3.362 \(\int x^m \tanh ^2(a+b x) \, dx\)

Optimal. Leaf size=15 \[ \text {Int}\left (x^m \tanh ^2(a+b x),x\right ) \]

[Out]

Unintegrable(x^m*tanh(b*x+a)^2,x)

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Rubi [A]  time = 0.03, 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 = {} \[ \int x^m \tanh ^2(a+b x) \, dx \]

Verification is Not applicable to the result.

[In]

Int[x^m*Tanh[a + b*x]^2,x]

[Out]

Defer[Int][x^m*Tanh[a + b*x]^2, x]

Rubi steps

\begin {align*} \int x^m \tanh ^2(a+b x) \, dx &=\int x^m \tanh ^2(a+b x) \, dx\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 0, normalized size = 0.00 \[ \int x^m \tanh ^2(a+b x) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x^m*Tanh[a + b*x]^2,x]

[Out]

Integrate[x^m*Tanh[a + b*x]^2, x]

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fricas [A]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{m} \operatorname {sech}\left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.28, size = 0, normalized size = 0.00 \[ \int x^{m} \mathrm {sech}\left (b x +a \right )^{2} \left (\sinh ^{2}\left (b x +a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {x e^{\left (4 \, b x + m \log \relax (x) + 4 \, a\right )}}{{\left (m + 1\right )} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, {\left (m + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + m + 1} - \int \frac {{\left (2 \, {\left (2 \, b x e^{\left (4 \, a\right )} + {\left (m + 1\right )} e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )} + {\left (m + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} - m - 1\right )} x^{m}}{{\left (m + 1\right )} e^{\left (6 \, b x + 6 \, a\right )} + 3 \, {\left (m + 1\right )} e^{\left (4 \, b x + 4 \, a\right )} + 3 \, {\left (m + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + m + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [A]  time = 0.00, size = -1, normalized size = -0.07 \[ \int \frac {x^m\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{{\mathrm {cosh}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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