Integrand size = 12, antiderivative size = 95 \[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{a^{5/2}}-\frac {b \tanh (x)}{3 a (a+b) \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {b (5 a+3 b) \tanh (x)}{3 a^2 (a+b)^2 \sqrt {a+b-b \tanh ^2(x)}} \]
arctanh(a^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/a^(5/2)-1/3*b*(5*a+3*b)*t anh(x)/a^2/(a+b)^2/(a+b-b*tanh(x)^2)^(1/2)-1/3*b*tanh(x)/a/(a+b)/(a+b-b*ta nh(x)^2)^(3/2)
Time = 0.34 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.37 \[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\text {sech}^5(x) \left (\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right ) (a+2 b+a \cosh (2 x))^{5/2}}{a^{5/2}}-\frac {4 b (a+2 b+a \cosh (2 x)) \left (3 a^2+7 a b+3 b^2+a (3 a+2 b) \cosh (2 x)\right ) \sinh (x)}{3 a^2 (a+b)^2}\right )}{8 \left (a+b \text {sech}^2(x)\right )^{5/2}} \]
(Sech[x]^5*((Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sinh[x])/Sqrt[a + 2*b + a*Co sh[2*x]]]*(a + 2*b + a*Cosh[2*x])^(5/2))/a^(5/2) - (4*b*(a + 2*b + a*Cosh[ 2*x])*(3*a^2 + 7*a*b + 3*b^2 + a*(3*a + 2*b)*Cosh[2*x])*Sinh[x])/(3*a^2*(a + b)^2)))/(8*(a + b*Sech[x]^2)^(5/2))
Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.16, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4616, 316, 25, 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.
\(\displaystyle \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a+b \sec (i x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4616 |
\(\displaystyle \int \frac {1}{\left (1-\tanh ^2(x)\right ) \left (a-b \tanh ^2(x)+b\right )^{5/2}}d\tanh (x)\) |
\(\Big \downarrow \) 316 |
\(\displaystyle -\frac {\int -\frac {2 b \tanh ^2(x)+3 a+b}{\left (1-\tanh ^2(x)\right ) \left (-b \tanh ^2(x)+a+b\right )^{3/2}}d\tanh (x)}{3 a (a+b)}-\frac {b \tanh (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {2 b \tanh ^2(x)+3 a+b}{\left (1-\tanh ^2(x)\right ) \left (-b \tanh ^2(x)+a+b\right )^{3/2}}d\tanh (x)}{3 a (a+b)}-\frac {b \tanh (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {-\frac {\int -\frac {3 (a+b)^2}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a (a+b)}-\frac {b (5 a+3 b) \tanh (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}}{3 a (a+b)}-\frac {b \tanh (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3 (a+b) \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a}-\frac {b (5 a+3 b) \tanh (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}}{3 a (a+b)}-\frac {b \tanh (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {3 (a+b) \int \frac {1}{1-\frac {a \tanh ^2(x)}{-b \tanh ^2(x)+a+b}}d\frac {\tanh (x)}{\sqrt {-b \tanh ^2(x)+a+b}}}{a}-\frac {b (5 a+3 b) \tanh (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}}{3 a (a+b)}-\frac {b \tanh (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {3 (a+b) \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{a^{3/2}}-\frac {b (5 a+3 b) \tanh (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}}{3 a (a+b)}-\frac {b \tanh (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\) |
-1/3*(b*Tanh[x])/(a*(a + b)*(a + b - b*Tanh[x]^2)^(3/2)) + ((3*(a + b)*Arc Tanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]])/a^(3/2) - (b*(5*a + 3*b )*Tanh[x])/(a*(a + b)*Sqrt[a + b - b*Tanh[x]^2]))/(3*a*(a + b))
3.3.17.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 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[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b + b*ff^2*x^2)^p /(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && NeQ[a + b, 0] && NeQ[p, -1]
\[\int \frac {1}{\left (a +\operatorname {sech}\left (x \right )^{2} b \right )^{\frac {5}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 2870 vs. \(2 (81) = 162\).
Time = 0.52 (sec) , antiderivative size = 6299, normalized size of antiderivative = 66.31 \[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {sech}\left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (81) = 162\).
Time = 0.41 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.44 \[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=-\frac {2 \, {\left ({\left ({\left (\frac {{\left (3 \, a^{6} b^{3} + 2 \, a^{5} b^{4}\right )} e^{\left (2 \, x\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}} + \frac {3 \, {\left (a^{6} b^{3} + 4 \, a^{5} b^{4} + 2 \, a^{4} b^{5}\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )} e^{\left (2 \, x\right )} - \frac {3 \, {\left (a^{6} b^{3} + 4 \, a^{5} b^{4} + 2 \, a^{4} b^{5}\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )} e^{\left (2 \, x\right )} - \frac {3 \, a^{6} b^{3} + 2 \, a^{5} b^{4}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )}}{3 \, {\left (a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + 4 \, b e^{\left (2 \, x\right )} + a\right )}^{\frac {3}{2}}} \]
-2/3*((((3*a^6*b^3 + 2*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 + 4*a^5*b^4 + 2*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 + 4*a^5*b^4 + 2*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^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. \[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{5/2}} \,d x \]