3.23 \(\int \frac{1}{(a g+b g x)^3 (A+B \log (e (\frac{a+b x}{c+d x})^n))} \, dx\)

Optimal. Leaf size=197 \[ \frac{b e^{\frac{2 A}{B n}} (c+d x)^2 \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )^{2/n} \text{Ei}\left (-\frac{2 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )}{B g^3 n (a+b x)^2 (b c-a d)^2}-\frac{d e^{\frac{A}{B n}} (c+d x) \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )^{\frac{1}{n}} \text{Ei}\left (-\frac{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{B n}\right )}{B g^3 n (a+b x) (b c-a d)^2} \]

[Out]

(b*E^((2*A)/(B*n))*(e*((a + b*x)/(c + d*x))^n)^(2/n)*(c + d*x)^2*ExpIntegralEi[(-2*(A + B*Log[e*((a + b*x)/(c
+ d*x))^n]))/(B*n)])/(B*(b*c - a*d)^2*g^3*n*(a + b*x)^2) - (d*E^(A/(B*n))*(e*((a + b*x)/(c + d*x))^n)^n^(-1)*(
c + d*x)*ExpIntegralEi[-((A + B*Log[e*((a + b*x)/(c + d*x))^n])/(B*n))])/(B*(b*c - a*d)^2*g^3*n*(a + b*x))

________________________________________________________________________________________

Rubi [F]  time = 0.0821707, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{1}{(a g+b g x)^3 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Int[1/((a*g + b*g*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]

[Out]

Defer[Int][1/((a*g + b*g*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n])), x]

Rubi steps

\begin{align*} \int \frac{1}{(a g+b g x)^3 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )} \, dx &=\int \frac{1}{(a g+b g x)^3 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )} \, dx\\ \end{align*}

Mathematica [A]  time = 0.2969, size = 172, normalized size = 0.87 \[ \frac{e^{\frac{A}{B n}} (c+d x) \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )^{\frac{1}{n}} \left (b e^{\frac{A}{B n}} (c+d x) \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )^{\frac{1}{n}} \text{Ei}\left (-\frac{2 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{B n}\right )-d (a+b x) \text{Ei}\left (-\frac{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{B n}\right )\right )}{B g^3 n (a+b x)^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a*g + b*g*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n])),x]

[Out]

(E^(A/(B*n))*(e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c + d*x)*(b*E^(A/(B*n))*(e*((a + b*x)/(c + d*x))^n)^n^(-1)*(c
+ d*x)*ExpIntegralEi[(-2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(B*n)] - d*(a + b*x)*ExpIntegralEi[-((A + B*
Log[e*((a + b*x)/(c + d*x))^n])/(B*n))]))/(B*(b*c - a*d)^2*g^3*n*(a + b*x)^2)

________________________________________________________________________________________

Maple [F]  time = 0.437, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( bgx+ag \right ) ^{3}} \left ( A+B\ln \left ( e \left ({\frac{bx+a}{dx+c}} \right ) ^{n} \right ) \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*g*x+a*g)^3/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)

[Out]

int(1/(b*g*x+a*g)^3/(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b g x + a g\right )}^{3}{\left (B \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right ) + A\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="maxima")

[Out]

integrate(1/((b*g*x + a*g)^3*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)

________________________________________________________________________________________

Fricas [A]  time = 0.936258, size = 354, normalized size = 1.8 \begin{align*} -\frac{d e^{\left (\frac{B \log \left (e\right ) + A}{B n}\right )} \logintegral \left (\frac{{\left (d x + c\right )} e^{\left (-\frac{B \log \left (e\right ) + A}{B n}\right )}}{b x + a}\right ) - b e^{\left (\frac{2 \,{\left (B \log \left (e\right ) + A\right )}}{B n}\right )} \logintegral \left (\frac{{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} e^{\left (-\frac{2 \,{\left (B \log \left (e\right ) + A\right )}}{B n}\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{{\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} g^{3} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="fricas")

[Out]

-(d*e^((B*log(e) + A)/(B*n))*log_integral((d*x + c)*e^(-(B*log(e) + A)/(B*n))/(b*x + a)) - b*e^(2*(B*log(e) +
A)/(B*n))*log_integral((d^2*x^2 + 2*c*d*x + c^2)*e^(-2*(B*log(e) + A)/(B*n))/(b^2*x^2 + 2*a*b*x + a^2)))/((B*b
^2*c^2 - 2*B*a*b*c*d + B*a^2*d^2)*g^3*n)

________________________________________________________________________________________

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(1/(b*g*x+a*g)**3/(A+B*ln(e*((b*x+a)/(d*x+c))**n)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b g x + a g\right )}^{3}{\left (B \log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right ) + A\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="giac")

[Out]

integrate(1/((b*g*x + a*g)^3*(B*log(e*((b*x + a)/(d*x + c))^n) + A)), x)