3.63 \(\int \frac{\sqrt{x}}{a+b \text{sech}(c+d \sqrt{x})} \, dx\)

Optimal. Leaf size=361 \[ -\frac{4 b \sqrt{x} \text{PolyLog}\left (2,-\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{b^2-a^2}}\right )}{a d^2 \sqrt{b^2-a^2}}+\frac{4 b \sqrt{x} \text{PolyLog}\left (2,-\frac{a e^{c+d \sqrt{x}}}{\sqrt{b^2-a^2}+b}\right )}{a d^2 \sqrt{b^2-a^2}}+\frac{4 b \text{PolyLog}\left (3,-\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{b^2-a^2}}\right )}{a d^3 \sqrt{b^2-a^2}}-\frac{4 b \text{PolyLog}\left (3,-\frac{a e^{c+d \sqrt{x}}}{\sqrt{b^2-a^2}+b}\right )}{a d^3 \sqrt{b^2-a^2}}-\frac{2 b x \log \left (\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{b^2-a^2}}+1\right )}{a d \sqrt{b^2-a^2}}+\frac{2 b x \log \left (\frac{a e^{c+d \sqrt{x}}}{\sqrt{b^2-a^2}+b}+1\right )}{a d \sqrt{b^2-a^2}}+\frac{2 x^{3/2}}{3 a} \]

[Out]

(2*x^(3/2))/(3*a) - (2*b*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (2*
b*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) - (4*b*Sqrt[x]*PolyLog[2, -(
(a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (4*b*Sqrt[x]*PolyLog[2, -((a*E^(c +
d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (4*b*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b -
Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - (4*b*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]
))])/(a*Sqrt[-a^2 + b^2]*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.776434, antiderivative size = 361, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {5436, 4191, 3320, 2264, 2190, 2531, 2282, 6589} \[ -\frac{4 b \sqrt{x} \text{PolyLog}\left (2,-\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{b^2-a^2}}\right )}{a d^2 \sqrt{b^2-a^2}}+\frac{4 b \sqrt{x} \text{PolyLog}\left (2,-\frac{a e^{c+d \sqrt{x}}}{\sqrt{b^2-a^2}+b}\right )}{a d^2 \sqrt{b^2-a^2}}+\frac{4 b \text{PolyLog}\left (3,-\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{b^2-a^2}}\right )}{a d^3 \sqrt{b^2-a^2}}-\frac{4 b \text{PolyLog}\left (3,-\frac{a e^{c+d \sqrt{x}}}{\sqrt{b^2-a^2}+b}\right )}{a d^3 \sqrt{b^2-a^2}}-\frac{2 b x \log \left (\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{b^2-a^2}}+1\right )}{a d \sqrt{b^2-a^2}}+\frac{2 b x \log \left (\frac{a e^{c+d \sqrt{x}}}{\sqrt{b^2-a^2}+b}+1\right )}{a d \sqrt{b^2-a^2}}+\frac{2 x^{3/2}}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2))/(3*a) - (2*b*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (2*
b*x*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) - (4*b*Sqrt[x]*PolyLog[2, -(
(a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (4*b*Sqrt[x]*PolyLog[2, -((a*E^(c +
d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (4*b*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b -
Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - (4*b*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]
))])/(a*Sqrt[-a^2 + b^2]*d^3)

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 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 3320

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol]
:> Dist[2, Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(E^(I*Pi*(k - 1/2))*(b + (2*a*E^(-(I*e) + f*fz*x))/E^(I*Pi*(k
- 1/2)) - (b*E^(2*(-(I*e) + f*fz*x)))/E^(2*I*k*Pi))), x], x] /; FreeQ[{a, b, c, d, e, f, fz}, x] && IntegerQ[
2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{\sqrt{x}}{a+b \text{sech}\left (c+d \sqrt{x}\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^2}{a+b \text{sech}(c+d x)} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{x^2}{a}-\frac{b x^2}{a (b+a \cosh (c+d x))}\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{2 x^{3/2}}{3 a}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{x^2}{b+a \cosh (c+d x)} \, dx,x,\sqrt{x}\right )}{a}\\ &=\frac{2 x^{3/2}}{3 a}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{e^{c+d x} x^2}{a+2 b e^{c+d x}+a e^{2 (c+d x)}} \, dx,x,\sqrt{x}\right )}{a}\\ &=\frac{2 x^{3/2}}{3 a}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{e^{c+d x} x^2}{2 b-2 \sqrt{-a^2+b^2}+2 a e^{c+d x}} \, dx,x,\sqrt{x}\right )}{\sqrt{-a^2+b^2}}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{e^{c+d x} x^2}{2 b+2 \sqrt{-a^2+b^2}+2 a e^{c+d x}} \, dx,x,\sqrt{x}\right )}{\sqrt{-a^2+b^2}}\\ &=\frac{2 x^{3/2}}{3 a}-\frac{2 b x \log \left (1+\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d}+\frac{2 b x \log \left (1+\frac{a e^{c+d \sqrt{x}}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d}+\frac{(4 b) \operatorname{Subst}\left (\int x \log \left (1+\frac{2 a e^{c+d x}}{2 b-2 \sqrt{-a^2+b^2}}\right ) \, dx,x,\sqrt{x}\right )}{a \sqrt{-a^2+b^2} d}-\frac{(4 b) \operatorname{Subst}\left (\int x \log \left (1+\frac{2 a e^{c+d x}}{2 b+2 \sqrt{-a^2+b^2}}\right ) \, dx,x,\sqrt{x}\right )}{a \sqrt{-a^2+b^2} d}\\ &=\frac{2 x^{3/2}}{3 a}-\frac{2 b x \log \left (1+\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d}+\frac{2 b x \log \left (1+\frac{a e^{c+d \sqrt{x}}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d}-\frac{4 b \sqrt{x} \text{Li}_2\left (-\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}+\frac{4 b \sqrt{x} \text{Li}_2\left (-\frac{a e^{c+d \sqrt{x}}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}+\frac{(4 b) \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{2 a e^{c+d x}}{2 b-2 \sqrt{-a^2+b^2}}\right ) \, dx,x,\sqrt{x}\right )}{a \sqrt{-a^2+b^2} d^2}-\frac{(4 b) \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{2 a e^{c+d x}}{2 b+2 \sqrt{-a^2+b^2}}\right ) \, dx,x,\sqrt{x}\right )}{a \sqrt{-a^2+b^2} d^2}\\ &=\frac{2 x^{3/2}}{3 a}-\frac{2 b x \log \left (1+\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d}+\frac{2 b x \log \left (1+\frac{a e^{c+d \sqrt{x}}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d}-\frac{4 b \sqrt{x} \text{Li}_2\left (-\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}+\frac{4 b \sqrt{x} \text{Li}_2\left (-\frac{a e^{c+d \sqrt{x}}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{a x}{-b+\sqrt{-a^2+b^2}}\right )}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{a \sqrt{-a^2+b^2} d^3}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{a x}{b+\sqrt{-a^2+b^2}}\right )}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{a \sqrt{-a^2+b^2} d^3}\\ &=\frac{2 x^{3/2}}{3 a}-\frac{2 b x \log \left (1+\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d}+\frac{2 b x \log \left (1+\frac{a e^{c+d \sqrt{x}}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d}-\frac{4 b \sqrt{x} \text{Li}_2\left (-\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}+\frac{4 b \sqrt{x} \text{Li}_2\left (-\frac{a e^{c+d \sqrt{x}}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}+\frac{4 b \text{Li}_3\left (-\frac{a e^{c+d \sqrt{x}}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^3}-\frac{4 b \text{Li}_3\left (-\frac{a e^{c+d \sqrt{x}}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^3}\\ \end{align*}

Mathematica [A]  time = 7.68598, size = 390, normalized size = 1.08 \[ \frac{2 \left (-6 b e^c d \sqrt{x} \text{PolyLog}\left (2,-\frac{a e^{2 c+d \sqrt{x}}}{b e^c-\sqrt{e^{2 c} \left (b^2-a^2\right )}}\right )+6 b e^c d \sqrt{x} \text{PolyLog}\left (2,-\frac{a e^{2 c+d \sqrt{x}}}{\sqrt{e^{2 c} \left (b^2-a^2\right )}+b e^c}\right )+6 b e^c \text{PolyLog}\left (3,-\frac{a e^{2 c+d \sqrt{x}}}{b e^c-\sqrt{e^{2 c} \left (b^2-a^2\right )}}\right )-6 b e^c \text{PolyLog}\left (3,-\frac{a e^{2 c+d \sqrt{x}}}{\sqrt{e^{2 c} \left (b^2-a^2\right )}+b e^c}\right )+d^3 x^{3/2} \sqrt{e^{2 c} \left (b^2-a^2\right )}-3 b e^c d^2 x \log \left (\frac{a e^{2 c+d \sqrt{x}}}{b e^c-\sqrt{e^{2 c} \left (b^2-a^2\right )}}+1\right )+3 b e^c d^2 x \log \left (\frac{a e^{2 c+d \sqrt{x}}}{\sqrt{e^{2 c} \left (b^2-a^2\right )}+b e^c}+1\right )\right )}{3 a d^3 \sqrt{e^{2 c} \left (b^2-a^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d^3*Sqrt[(-a^2 + b^2)*E^(2*c)]*x^(3/2) - 3*b*d^2*E^c*x*Log[1 + (a*E^(2*c + d*Sqrt[x]))/(b*E^c - Sqrt[(-a^2
+ b^2)*E^(2*c)])] + 3*b*d^2*E^c*x*Log[1 + (a*E^(2*c + d*Sqrt[x]))/(b*E^c + Sqrt[(-a^2 + b^2)*E^(2*c)])] - 6*b
*d*E^c*Sqrt[x]*PolyLog[2, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c - Sqrt[(-a^2 + b^2)*E^(2*c)]))] + 6*b*d*E^c*Sqrt[x]
*PolyLog[2, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c + Sqrt[(-a^2 + b^2)*E^(2*c)]))] + 6*b*E^c*PolyLog[3, -((a*E^(2*c
+ d*Sqrt[x]))/(b*E^c - Sqrt[(-a^2 + b^2)*E^(2*c)]))] - 6*b*E^c*PolyLog[3, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c + S
qrt[(-a^2 + b^2)*E^(2*c)]))]))/(3*a*d^3*Sqrt[(-a^2 + b^2)*E^(2*c)])

________________________________________________________________________________________

Maple [F]  time = 0.082, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{x} \left ( a+b{\rm sech} \left (c+d\sqrt{x}\right ) \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x}}{b \operatorname{sech}\left (d \sqrt{x} + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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