\(\int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log (e (a+b x)^n (c+d x)^{-n})} \, dx\) [240]

Optimal result

Integrand size = 40, antiderivative size = 88 \[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (e (a+b x)^n (c+d x)^{-n}\right )^{-\frac {1+m}{n}} \operatorname {ExpIntegralEi}\left (\frac {(1+m) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{n}\right )}{(b c-a d) n} \]

[Out]

(b*x+a)^(1+m)*(d*x+c)^(-1-m)*Ei((1+m)*ln(e*(b*x+a)^n/((d*x+c)^n))/n)/(-a*d+b*c)/n/((e*(b*x+a)^n/((d*x+c)^n))^(
(1+m)/n))

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2573, 2563, 2347, 2209} \[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\frac {(a+b x)^{m+1} (c+d x)^{-m-1} \left (e (a+b x)^n (c+d x)^{-n}\right )^{-\frac {m+1}{n}} \operatorname {ExpIntegralEi}\left (\frac {(m+1) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{n}\right )}{n (b c-a d)} \]

[In]

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

[Out]

((a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*ExpIntegralEi[((1 + m)*Log[(e*(a + b*x)^n)/(c + d*x)^n])/n])/((b*c - a*d
)*n*((e*(a + b*x)^n)/(c + d*x)^n)^((1 + m)/n))

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2347

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2563

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[d^2*((g*((a + b*x)/b))^m/(i^2*(b*c - a*d)*(i*((c + d*x)/d))^
m*((a + b*x)/(c + d*x))^m)), Subst[Int[x^m*(A + B*Log[e*x^n])^p, x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a,
b, c, d, e, f, g, h, i, A, B, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] &
& EqQ[m + q + 2, 0]

Rule 2573

Int[((A_.) + Log[(e_.)*(u_)^(n_.)*(v_)^(mn_)]*(B_.))^(p_.)*(w_.), x_Symbol] :> Subst[Int[w*(A + B*Log[e*(u/v)^
n])^p, x], e*(u/v)^n, e*(u^n/v^n)] /; FreeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] &&  !I
ntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx,e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = \text {Subst}\left (\frac {\left ((a+b x)^m \left (\frac {a+b x}{c+d x}\right )^{-m} (c+d x)^{-m}\right ) \text {Subst}\left (\int \frac {x^m}{\log \left (e x^n\right )} \, dx,x,\frac {a+b x}{c+d x}\right )}{b c-a d},e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = \text {Subst}\left (\frac {\left ((a+b x)^{1+m} \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-\frac {1+m}{n}} (c+d x)^{-1-m}\right ) \text {Subst}\left (\int \frac {e^{\frac {(1+m) x}{n}}}{x} \, dx,x,\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) n},e \left (\frac {a+b x}{c+d x}\right )^n,e (a+b x)^n (c+d x)^{-n}\right ) \\ & = \frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (e (a+b x)^n (c+d x)^{-n}\right )^{-\frac {1+m}{n}} \text {Ei}\left (\frac {(1+m) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{n}\right )}{(b c-a d) n} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-2-m}}{\ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )}d x\]

[In]

int((b*x+a)^m*(d*x+c)^(-2-m)/ln(e*(b*x+a)^n/((d*x+c)^n)),x)

[Out]

int((b*x+a)^m*(d*x+c)^(-2-m)/ln(e*(b*x+a)^n/((d*x+c)^n)),x)

Fricas [F]

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

[In]

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

[Out]

integral((b*x + a)^m*(d*x + c)^(-m - 2)/log((b*x + a)^n*e/(d*x + c)^n), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\,{\left (c+d\,x\right )}^{m+2}} \,d x \]

[In]

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

[Out]

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