\(\int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\) [134]

Optimal result

Integrand size = 23, antiderivative size = 79 \[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {2 a \sqrt {a+b \text {sech}(c+d x)}}{b^2 d}+\frac {2 (a+b \text {sech}(c+d x))^{3/2}}{3 b^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3970, 912, 1167, 213} \[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {2 (a+b \text {sech}(c+d x))^{3/2}}{3 b^2 d}-\frac {2 a \sqrt {a+b \text {sech}(c+d x)}}{b^2 d} \]

[In]

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

Rule 213

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

Rule 912

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 1167

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}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 3970

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

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \left (\frac {3 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {(-2 a+b \text {sech}(c+d x)) \sqrt {a+b \text {sech}(c+d x)}}{b^2}\right )}{3 d} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\tanh \left (d x +c \right )^{3}}{\sqrt {a +b \,\operatorname {sech}\left (d x +c \right )}}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (67) = 134\).

Time = 0.68 (sec) , antiderivative size = 925, normalized size of antiderivative = 11.71 \[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/6*(3*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 + b^2)*sqrt(a)*log(-(2*
a^2*cosh(d*x + c)^4 + 2*a^2*sinh(d*x + c)^4 + 4*a*b*cosh(d*x + c)^3 + 4*(2*a^2*cosh(d*x + c) + a*b)*sinh(d*x +
c)^3 + 4*a*b*cosh(d*x + c) + (4*a^2 + b^2)*cosh(d*x + c)^2 + (12*a^2*cosh(d*x + c)^2 + 12*a*b*cosh(d*x + c) +
4*a^2 + b^2)*sinh(d*x + c)^2 + 2*a^2 + 2*(a*cosh(d*x + c)^4 + a*sinh(d*x + c)^4 + b*cosh(d*x + c)^3 + (4*a*co
sh(d*x + c) + b)*sinh(d*x + c)^3 + 2*a*cosh(d*x + c)^2 + (6*a*cosh(d*x + c)^2 + 3*b*cosh(d*x + c) + 2*a)*sinh(
d*x + c)^2 + b*cosh(d*x + c) + (4*a*cosh(d*x + c)^3 + 3*b*cosh(d*x + c)^2 + 4*a*cosh(d*x + c) + b)*sinh(d*x +
c) + a)*sqrt(a)*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)) + 2*(4*a^2*cosh(d*x + c)^3 + 6*a*b*cosh(d*x + c)^2 +
2*a*b + (4*a^2 + b^2)*cosh(d*x + c))*sinh(d*x + c))/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d
*x + c)^2)) - 8*(a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 - a*b*cosh(d*x + c) + a^2 + (2*a^2*cosh(d*x + c) -
a*b)*sinh(d*x + c))*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)))/(a*b^2*d*cosh(d*x + c)^2 + 2*a*b^2*d*cosh(d*x +
c)*sinh(d*x + c) + a*b^2*d*sinh(d*x + c)^2 + a*b^2*d), -1/3*(3*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sin
h(d*x + c) + b^2*sinh(d*x + c)^2 + b^2)*sqrt(-a)*arctan((a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + b*cosh(d*x +
c) + (2*a*cosh(d*x + c) + b)*sinh(d*x + c) + a)*sqrt(-a)*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c))/(a^2*cosh(d
*x + c)^2 + a^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + a^2 + 2*(a^2*cosh(d*x + c) + a*b)*sinh(d*x + c))) + 4*
(a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 - a*b*cosh(d*x + c) + a^2 + (2*a^2*cosh(d*x + c) - a*b)*sinh(d*x +
c))*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)))/(a*b^2*d*cosh(d*x + c)^2 + 2*a*b^2*d*cosh(d*x + c)*sinh(d*x + c
) + a*b^2*d*sinh(d*x + c)^2 + a*b^2*d)]

Sympy [F]

\[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {\tanh ^{3}{\left (c + d x \right )}}{\sqrt {a + b \operatorname {sech}{\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\tanh \left (d x + c\right )^{3}}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\tanh \left (d x + c\right )^{3}}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {{\mathrm {tanh}\left (c+d\,x\right )}^3}{\sqrt {a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}}} \,d x \]

[In]

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

[Out]

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