right ) + n \log \left (\log \left (d\right ) + \log \left (x\right )\right )\right )}}{x^{2}} \]

[In]

integrate(F^(c*(b*x+a))*log(d*x)^n*(e+e*n+e*(-2+b*c*x*log(F))*log(d*x))/x^3,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {F^{c (a+b x)} \log ^n(d x) (e+e n+e (-2+b c x \log (F)) \log (d x))}{x^3} \, dx=\int { \frac {{\left ({\left (b c x \log \left (F\right ) - 2\right )} e \log \left (d x\right ) + e n + e\right )} F^{{\left (b x + a\right )} c} \log \left (d x\right )^{n}}{x^{3}} \,d x } \]

[In]

integrate(F^(c*(b*x+a))*log(d*x)^n*(e+e*n+e*(-2+b*c*x*log(F))*log(d*x))/x^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {F^{c (a+b x)} \log ^n(d x) (e+e n+e (-2+b c x \log (F)) \log (d x))}{x^3} \, dx=\frac {F^{a\,c+b\,c\,x}\,e\,{\ln \left (d\,x\right )}^{n+1}}{x^2} \]

[In]

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

[Out]

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