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

Optimal. Leaf size=362 \[ -\frac{\coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{d \sqrt{a+b}}-\frac{b^2 \tanh (c+d x)}{d \left (a^2-b^2\right ) \sqrt{a+b \text{sech}(c+d x)}}-\frac{\coth (c+d x)}{d \sqrt{a+b \text{sech}(c+d x)}}+\frac{\coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d \sqrt{a+b}}+\frac{2 \sqrt{a+b} \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a d} \]

[Out]

(Coth[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c +
d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(Sqrt[a + b]*d) - (Coth[c + d*x]*EllipticF[ArcSin[S
qrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + S
ech[c + d*x]))/(a - b))])/(Sqrt[a + b]*d) + (2*Sqrt[a + b]*Coth[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a +
b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c +
d*x]))/(a - b))])/(a*d) - Coth[c + d*x]/(d*Sqrt[a + b*Sech[c + d*x]]) - (b^2*Tanh[c + d*x])/((a^2 - b^2)*d*Sq
rt[a + b*Sech[c + d*x]])

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Rubi [A]  time = 0.436081, antiderivative size = 362, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {3896, 3784, 3875, 3833, 21, 3829, 3832, 4004} \[ -\frac{b^2 \tanh (c+d x)}{d \left (a^2-b^2\right ) \sqrt{a+b \text{sech}(c+d x)}}-\frac{\coth (c+d x)}{d \sqrt{a+b \text{sech}(c+d x)}}-\frac{\coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d \sqrt{a+b}}+\frac{\coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d \sqrt{a+b}}+\frac{2 \sqrt{a+b} \coth (c+d x) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (\text{sech}(c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Coth[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c +
d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(Sqrt[a + b]*d) - (Coth[c + d*x]*EllipticF[ArcSin[S
qrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + S
ech[c + d*x]))/(a - b))])/(Sqrt[a + b]*d) + (2*Sqrt[a + b]*Coth[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a +
b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c +
d*x]))/(a - b))])/(a*d) - Coth[c + d*x]/(d*Sqrt[a + b*Sech[c + d*x]]) - (b^2*Tanh[c + d*x])/((a^2 - b^2)*d*Sq
rt[a + b*Sech[c + d*x]])

Rule 3896

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

Rule 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3875

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)/cos[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[(Tan[e + f*x]*(a
+ b*Csc[e + f*x])^m)/f, x] + Dist[b*m, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e
, f, m}, x]

Rule 3833

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b*Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^
2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3829

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[a - b, Int[Csc[e + f
*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[b, Int[(Csc[e + f*x]*(1 + Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]],
x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

\begin{align*} \int \frac{\coth ^2(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}} \, dx &=-\int \left (-\frac{1}{\sqrt{a+b \text{sech}(c+d x)}}-\frac{\text{csch}^2(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}}\right ) \, dx\\ &=\int \frac{1}{\sqrt{a+b \text{sech}(c+d x)}} \, dx+\int \frac{\text{csch}^2(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}} \, dx\\ &=\frac{2 \sqrt{a+b} \coth (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{a d}-\frac{\coth (c+d x)}{d \sqrt{a+b \text{sech}(c+d x)}}+\frac{1}{2} b \int \frac{\text{sech}(c+d x)}{(a+b \text{sech}(c+d x))^{3/2}} \, dx\\ &=\frac{2 \sqrt{a+b} \coth (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{a d}-\frac{\coth (c+d x)}{d \sqrt{a+b \text{sech}(c+d x)}}-\frac{b^2 \tanh (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \text{sech}(c+d x)}}-\frac{b \int \frac{\text{sech}(c+d x) \left (-\frac{a}{2}-\frac{1}{2} b \text{sech}(c+d x)\right )}{\sqrt{a+b \text{sech}(c+d x)}} \, dx}{a^2-b^2}\\ &=\frac{2 \sqrt{a+b} \coth (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{a d}-\frac{\coth (c+d x)}{d \sqrt{a+b \text{sech}(c+d x)}}-\frac{b^2 \tanh (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \text{sech}(c+d x)}}+\frac{b \int \text{sech}(c+d x) \sqrt{a+b \text{sech}(c+d x)} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac{2 \sqrt{a+b} \coth (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{a d}-\frac{\coth (c+d x)}{d \sqrt{a+b \text{sech}(c+d x)}}-\frac{b^2 \tanh (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \text{sech}(c+d x)}}+\frac{b \int \frac{\text{sech}(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}} \, dx}{2 (a+b)}+\frac{b^2 \int \frac{\text{sech}(c+d x) (1+\text{sech}(c+d x))}{\sqrt{a+b \text{sech}(c+d x)}} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac{\coth (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{\sqrt{a+b} d}-\frac{\coth (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{\sqrt{a+b} d}+\frac{2 \sqrt{a+b} \coth (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\text{sech}(c+d x))}{a+b}} \sqrt{-\frac{b (1+\text{sech}(c+d x))}{a-b}}}{a d}-\frac{\coth (c+d x)}{d \sqrt{a+b \text{sech}(c+d x)}}-\frac{b^2 \tanh (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \text{sech}(c+d x)}}\\ \end{align*}

Mathematica [F]  time = 86.4917, size = 0, normalized size = 0. \[ \int \frac{\coth ^2(c+d x)}{\sqrt{a+b \text{sech}(c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.215, size = 0, normalized size = 0. \begin{align*} \int{ \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}{\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)^2/(a+b*sech(d*x+c))^(1/2),x)

[Out]

int(coth(d*x+c)^2/(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 )^{2}}{\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)^2/(a+b*sech(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(coth(d*x + c)^2/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)^2/(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 ^{2}{\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)**2/(a+b*sech(d*x+c))**(1/2),x)

[Out]

Integral(coth(c + d*x)**2/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 )^{2}}{\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)^2/(a+b*sech(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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