\(\int \sqrt {a+b \text {sech}^2(x)} \tanh ^4(x) \, dx\) [177]

Optimal result

Integrand size = 17, antiderivative size = 125 \[ \int \sqrt {a+b \text {sech}^2(x)} \tanh ^4(x) \, dx=-\frac {\left (a^2+6 a b-3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{8 b^{3/2}}+\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )+\frac {(a-3 b) \tanh (x) \sqrt {a+b-b \tanh ^2(x)}}{8 b}-\frac {1}{4} \tanh ^3(x) \sqrt {a+b-b \tanh ^2(x)} \] Output:

-1/8*(a^2+6*a*b-3*b^2)*arctan(b^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/b^( 
3/2)+a^(1/2)*arctanh(a^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))+1/8*(a-3*b)* 
tanh(x)*(a+b-b*tanh(x)^2)^(1/2)/b-1/4*tanh(x)^3*(a+b-b*tanh(x)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.54 \[ \int \sqrt {a+b \text {sech}^2(x)} \tanh ^4(x) \, dx=-\frac {\cosh (x) \sqrt {a+b \text {sech}^2(x)} \left (\sqrt {2} \left (a^2+6 a b-3 b^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {b} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right )-8 \sqrt {2} \sqrt {a} b^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right )-(a-5 b) \sqrt {b} \sqrt {a+2 b+a \cosh (2 x)} \text {sech}(x) \tanh (x)-2 b^{3/2} \sqrt {a+2 b+a \cosh (2 x)} \text {sech}^3(x) \tanh (x)\right )}{8 b^{3/2} \sqrt {a+2 b+a \cosh (2 x)}} \] Input:

Integrate[Sqrt[a + b*Sech[x]^2]*Tanh[x]^4,x]
 

Output:

-1/8*(Cosh[x]*Sqrt[a + b*Sech[x]^2]*(Sqrt[2]*(a^2 + 6*a*b - 3*b^2)*ArcTan[ 
(Sqrt[2]*Sqrt[b]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]] - 8*Sqrt[2]*Sqrt[a] 
*b^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]] - 
(a - 5*b)*Sqrt[b]*Sqrt[a + 2*b + a*Cosh[2*x]]*Sech[x]*Tanh[x] - 2*b^(3/2)* 
Sqrt[a + 2*b + a*Cosh[2*x]]*Sech[x]^3*Tanh[x]))/(b^(3/2)*Sqrt[a + 2*b + a* 
Cosh[2*x]])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.588, Rules used = {3042, 4629, 2075, 380, 444, 398, 224, 216, 291, 219}

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.

\(\displaystyle \int \tanh ^4(x) \sqrt {a+b \text {sech}^2(x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (i x)^4 \sqrt {a+b \sec (i x)^2}dx\)

\(\Big \downarrow \) 4629

\(\displaystyle \int \frac {\tanh ^4(x) \sqrt {a+b \left (1-\tanh ^2(x)\right )}}{1-\tanh ^2(x)}d\tanh (x)\)

\(\Big \downarrow \) 2075

\(\displaystyle \int \frac {\tanh ^4(x) \sqrt {a-b \tanh ^2(x)+b}}{1-\tanh ^2(x)}d\tanh (x)\)

\(\Big \downarrow \) 380

\(\displaystyle \frac {1}{4} \int \frac {\tanh ^2(x) \left ((a-3 b) \tanh ^2(x)+3 (a+b)\right )}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)-\frac {1}{4} \tanh ^3(x) \sqrt {a-b \tanh ^2(x)+b}\)

\(\Big \downarrow \) 444

\(\displaystyle \frac {1}{4} \left (\frac {(a-3 b) \tanh (x) \sqrt {a-b \tanh ^2(x)+b}}{2 b}-\frac {\int \frac {(a-3 b) (a+b)-\left (a^2+6 b a-3 b^2\right ) \tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{2 b}\right )-\frac {1}{4} \tanh ^3(x) \sqrt {a-b \tanh ^2(x)+b}\)

\(\Big \downarrow \) 398

\(\displaystyle \frac {1}{4} \left (\frac {(a-3 b) \tanh (x) \sqrt {a-b \tanh ^2(x)+b}}{2 b}-\frac {\left (a^2+6 a b-3 b^2\right ) \int \frac {1}{\sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)-8 a b \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{2 b}\right )-\frac {1}{4} \tanh ^3(x) \sqrt {a-b \tanh ^2(x)+b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {1}{4} \left (\frac {(a-3 b) \tanh (x) \sqrt {a-b \tanh ^2(x)+b}}{2 b}-\frac {\left (a^2+6 a b-3 b^2\right ) \int \frac {1}{\frac {b \tanh ^2(x)}{-b \tanh ^2(x)+a+b}+1}d\frac {\tanh (x)}{\sqrt {-b \tanh ^2(x)+a+b}}-8 a b \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{2 b}\right )-\frac {1}{4} \tanh ^3(x) \sqrt {a-b \tanh ^2(x)+b}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{4} \left (\frac {(a-3 b) \tanh (x) \sqrt {a-b \tanh ^2(x)+b}}{2 b}-\frac {\frac {\left (a^2+6 a b-3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{\sqrt {b}}-8 a b \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{2 b}\right )-\frac {1}{4} \tanh ^3(x) \sqrt {a-b \tanh ^2(x)+b}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {1}{4} \left (\frac {(a-3 b) \tanh (x) \sqrt {a-b \tanh ^2(x)+b}}{2 b}-\frac {\frac {\left (a^2+6 a b-3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{\sqrt {b}}-8 a b \int \frac {1}{1-\frac {a \tanh ^2(x)}{-b \tanh ^2(x)+a+b}}d\frac {\tanh (x)}{\sqrt {-b \tanh ^2(x)+a+b}}}{2 b}\right )-\frac {1}{4} \tanh ^3(x) \sqrt {a-b \tanh ^2(x)+b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} \left (\frac {(a-3 b) \tanh (x) \sqrt {a-b \tanh ^2(x)+b}}{2 b}-\frac {\frac {\left (a^2+6 a b-3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{\sqrt {b}}-8 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{2 b}\right )-\frac {1}{4} \tanh ^3(x) \sqrt {a-b \tanh ^2(x)+b}\)

Input:

Int[Sqrt[a + b*Sech[x]^2]*Tanh[x]^4,x]
 

Output:

-1/4*(Tanh[x]^3*Sqrt[a + b - b*Tanh[x]^2]) + (-1/2*(((a^2 + 6*a*b - 3*b^2) 
*ArcTan[(Sqrt[b]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]])/Sqrt[b] - 8*Sqrt[a]* 
b*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]])/b + ((a - 3*b)*Tan 
h[x]*Sqrt[a + b - b*Tanh[x]^2])/(2*b))/4
 

Defintions of rubi rules used

rule 216

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

rule 219

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

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 291

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

rule 380

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

rule 398

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

rule 444

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

rule 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

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

rule 4629

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [F]

\[\int \sqrt {a +\operatorname {sech}\left (x \right )^{2} b}\, \tanh \left (x \right )^{4}d x\]

Input:

int((a+sech(x)^2*b)^(1/2)*tanh(x)^4,x)
 

Output:

int((a+sech(x)^2*b)^(1/2)*tanh(x)^4,x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1861 vs. \(2 (103) = 206\).

Time = 0.38 (sec) , antiderivative size = 8720, normalized size of antiderivative = 69.76 \[ \int \sqrt {a+b \text {sech}^2(x)} \tanh ^4(x) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sech(x)^2)^(1/2)*tanh(x)^4,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \sqrt {a+b \text {sech}^2(x)} \tanh ^4(x) \, dx=\int \sqrt {a + b \operatorname {sech}^{2}{\left (x \right )}} \tanh ^{4}{\left (x \right )}\, dx \] Input:

integrate((a+b*sech(x)**2)**(1/2)*tanh(x)**4,x)
 

Output:

Integral(sqrt(a + b*sech(x)**2)*tanh(x)**4, x)
 

Maxima [F]

\[ \int \sqrt {a+b \text {sech}^2(x)} \tanh ^4(x) \, dx=\int { \sqrt {b \operatorname {sech}\left (x\right )^{2} + a} \tanh \left (x\right )^{4} \,d x } \] Input:

integrate((a+b*sech(x)^2)^(1/2)*tanh(x)^4,x, algorithm="maxima")
 

Output:

integrate(sqrt(b*sech(x)^2 + a)*tanh(x)^4, x)
 

Giac [F(-2)]

Exception generated. \[ \int \sqrt {a+b \text {sech}^2(x)} \tanh ^4(x) \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*sech(x)^2)^(1/2)*tanh(x)^4,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b \text {sech}^2(x)} \tanh ^4(x) \, dx=\int {\mathrm {tanh}\left (x\right )}^4\,\sqrt {a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}} \,d x \] Input:

int(tanh(x)^4*(a + b/cosh(x)^2)^(1/2),x)
 

Output:

int(tanh(x)^4*(a + b/cosh(x)^2)^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {a+b \text {sech}^2(x)} \tanh ^4(x) \, dx=\int \sqrt {\mathrm {sech}\left (x \right )^{2} b +a}\, \tanh \left (x \right )^{4}d x \] Input:

int((a+b*sech(x)^2)^(1/2)*tanh(x)^4,x)
 

Output:

int(sqrt(sech(x)**2*b + a)*tanh(x)**4,x)