3.222 \(\int \frac {1}{(a g+b g x)^2 (A+B \log (\frac {e (c+d x)^2}{(a+b x)^2}))} \, dx\)

Optimal. Leaf size=91 \[ -\frac {e^{-\frac {A}{2 B}} (c+d x) \text {Ei}\left (\frac {A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )}{2 B}\right )}{2 B g^2 (a+b x) (b c-a d) \sqrt {\frac {e (c+d x)^2}{(a+b x)^2}}} \]

[Out]

-1/2*(d*x+c)*Ei(1/2*(A+B*ln(e*(d*x+c)^2/(b*x+a)^2))/B)/B/(-a*d+b*c)/exp(1/2*A/B)/g^2/(b*x+a)/(e*(d*x+c)^2/(b*x
+a)^2)^(1/2)

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Rubi [F]  time = 0.09, 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 = {} \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Int[1/((a*g + b*g*x)^2*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])),x]

[Out]

Defer[Int][1/((a*g + b*g*x)^2*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])), x]

Rubi steps

\begin {align*} \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )} \, dx &=\int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )} \, dx\\ \end {align*}

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Mathematica [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {1}{(a g+b g x)^2 \left (A+B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[1/((a*g + b*g*x)^2*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])),x]

[Out]

Integrate[1/((a*g + b*g*x)^2*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])), x]

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{A b^{2} g^{2} x^{2} + 2 \, A a b g^{2} x + A a^{2} g^{2} + {\left (B b^{2} g^{2} x^{2} + 2 \, B a b g^{2} x + B a^{2} g^{2}\right )} \log \left (\frac {d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^2/(A+B*log(e*(d*x+c)^2/(b*x+a)^2)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^2/(A+B*log(e*(d*x+c)^2/(b*x+a)^2)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.87, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b g x +a g \right )^{2} \left (B \ln \left (\frac {\left (d x +c \right )^{2} e}{\left (b x +a \right )^{2}}\right )+A \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*g*x+a*g)^2/(B*ln((d*x+c)^2/(b*x+a)^2*e)+A),x)

[Out]

int(1/(b*g*x+a*g)^2/(B*ln((d*x+c)^2/(b*x+a)^2*e)+A),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^2/(A+B*log(e*(d*x+c)^2/(b*x+a)^2)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a\,g+b\,g\,x\right )}^2\,\left (A+B\,\ln \left (\frac {e\,{\left (c+d\,x\right )}^2}{{\left (a+b\,x\right )}^2}\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)**2/(A+B*ln(e*(d*x+c)**2/(b*x+a)**2)),x)

[Out]

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

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