3.46 \(\int \frac {\tanh (x)}{(a+b \coth ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=108 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \coth ^2(x)}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {b (2 a+b)}{a^2 (a+b)^2 \sqrt {a+b \coth ^2(x)}}+\frac {b}{3 a (a+b) \left (a+b \coth ^2(x)\right )^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \coth ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}} \]

[Out]

-arctanh((a+b*coth(x)^2)^(1/2)/a^(1/2))/a^(5/2)+arctanh((a+b*coth(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)+1/3*b/a
/(a+b)/(a+b*coth(x)^2)^(3/2)+b*(2*a+b)/a^2/(a+b)^2/(a+b*coth(x)^2)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3670, 446, 85, 152, 156, 63, 208} \[ \frac {b (2 a+b)}{a^2 (a+b)^2 \sqrt {a+b \coth ^2(x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \coth ^2(x)}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {b}{3 a (a+b) \left (a+b \coth ^2(x)\right )^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \coth ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/(a + b*Coth[x]^2)^(5/2),x]

[Out]

-(ArcTanh[Sqrt[a + b*Coth[x]^2]/Sqrt[a]]/a^(5/2)) + ArcTanh[Sqrt[a + b*Coth[x]^2]/Sqrt[a + b]]/(a + b)^(5/2) +
b/(3*a*(a + b)*(a + b*Coth[x]^2)^(3/2)) + (b*(2*a + b))/(a^2*(a + b)^2*Sqrt[a + b*Coth[x]^2])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 85

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[(f*(e + f*x)^(p +
1))/((p + 1)*(b*e - a*f)*(d*e - c*f)), x] + Dist[1/((b*e - a*f)*(d*e - c*f)), Int[((b*d*e - b*c*f - a*d*f - b
*d*f*x)*(e + f*x)^(p + 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[p, -1]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

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]

Rule 446

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \frac {\tanh (x)}{\left (a+b \coth ^2(x)\right )^{5/2}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{x \left (1-x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\coth (x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(1-x) x (a+b x)^{5/2}} \, dx,x,\coth ^2(x)\right )\\ &=\frac {b}{3 a (a+b) \left (a+b \coth ^2(x)\right )^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {-a-b+b x}{(1-x) x (a+b x)^{3/2}} \, dx,x,\coth ^2(x)\right )}{2 a (a+b)}\\ &=\frac {b}{3 a (a+b) \left (a+b \coth ^2(x)\right )^{3/2}}+\frac {b (2 a+b)}{a^2 (a+b)^2 \sqrt {a+b \coth ^2(x)}}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{2} (a+b)^2+\frac {1}{2} b (2 a+b) x}{(1-x) x \sqrt {a+b x}} \, dx,x,\coth ^2(x)\right )}{a^2 (a+b)^2}\\ &=\frac {b}{3 a (a+b) \left (a+b \coth ^2(x)\right )^{3/2}}+\frac {b (2 a+b)}{a^2 (a+b)^2 \sqrt {a+b \coth ^2(x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\coth ^2(x)\right )}{2 a^2}+\frac {\operatorname {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\coth ^2(x)\right )}{2 (a+b)^2}\\ &=\frac {b}{3 a (a+b) \left (a+b \coth ^2(x)\right )^{3/2}}+\frac {b (2 a+b)}{a^2 (a+b)^2 \sqrt {a+b \coth ^2(x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \coth ^2(x)}\right )}{a^2 b}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \coth ^2(x)}\right )}{b (a+b)^2}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \coth ^2(x)}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \coth ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}+\frac {b}{3 a (a+b) \left (a+b \coth ^2(x)\right )^{3/2}}+\frac {b (2 a+b)}{a^2 (a+b)^2 \sqrt {a+b \coth ^2(x)}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 73, normalized size = 0.68 \[ \frac {(a+b) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b \coth ^2(x)}{a}+1\right )-a \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b \coth ^2(x)+a}{a+b}\right )}{3 a (a+b) \left (a+b \coth ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/(a + b*Coth[x]^2)^(5/2),x]

[Out]

(-(a*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Coth[x]^2)/(a + b)]) + (a + b)*Hypergeometric2F1[-3/2, 1, -1/2, 1
+ (b*Coth[x]^2)/a])/(3*a*(a + b)*(a + b*Coth[x]^2)^(3/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*coth(x)^2)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*coth(x)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep-1)]Evaluation time: 0.72Error: Bad Argument Type

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \relax (x )}{\left (a +b \left (\coth ^{2}\relax (x )\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a+b*coth(x)^2)^(5/2),x)

[Out]

int(tanh(x)/(a+b*coth(x)^2)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \relax (x)}{{\left (b \coth \relax (x)^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*coth(x)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(tanh(x)/(b*coth(x)^2 + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {tanh}\relax (x)}{{\left (b\,{\mathrm {coth}\relax (x)}^2+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a + b*coth(x)^2)^(5/2),x)

[Out]

int(tanh(x)/(a + b*coth(x)^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh {\relax (x )}}{\left (a + b \coth ^{2}{\relax (x )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*coth(x)**2)**(5/2),x)

[Out]

Integral(tanh(x)/(a + b*coth(x)**2)**(5/2), x)

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