3.1.59 \(\int \frac {(a+b \text {sech}(c+d \sqrt {x}))^2}{\sqrt {x}} \, dx\) [59]

Optimal. Leaf size=47 \[ 2 a^2 \sqrt {x}+\frac {4 a b \text {ArcTan}\left (\sinh \left (c+d \sqrt {x}\right )\right )}{d}+\frac {2 b^2 \tanh \left (c+d \sqrt {x}\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5544, 3858, 3855, 3852, 8} \begin {gather*} 2 a^2 \sqrt {x}+\frac {4 a b \text {ArcTan}\left (\sinh \left (c+d \sqrt {x}\right )\right )}{d}+\frac {2 b^2 \tanh \left (c+d \sqrt {x}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*Sqrt[x]])^2/Sqrt[x],x]

[Out]

2*a^2*Sqrt[x] + (4*a*b*ArcTan[Sinh[c + d*Sqrt[x]]])/d + (2*b^2*Tanh[c + d*Sqrt[x]])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

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

Rule 3858

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[a^2*x, x] + (Dist[2*a*b, Int[Csc[c + d*x], x],
x] + Dist[b^2, Int[Csc[c + d*x]^2, x], x]) /; FreeQ[{a, b, c, d}, x]

Rule 5544

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2}{\sqrt {x}} \, dx &=2 \text {Subst}\left (\int (a+b \text {sech}(c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 a^2 \sqrt {x}+(4 a b) \text {Subst}\left (\int \text {sech}(c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \text {Subst}\left (\int \text {sech}^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=2 a^2 \sqrt {x}+\frac {4 a b \tan ^{-1}\left (\sinh \left (c+d \sqrt {x}\right )\right )}{d}+\frac {\left (2 i b^2\right ) \text {Subst}\left (\int 1 \, dx,x,-i \tanh \left (c+d \sqrt {x}\right )\right )}{d}\\ &=2 a^2 \sqrt {x}+\frac {4 a b \tan ^{-1}\left (\sinh \left (c+d \sqrt {x}\right )\right )}{d}+\frac {2 b^2 \tanh \left (c+d \sqrt {x}\right )}{d}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 48, normalized size = 1.02 \begin {gather*} \frac {2 \left (a \left (a \left (c+d \sqrt {x}\right )+2 b \text {ArcTan}\left (\sinh \left (c+d \sqrt {x}\right )\right )\right )+b^2 \tanh \left (c+d \sqrt {x}\right )\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \,\mathrm {sech}\left (c +d \sqrt {x}\right )\right )^{2}}{\sqrt {x}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sech(c+d*x^(1/2)))^2/x^(1/2),x)

[Out]

int((a+b*sech(c+d*x^(1/2)))^2/x^(1/2),x)

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Maxima [A]
time = 0.27, size = 48, normalized size = 1.02 \begin {gather*} 2 \, a^{2} \sqrt {x} + \frac {4 \, a b \arctan \left (\sinh \left (d \sqrt {x} + c\right )\right )}{d} + \frac {4 \, b^{2}}{d {\left (e^{\left (-2 \, d \sqrt {x} - 2 \, c\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^2*sqrt(x) + 4*a*b*arctan(sinh(d*sqrt(x) + c))/d + 4*b^2/(d*(e^(-2*d*sqrt(x) - 2*c) + 1))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (41) = 82\).
time = 0.37, size = 194, normalized size = 4.13 \begin {gather*} \frac {2 \, {\left (a^{2} d \sqrt {x} \cosh \left (d \sqrt {x} + c\right )^{2} + 2 \, a^{2} d \sqrt {x} \cosh \left (d \sqrt {x} + c\right ) \sinh \left (d \sqrt {x} + c\right ) + a^{2} d \sqrt {x} \sinh \left (d \sqrt {x} + c\right )^{2} + a^{2} d \sqrt {x} - 2 \, b^{2} + 4 \, {\left (a b \cosh \left (d \sqrt {x} + c\right )^{2} + 2 \, a b \cosh \left (d \sqrt {x} + c\right ) \sinh \left (d \sqrt {x} + c\right ) + a b \sinh \left (d \sqrt {x} + c\right )^{2} + a b\right )} \arctan \left (\cosh \left (d \sqrt {x} + c\right ) + \sinh \left (d \sqrt {x} + c\right )\right )\right )}}{d \cosh \left (d \sqrt {x} + c\right )^{2} + 2 \, d \cosh \left (d \sqrt {x} + c\right ) \sinh \left (d \sqrt {x} + c\right ) + d \sinh \left (d \sqrt {x} + c\right )^{2} + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a^2*d*sqrt(x)*cosh(d*sqrt(x) + c)^2 + 2*a^2*d*sqrt(x)*cosh(d*sqrt(x) + c)*sinh(d*sqrt(x) + c) + a^2*d*sqrt(
x)*sinh(d*sqrt(x) + c)^2 + a^2*d*sqrt(x) - 2*b^2 + 4*(a*b*cosh(d*sqrt(x) + c)^2 + 2*a*b*cosh(d*sqrt(x) + c)*si
nh(d*sqrt(x) + c) + a*b*sinh(d*sqrt(x) + c)^2 + a*b)*arctan(cosh(d*sqrt(x) + c) + sinh(d*sqrt(x) + c)))/(d*cos
h(d*sqrt(x) + c)^2 + 2*d*cosh(d*sqrt(x) + c)*sinh(d*sqrt(x) + c) + d*sinh(d*sqrt(x) + c)^2 + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )^{2}}{\sqrt {x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sech(c+d*x**(1/2)))**2/x**(1/2),x)

[Out]

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

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Giac [A]
time = 0.40, size = 55, normalized size = 1.17 \begin {gather*} \frac {2 \, {\left (d \sqrt {x} + c\right )} a^{2}}{d} + \frac {8 \, a b \arctan \left (e^{\left (d \sqrt {x} + c\right )}\right )}{d} - \frac {4 \, b^{2}}{d {\left (e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(d*sqrt(x) + c)*a^2/d + 8*a*b*arctan(e^(d*sqrt(x) + c))/d - 4*b^2/(d*(e^(2*d*sqrt(x) + 2*c) + 1))

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Mupad [B]
time = 1.32, size = 77, normalized size = 1.64 \begin {gather*} 2\,a^2\,\sqrt {x}+\frac {8\,\mathrm {atan}\left (\frac {a\,b\,{\mathrm {e}}^{d\,\sqrt {x}}\,{\mathrm {e}}^c\,\sqrt {d^2}}{d\,\sqrt {a^2\,b^2}}\right )\,\sqrt {a^2\,b^2}}{\sqrt {d^2}}-\frac {4\,b^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,\sqrt {x}}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cosh(c + d*x^(1/2)))^2/x^(1/2),x)

[Out]

2*a^2*x^(1/2) + (8*atan((a*b*exp(d*x^(1/2))*exp(c)*(d^2)^(1/2))/(d*(a^2*b^2)^(1/2)))*(a^2*b^2)^(1/2))/(d^2)^(1
/2) - (4*b^2)/(d*(exp(2*c + 2*d*x^(1/2)) + 1))

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