3.375 \(\int \frac {\text {sech}(a+b x) \tanh ^2(a+b x)}{x^2} \, dx\)

Optimal. Leaf size=31 \[ \text {Int}\left (\frac {\text {sech}(a+b x)}{x^2},x\right )-\text {Int}\left (\frac {\text {sech}^3(a+b x)}{x^2},x\right ) \]

[Out]

Unintegrable(sech(b*x+a)/x^2,x)-Unintegrable(sech(b*x+a)^3/x^2,x)

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Rubi [A]  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 = {} \[ \int \frac {\text {sech}(a+b x) \tanh ^2(a+b x)}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(Sech[a + b*x]*Tanh[a + b*x]^2)/x^2,x]

[Out]

Defer[Int][Sech[a + b*x]/x^2, x] - Defer[Int][Sech[a + b*x]^3/x^2, x]

Rubi steps

\begin {align*} \int \frac {\text {sech}(a+b x) \tanh ^2(a+b x)}{x^2} \, dx &=\int \frac {\text {sech}(a+b x)}{x^2} \, dx-\int \frac {\text {sech}^3(a+b x)}{x^2} \, dx\\ \end {align*}

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Mathematica [A]  time = 13.40, size = 0, normalized size = 0.00 \[ \int \frac {\text {sech}(a+b x) \tanh ^2(a+b x)}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(Sech[a + b*x]*Tanh[a + b*x]^2)/x^2,x]

[Out]

Integrate[(Sech[a + b*x]*Tanh[a + b*x]^2)/x^2, x]

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fricas [A]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {sech}\left (b x + a\right )^{3} \sinh \left (b x + a\right )^{2}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*sinh(b*x+a)^2/x^2,x, algorithm="fricas")

[Out]

integral(sech(b*x + a)^3*sinh(b*x + a)^2/x^2, x)

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}\left (b x + a\right )^{3} \sinh \left (b x + a\right )^{2}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*sinh(b*x+a)^2/x^2,x, algorithm="giac")

[Out]

integrate(sech(b*x + a)^3*sinh(b*x + a)^2/x^2, x)

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maple [A]  time = 0.93, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {sech}\left (b x +a \right )^{3} \left (\sinh ^{2}\left (b x +a \right )\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^3*sinh(b*x+a)^2/x^2,x)

[Out]

int(sech(b*x+a)^3*sinh(b*x+a)^2/x^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*sinh(b*x+a)^2/x^2,x, algorithm="maxima")

[Out]

-((b*x*e^(3*a) - 2*e^(3*a))*e^(3*b*x) - (b*x*e^a + 2*e^a)*e^(b*x))/(b^2*x^3*e^(4*b*x + 4*a) + 2*b^2*x^3*e^(2*b
*x + 2*a) + b^2*x^3) + 2*integrate(1/2*(b^2*x^2*e^a + 6*e^a)*e^(b*x)/(b^2*x^4*e^(2*b*x + 2*a) + b^2*x^4), x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {{\mathrm {sinh}\left (a+b\,x\right )}^2}{x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^2/(x^2*cosh(a + b*x)^3),x)

[Out]

int(sinh(a + b*x)^2/(x^2*cosh(a + b*x)^3), x)

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{2}{\left (a + b x \right )} \operatorname {sech}^{3}{\left (a + b x \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**3*sinh(b*x+a)**2/x**2,x)

[Out]

Integral(sinh(a + b*x)**2*sech(a + b*x)**3/x**2, x)

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