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

Optimal. Leaf size=88 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-1} \left (e (a+b x)^n (c+d x)^{-n}\right )^{-\frac{m+1}{n}} \text{Ei}\left (\frac{(m+1) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{n}\right )}{n (b c-a d)} \]

[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))

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Rubi [A]  time = 0.10988, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.025, Rules used = {2510} \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-1} \left (e (a+b x)^n (c+d x)^{-n}\right )^{-\frac{m+1}{n}} \text{Ei}\left (\frac{(m+1) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{n}\right )}{n (b c-a d)} \]

Antiderivative was successfully verified.

[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 2510

Int[(((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.))/Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.)
+ (d_.)*(x_))^(q_.))^(r_.)], x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1)*ExpIntegralEi[((m + 1)*Lo
g[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(p*r)])/(p*r*(b*c - a*d)*(e*(f*(a + b*x)^p*(c + d*x)^q)^r)^((m + 1)/(p*r))
), x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q, 0] && EqQ[m + n + 2, 0
] && NeQ[m, -1]

Rubi steps

\begin{align*} \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}} \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]  time = 0.359671, size = 0, normalized size = 0. \[ \int \frac{(a+b x)^m (c+d x)^{-2-m}}{\log \left (e (a+b x)^n (c+d x)^{-n}\right )} \, dx \]

Verification is Not applicable to the result.

[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]

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Maple [F]  time = 1.237, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-2-m} \left ( \ln \left ({\frac{e \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ) \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\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 )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)