Integrand size = 17, antiderivative size = 118 \[ \int \frac {\tanh ^6(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{b^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{a^{5/2}}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {\left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}} \]
-arctan(b^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/b^(5/2)+arctanh(a^(1/2)*t anh(x)/(a+b-b*tanh(x)^2)^(1/2))/a^(5/2)-(1/a^2-1/b^2)*tanh(x)/(a+b-b*tanh( x)^2)^(1/2)-1/3*(a+b)*tanh(x)^3/a/b/(a+b-b*tanh(x)^2)^(3/2)
Time = 0.58 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.51 \[ \int \frac {\tanh ^6(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\text {sech}^5(x) \left (\frac {\sqrt {2} \left (-a^{5/2} \arctan \left (\frac {\sqrt {2} \sqrt {b} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right )+b^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right )\right ) (a+2 b+a \cosh (2 x))^{5/2}}{a^{5/2} b^{5/2}}+\frac {2 (a+b) (a+2 b+a \cosh (2 x)) \left (3 a^2+4 a b-6 b^2+a (3 a-4 b) \cosh (2 x)\right ) \sinh (x)}{3 a^2 b^2}\right )}{8 \left (a+b \text {sech}^2(x)\right )^{5/2}} \]
(Sech[x]^5*((Sqrt[2]*(-(a^(5/2)*ArcTan[(Sqrt[2]*Sqrt[b]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]]) + b^(5/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]])*(a + 2*b + a*Cosh[2*x])^(5/2))/(a^(5/2)*b^(5/2)) + (2 *(a + b)*(a + 2*b + a*Cosh[2*x])*(3*a^2 + 4*a*b - 6*b^2 + a*(3*a - 4*b)*Co sh[2*x])*Sinh[x])/(3*a^2*b^2)))/(8*(a + b*Sech[x]^2)^(5/2))
Time = 0.49 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.24, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.765, Rules used = {3042, 25, 4629, 25, 2075, 372, 27, 440, 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.
\(\displaystyle \int \frac {\tanh ^6(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\tan (i x)^6}{\left (a+b \sec (i x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\tan (i x)^6}{\left (b \sec (i x)^2+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle -\int -\frac {\tanh ^6(x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \left (1-\tanh ^2(x)\right )\right )^{5/2}}d\tanh (x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\tanh ^6(x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \left (1-\tanh ^2(x)\right )\right )^{5/2}}d\tanh (x)\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \int \frac {\tanh ^6(x)}{\left (1-\tanh ^2(x)\right ) \left (a-b \tanh ^2(x)+b\right )^{5/2}}d\tanh (x)\) |
\(\Big \downarrow \) 372 |
\(\displaystyle \frac {\int \frac {3 \tanh ^2(x) \left (-a \tanh ^2(x)+a+b\right )}{\left (1-\tanh ^2(x)\right ) \left (-b \tanh ^2(x)+a+b\right )^{3/2}}d\tanh (x)}{3 a b}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\tanh ^2(x) \left (-a \tanh ^2(x)+a+b\right )}{\left (1-\tanh ^2(x)\right ) \left (-b \tanh ^2(x)+a+b\right )^{3/2}}d\tanh (x)}{a b}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 440 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \tanh (x)}{a b \sqrt {a-b \tanh ^2(x)+b}}-\frac {\int \frac {-\tanh ^2(x) a^2+a^2-b^2}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a b}}{a b}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 398 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \tanh (x)}{a b \sqrt {a-b \tanh ^2(x)+b}}-\frac {a^2 \int \frac {1}{\sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)-b^2 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a b}}{a b}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \tanh (x)}{a b \sqrt {a-b \tanh ^2(x)+b}}-\frac {a^2 \int \frac {1}{\frac {b \tanh ^2(x)}{-b \tanh ^2(x)+a+b}+1}d\frac {\tanh (x)}{\sqrt {-b \tanh ^2(x)+a+b}}-b^2 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a b}}{a b}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \tanh (x)}{a b \sqrt {a-b \tanh ^2(x)+b}}-\frac {\frac {a^2 \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{\sqrt {b}}-b^2 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a b}}{a b}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \tanh (x)}{a b \sqrt {a-b \tanh ^2(x)+b}}-\frac {\frac {a^2 \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{\sqrt {b}}-b^2 \int \frac {1}{1-\frac {a \tanh ^2(x)}{-b \tanh ^2(x)+a+b}}d\frac {\tanh (x)}{\sqrt {-b \tanh ^2(x)+a+b}}}{a b}}{a b}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \tanh (x)}{a b \sqrt {a-b \tanh ^2(x)+b}}-\frac {\frac {a^2 \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{\sqrt {b}}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{\sqrt {a}}}{a b}}{a b}-\frac {(a+b) \tanh ^3(x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
-1/3*((a + b)*Tanh[x]^3)/(a*b*(a + b - b*Tanh[x]^2)^(3/2)) + (-(((a^2*ArcT an[(Sqrt[b]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]])/Sqrt[b] - (b^2*ArcTanh[(S qrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]])/Sqrt[a])/(a*b)) + ((a^2 - b^2) *Tanh[x])/(a*b*Sqrt[a + b - b*Tanh[x]^2]))/(a*b)
3.3.11.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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])
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]
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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
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]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ g^2/(2*b*(b*c - a*d)*(p + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c *f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && LtQ[p, -1] && GtQ[m, 1]
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]
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])
\[\int \frac {\tanh \left (x \right )^{6}}{\left (a +\operatorname {sech}\left (x \right )^{2} b \right )^{\frac {5}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 2666 vs. \(2 (100) = 200\).
Time = 0.71 (sec) , antiderivative size = 11939, normalized size of antiderivative = 101.18 \[ \int \frac {\tanh ^6(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tanh ^6(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {\tanh ^{6}{\left (x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\tanh ^6(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )^{6}}{{\left (b \operatorname {sech}\left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int \frac {\tanh ^6(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {\tanh ^6(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^6}{{\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{5/2}} \,d x \]