3.5 \(\int x (a+b \text{sech}(c+d x^2)) \, dx\)

Optimal. Leaf size=26 \[ \frac{a x^2}{2}+\frac{b \tan ^{-1}\left (\sinh \left (c+d x^2\right )\right )}{2 d} \]

[Out]

(a*x^2)/2 + (b*ArcTan[Sinh[c + d*x^2]])/(2*d)

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Rubi [A]  time = 0.024407, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {14, 5436, 3770} \[ \frac{a x^2}{2}+\frac{b \tan ^{-1}\left (\sinh \left (c+d x^2\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Sech[c + d*x^2]),x]

[Out]

(a*x^2)/2 + (b*ArcTan[Sinh[c + d*x^2]])/(2*d)

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]]

Rule 5436

Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sech[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x \left (a+b \text{sech}\left (c+d x^2\right )\right ) \, dx &=\int \left (a x+b x \text{sech}\left (c+d x^2\right )\right ) \, dx\\ &=\frac{a x^2}{2}+b \int x \text{sech}\left (c+d x^2\right ) \, dx\\ &=\frac{a x^2}{2}+\frac{1}{2} b \operatorname{Subst}\left (\int \text{sech}(c+d x) \, dx,x,x^2\right )\\ &=\frac{a x^2}{2}+\frac{b \tan ^{-1}\left (\sinh \left (c+d x^2\right )\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0138505, size = 26, normalized size = 1. \[ \frac{a x^2}{2}+\frac{b \tan ^{-1}\left (\sinh \left (c+d x^2\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*Sech[c + d*x^2]),x]

[Out]

(a*x^2)/2 + (b*ArcTan[Sinh[c + d*x^2]])/(2*d)

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Maple [A]  time = 0.013, size = 30, normalized size = 1.2 \begin{align*}{\frac{a{x}^{2}}{2}}+{\frac{b\arctan \left ( \sinh \left ( d{x}^{2}+c \right ) \right ) }{2\,d}}+{\frac{ac}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*sech(d*x^2+c)),x)

[Out]

1/2*a*x^2+1/2*b*arctan(sinh(d*x^2+c))/d+1/2/d*a*c

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Maxima [A]  time = 1.17881, size = 30, normalized size = 1.15 \begin{align*} \frac{1}{2} \, a x^{2} + \frac{b \arctan \left (\sinh \left (d x^{2} + c\right )\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 + 1/2*b*arctan(sinh(d*x^2 + c))/d

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Fricas [A]  time = 2.40488, size = 88, normalized size = 3.38 \begin{align*} \frac{a d x^{2} + 2 \, b \arctan \left (\cosh \left (d x^{2} + c\right ) + \sinh \left (d x^{2} + c\right )\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*d*x^2 + 2*b*arctan(cosh(d*x^2 + c) + sinh(d*x^2 + c)))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b \operatorname{sech}{\left (c + d x^{2} \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*sech(d*x**2+c)),x)

[Out]

Integral(x*(a + b*sech(c + d*x**2)), x)

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Giac [A]  time = 1.13095, size = 38, normalized size = 1.46 \begin{align*} \frac{{\left (d x^{2} + c\right )} a}{2 \, d} + \frac{b \arctan \left (e^{\left (d x^{2} + c\right )}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d*x^2 + c)*a/d + b*arctan(e^(d*x^2 + c))/d