3.69 \(\int \frac{1}{\sqrt{x} (a+b \text{sech}(c+d \sqrt{x}))^2} \, dx\)

Optimal. Leaf size=127 \[ -\frac{4 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{2 b^2 \tanh \left (c+d \sqrt{x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right )}+\frac{2 \sqrt{x}}{a^2} \]

[Out]

(2*Sqrt[x])/a^2 - (4*b*(2*a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tanh[(c + d*Sqrt[x])/2])/Sqrt[a + b]])/(a^2*(a - b)^(
3/2)*(a + b)^(3/2)*d) + (2*b^2*Tanh[c + d*Sqrt[x]])/(a*(a^2 - b^2)*d*(a + b*Sech[c + d*Sqrt[x]]))

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Rubi [A]  time = 0.200421, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {5436, 3785, 3919, 3831, 2659, 208} \[ -\frac{4 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{2 b^2 \tanh \left (c+d \sqrt{x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right )}+\frac{2 \sqrt{x}}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(a + b*Sech[c + d*Sqrt[x]])^2),x]

[Out]

(2*Sqrt[x])/a^2 - (4*b*(2*a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tanh[(c + d*Sqrt[x])/2])/Sqrt[a + b]])/(a^2*(a - b)^(
3/2)*(a + b)^(3/2)*d) + (2*b^2*Tanh[c + d*Sqrt[x]])/(a*(a^2 - b^2)*d*(a + b*Sech[c + d*Sqrt[x]]))

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 3785

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n +
1))/(a*d*(n + 1)*(a^2 - b^2)), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.482957, size = 232, normalized size = 1.83 \[ \frac{2 \left (b \left (\left (a^2-b^2\right )^{3/2} \left (c+d \sqrt{x}\right )+a b \sqrt{a^2-b^2} \sinh \left (c+d \sqrt{x}\right )+\left (4 a^2 b-2 b^3\right ) \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a^2-b^2}}\right )\right )+a \cosh \left (c+d \sqrt{x}\right ) \left (\left (a^2-b^2\right )^{3/2} \left (c+d \sqrt{x}\right )+\left (4 a^2 b-2 b^3\right ) \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a^2-b^2}}\right )\right )\right )}{a^2 d (a-b) (a+b) \sqrt{a^2-b^2} \left (a \cosh \left (c+d \sqrt{x}\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(a + b*Sech[c + d*Sqrt[x]])^2),x]

[Out]

(2*(a*((a^2 - b^2)^(3/2)*(c + d*Sqrt[x]) + (4*a^2*b - 2*b^3)*ArcTan[((-a + b)*Tanh[(c + d*Sqrt[x])/2])/Sqrt[a^
2 - b^2]])*Cosh[c + d*Sqrt[x]] + b*((a^2 - b^2)^(3/2)*(c + d*Sqrt[x]) + (4*a^2*b - 2*b^3)*ArcTan[((-a + b)*Tan
h[(c + d*Sqrt[x])/2])/Sqrt[a^2 - b^2]] + a*b*Sqrt[a^2 - b^2]*Sinh[c + d*Sqrt[x]])))/(a^2*(a - b)*(a + b)*Sqrt[
a^2 - b^2]*d*(b + a*Cosh[c + d*Sqrt[x]]))

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Maple [B]  time = 0.066, size = 236, normalized size = 1.9 \begin{align*} -2\,{\frac{\ln \left ( \tanh \left ( c/2+1/2\,d\sqrt{x} \right ) -1 \right ) }{d{a}^{2}}}+4\,{\frac{{b}^{2}\tanh \left ( c/2+1/2\,d\sqrt{x} \right ) }{ad \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tanh \left ( c/2+1/2\,d\sqrt{x} \right ) \right ) ^{2}a- \left ( \tanh \left ( c/2+1/2\,d\sqrt{x} \right ) \right ) ^{2}b+a+b \right ) }}-8\,{\frac{b}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tanh \left ( c/2+1/2\,d\sqrt{x} \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+4\,{\frac{{b}^{3}}{d{a}^{2} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tanh \left ( c/2+1/2\,d\sqrt{x} \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{\ln \left ( \tanh \left ( c/2+1/2\,d\sqrt{x} \right ) +1 \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/a^2*ln(tanh(1/2*c+1/2*d*x^(1/2))-1)+4/d*b^2/a/(a^2-b^2)*tanh(1/2*c+1/2*d*x^(1/2))/(tanh(1/2*c+1/2*d*x^(1/
2))^2*a-tanh(1/2*c+1/2*d*x^(1/2))^2*b+a+b)-8/d*b/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*c+1/2*d
*x^(1/2))/((a+b)*(a-b))^(1/2))+4/d*b^3/a^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*c+1/2*d*x^(1/
2))/((a+b)*(a-b))^(1/2))+2/d/a^2*ln(tanh(1/2*c+1/2*d*x^(1/2))+1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.64354, size = 3187, normalized size = 25.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-2*(2*a^3*b^2 - 2*a*b^4 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*cosh(d*sqrt(x) + c)^2 - (a^5 - 2*a^3*b^2 + a*b^
4)*d*sqrt(x)*sinh(d*sqrt(x) + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x) + 2*(a^2*b^3 - b^5 - (a^4*b - 2*a^2*b
^3 + b^5)*d*sqrt(x))*cosh(d*sqrt(x) + c) + ((2*a^3*b - a*b^3)*sqrt(-a^2 + b^2)*cosh(d*sqrt(x) + c)^2 + (2*a^3*
b - a*b^3)*sqrt(-a^2 + b^2)*sinh(d*sqrt(x) + c)^2 + 2*(2*a^2*b^2 - b^4)*sqrt(-a^2 + b^2)*cosh(d*sqrt(x) + c) +
2*((2*a^3*b - a*b^3)*sqrt(-a^2 + b^2)*cosh(d*sqrt(x) + c) + (2*a^2*b^2 - b^4)*sqrt(-a^2 + b^2))*sinh(d*sqrt(x
) + c) + (2*a^3*b - a*b^3)*sqrt(-a^2 + b^2))*log((a*b + (b^2 + sqrt(-a^2 + b^2)*b)*cosh(d*sqrt(x) + c) + (a^2
- b^2 - sqrt(-a^2 + b^2)*b)*sinh(d*sqrt(x) + c) + sqrt(-a^2 + b^2)*a)/(a*cosh(d*sqrt(x) + c) + b)) + 2*(a^2*b^
3 - b^5 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*cosh(d*sqrt(x) + c) - (a^4*b - 2*a^2*b^3 + b^5)*d*sqrt(x))*sinh(
d*sqrt(x) + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cosh(d*sqrt(x) + c)^2 + (a^7 - 2*a^5*b^2 + a^3*b^4)*d*sinh(d*sq
rt(x) + c)^2 + 2*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cosh(d*sqrt(x) + c) + (a^7 - 2*a^5*b^2 + a^3*b^4)*d + 2*((a^7
- 2*a^5*b^2 + a^3*b^4)*d*cosh(d*sqrt(x) + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d)*sinh(d*sqrt(x) + c)), -2*(2*a
^3*b^2 - 2*a*b^4 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*cosh(d*sqrt(x) + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqr
t(x)*sinh(d*sqrt(x) + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x) - 2*((2*a^3*b - a*b^3)*sqrt(a^2 - b^2)*cosh(d
*sqrt(x) + c)^2 + (2*a^3*b - a*b^3)*sqrt(a^2 - b^2)*sinh(d*sqrt(x) + c)^2 + 2*(2*a^2*b^2 - b^4)*sqrt(a^2 - b^2
)*cosh(d*sqrt(x) + c) + 2*((2*a^3*b - a*b^3)*sqrt(a^2 - b^2)*cosh(d*sqrt(x) + c) + (2*a^2*b^2 - b^4)*sqrt(a^2
- b^2))*sinh(d*sqrt(x) + c) + (2*a^3*b - a*b^3)*sqrt(a^2 - b^2))*arctan(-(sqrt(a^2 - b^2)*a*cosh(d*sqrt(x) + c
) + sqrt(a^2 - b^2)*a*sinh(d*sqrt(x) + c) + sqrt(a^2 - b^2)*b)/(a^2 - b^2)) + 2*(a^2*b^3 - b^5 - (a^4*b - 2*a^
2*b^3 + b^5)*d*sqrt(x))*cosh(d*sqrt(x) + c) + 2*(a^2*b^3 - b^5 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*cosh(d*sq
rt(x) + c) - (a^4*b - 2*a^2*b^3 + b^5)*d*sqrt(x))*sinh(d*sqrt(x) + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cosh(d*s
qrt(x) + c)^2 + (a^7 - 2*a^5*b^2 + a^3*b^4)*d*sinh(d*sqrt(x) + c)^2 + 2*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cosh(d
*sqrt(x) + c) + (a^7 - 2*a^5*b^2 + a^3*b^4)*d + 2*((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cosh(d*sqrt(x) + c) + (a^6*b
- 2*a^4*b^3 + a^2*b^5)*d)*sinh(d*sqrt(x) + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x} \left (a + b \operatorname{sech}{\left (c + d \sqrt{x} \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12855, size = 200, normalized size = 1.57 \begin{align*} -\frac{4 \,{\left (2 \, a^{2} b - b^{3}\right )} \arctan \left (\frac{a e^{\left (d \sqrt{x} + c\right )} + b}{\sqrt{a^{2} - b^{2}}}\right )}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt{a^{2} - b^{2}}} - \frac{4 \,{\left (b^{3} e^{\left (d \sqrt{x} + c\right )} + a b^{2}\right )}}{{\left (a^{4} d - a^{2} b^{2} d\right )}{\left (a e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + 2 \, b e^{\left (d \sqrt{x} + c\right )} + a\right )}} + \frac{2 \,{\left (d \sqrt{x} + c\right )}}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(2*a^2*b - b^3)*arctan((a*e^(d*sqrt(x) + c) + b)/sqrt(a^2 - b^2))/((a^4*d - a^2*b^2*d)*sqrt(a^2 - b^2)) - 4
*(b^3*e^(d*sqrt(x) + c) + a*b^2)/((a^4*d - a^2*b^2*d)*(a*e^(2*d*sqrt(x) + 2*c) + 2*b*e^(d*sqrt(x) + c) + a)) +
2*(d*sqrt(x) + c)/(a^2*d)