3.61 \(\int \frac {(a+b \tanh (e+f x))^2}{c+d x} \, dx\)

Optimal. Leaf size=23 \[ \text {Int}\left (\frac {(a+b \tanh (e+f x))^2}{c+d x},x\right ) \]

[Out]

Unintegrable((a+b*tanh(f*x+e))^2/(d*x+c),x)

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Rubi [A]  time = 0.06, 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 = {} \[ \int \frac {(a+b \tanh (e+f x))^2}{c+d x} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(a + b*Tanh[e + f*x])^2/(c + d*x),x]

[Out]

Defer[Int][(a + b*Tanh[e + f*x])^2/(c + d*x), x]

Rubi steps

\begin {align*} \int \frac {(a+b \tanh (e+f x))^2}{c+d x} \, dx &=\int \frac {(a+b \tanh (e+f x))^2}{c+d x} \, dx\\ \end {align*}

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Mathematica [A]  time = 41.59, size = 0, normalized size = 0.00 \[ \int \frac {(a+b \tanh (e+f x))^2}{c+d x} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Tanh[e + f*x])^2/(c + d*x),x]

[Out]

Integrate[(a + b*Tanh[e + f*x])^2/(c + d*x), x]

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fricas [A]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \tanh \left (f x + e\right )^{2} + 2 \, a b \tanh \left (f x + e\right ) + a^{2}}{d x + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tanh(f*x+e))^2/(d*x+c),x, algorithm="fricas")

[Out]

integral((b^2*tanh(f*x + e)^2 + 2*a*b*tanh(f*x + e) + a^2)/(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tanh(f*x+e))^2/(d*x+c),x, algorithm="giac")

[Out]

integrate((b*tanh(f*x + e) + a)^2/(d*x + c), x)

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maple [A]  time = 0.81, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \tanh \left (f x +e \right )\right )^{2}}{d x +c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tanh(f*x+e))^2/(d*x+c),x)

[Out]

int((a+b*tanh(f*x+e))^2/(d*x+c),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tanh(f*x+e))^2/(d*x+c),x, algorithm="maxima")

[Out]

a^2*log(d*x + c)/d + 2*b^2/(d*f*x + c*f + (d*f*x*e^(2*e) + c*f*e^(2*e))*e^(2*f*x)) + (2*a*b + b^2)*log(d*x + c
)/d - integrate(2*(2*a*b*d*f*x + 2*a*b*c*f - b^2*d)/(d^2*f*x^2 + 2*c*d*f*x + c^2*f + (d^2*f*x^2*e^(2*e) + 2*c*
d*f*x*e^(2*e) + c^2*f*e^(2*e))*e^(2*f*x)), x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {{\left (a+b\,\mathrm {tanh}\left (e+f\,x\right )\right )}^2}{c+d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tanh(e + f*x))^2/(c + d*x),x)

[Out]

int((a + b*tanh(e + f*x))^2/(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tanh(f*x+e))**2/(d*x+c),x)

[Out]

Integral((a + b*tanh(e + f*x))**2/(c + d*x), x)

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