Integrand size = 12, antiderivative size = 57 \[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{a^{3/2}}-\frac {b \tanh (x)}{a (a+b) \sqrt {a+b-b \tanh ^2(x)}} \] Output:
arctanh(a^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/a^(3/2)-b*tanh(x)/a/(a+b) /(a+b-b*tanh(x)^2)^(1/2)
Time = 0.56 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.88 \[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\frac {(a+2 b+a \cosh (2 x)) \text {sech}^3(x) \left ((a+b)^{3/2} \text {arcsinh}\left (\frac {\sqrt {a} \sinh (x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+2 b+a \cosh (2 x)}{a+b}}-\sqrt {2} \sqrt {a} b \sinh (x)\right )}{2 \sqrt {2} a^{3/2} (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}} \] Input:
Integrate[(a + b*Sech[x]^2)^(-3/2),x]
Output:
((a + 2*b + a*Cosh[2*x])*Sech[x]^3*((a + b)^(3/2)*ArcSinh[(Sqrt[a]*Sinh[x] )/Sqrt[a + b]]*Sqrt[(a + 2*b + a*Cosh[2*x])/(a + b)] - Sqrt[2]*Sqrt[a]*b*S inh[x]))/(2*Sqrt[2]*a^(3/2)*(a + b)*(a + b*Sech[x]^2)^(3/2))
Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4616, 296, 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.
\(\displaystyle \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a+b \sec (i x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4616 |
\(\displaystyle \int \frac {1}{\left (1-\tanh ^2(x)\right ) \left (a-b \tanh ^2(x)+b\right )^{3/2}}d\tanh (x)\) |
\(\Big \downarrow \) 296 |
\(\displaystyle \frac {\int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a}-\frac {b \tanh (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\int \frac {1}{1-\frac {a \tanh ^2(x)}{-b \tanh ^2(x)+a+b}}d\frac {\tanh (x)}{\sqrt {-b \tanh ^2(x)+a+b}}}{a}-\frac {b \tanh (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{a^{3/2}}-\frac {b \tanh (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}\) |
Input:
Int[(a + b*Sech[x]^2)^(-3/2),x]
Output:
ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]]/a^(3/2) - (b*Tanh[x]) /(a*(a + b)*Sqrt[a + b - b*Tanh[x]^2])
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[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[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]
\[\int \frac {1}{\left (a +\operatorname {sech}\left (x \right )^{2} b \right )^{\frac {3}{2}}}d x\]
Input:
int(1/(a+sech(x)^2*b)^(3/2),x)
Output:
int(1/(a+sech(x)^2*b)^(3/2),x)
Leaf count of result is larger than twice the leaf count of optimal. 735 vs. \(2 (49) = 98\).
Time = 0.16 (sec) , antiderivative size = 2029, normalized size of antiderivative = 35.60 \[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b*sech(x)^2)^(3/2),x, algorithm="fricas")
Output:
[1/4*(((a^2 + a*b)*cosh(x)^4 + 4*(a^2 + a*b)*cosh(x)*sinh(x)^3 + (a^2 + a* b)*sinh(x)^4 + 2*(a^2 + 3*a*b + 2*b^2)*cosh(x)^2 + 2*(3*(a^2 + a*b)*cosh(x )^2 + a^2 + 3*a*b + 2*b^2)*sinh(x)^2 + a^2 + a*b + 4*((a^2 + a*b)*cosh(x)^ 3 + (a^2 + 3*a*b + 2*b^2)*cosh(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*cosh(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)*cosh(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 + (15*b^2*cosh(x)^4 - 18*b^2*cosh(x)^2 - a^2 - 4*a*b )*sinh(x)^2 - a^2 + 2*(3*b^2*cosh(x)^5 - 6*b^2*cosh(x)^3 - (a^2 + 4*a*b)*c osh(x))*sinh(x))*sqrt(a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh( x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(2*a*b^2*cosh(x)^7 - 3*(a*b^2 - b^3)*cosh(x)^5 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x)^3 + (a^3 + 3*a^2*b)*co sh(x))*sinh(x))/(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)...
\[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(a+b*sech(x)**2)**(3/2),x)
Output:
Integral((a + b*sech(x)**2)**(-3/2), x)
\[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {sech}\left (x\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a+b*sech(x)^2)^(3/2),x, algorithm="maxima")
Output:
integrate((b*sech(x)^2 + a)^(-3/2), x)
Time = 0.17 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=-\frac {\frac {a b^{2} e^{\left (2 \, x\right )}}{a^{3} b + a^{2} b^{2}} - \frac {a b^{2}}{a^{3} b + a^{2} b^{2}}}{\sqrt {a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + 4 \, b e^{\left (2 \, x\right )} + a}} \] Input:
integrate(1/(a+b*sech(x)^2)^(3/2),x, algorithm="giac")
Output:
-(a*b^2*e^(2*x)/(a^3*b + a^2*b^2) - a*b^2/(a^3*b + a^2*b^2))/sqrt(a*e^(4*x ) + 2*a*e^(2*x) + 4*b*e^(2*x) + a)
Timed out. \[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{3/2}} \,d x \] Input:
int(1/(a + b/cosh(x)^2)^(3/2),x)
Output:
int(1/(a + b/cosh(x)^2)^(3/2), x)
\[ \int \frac {1}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (x \right )^{2} b +a}}{\mathrm {sech}\left (x \right )^{4} b^{2}+2 \mathrm {sech}\left (x \right )^{2} a b +a^{2}}d x \] Input:
int(1/(a+b*sech(x)^2)^(3/2),x)
Output:
int(sqrt(sech(x)**2*b + a)/(sech(x)**4*b**2 + 2*sech(x)**2*a*b + a**2),x)