Integrand size = 12, antiderivative size = 88 \[ \int \left (a+b \text {sech}^2(x)\right )^{3/2} \, dx=\frac {1}{2} \sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )+a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )+\frac {1}{2} b \tanh (x) \sqrt {a+b-b \tanh ^2(x)} \] Output:
1/2*b^(1/2)*(3*a+b)*arctan(b^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))+a^(3/2 )*arctanh(a^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))+1/2*b*tanh(x)*(a+b-b*ta nh(x)^2)^(1/2)
Time = 0.18 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.73 \[ \int \left (a+b \text {sech}^2(x)\right )^{3/2} \, dx=\frac {\left (b+a \cosh ^2(x)\right ) \text {sech}(x) \sqrt {a+b \text {sech}^2(x)} \left (\sqrt {2} \sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {2} \sqrt {b} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right ) \cosh ^2(x)+2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right ) \cosh ^2(x)+b \sqrt {a+2 b+a \cosh (2 x)} \sinh (x)\right )}{(a+2 b+a \cosh (2 x))^{3/2}} \] Input:
Integrate[(a + b*Sech[x]^2)^(3/2),x]
Output:
((b + a*Cosh[x]^2)*Sech[x]*Sqrt[a + b*Sech[x]^2]*(Sqrt[2]*Sqrt[b]*(3*a + b )*ArcTan[(Sqrt[2]*Sqrt[b]*Sinh[x])/Sqrt[a + 2*b + a*Cosh[2*x]]]*Cosh[x]^2 + 2*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sinh[x])/Sqrt[a + 2*b + a*Cos h[2*x]]]*Cosh[x]^2 + b*Sqrt[a + 2*b + a*Cosh[2*x]]*Sinh[x]))/(a + 2*b + a* Cosh[2*x])^(3/2)
Time = 0.33 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 4616, 318, 25, 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.
\(\displaystyle \int \left (a+b \text {sech}^2(x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \sec (i x)^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 4616 |
\(\displaystyle \int \frac {\left (a-b \tanh ^2(x)+b\right )^{3/2}}{1-\tanh ^2(x)}d\tanh (x)\) |
\(\Big \downarrow \) 318 |
\(\displaystyle \frac {1}{2} b \tanh (x) \sqrt {a-b \tanh ^2(x)+b}-\frac {1}{2} \int -\frac {(a+b) (2 a+b)-b (3 a+b) \tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \int \frac {(a+b) (2 a+b)-b (3 a+b) \tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)+\frac {1}{2} b \tanh (x) \sqrt {a-b \tanh ^2(x)+b}\) |
\(\Big \downarrow \) 398 |
\(\displaystyle \frac {1}{2} \left (2 a^2 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)+b (3 a+b) \int \frac {1}{\sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)\right )+\frac {1}{2} b \tanh (x) \sqrt {a-b \tanh ^2(x)+b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} \left (2 a^2 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)+b (3 a+b) \int \frac {1}{\frac {b \tanh ^2(x)}{-b \tanh ^2(x)+a+b}+1}d\frac {\tanh (x)}{\sqrt {-b \tanh ^2(x)+a+b}}\right )+\frac {1}{2} b \tanh (x) \sqrt {a-b \tanh ^2(x)+b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (2 a^2 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)+\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )\right )+\frac {1}{2} b \tanh (x) \sqrt {a-b \tanh ^2(x)+b}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {1}{2} \left (2 a^2 \int \frac {1}{1-\frac {a \tanh ^2(x)}{-b \tanh ^2(x)+a+b}}d\frac {\tanh (x)}{\sqrt {-b \tanh ^2(x)+a+b}}+\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )\right )+\frac {1}{2} b \tanh (x) \sqrt {a-b \tanh ^2(x)+b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )+\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )\right )+\frac {1}{2} b \tanh (x) \sqrt {a-b \tanh ^2(x)+b}\) |
Input:
Int[(a + b*Sech[x]^2)^(3/2),x]
Output:
(Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[b]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]] + 2 *a^(3/2)*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]])/2 + (b*Tanh [x]*Sqrt[a + b - b*Tanh[x]^2])/2
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 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[((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]
\[\int \left (a +\operatorname {sech}\left (x \right )^{2} b \right )^{\frac {3}{2}}d x\]
Input:
int((a+sech(x)^2*b)^(3/2),x)
Output:
int((a+sech(x)^2*b)^(3/2),x)
Leaf count of result is larger than twice the leaf count of optimal. 716 vs. \(2 (70) = 140\).
Time = 0.23 (sec) , antiderivative size = 4140, normalized size of antiderivative = 47.05 \[ \int \left (a+b \text {sech}^2(x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:
integrate((a+b*sech(x)^2)^(3/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \left (a+b \text {sech}^2(x)\right )^{3/2} \, dx=\int \left (a + b \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((a+b*sech(x)**2)**(3/2),x)
Output:
Integral((a + b*sech(x)**2)**(3/2), x)
\[ \int \left (a+b \text {sech}^2(x)\right )^{3/2} \, dx=\int { {\left (b \operatorname {sech}\left (x\right )^{2} + a\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((a+b*sech(x)^2)^(3/2),x, algorithm="maxima")
Output:
integrate((b*sech(x)^2 + a)^(3/2), x)
Exception generated. \[ \int \left (a+b \text {sech}^2(x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*sech(x)^2)^(3/2),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
Timed out. \[ \int \left (a+b \text {sech}^2(x)\right )^{3/2} \, dx=\int {\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{3/2} \,d x \] Input:
int((a + b/cosh(x)^2)^(3/2),x)
Output:
int((a + b/cosh(x)^2)^(3/2), x)
\[ \int \left (a+b \text {sech}^2(x)\right )^{3/2} \, dx=\left (\int \sqrt {\mathrm {sech}\left (x \right )^{2} b +a}d x \right ) a +\left (\int \sqrt {\mathrm {sech}\left (x \right )^{2} b +a}\, \mathrm {sech}\left (x \right )^{2}d x \right ) b \] Input:
int((a+b*sech(x)^2)^(3/2),x)
Output:
int(sqrt(sech(x)**2*b + a),x)*a + int(sqrt(sech(x)**2*b + a)*sech(x)**2,x) *b