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

Optimal. Leaf size=107 \[ \frac {b e^2 e^{\frac {2 A}{B}} \text {Ei}\left (-\frac {2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{B}\right )}{B g^3 (b c-a d)^2}-\frac {d e e^{A/B} \text {Ei}\left (-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{B}\right )}{B g^3 (b c-a d)^2} \]

[Out]

b*e^2*exp(2*A/B)*Ei(-2*(A+B*ln(e*(b*x+a)/(d*x+c)))/B)/B/(-a*d+b*c)^2/g^3-d*e*exp(A/B)*Ei((-A-B*ln(e*(b*x+a)/(d
*x+c)))/B)/B/(-a*d+b*c)^2/g^3

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Rubi [F]  time = 0.07, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx \]

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)])),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 89, normalized size = 0.83 \[ \frac {e e^{A/B} \left (b e e^{A/B} \text {Ei}\left (-\frac {2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{B}\right )-d \text {Ei}\left (-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{B}\right )\right )}{B g^3 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 130, normalized size = 1.21 \[ \frac {b e^{2} e^{\left (\frac {2 \, A}{B}\right )} \operatorname {log\_integral}\left (\frac {{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} e^{\left (-\frac {2 \, A}{B}\right )}}{b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}}\right ) - d e e^{\frac {A}{B}} \operatorname {log\_integral}\left (\frac {{\left (d x + c\right )} e^{\left (-\frac {A}{B}\right )}}{b e x + a e}\right )}{{\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} g^{3}} \]

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))),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 1.32, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b g x +a g \right )^{3} \left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b g x + a g\right )}^{3} {\left (B \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}}\,{d x} \]

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))),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a\,g+b\,g\,x\right )}^3\,\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{A a^{3} + 3 A a^{2} b x + 3 A a b^{2} x^{2} + A b^{3} x^{3} + B a^{3} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )} + 3 B a^{2} b x \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )} + 3 B a b^{2} x^{2} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )} + B b^{3} x^{3} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx}{g^{3}} \]

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))),x)

[Out]

Integral(1/(A*a**3 + 3*A*a**2*b*x + 3*A*a*b**2*x**2 + A*b**3*x**3 + B*a**3*log(a*e/(c + d*x) + b*e*x/(c + d*x)
) + 3*B*a**2*b*x*log(a*e/(c + d*x) + b*e*x/(c + d*x)) + 3*B*a*b**2*x**2*log(a*e/(c + d*x) + b*e*x/(c + d*x)) +
B*b**3*x**3*log(a*e/(c + d*x) + b*e*x/(c + d*x))), x)/g**3

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