3.91 \(\int \text{sech}^4(a+b x) \sqrt{\tanh (a+b x)} \, dx\)

Optimal. Leaf size=35 \[ \frac{2 \tanh ^{\frac{3}{2}}(a+b x)}{3 b}-\frac{2 \tanh ^{\frac{7}{2}}(a+b x)}{7 b} \]

[Out]

(2*Tanh[a + b*x]^(3/2))/(3*b) - (2*Tanh[a + b*x]^(7/2))/(7*b)

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Rubi [A]  time = 0.0366913, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2607, 14} \[ \frac{2 \tanh ^{\frac{3}{2}}(a+b x)}{3 b}-\frac{2 \tanh ^{\frac{7}{2}}(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*x]^4*Sqrt[Tanh[a + b*x]],x]

[Out]

(2*Tanh[a + b*x]^(3/2))/(3*b) - (2*Tanh[a + b*x]^(7/2))/(7*b)

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \text{sech}^4(a+b x) \sqrt{\tanh (a+b x)} \, dx &=-\frac{i \operatorname{Subst}\left (\int \sqrt{-i x} \left (1+x^2\right ) \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (\sqrt{-i x}-(-i x)^{5/2}\right ) \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac{2 \tanh ^{\frac{3}{2}}(a+b x)}{3 b}-\frac{2 \tanh ^{\frac{7}{2}}(a+b x)}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.105289, size = 29, normalized size = 0.83 \[ \frac{2 \tanh ^{\frac{3}{2}}(a+b x) \left (3 \text{sech}^2(a+b x)+4\right )}{21 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[a + b*x]^4*Sqrt[Tanh[a + b*x]],x]

[Out]

(2*(4 + 3*Sech[a + b*x]^2)*Tanh[a + b*x]^(3/2))/(21*b)

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Maple [F]  time = 0.194, size = 0, normalized size = 0. \begin{align*} \int \left ({\rm sech} \left (bx+a\right ) \right ) ^{4}\sqrt{\tanh \left ( bx+a \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^4*tanh(b*x+a)^(1/2),x)

[Out]

int(sech(b*x+a)^4*tanh(b*x+a)^(1/2),x)

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Maxima [B]  time = 1.80421, size = 475, normalized size = 13.57 \begin{align*} \frac{32 \, \sqrt{e^{\left (-b x - a\right )} + 1} \sqrt{-e^{\left (-b x - a\right )} + 1} e^{\left (-2 \, b x - 2 \, a\right )}}{21 \, b{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 3 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )} + 1\right )} \sqrt{e^{\left (-2 \, b x - 2 \, a\right )} + 1}} - \frac{32 \, \sqrt{e^{\left (-b x - a\right )} + 1} \sqrt{-e^{\left (-b x - a\right )} + 1} e^{\left (-4 \, b x - 4 \, a\right )}}{21 \, b{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 3 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )} + 1\right )} \sqrt{e^{\left (-2 \, b x - 2 \, a\right )} + 1}} - \frac{8 \, \sqrt{e^{\left (-b x - a\right )} + 1} \sqrt{-e^{\left (-b x - a\right )} + 1} e^{\left (-6 \, b x - 6 \, a\right )}}{21 \, b{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 3 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )} + 1\right )} \sqrt{e^{\left (-2 \, b x - 2 \, a\right )} + 1}} + \frac{8 \, \sqrt{e^{\left (-b x - a\right )} + 1} \sqrt{-e^{\left (-b x - a\right )} + 1}}{21 \, b{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 3 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )} + 1\right )} \sqrt{e^{\left (-2 \, b x - 2 \, a\right )} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^4*tanh(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

32/21*sqrt(e^(-b*x - a) + 1)*sqrt(-e^(-b*x - a) + 1)*e^(-2*b*x - 2*a)/(b*(3*e^(-2*b*x - 2*a) + 3*e^(-4*b*x - 4
*a) + e^(-6*b*x - 6*a) + 1)*sqrt(e^(-2*b*x - 2*a) + 1)) - 32/21*sqrt(e^(-b*x - a) + 1)*sqrt(-e^(-b*x - a) + 1)
*e^(-4*b*x - 4*a)/(b*(3*e^(-2*b*x - 2*a) + 3*e^(-4*b*x - 4*a) + e^(-6*b*x - 6*a) + 1)*sqrt(e^(-2*b*x - 2*a) +
1)) - 8/21*sqrt(e^(-b*x - a) + 1)*sqrt(-e^(-b*x - a) + 1)*e^(-6*b*x - 6*a)/(b*(3*e^(-2*b*x - 2*a) + 3*e^(-4*b*
x - 4*a) + e^(-6*b*x - 6*a) + 1)*sqrt(e^(-2*b*x - 2*a) + 1)) + 8/21*sqrt(e^(-b*x - a) + 1)*sqrt(-e^(-b*x - a)
+ 1)/(b*(3*e^(-2*b*x - 2*a) + 3*e^(-4*b*x - 4*a) + e^(-6*b*x - 6*a) + 1)*sqrt(e^(-2*b*x - 2*a) + 1))

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Fricas [B]  time = 1.8987, size = 1521, normalized size = 43.46 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^4*tanh(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

8/21*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 + 1)*sinh(b*x
+ a)^4 + 3*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 +
6*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 3*cosh(b*x + a)^2 + 6*(cosh(b*x + a)^5 + 2*cosh(b*x + a)^3 + cosh(b*
x + a))*sinh(b*x + a) + (cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + (15*cosh(b*x +
a)^2 + 4)*sinh(b*x + a)^4 + 4*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 + 4*cosh(b*x + a))*sinh(b*x + a)^3 + (15*
cosh(b*x + a)^4 + 24*cosh(b*x + a)^2 - 4)*sinh(b*x + a)^2 - 4*cosh(b*x + a)^2 + 2*(3*cosh(b*x + a)^5 + 8*cosh(
b*x + a)^3 - 4*cosh(b*x + a))*sinh(b*x + a) - 1)*sqrt(sinh(b*x + a)/cosh(b*x + a)) + 1)/(b*cosh(b*x + a)^6 + 6
*b*cosh(b*x + a)*sinh(b*x + a)^5 + b*sinh(b*x + a)^6 + 3*b*cosh(b*x + a)^4 + 3*(5*b*cosh(b*x + a)^2 + b)*sinh(
b*x + a)^4 + 4*(5*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^3 + 3*b*cosh(b*x + a)^2 + 3*(5*b*cosh(b
*x + a)^4 + 6*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 6*(b*cosh(b*x + a)^5 + 2*b*cosh(b*x + a)^3 + b*cosh(b*x
+ a))*sinh(b*x + a) + b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\tanh{\left (a + b x \right )}} \operatorname{sech}^{4}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**4*tanh(b*x+a)**(1/2),x)

[Out]

Integral(sqrt(tanh(a + b*x))*sech(a + b*x)**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{sech}\left (b x + a\right )^{4} \sqrt{\tanh \left (b x + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^4*tanh(b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate(sech(b*x + a)^4*sqrt(tanh(b*x + a)), x)