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

Optimal. Leaf size=262 \[ -\frac{\sqrt{a+b \text{sech}(c+d x)}}{4 d (a+b) (1-\text{sech}(c+d x))}-\frac{\sqrt{a+b \text{sech}(c+d x)}}{4 d (a-b) (\text{sech}(c+d x)+1)}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{4 d (a-b)^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{d \sqrt{a-b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{d \sqrt{a+b}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{4 d (a+b)^{3/2}} \]

[Out]

(2*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) - ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]]/(S
qrt[a - b]*d) + (b*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]])/(4*(a - b)^(3/2)*d) - (b*ArcTanh[Sqrt[a + b
*Sech[c + d*x]]/Sqrt[a + b]])/(4*(a + b)^(3/2)*d) - ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]]/(Sqrt[a + b
]*d) - Sqrt[a + b*Sech[c + d*x]]/(4*(a + b)*d*(1 - Sech[c + d*x])) - Sqrt[a + b*Sech[c + d*x]]/(4*(a - b)*d*(1
+ Sech[c + d*x]))

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Rubi [A]  time = 0.296669, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3885, 898, 1238, 206, 199, 207} \[ -\frac{\sqrt{a+b \text{sech}(c+d x)}}{4 d (a+b) (1-\text{sech}(c+d x))}-\frac{\sqrt{a+b \text{sech}(c+d x)}}{4 d (a-b) (\text{sech}(c+d x)+1)}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{4 d (a-b)^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{d \sqrt{a-b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{d \sqrt{a+b}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{4 d (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) - ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]]/(S
qrt[a - b]*d) + (b*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]])/(4*(a - b)^(3/2)*d) - (b*ArcTanh[Sqrt[a + b
*Sech[c + d*x]]/Sqrt[a + b]])/(4*(a + b)^(3/2)*d) - ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]]/(Sqrt[a + b
]*d) - Sqrt[a + b*Sech[c + d*x]]/(4*(a + b)*d*(1 - Sech[c + d*x])) - Sqrt[a + b*Sech[c + d*x]]/(4*(a - b)*d*(1
+ Sech[c + d*x]))

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]

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 1238

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d
+ e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b^2 - 4*a*c, 0] && ((Intege
rQ[p] && IntegerQ[q]) || IGtQ[p, 0] || IGtQ[q, 0])

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 207

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

Rubi steps

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

Mathematica [B]  time = 7.28054, size = 902, normalized size = 3.44 \[ \frac{\sqrt{b+a \cosh (c+d x)} \sqrt{\text{sech}(c+d x)} \left (\frac{\left (2 a^2-2 b^2\right ) \left (\sqrt{a} \left (\sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{b+a \cosh (c+d x)}}{\sqrt{a-b} \sqrt{-a \cosh (c+d x)}}\right )+\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{b+a \cosh (c+d x)}}{\sqrt{a+b} \sqrt{-a \cosh (c+d x)}}\right )\right )-4 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{b+a \cosh (c+d x)}}{\sqrt{-a \cosh (c+d x)}}\right )\right ) \sqrt{-a \cosh (c+d x)} \sqrt{\frac{a \cosh (c+d x)-a}{\cosh (c+d x) a+a}} \cosh (2 (c+d x)) \sqrt{\text{sech}(c+d x)} (\cosh (c+d x) a+a)}{\sqrt{a-b} \sqrt{a+b} \sqrt{\cosh (c+d x)-1} \sqrt{\cosh (c+d x)+1} \left (a^2-2 b^2-2 (b+a \cosh (c+d x))^2+4 b (b+a \cosh (c+d x))\right )}-\frac{\left (2 a^2-3 b^2\right ) \left (\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{b+a \cosh (c+d x)}}{\sqrt{a-b} \sqrt{a \cosh (c+d x)}}\right )+\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{b+a \cosh (c+d x)}}{\sqrt{a+b} \sqrt{a \cosh (c+d x)}}\right )\right ) \sqrt{a \cosh (c+d x)} \sqrt{\frac{a \cosh (c+d x)-a}{\cosh (c+d x) a+a}} \sqrt{\text{sech}(c+d x)} (\cosh (c+d x) a+a)}{a^{3/2} \sqrt{a-b} \sqrt{a+b} \sqrt{\cosh (c+d x)-1} \sqrt{\cosh (c+d x)+1}}+\frac{\sqrt{a} b \left (\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{b+a \cosh (c+d x)}}{\sqrt{-a-b} \sqrt{a \cosh (c+d x)}}\right )+\sqrt{-a-b} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{b+a \cosh (c+d x)}}{\sqrt{a-b} \sqrt{a \cosh (c+d x)}}\right )\right ) \sqrt{\frac{a \cosh (c+d x)-a}{\cosh (c+d x) a+a}} (\cosh (c+d x) a+a)}{\sqrt{-a-b} \sqrt{a-b} \sqrt{\cosh (c+d x)-1} \sqrt{a \cosh (c+d x)} \sqrt{\cosh (c+d x)+1} \sqrt{\text{sech}(c+d x)}}\right )}{4 (a-b) (a+b) d \sqrt{a+b \text{sech}(c+d x)}}+\frac{(b+a \cosh (c+d x)) \left (\frac{(a-b \cosh (c+d x)) \text{csch}^2(c+d x)}{2 \left (b^2-a^2\right )}-\frac{a}{2 \left (a^2-b^2\right )}\right ) \text{sech}(c+d x)}{d \sqrt{a+b \text{sech}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b + a*Cosh[c + d*x]]*((Sqrt[a]*b*(Sqrt[a - b]*ArcTan[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(Sqrt[-a - b]*S
qrt[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]*Sqrt[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]]) - ((2*a^2 - 3*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]*A
rcTanh[(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])*Sqrt[Sech[c + d*x]])/(a^(3/2)*Sqrt[a - b]*Sq
rt[a + b]*Sqrt[-1 + Cosh[c + d*x]]*Sqrt[1 + Cosh[c + d*x]]) + ((2*a^2 - 2*b^2)*(-4*Sqrt[a - b]*Sqrt[a + b]*Arc
Tan[Sqrt[b + a*Cosh[c + d*x]]/Sqrt[-(a*Cosh[c + d*x])]] + Sqrt[a]*(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]*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)))*Sqrt[S
ech[c + d*x]])/(4*(a - b)*(a + b)*d*Sqrt[a + b*Sech[c + d*x]]) + ((b + a*Cosh[c + d*x])*(-a/(2*(a^2 - b^2)) +
((a - b*Cosh[c + d*x])*Csch[c + d*x]^2)/(2*(-a^2 + b^2)))*Sech[c + d*x])/(d*Sqrt[a + b*Sech[c + d*x]])

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Maple [F]  time = 0.404, size = 0, normalized size = 0. \begin{align*} \int{ \left ({\rm coth} \left (dx+c\right ) \right ) ^{3}{\frac{1}{\sqrt{a+b{\rm sech} \left (dx+c\right )}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x)

[Out]

int(coth(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth \left (d x + c\right )^{3}}{\sqrt{b \operatorname{sech}\left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{3}{\left (c + d x \right )}}{\sqrt{a + b \operatorname{sech}{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**3/(a+b*sech(d*x+c))**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth \left (d x + c\right )^{3}}{\sqrt{b \operatorname{sech}\left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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