3.1.62 \(\int \frac {\sqrt {x}}{a+b \text {csch}(c+d \sqrt {x})} \, dx\) [62]

Optimal. Leaf size=337 \[ \frac {2 x^{3/2}}{3 a}-\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {2 b x \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {4 b \sqrt {x} \text {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {4 b \sqrt {x} \text {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {4 b \text {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {4 b \text {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3} \]

[Out]

2/3*x^(3/2)/a-2*b*x*ln(1+a*exp(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))/a/d/(a^2+b^2)^(1/2)+2*b*x*ln(1+a*exp(c+d*x^(1
/2))/(b+(a^2+b^2)^(1/2)))/a/d/(a^2+b^2)^(1/2)+4*b*polylog(3,-a*exp(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))/a/d^3/(a^
2+b^2)^(1/2)-4*b*polylog(3,-a*exp(c+d*x^(1/2))/(b+(a^2+b^2)^(1/2)))/a/d^3/(a^2+b^2)^(1/2)-4*b*polylog(2,-a*exp
(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))*x^(1/2)/a/d^2/(a^2+b^2)^(1/2)+4*b*polylog(2,-a*exp(c+d*x^(1/2))/(b+(a^2+b^2
)^(1/2)))*x^(1/2)/a/d^2/(a^2+b^2)^(1/2)

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Rubi [A]
time = 0.55, antiderivative size = 337, normalized size of antiderivative = 1.00, 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 = {5545, 4276, 3403, 2296, 2221, 2611, 2320, 6724} \begin {gather*} \frac {4 b \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {4 b \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^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 d^2 \sqrt {a^2+b^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 d^2 \sqrt {a^2+b^2}}-\frac {2 b x \log \left (\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {2 b x \log \left (\frac {a e^{c+d \sqrt {x}}}{\sqrt {a^2+b^2}+b}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {2 x^{3/2}}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a + b*Csch[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*Sqr
t[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*Sqr
t[a^2 + b^2]*d^3)

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2320

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 2611

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 3403

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

Rule 4276

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 5545

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Csch[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 6724

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 {csch}\left (c+d \sqrt {x}\right )} \, dx &=2 \text {Subst}\left (\int \frac {x^2}{a+b \text {csch}(c+d x)} \, dx,x,\sqrt {x}\right )\\ &=2 \text {Subst}\left (\int \left (\frac {x^2}{a}-\frac {b x^2}{a (b+a \sinh (c+d x))}\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {2 x^{3/2}}{3 a}-\frac {(2 b) \text {Subst}\left (\int \frac {x^2}{b+a \sinh (c+d x)} \, dx,x,\sqrt {x}\right )}{a}\\ &=\frac {2 x^{3/2}}{3 a}-\frac {(4 b) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 6.38, size = 374, normalized size = 1.11 \begin {gather*} \frac {2 \left (d^3 \sqrt {\left (a^2+b^2\right ) e^{2 c}} x^{3/2}-3 b d^2 e^c x \log \left (1+\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+3 b d^2 e^c x \log \left (1+\frac {a e^{2 c+d \sqrt {x}}}{b e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-6 b d e^c \sqrt {x} \text {PolyLog}\left (2,-\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+6 b d e^c \sqrt {x} \text {PolyLog}\left (2,-\frac {a e^{2 c+d \sqrt {x}}}{b e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+6 b e^c \text {PolyLog}\left (3,-\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-6 b e^c \text {PolyLog}\left (3,-\frac {a e^{2 c+d \sqrt {x}}}{b e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )\right )}{3 a d^3 \sqrt {\left (a^2+b^2\right ) e^{2 c}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(a + b*Csch[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]*Pol
yLog[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*S
qrt[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)]))]))/(3*a*d^3*Sqrt[(a^2 + b^2)*E^(2*c)])

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Maple [F]
time = 3.96, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {x}}{a +b \,\mathrm {csch}\left (c +d \sqrt {x}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {x}}{a+\frac {b}{\mathrm {sinh}\left (c+d\,\sqrt {x}\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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