\(\int \sqrt {a+b \text {sech}(c+d x)} \tanh ^3(c+d x) \, dx\) [126]

Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^3(c+d x) \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d}-\frac {2 a (a+b \text {sech}(c+d x))^{3/2}}{3 b^2 d}+\frac {2 (a+b \text {sech}(c+d x))^{5/2}}{5 b^2 d} \]

[Out]

-2/3*a*(a+b*sech(d*x+c))^(3/2)/b^2/d+2/5*(a+b*sech(d*x+c))^(5/2)/b^2/d+2*arctanh((a+b*sech(d*x+c))^(1/2)/a^(1/
2))*a^(1/2)/d-2*(a+b*sech(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 100, 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, 1275, 213} \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^3(c+d x) \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 (a+b \text {sech}(c+d x))^{5/2}}{5 b^2 d}-\frac {2 a (a+b \text {sech}(c+d x))^{3/2}}{3 b^2 d}-\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d} \]

[In]

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

[Out]

(2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/d - (2*Sqrt[a + b*Sech[c + d*x]])/d - (2*a*(a + b*Sech[
c + d*x])^(3/2))/(3*b^2*d) + (2*(a + b*Sech[c + d*x])^(5/2))/(5*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 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 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 {\sqrt {a+x} \left (b^2-x^2\right )}{x} \, dx,x,b \text {sech}(c+d x)\right )}{b^2 d} \\ & = -\frac {2 \text {Subst}\left (\int \frac {x^2 \left (-a^2+b^2+2 a x^2-x^4\right )}{-a+x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{b^2 d} \\ & = -\frac {2 \text {Subst}\left (\int \left (b^2+a x^2-x^4+\frac {a b^2}{-a+x^2}\right ) \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{b^2 d} \\ & = -\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d}-\frac {2 a (a+b \text {sech}(c+d x))^{3/2}}{3 b^2 d}+\frac {2 (a+b \text {sech}(c+d x))^{5/2}}{5 b^2 d}-\frac {(2 a) \text {Subst}\left (\int \frac {1}{-a+x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d} \\ & = \frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d}-\frac {2 a (a+b \text {sech}(c+d x))^{3/2}}{3 b^2 d}+\frac {2 (a+b \text {sech}(c+d x))^{5/2}}{5 b^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^3(c+d x) \, dx=\frac {2 \left (15 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )+\frac {\sqrt {a+b \text {sech}(c+d x)} \left (-2 a^2-15 b^2+a b \text {sech}(c+d x)+3 b^2 \text {sech}^2(c+d x)\right )}{b^2}\right )}{15 d} \]

[In]

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

[Out]

(2*(15*Sqrt[a]*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]] + (Sqrt[a + b*Sech[c + d*x]]*(-2*a^2 - 15*b^2 + a*b*
Sech[c + d*x] + 3*b^2*Sech[c + d*x]^2))/b^2))/(15*d)

Maple [F]

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 665 vs. \(2 (84) = 168\).

Time = 0.69 (sec) , antiderivative size = 1589, normalized size of antiderivative = 15.89 \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^3(c+d x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/30*(15*(b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 + 2*b^2*cosh(d*x +
c)^2 + 2*(3*b^2*cosh(d*x + c)^2 + b^2)*sinh(d*x + c)^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 + b^2*cosh(d*x + c))*sin
h(d*x + c))*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*cos
h(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*cosh(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*co
sh(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)) + 4*(2*a*b*cosh(d*x + c)^3 - (2*a^2 + 15*b^2)*cosh(d*x + c)^4 - (2*a^2
+ 15*b^2)*sinh(d*x + c)^4 + 2*(a*b - 2*(2*a^2 + 15*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 2*a*b*cosh(d*x + c)
- 2*(2*a^2 + 9*b^2)*cosh(d*x + c)^2 + 2*(3*a*b*cosh(d*x + c) - 3*(2*a^2 + 15*b^2)*cosh(d*x + c)^2 - 2*a^2 - 9*
b^2)*sinh(d*x + c)^2 - 2*a^2 - 15*b^2 + 2*(3*a*b*cosh(d*x + c)^2 - 2*(2*a^2 + 15*b^2)*cosh(d*x + c)^3 + a*b -
2*(2*a^2 + 9*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)))/(b^2*d*cosh(d*x + c
)^4 + 4*b^2*d*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*d*sinh(d*x + c)^4 + 2*b^2*d*cosh(d*x + c)^2 + b^2*d + 2*(3*b
^2*d*cosh(d*x + c)^2 + b^2*d)*sinh(d*x + c)^2 + 4*(b^2*d*cosh(d*x + c)^3 + b^2*d*cosh(d*x + c))*sinh(d*x + c))
, -1/15*(15*(b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 + 2*b^2*cosh(d*x
+ c)^2 + 2*(3*b^2*cosh(d*x + c)^2 + b^2)*sinh(d*x + c)^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 + b^2*cosh(d*x + c))*s
inh(d*x + c))*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))) - 2*(2*a*b*cosh(d*x + c)^3 - (
2*a^2 + 15*b^2)*cosh(d*x + c)^4 - (2*a^2 + 15*b^2)*sinh(d*x + c)^4 + 2*(a*b - 2*(2*a^2 + 15*b^2)*cosh(d*x + c)
)*sinh(d*x + c)^3 + 2*a*b*cosh(d*x + c) - 2*(2*a^2 + 9*b^2)*cosh(d*x + c)^2 + 2*(3*a*b*cosh(d*x + c) - 3*(2*a^
2 + 15*b^2)*cosh(d*x + c)^2 - 2*a^2 - 9*b^2)*sinh(d*x + c)^2 - 2*a^2 - 15*b^2 + 2*(3*a*b*cosh(d*x + c)^2 - 2*(
2*a^2 + 15*b^2)*cosh(d*x + c)^3 + a*b - 2*(2*a^2 + 9*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a*cosh(d*x + c)
+ b)/cosh(d*x + c)))/(b^2*d*cosh(d*x + c)^4 + 4*b^2*d*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*d*sinh(d*x + c)^4 +
2*b^2*d*cosh(d*x + c)^2 + b^2*d + 2*(3*b^2*d*cosh(d*x + c)^2 + b^2*d)*sinh(d*x + c)^2 + 4*(b^2*d*cosh(d*x + c)
^3 + b^2*d*cosh(d*x + c))*sinh(d*x + c))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)