### 3.1047 $$\int \text{sech}(x) \sqrt{1+\text{sech}(x)} \tanh ^3(x) \, dx$$

Optimal. Leaf size=25 $\frac{2}{7} (\text{sech}(x)+1)^{7/2}-\frac{4}{5} (\text{sech}(x)+1)^{5/2}$

[Out]

(-4*(1 + Sech[x])^(5/2))/5 + (2*(1 + Sech[x])^(7/2))/7

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Rubi [A]  time = 0.100077, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {4373, 1570, 1469, 627, 43} $\frac{2}{7} (\text{sech}(x)+1)^{7/2}-\frac{4}{5} (\text{sech}(x)+1)^{5/2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[x]*Sqrt[1 + Sech[x]]*Tanh[x]^3,x]

[Out]

(-4*(1 + Sech[x])^(5/2))/5 + (2*(1 + Sech[x])^(7/2))/7

Rule 4373

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dis
t[(b*c*d^(n - 1))^(-1), Subst[Int[SubstFor[(1 - d^2*x^2)^((n - 1)/2)/x^n, Cos[c*(a + b*x)]/d, u, x], x], x, Co
s[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && IntegerQ[(n - 1)/2]
&& NonsumQ[u] && (EqQ[F, Tan] || EqQ[F, tan])

Rule 1570

Int[(x_)^(m_.)*((a_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Int[x^(m - 2*n
*p)*(d + e*x^n)^q*(c + a*x^(2*n))^p, x] /; FreeQ[{a, c, d, e, m, n, q}, x] && EqQ[mn2, -2*n] && IntegerQ[p]

Rule 1469

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[Sim
plify[m - n + 1], 0]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.200845, size = 30, normalized size = 1.2 $-\frac{8}{35} \cosh ^4\left (\frac{x}{2}\right ) (9 \cosh (x)-5) \text{sech}^3(x) \sqrt{\text{sech}(x)+1}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[x]*Sqrt[1 + Sech[x]]*Tanh[x]^3,x]

[Out]

(-8*Cosh[x/2]^4*(-5 + 9*Cosh[x])*Sech[x]^3*Sqrt[1 + Sech[x]])/35

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Maple [F]  time = 0.102, size = 0, normalized size = 0. \begin{align*} \int{\rm sech} \left (x\right )\sqrt{1+{\rm sech} \left (x\right )} \left ( \tanh \left ( x \right ) \right ) ^{3}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)*(1+sech(x))^(1/2)*tanh(x)^3,x)

[Out]

int(sech(x)*(1+sech(x))^(1/2)*tanh(x)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\operatorname{sech}\left (x\right ) + 1} \operatorname{sech}\left (x\right ) \tanh \left (x\right )^{3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)*(1+sech(x))^(1/2)*tanh(x)^3,x, algorithm="maxima")

[Out]

integrate(sqrt(sech(x) + 1)*sech(x)*tanh(x)^3, x)

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Fricas [B]  time = 2.07246, size = 1465, normalized size = 58.6 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)*(1+sech(x))^(1/2)*tanh(x)^3,x, algorithm="fricas")

[Out]

-2/35*(9*cosh(x)^6 + 54*cosh(x)*sinh(x)^5 + 9*sinh(x)^6 + 27*(5*cosh(x)^2 + 1)*sinh(x)^4 + 27*cosh(x)^4 + 36*(
5*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 27*(5*cosh(x)^4 + 6*cosh(x)^2 + 1)*sinh(x)^2 + 27*cosh(x)^2 + 54*(cosh(x)
^5 + 2*cosh(x)^3 + cosh(x))*sinh(x) + (9*cosh(x)^7 + 7*(9*cosh(x) + 5)*sinh(x)^6 + 9*sinh(x)^7 + 35*cosh(x)^6
+ 7*(27*cosh(x)^2 + 30*cosh(x) + 7)*sinh(x)^5 + 49*cosh(x)^5 + 35*(9*cosh(x)^3 + 15*cosh(x)^2 + 7*cosh(x) + 1)
*sinh(x)^4 + 35*cosh(x)^4 + 35*(9*cosh(x)^4 + 20*cosh(x)^3 + 14*cosh(x)^2 + 4*cosh(x) + 1)*sinh(x)^3 + 35*cosh
(x)^3 + 7*(27*cosh(x)^5 + 75*cosh(x)^4 + 70*cosh(x)^3 + 30*cosh(x)^2 + 15*cosh(x) + 7)*sinh(x)^2 + 49*cosh(x)^
2 + 7*(9*cosh(x)^6 + 30*cosh(x)^5 + 35*cosh(x)^4 + 20*cosh(x)^3 + 15*cosh(x)^2 + 14*cosh(x) + 5)*sinh(x) + 35*
cosh(x) + 9)/sqrt(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1) + 9)/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(
x)^6 + 3*(5*cosh(x)^2 + 1)*sinh(x)^4 + 3*cosh(x)^4 + 4*(5*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 3*(5*cosh(x)^4 +
6*cosh(x)^2 + 1)*sinh(x)^2 + 3*cosh(x)^2 + 6*(cosh(x)^5 + 2*cosh(x)^3 + cosh(x))*sinh(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\operatorname{sech}{\left (x \right )} + 1} \tanh ^{3}{\left (x \right )} \operatorname{sech}{\left (x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)*(1+sech(x))**(1/2)*tanh(x)**3,x)

[Out]

Integral(sqrt(sech(x) + 1)*tanh(x)**3*sech(x), x)

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Giac [B]  time = 1.1923, size = 62, normalized size = 2.48 \begin{align*} -\frac{2 \,{\left ({\left ({\left ({\left ({\left ({\left ({\left (9 \, e^{x} + 35\right )} e^{x} + 49\right )} e^{x} + 35\right )} e^{x} + 35\right )} e^{x} + 49\right )} e^{x} + 35\right )} e^{x} + 9\right )}}{35 \,{\left (e^{\left (2 \, x\right )} + 1\right )}^{\frac{7}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)*(1+sech(x))^(1/2)*tanh(x)^3,x, algorithm="giac")

[Out]

-2/35*(((((((9*e^x + 35)*e^x + 49)*e^x + 35)*e^x + 35)*e^x + 49)*e^x + 35)*e^x + 9)/(e^(2*x) + 1)^(7/2)