3.2.41 \(\int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\) [141]

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

[Out]

coth(d*x+c)*EllipticE((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sech(d*x+c))/(a+b))^(1/2)
*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/d/(a+b)^(1/2)-coth(d*x+c)*EllipticF((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2),((a+
b)/(a-b))^(1/2))*(b*(1-sech(d*x+c))/(a+b))^(1/2)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/d/(a+b)^(1/2)+2*coth(d*x+c)*
EllipticPi((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sech(d*x+c))/(a+
b))^(1/2)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/a/d-coth(d*x+c)/d/(a+b*sech(d*x+c))^(1/2)-b^2*tanh(d*x+c)/(a^2-b^2)
/d/(a+b*sech(d*x+c))^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 362, normalized size of antiderivative = 1.00, 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 = {3981, 3869, 3960, 3918, 21, 3914, 3917, 4089} \begin {gather*} -\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) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} F\left (\text {ArcSin}\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 (\text {ArcSin}\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};\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d}-\frac {\coth (c+d x)}{d \sqrt {a+b \text {sech}(c+d x)}} \end {gather*}

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 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 3869

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

Rule 3914

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 3917

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

Rule 3918

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 3960

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 3981

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 4089

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 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], 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*}

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Mathematica [F]
time = 90.93, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx \end {gather*}

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 = 2.86, size = 0, normalized size = 0.00 \[\int \frac {\coth ^{2}\left (d x +c \right )}{\sqrt {a +b \,\mathrm {sech}\left (d x +c \right )}}\, dx\]

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 = 0.00, 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(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]
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(coth(d*x+c)^2/(a+b*sech(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\coth ^{2}{\left (c + d x \right )}}{\sqrt {a + b \operatorname {sech}{\left (c + d x \right )}}}\, dx \end {gather*}

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 = 0.00, 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(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {coth}\left (c+d\,x\right )}^2}{\sqrt {a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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