3.145 \(\int \frac {\coth (c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=142 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 b^2}{a d \left (a^2-b^2\right ) \sqrt {a+b \text {sech}(c+d x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{d (a-b)^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{d (a+b)^{3/2}} \]

[Out]

2*arctanh((a+b*sech(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d-arctanh((a+b*sech(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(3/2)/
d-arctanh((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)/d+2*b^2/a/(a^2-b^2)/d/(a+b*sech(d*x+c))^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3885, 898, 1287, 206} \[ \frac {2 b^2}{a d \left (a^2-b^2\right ) \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{d (a-b)^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{d (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]/(a + b*Sech[c + d*x])^(3/2),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 898

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 + a*e^2)/e^2 - (2*c
*d*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1287

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[Ex
pandIntegrand[((f*x)^m*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^
2 - 4*a*c, 0] && IntegerQ[q] && IntegerQ[m]

Rule 3885

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

Rubi steps

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

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Mathematica [B]  time = 7.37, size = 904, normalized size = 6.37 \[ \frac {(b+a \cosh (c+d x))^2 \left (-\frac {2 b^3}{a^2 \left (a^2-b^2\right ) (b+a \cosh (c+d x))}-\frac {2 b^2}{a^2 \left (b^2-a^2\right )}\right ) \text {sech}^2(c+d x)}{d (a+b \text {sech}(c+d x))^{3/2}}-\frac {(b+a \cosh (c+d x))^{3/2} \left (\frac {\left (a^2-b^2\right ) \left (\sqrt {a} \left (\sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {a-b} \sqrt {-a \cosh (c+d x)}}\right )+\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {a+b} \sqrt {-a \cosh (c+d x)}}\right )\right )-4 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {b+a \cosh (c+d x)}}{\sqrt {-a \cosh (c+d x)}}\right )\right ) \sqrt {-a \cosh (c+d x)} \sqrt {\frac {a \cosh (c+d x)-a}{\cosh (c+d x) a+a}} \cosh (2 (c+d x)) \sqrt {\text {sech}(c+d x)} (\cosh (c+d x) a+a)}{\sqrt {a-b} \sqrt {a+b} \sqrt {\cosh (c+d x)-1} \sqrt {\cosh (c+d x)+1} \left (a^2-2 b^2-2 (b+a \cosh (c+d x))^2+4 b (b+a \cosh (c+d x))\right )}-\frac {\left (a^2+b^2\right ) \left (\sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {a-b} \sqrt {a \cosh (c+d x)}}\right )+\sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {a+b} \sqrt {a \cosh (c+d x)}}\right )\right ) \sqrt {a \cosh (c+d x)} \sqrt {\frac {a \cosh (c+d x)-a}{\cosh (c+d x) a+a}} \sqrt {\text {sech}(c+d x)} (\cosh (c+d x) a+a)}{a^{3/2} \sqrt {a-b} \sqrt {a+b} \sqrt {\cosh (c+d x)-1} \sqrt {\cosh (c+d x)+1}}-\frac {2 \sqrt {a} b \left (\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {-a-b} \sqrt {a \cosh (c+d x)}}\right )+\sqrt {-a-b} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {a-b} \sqrt {a \cosh (c+d x)}}\right )\right ) \sqrt {\frac {a \cosh (c+d x)-a}{\cosh (c+d x) a+a}} (\cosh (c+d x) a+a)}{\sqrt {-a-b} \sqrt {a-b} \sqrt {\cosh (c+d x)-1} \sqrt {a \cosh (c+d x)} \sqrt {\cosh (c+d x)+1} \sqrt {\text {sech}(c+d x)}}\right ) \text {sech}^{\frac {3}{2}}(c+d x)}{2 a (b-a) (a+b) d (a+b \text {sech}(c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]/(a + b*Sech[c + d*x])^(3/2),x]

[Out]

-1/2*((b + a*Cosh[c + d*x])^(3/2)*((-2*Sqrt[a]*b*(Sqrt[a - b]*ArcTan[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(Sqrt
[-a - b]*Sqrt[a*Cosh[c + d*x]])] + Sqrt[-a - b]*ArcTanh[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(Sqrt[a - b]*Sqrt[
a*Cosh[c + d*x]])])*Sqrt[(-a + a*Cosh[c + d*x])/(a + a*Cosh[c + d*x])]*(a + a*Cosh[c + d*x]))/(Sqrt[-a - b]*Sq
rt[a - b]*Sqrt[-1 + Cosh[c + d*x]]*Sqrt[a*Cosh[c + d*x]]*Sqrt[1 + Cosh[c + d*x]]*Sqrt[Sech[c + d*x]]) - ((a^2
+ b^2)*(Sqrt[a + b]*ArcTanh[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(Sqrt[a - b]*Sqrt[a*Cosh[c + d*x]])] + Sqrt[a
- b]*ArcTanh[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(Sqrt[a + b]*Sqrt[a*Cosh[c + d*x]])])*Sqrt[a*Cosh[c + d*x]]*S
qrt[(-a + a*Cosh[c + d*x])/(a + a*Cosh[c + d*x])]*(a + a*Cosh[c + d*x])*Sqrt[Sech[c + d*x]])/(a^(3/2)*Sqrt[a -
b]*Sqrt[a + b]*Sqrt[-1 + Cosh[c + d*x]]*Sqrt[1 + Cosh[c + d*x]]) + ((a^2 - b^2)*(-4*Sqrt[a - b]*Sqrt[a + b]*A
rcTan[Sqrt[b + a*Cosh[c + d*x]]/Sqrt[-(a*Cosh[c + d*x])]] + Sqrt[a]*(Sqrt[a + b]*ArcTan[(Sqrt[a]*Sqrt[b + a*Co
sh[c + d*x]])/(Sqrt[a - b]*Sqrt[-(a*Cosh[c + d*x])])] + Sqrt[a - b]*ArcTan[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])
/(Sqrt[a + b]*Sqrt[-(a*Cosh[c + d*x])])]))*Sqrt[-(a*Cosh[c + d*x])]*Sqrt[(-a + a*Cosh[c + d*x])/(a + a*Cosh[c
+ d*x])]*(a + a*Cosh[c + d*x])*Cosh[2*(c + d*x)]*Sqrt[Sech[c + d*x]])/(Sqrt[a - b]*Sqrt[a + b]*Sqrt[-1 + Cosh[
c + d*x]]*Sqrt[1 + Cosh[c + d*x]]*(a^2 - 2*b^2 + 4*b*(b + a*Cosh[c + d*x]) - 2*(b + a*Cosh[c + d*x])^2)))*Sech
[c + d*x]^(3/2))/(a*(-a + b)*(a + b)*d*(a + b*Sech[c + d*x])^(3/2)) + ((b + a*Cosh[c + d*x])^2*((-2*b^2)/(a^2*
(-a^2 + b^2)) - (2*b^3)/(a^2*(a^2 - b^2)*(b + a*Cosh[c + d*x])))*Sech[c + d*x]^2)/(d*(a + b*Sech[c + d*x])^(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(coth(d*x+c)/(a+b*sech(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth \left (d x + c\right )}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)/(a+b*sech(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(coth(d*x + c)/(b*sech(d*x + c) + a)^(3/2), x)

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maple [F]  time = 0.57, size = 0, normalized size = 0.00 \[ \int \frac {\coth \left (d x +c \right )}{\left (a +b \,\mathrm {sech}\left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)/(a+b*sech(d*x+c))^(3/2),x)

[Out]

int(coth(d*x+c)/(a+b*sech(d*x+c))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth \left (d x + c\right )}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)/(a+b*sech(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(coth(d*x + c)/(b*sech(d*x + c) + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {coth}\left (c+d\,x\right )}{{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + d*x)/(a + b/cosh(c + d*x))^(3/2),x)

[Out]

int(coth(c + d*x)/(a + b/cosh(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth {\left (c + d x \right )}}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)/(a+b*sech(d*x+c))**(3/2),x)

[Out]

Integral(coth(c + d*x)/(a + b*sech(c + d*x))**(3/2), x)

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