\(\int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\) [143]

Optimal result

Integrand size = 23, antiderivative size = 88 \[ \int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (a^2-b^2\right )}{a b^2 d \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{b^2 d} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.32 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.75 \[ \int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\frac {2 \left (-b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \text {sech}(c+d x)}{a}\right )+a (2 a+b \text {sech}(c+d x))\right )}{a b^2 d \sqrt {a+b \text {sech}(c+d x)}} \] Input:

Integrate[Tanh[c + d*x]^3/(a + b*Sech[c + d*x])^(3/2),x]
 

Output:

(2*(-(b^2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Sech[c + d*x])/a]) + a*(2 
*a + b*Sech[c + d*x])))/(a*b^2*d*Sqrt[a + b*Sech[c + d*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.59 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 26, 4373, 517, 1584, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \cot \left (i c+i d x+\frac {\pi }{2}\right )^3}{\left (a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\cot \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^3}{\left (a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4373

\(\displaystyle -\frac {\int \frac {\cosh (c+d x) \left (b^2-b^2 \text {sech}^2(c+d x)\right )}{b (a+b \text {sech}(c+d x))^{3/2}}d(b \text {sech}(c+d x))}{b^2 d}\)

\(\Big \downarrow \) 517

\(\displaystyle -\frac {2 \int \frac {\cosh ^2(c+d x) \left (b^4 \text {sech}^4(c+d x)-2 a b^2 \text {sech}^2(c+d x)+a^2-b^2\right )}{b^2 \left (a-b^2 \text {sech}^2(c+d x)\right )}d\sqrt {a+b \text {sech}(c+d x)}}{b^2 d}\)

\(\Big \downarrow \) 1584

\(\displaystyle -\frac {2 \int \left (-\frac {b^2}{a \left (a-b^2 \text {sech}^2(c+d x)\right )}-1+\frac {\left (a^2-b^2\right ) \cosh ^2(c+d x)}{a b^2}\right )d\sqrt {a+b \text {sech}(c+d x)}}{b^2 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\left (a^2-b^2\right ) \cosh (c+d x)}{a b}-\sqrt {a+b \text {sech}(c+d x)}\right )}{b^2 d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 517

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1))   Subst[Int[x^(2*n + 1)*(-c + x^ 
2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr 
eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
 

rule 1584

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4373

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_), x_Symbol] :> Simp[-(-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]
 

Maple [F]

\[\int \frac {\tanh \left (d x +c \right )^{3}}{\left (a +b \,\operatorname {sech}\left (d x +c \right )\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (78) = 156\).

Time = 0.49 (sec) , antiderivative size = 1107, normalized size of antiderivative = 12.58 \[ \int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/2*((a*b^2*cosh(d*x + c)^2 + a*b^2*sinh(d*x + c)^2 + 2*b^3*cosh(d*x + c) 
 + a*b^2 + 2*(a*b^2*cosh(d*x + c) + b^3)*sinh(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*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*cosh(d*x + c) + b)*sinh(d*x + c)^3 + 2*a*cos 
h(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)/c 
osh(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^2*b*cosh(d*x + c) + 2*a^3 
 - a*b^2 + (2*a^3 - a*b^2)*cosh(d*x + c)^2 + (2*a^3 - a*b^2)*sinh(d*x + c) 
^2 + 2*(a^2*b + (2*a^3 - a*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a*cosh 
(d*x + c) + b)/cosh(d*x + c)))/(a^3*b^2*d*cosh(d*x + c)^2 + a^3*b^2*d*sinh 
(d*x + c)^2 + 2*a^2*b^3*d*cosh(d*x + c) + a^3*b^2*d + 2*(a^3*b^2*d*cosh(d* 
x + c) + a^2*b^3*d)*sinh(d*x + c)), -((a*b^2*cosh(d*x + c)^2 + a*b^2*sinh( 
d*x + c)^2 + 2*b^3*cosh(d*x + c) + a*b^2 + 2*(a*b^2*cosh(d*x + c) + b^3)*s 
inh(d*x + c))*sqrt(-a)*arctan((a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + ...
 

Sympy [F]

\[ \int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int \frac {\tanh ^{3}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int { \frac {\tanh \left (d x + c\right )^{3}}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int { \frac {\tanh \left (d x + c\right )^{3}}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tanh}\left (c+d\,x\right )}^3}{{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\tanh ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (d x +c \right ) b +a}\, \tanh \left (d x +c \right )^{3}}{\mathrm {sech}\left (d x +c \right )^{2} b^{2}+2 \,\mathrm {sech}\left (d x +c \right ) a b +a^{2}}d x \] Input:

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

Output:

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