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

Optimal. Leaf size=35 \[ \text {Int}\left (\frac {(a g+b g x)^2}{\left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )^2},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.21, 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 = {} \[ \int \frac {(a g+b g x)^2}{\left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 1.40, size = 0, normalized size = 0.00 \[ \int \frac {(a g+b g x)^2}{\left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 2.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} g^{2} x^{2} + 2 \, a b g^{2} x + a^{2} g^{2}}{B^{2} \log \left (\frac {d e x + c e}{b x + a}\right )^{2} + 2 \, A B \log \left (\frac {d e x + c e}{b x + a}\right ) + A^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [A]  time = 1.07, size = 0, normalized size = 0.00 \[ \int \frac {\left (b g x +a g \right )^{2}}{\left (B \ln \left (\frac {\left (d x +c \right ) e}{b x +a}\right )+A \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b^{3} d g^{2} x^{4} + a^{3} c g^{2} + {\left (b^{3} c g^{2} + 3 \, a b^{2} d g^{2}\right )} x^{3} + 3 \, {\left (a b^{2} c g^{2} + a^{2} b d g^{2}\right )} x^{2} + {\left (3 \, a^{2} b c g^{2} + a^{3} d g^{2}\right )} x}{{\left (b c - a d\right )} B^{2} \log \left (b x + a\right ) - {\left (b c - a d\right )} B^{2} \log \left (d x + c\right ) - {\left (b c - a d\right )} A B - {\left (b c \log \relax (e) - a d \log \relax (e)\right )} B^{2}} + \int \frac {4 \, b^{3} d g^{2} x^{3} + 3 \, a^{2} b c g^{2} + a^{3} d g^{2} + 3 \, {\left (b^{3} c g^{2} + 3 \, a b^{2} d g^{2}\right )} x^{2} + 6 \, {\left (a b^{2} c g^{2} + a^{2} b d g^{2}\right )} x}{{\left (b c - a d\right )} B^{2} \log \left (b x + a\right ) - {\left (b c - a d\right )} B^{2} \log \left (d x + c\right ) - {\left (b c - a d\right )} A B - {\left (b c \log \relax (e) - a d \log \relax (e)\right )} B^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^3*d*g^2*x^4 + a^3*c*g^2 + (b^3*c*g^2 + 3*a*b^2*d*g^2)*x^3 + 3*(a*b^2*c*g^2 + a^2*b*d*g^2)*x^2 + (3*a^2*b*c
*g^2 + a^3*d*g^2)*x)/((b*c - a*d)*B^2*log(b*x + a) - (b*c - a*d)*B^2*log(d*x + c) - (b*c - a*d)*A*B - (b*c*log
(e) - a*d*log(e))*B^2) + integrate((4*b^3*d*g^2*x^3 + 3*a^2*b*c*g^2 + a^3*d*g^2 + 3*(b^3*c*g^2 + 3*a*b^2*d*g^2
)*x^2 + 6*(a*b^2*c*g^2 + a^2*b*d*g^2)*x)/((b*c - a*d)*B^2*log(b*x + a) - (b*c - a*d)*B^2*log(d*x + c) - (b*c -
a*d)*A*B - (b*c*log(e) - a*d*log(e))*B^2), x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {{\left (a\,g+b\,g\,x\right )}^2}{{\left (A+B\,\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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