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

Optimal. Leaf size=217 \[ -\frac{\coth ^2(c+d x) \sqrt{a+b \text{sech}(c+d x)}}{2 d}+\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a}}\right )}{d}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{4 d \sqrt{a-b}}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{d \sqrt{a-b}}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{d \sqrt{a+b}}-\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{4 d \sqrt{a+b}} \]

[Out]

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

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Rubi [A]  time = 0.327092, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {3885, 898, 1315, 1178, 12, 1093, 206, 1170, 207} \[ -\frac{\coth ^2(c+d x) \sqrt{a+b \text{sech}(c+d x)}}{2 d}+\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a}}\right )}{d}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{4 d \sqrt{a-b}}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{d \sqrt{a-b}}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{d \sqrt{a+b}}-\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{4 d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1315

Int[(((f_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[
f^2/(c*d^2 - b*d*e + a*e^2), Int[(f*x)^(m - 2)*(a*e + c*d*x^2)*(a + b*x^2 + c*x^4)^p, x], x] - Dist[(d*e*f^2)/
(c*d^2 - b*d*e + a*e^2), Int[((f*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p + 1))/(d + e*x^2), x], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 0]

Rule 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e
^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1093

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/
2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && PosQ[b^2 - 4*a*c]

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 1170

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

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 \coth ^3(c+d x) \sqrt{a+b \text{sech}(c+d x)} \, dx &=-\frac{b^4 \operatorname{Subst}\left (\int \frac{\sqrt{a+x}}{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{x^2}{\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^2\right ) \operatorname{Subst}\left (\int \frac{-a^2+b^2+a x^2}{\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 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\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{b^2 \sqrt{a+b \text{sech}(c+d x)}}{2 d \left (a^2-b^2-2 a (a+b \text{sech}(c+d x))+(a+b \text{sech}(c+d x))^2\right )}-\frac{\left (2 a b^2\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{b^2 \left (a-x^2\right )}+\frac{1}{2 b^2 \left (a+b-x^2\right )}-\frac{1}{2 b^2 \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{6 b^2 \left (a^2-b^2\right )}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt{a+b \text{sech}(c+d x)}\right )}{4 \left (a^2-b^2\right ) d}\\ &=\frac{b^2 \sqrt{a+b \text{sech}(c+d x)}}{2 d \left (a^2-b^2-2 a (a+b \text{sech}(c+d x))+(a+b \text{sech}(c+d x))^2\right )}-\frac{a \operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\sqrt{a+b \text{sech}(c+d x)}\right )}{d}+\frac{a \operatorname{Subst}\left (\int \frac{1}{-a+b+x^2} \, dx,x,\sqrt{a+b \text{sech}(c+d x)}\right )}{d}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\sqrt{a+b \text{sech}(c+d x)}\right )}{d}-\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt{a+b \text{sech}(c+d x)}\right )}{2 d}\\ &=\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a}}\right )}{d}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{\sqrt{a-b} d}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{\sqrt{a+b} d}+\frac{b^2 \sqrt{a+b \text{sech}(c+d x)}}{2 d \left (a^2-b^2-2 a (a+b \text{sech}(c+d x))+(a+b \text{sech}(c+d x))^2\right )}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{a-b-x^2} \, dx,x,\sqrt{a+b \text{sech}(c+d x)}\right )}{4 d}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\sqrt{a+b \text{sech}(c+d x)}\right )}{4 d}\\ &=\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a}}\right )}{d}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{\sqrt{a-b} d}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a-b}}\right )}{4 \sqrt{a-b} d}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{\sqrt{a+b} d}-\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b \text{sech}(c+d x)}}{\sqrt{a+b}}\right )}{4 \sqrt{a+b} d}+\frac{b^2 \sqrt{a+b \text{sech}(c+d x)}}{2 d \left (a^2-b^2-2 a (a+b \text{sech}(c+d x))+(a+b \text{sech}(c+d x))^2\right )}\\ \end{align*}

Mathematica [B]  time = 20.391, size = 518, normalized size = 2.39 \[ \frac{\sqrt{a+b \text{sech}(c+d x)} \left (\frac{8 \sqrt{-a \cosh (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a \cosh (c+d x)+b}}{\sqrt{-a \cosh (c+d x)}}\right )}{\sqrt{a \cosh (c+d x)+b}}-\frac{2 \sqrt{a} \sqrt{-a \cosh (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{a \cosh (c+d x)+b}}{\sqrt{a-b} \sqrt{-a \cosh (c+d x)}}\right )}{\sqrt{a-b} \sqrt{a \cosh (c+d x)+b}}-\frac{2 \sqrt{a} \sqrt{-a \cosh (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{a \cosh (c+d x)+b}}{\sqrt{a+b} \sqrt{-a \cosh (c+d x)}}\right )}{\sqrt{a+b} \sqrt{a \cosh (c+d x)+b}}+\frac{3 b \sqrt{a \cosh (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{a \cosh (c+d x)+b}}{\sqrt{-a-b} \sqrt{a \cosh (c+d x)}}\right )}{\sqrt{a} \sqrt{-a-b} \sqrt{a \cosh (c+d x)+b}}-\frac{(2 a-3 b) \sqrt{a \cosh (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{a \cosh (c+d x)+b}}{\sqrt{a-b} \sqrt{a \cosh (c+d x)}}\right )}{\sqrt{a} \sqrt{a-b} \sqrt{a \cosh (c+d x)+b}}-\frac{2 \sqrt{a} \sqrt{a \cosh (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{a \cosh (c+d x)+b}}{\sqrt{a+b} \sqrt{a \cosh (c+d x)}}\right )}{\sqrt{a+b} \sqrt{a \cosh (c+d x)+b}}-2 \coth ^2(c+d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((8*ArcTan[Sqrt[b + a*Cosh[c + d*x]]/Sqrt[-(a*Cosh[c + d*x])]]*Sqrt[-(a*Cosh[c + d*x])])/Sqrt[b + a*Cosh[c +
d*x]] - (2*Sqrt[a]*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 - b]*Sqrt[b + a*Cosh[c + d*x]]) - (2*Sqrt[a]*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 + b]*Sqrt[b + a*Cosh[c + d*x]])
+ (3*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]*Sqrt[-a - b]*Sqrt[b + a*Cosh[c + d*x]]) - ((2*a - 3*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]])/(Sqrt[a]*Sqrt[a - b]*Sqrt[b + a*Cosh[c + d*x]]) - (
2*Sqrt[a]*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
]])/(Sqrt[a + b]*Sqrt[b + a*Cosh[c + d*x]]) - 2*Coth[c + d*x]^2)*Sqrt[a + b*Sech[c + d*x]])/(4*d)

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Maple [F]  time = 0.201, size = 0, normalized size = 0. \begin{align*} \int \left ({\rm coth} \left (dx+c\right ) \right ) ^{3}\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 \sqrt{b \operatorname{sech}\left (d x + c\right ) + a} \coth \left (d x + c\right )^{3}\,{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(sqrt(b*sech(d*x + c) + a)*coth(d*x + c)^3, 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(-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)

[Out]

Timed out

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