Integrand size = 17, antiderivative size = 90 \[ \int \frac {\tanh ^4(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{a^{5/2}}-\frac {(a+b) \tanh (x)}{3 a b \left (a+b-b \tanh ^2(x)\right )^{3/2}}+\frac {(a-3 b) \tanh (x)}{3 a^2 b \sqrt {a+b-b \tanh ^2(x)}} \]
arctanh(a^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/a^(5/2)+1/3*(a-3*b)*tanh( x)/a^2/b/(a+b-b*tanh(x)^2)^(1/2)-1/3*(a+b)*tanh(x)/a/b/(a+b-b*tanh(x)^2)^( 3/2)
Leaf count is larger than twice the leaf count of optimal. \(290\) vs. \(2(90)=180\).
Time = 1.46 (sec) , antiderivative size = 290, normalized size of antiderivative = 3.22 \[ \int \frac {\tanh ^4(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\text {sech}^4(x) \left (\frac {\sqrt {2} (a+2 b+a \cosh (2 x))^{5/2} \text {csch}(x) \text {sech}(x) \left (\frac {\sinh ^2(x)}{a+b}+\frac {12 \sinh ^4(x)}{a+b}+\frac {2 \sinh ^2(x) \left (a+b+a \sinh ^2(x)\right )}{(a+b)^2}-\frac {16 \left (a+b+a \sinh ^2(x)\right ) \left (1+\frac {a \sinh ^2(x)}{a+b}\right ) \left (\frac {a^2 (a+b) \sinh ^4(x)}{\left (a+b+a \sinh ^2(x)\right )^2}+\frac {3 a (a+b) \sinh ^2(x)}{a+b+a \sinh ^2(x)}-\frac {3 \sqrt {a} \sqrt {a+b} \text {arcsinh}\left (\frac {\sqrt {a} \sinh (x)}{\sqrt {a+b}}\right ) \sinh (x)}{\sqrt {\frac {a+b+a \sinh ^2(x)}{a+b}}}\right )}{a^3}\right )}{\left (a+b+a \sinh ^2(x)\right )^{3/2}}+\frac {8 (a+2 b+a \cosh (2 x)) (2 a+3 b+a \cosh (2 x)) \tanh (x)}{(a+b)^2}-\frac {12 (a+2 b+a \cosh (2 x)) (b+(3 a+2 b) \cosh (2 x)) \tanh (x)}{(a+b)^2}\right )}{384 \left (a+b \text {sech}^2(x)\right )^{5/2}} \]
(Sech[x]^4*((Sqrt[2]*(a + 2*b + a*Cosh[2*x])^(5/2)*Csch[x]*Sech[x]*(Sinh[x ]^2/(a + b) + (12*Sinh[x]^4)/(a + b) + (2*Sinh[x]^2*(a + b + a*Sinh[x]^2)) /(a + b)^2 - (16*(a + b + a*Sinh[x]^2)*(1 + (a*Sinh[x]^2)/(a + b))*((a^2*( a + b)*Sinh[x]^4)/(a + b + a*Sinh[x]^2)^2 + (3*a*(a + b)*Sinh[x]^2)/(a + b + a*Sinh[x]^2) - (3*Sqrt[a]*Sqrt[a + b]*ArcSinh[(Sqrt[a]*Sinh[x])/Sqrt[a + b]]*Sinh[x])/Sqrt[(a + b + a*Sinh[x]^2)/(a + b)]))/a^3))/(a + b + a*Sinh [x]^2)^(3/2) + (8*(a + 2*b + a*Cosh[2*x])*(2*a + 3*b + a*Cosh[2*x])*Tanh[x ])/(a + b)^2 - (12*(a + 2*b + a*Cosh[2*x])*(b + (3*a + 2*b)*Cosh[2*x])*Tan h[x])/(a + b)^2))/(384*(a + b*Sech[x]^2)^(5/2))
Time = 0.40 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 4629, 2075, 372, 402, 27, 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.
\(\displaystyle \int \frac {\tanh ^4(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (i x)^4}{\left (a+b \sec (i x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \int \frac {\tanh ^4(x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \left (1-\tanh ^2(x)\right )\right )^{5/2}}d\tanh (x)\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \int \frac {\tanh ^4(x)}{\left (1-\tanh ^2(x)\right ) \left (a-b \tanh ^2(x)+b\right )^{5/2}}d\tanh (x)\) |
\(\Big \downarrow \) 372 |
\(\displaystyle \frac {\int \frac {-\left ((a-2 b) \tanh ^2(x)\right )+a+b}{\left (1-\tanh ^2(x)\right ) \left (-b \tanh ^2(x)+a+b\right )^{3/2}}d\tanh (x)}{3 a b}-\frac {(a+b) \tanh (x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {(a-3 b) \tanh (x)}{a \sqrt {a-b \tanh ^2(x)+b}}-\frac {\int -\frac {3 b (a+b)}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a (a+b)}}{3 a b}-\frac {(a+b) \tanh (x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3 b \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a}+\frac {(a-3 b) \tanh (x)}{a \sqrt {a-b \tanh ^2(x)+b}}}{3 a b}-\frac {(a+b) \tanh (x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {3 b \int \frac {1}{1-\frac {a \tanh ^2(x)}{-b \tanh ^2(x)+a+b}}d\frac {\tanh (x)}{\sqrt {-b \tanh ^2(x)+a+b}}}{a}+\frac {(a-3 b) \tanh (x)}{a \sqrt {a-b \tanh ^2(x)+b}}}{3 a b}-\frac {(a+b) \tanh (x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {3 b \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{a^{3/2}}+\frac {(a-3 b) \tanh (x)}{a \sqrt {a-b \tanh ^2(x)+b}}}{3 a b}-\frac {(a+b) \tanh (x)}{3 a b \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
-1/3*((a + b)*Tanh[x])/(a*b*(a + b - b*Tanh[x]^2)^(3/2)) + ((3*b*ArcTanh[( Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]])/a^(3/2) + ((a - 3*b)*Tanh[x]) /(a*Sqrt[a + b - b*Tanh[x]^2]))/(3*a*b)
3.3.13.3.1 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]))* 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]*((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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -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])
\[\int \frac {\tanh \left (x \right )^{4}}{\left (a +\operatorname {sech}\left (x \right )^{2} b \right )^{\frac {5}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 1500 vs. \(2 (76) = 152\).
Time = 0.47 (sec) , antiderivative size = 3559, normalized size of antiderivative = 39.54 \[ \int \frac {\tanh ^4(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]
[1/12*(3*(a^2*cosh(x)^8 + 8*a^2*cosh(x)*sinh(x)^7 + a^2*sinh(x)^8 + 4*(a^2 + 2*a*b)*cosh(x)^6 + 4*(7*a^2*cosh(x)^2 + a^2 + 2*a*b)*sinh(x)^6 + 8*(7*a ^2*cosh(x)^3 + 3*(a^2 + 2*a*b)*cosh(x))*sinh(x)^5 + 2*(3*a^2 + 8*a*b + 8*b ^2)*cosh(x)^4 + 2*(35*a^2*cosh(x)^4 + 30*(a^2 + 2*a*b)*cosh(x)^2 + 3*a^2 + 8*a*b + 8*b^2)*sinh(x)^4 + 8*(7*a^2*cosh(x)^5 + 10*(a^2 + 2*a*b)*cosh(x)^ 3 + (3*a^2 + 8*a*b + 8*b^2)*cosh(x))*sinh(x)^3 + 4*(a^2 + 2*a*b)*cosh(x)^2 + 4*(7*a^2*cosh(x)^6 + 15*(a^2 + 2*a*b)*cosh(x)^4 + 3*(3*a^2 + 8*a*b + 8* b^2)*cosh(x)^2 + a^2 + 2*a*b)*sinh(x)^2 + a^2 + 8*(a^2*cosh(x)^7 + 3*(a^2 + 2*a*b)*cosh(x)^5 + (3*a^2 + 8*a*b + 8*b^2)*cosh(x)^3 + (a^2 + 2*a*b)*cos h(x))*sinh(x))*sqrt(a)*log((a*b^2*cosh(x)^8 + 8*a*b^2*cosh(x)*sinh(x)^7 + a*b^2*sinh(x)^8 - 2*(a*b^2 - b^3)*cosh(x)^6 + 2*(14*a*b^2*cosh(x)^2 - a*b^ 2 + b^3)*sinh(x)^6 + 4*(14*a*b^2*cosh(x)^3 - 3*(a*b^2 - b^3)*cosh(x))*sinh (x)^5 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x)^4 + (70*a*b^2*cosh(x)^4 + a^3 + 4*a^2*b + 9*a*b^2 - 30*(a*b^2 - b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*a*b^2*co sh(x)^5 - 10*(a*b^2 - b^3)*cosh(x)^3 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x))* sinh(x)^3 + a^3 + 2*(a^3 + 3*a^2*b)*cosh(x)^2 + 2*(14*a*b^2*cosh(x)^6 - 15 *(a*b^2 - b^3)*cosh(x)^4 + a^3 + 3*a^2*b + 3*(a^3 + 4*a^2*b + 9*a*b^2)*cos h(x)^2)*sinh(x)^2 + sqrt(2)*(b^2*cosh(x)^6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2 *sinh(x)^6 - 3*b^2*cosh(x)^4 + 3*(5*b^2*cosh(x)^2 - b^2)*sinh(x)^4 + 4*(5* b^2*cosh(x)^3 - 3*b^2*cosh(x))*sinh(x)^3 - (a^2 + 4*a*b)*cosh(x)^2 + (1...
\[ \int \frac {\tanh ^4(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {\tanh ^{4}{\left (x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\tanh ^4(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )^{4}}{{\left (b \operatorname {sech}\left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (76) = 152\).
Time = 0.41 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.69 \[ \int \frac {\tanh ^4(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=-\frac {4 \, {\left ({\left ({\left (\frac {{\left (a^{7} b^{2} + 2 \, a^{6} b^{3} + a^{5} b^{4}\right )} e^{\left (2 \, x\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}} + \frac {3 \, {\left (a^{6} b^{3} + 2 \, a^{5} b^{4} + a^{4} b^{5}\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )} e^{\left (2 \, x\right )} - \frac {3 \, {\left (a^{6} b^{3} + 2 \, a^{5} b^{4} + a^{4} b^{5}\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )} e^{\left (2 \, x\right )} - \frac {a^{7} b^{2} + 2 \, a^{6} b^{3} + a^{5} b^{4}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )}}{3 \, {\left (a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + 4 \, b e^{\left (2 \, x\right )} + a\right )}^{\frac {3}{2}}} \]
-4/3*((((a^7*b^2 + 2*a^6*b^3 + a^5*b^4)*e^(2*x)/(a^8*b^2 + 2*a^7*b^3 + a^6 *b^4) + 3*(a^6*b^3 + 2*a^5*b^4 + a^4*b^5)/(a^8*b^2 + 2*a^7*b^3 + a^6*b^4)) *e^(2*x) - 3*(a^6*b^3 + 2*a^5*b^4 + a^4*b^5)/(a^8*b^2 + 2*a^7*b^3 + a^6*b^ 4))*e^(2*x) - (a^7*b^2 + 2*a^6*b^3 + a^5*b^4)/(a^8*b^2 + 2*a^7*b^3 + a^6*b ^4))/(a*e^(4*x) + 2*a*e^(2*x) + 4*b*e^(2*x) + a)^(3/2)
Timed out. \[ \int \frac {\tanh ^4(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^4}{{\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{5/2}} \,d x \]