Integrand size = 15, antiderivative size = 90 \[ \int \frac {\text {sech}^5(x)}{a+b \cosh ^2(x)} \, dx=\frac {\left (3 a^2-4 a b+8 b^2\right ) \arctan (\sinh (x))}{8 a^3}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \sinh (x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b}}+\frac {(3 a-4 b) \text {sech}(x) \tanh (x)}{8 a^2}+\frac {\text {sech}^3(x) \tanh (x)}{4 a} \] Output:
1/8*(3*a^2-4*a*b+8*b^2)*arctan(sinh(x))/a^3-b^(5/2)*arctan(b^(1/2)*sinh(x) /(a+b)^(1/2))/a^3/(a+b)^(1/2)+1/8*(3*a-4*b)*sech(x)*tanh(x)/a^2+1/4*sech(x )^3*tanh(x)/a
Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96 \[ \int \frac {\text {sech}^5(x)}{a+b \cosh ^2(x)} \, dx=\frac {\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {a+b} \text {csch}(x)}{\sqrt {b}}\right )}{\sqrt {a+b}}+2 \left (3 a^2-4 a b+8 b^2\right ) \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+a (3 a-4 b) \text {sech}(x) \tanh (x)+2 a^2 \text {sech}^3(x) \tanh (x)}{8 a^3} \] Input:
Integrate[Sech[x]^5/(a + b*Cosh[x]^2),x]
Output:
((8*b^(5/2)*ArcTan[(Sqrt[a + b]*Csch[x])/Sqrt[b]])/Sqrt[a + b] + 2*(3*a^2 - 4*a*b + 8*b^2)*ArcTan[Tanh[x/2]] + a*(3*a - 4*b)*Sech[x]*Tanh[x] + 2*a^2 *Sech[x]^3*Tanh[x])/(8*a^3)
Time = 0.35 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.26, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3665, 316, 25, 402, 25, 397, 216, 218}
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.
| \(\displaystyle \int \frac {\text {sech}^5(x)}{a+b \cosh ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (\frac {\pi }{2}+i x\right )^5 \left (a+b \sin \left (\frac {\pi }{2}+i x\right )^2\right )}dx\) |
| \(\Big \downarrow \) 3665 |
| \(\displaystyle \int \frac {1}{\left (\sinh ^2(x)+1\right )^3 \left (a+b \sinh ^2(x)+b\right )}d\sinh (x)\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\sinh (x)}{4 a \left (\sinh ^2(x)+1\right )^2}-\frac {\int -\frac {3 b \sinh ^2(x)+3 a-b}{\left (\sinh ^2(x)+1\right )^2 \left (b \sinh ^2(x)+a+b\right )}d\sinh (x)}{4 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {3 b \sinh ^2(x)+3 a-b}{\left (\sinh ^2(x)+1\right )^2 \left (b \sinh ^2(x)+a+b\right )}d\sinh (x)}{4 a}+\frac {\sinh (x)}{4 a \left (\sinh ^2(x)+1\right )^2}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {(3 a-4 b) \sinh (x)}{2 a \left (\sinh ^2(x)+1\right )}-\frac {\int -\frac {3 a^2-b a+4 b^2+(3 a-4 b) b \sinh ^2(x)}{\left (\sinh ^2(x)+1\right ) \left (b \sinh ^2(x)+a+b\right )}d\sinh (x)}{2 a}}{4 a}+\frac {\sinh (x)}{4 a \left (\sinh ^2(x)+1\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {3 a^2-b a+4 b^2+(3 a-4 b) b \sinh ^2(x)}{\left (\sinh ^2(x)+1\right ) \left (b \sinh ^2(x)+a+b\right )}d\sinh (x)}{2 a}+\frac {(3 a-4 b) \sinh (x)}{2 a \left (\sinh ^2(x)+1\right )}}{4 a}+\frac {\sinh (x)}{4 a \left (\sinh ^2(x)+1\right )^2}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\left (3 a^2-4 a b+8 b^2\right ) \int \frac {1}{\sinh ^2(x)+1}d\sinh (x)}{a}-\frac {8 b^3 \int \frac {1}{b \sinh ^2(x)+a+b}d\sinh (x)}{a}}{2 a}+\frac {(3 a-4 b) \sinh (x)}{2 a \left (\sinh ^2(x)+1\right )}}{4 a}+\frac {\sinh (x)}{4 a \left (\sinh ^2(x)+1\right )^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\left (3 a^2-4 a b+8 b^2\right ) \arctan (\sinh (x))}{a}-\frac {8 b^3 \int \frac {1}{b \sinh ^2(x)+a+b}d\sinh (x)}{a}}{2 a}+\frac {(3 a-4 b) \sinh (x)}{2 a \left (\sinh ^2(x)+1\right )}}{4 a}+\frac {\sinh (x)}{4 a \left (\sinh ^2(x)+1\right )^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\left (3 a^2-4 a b+8 b^2\right ) \arctan (\sinh (x))}{a}-\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {b} \sinh (x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{2 a}+\frac {(3 a-4 b) \sinh (x)}{2 a \left (\sinh ^2(x)+1\right )}}{4 a}+\frac {\sinh (x)}{4 a \left (\sinh ^2(x)+1\right )^2}\) |
Input:
Int[Sech[x]^5/(a + b*Cosh[x]^2),x]
Output:
Sinh[x]/(4*a*(1 + Sinh[x]^2)^2) + ((((3*a^2 - 4*a*b + 8*b^2)*ArcTan[Sinh[x ]])/a - (8*b^(5/2)*ArcTan[(Sqrt[b]*Sinh[x])/Sqrt[a + b]])/(a*Sqrt[a + b])) /(2*a) + ((3*a - 4*b)*Sinh[x])/(2*a*(1 + Sinh[x]^2)))/(4*a)
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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, 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[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(182\) vs. \(2(76)=152\).
Time = 3.27 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.03
| method | result | size |
| default | \(\frac {\frac {2 \left (\left (-\frac {5}{8} a^{2}+\frac {1}{2} a b \right ) \tanh \left (\frac {x}{2}\right )^{7}+\left (\frac {3}{8} a^{2}+\frac {1}{2} a b \right ) \tanh \left (\frac {x}{2}\right )^{5}+\left (-\frac {3}{8} a^{2}-\frac {1}{2} a b \right ) \tanh \left (\frac {x}{2}\right )^{3}+\left (\frac {5}{8} a^{2}-\frac {1}{2} a b \right ) \tanh \left (\frac {x}{2}\right )\right )}{\left (\tanh \left (\frac {x}{2}\right )^{2}+1\right )^{4}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{4}}{a^{3}}-\frac {2 b^{3} \left (\frac {\arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {x}{2}\right )+2 \sqrt {a}}{2 \sqrt {b}}\right )}{2 \sqrt {a +b}\, \sqrt {b}}+\frac {\arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {x}{2}\right )-2 \sqrt {a}}{2 \sqrt {b}}\right )}{2 \sqrt {a +b}\, \sqrt {b}}\right )}{a^{3}}\) | \(183\) |
| risch | \(\frac {{\mathrm e}^{x} \left (3 \,{\mathrm e}^{6 x} a -4 \,{\mathrm e}^{6 x} b +11 \,{\mathrm e}^{4 x} a -4 \,{\mathrm e}^{4 x} b -11 \,{\mathrm e}^{2 x} a +4 \,{\mathrm e}^{2 x} b -3 a +4 b \right )}{4 \left ({\mathrm e}^{2 x}+1\right )^{4} a^{2}}+\frac {3 i \ln \left ({\mathrm e}^{x}+i\right )}{8 a}-\frac {i b \ln \left ({\mathrm e}^{x}+i\right )}{2 a^{2}}+\frac {i \ln \left ({\mathrm e}^{x}+i\right ) b^{2}}{a^{3}}-\frac {3 i \ln \left ({\mathrm e}^{x}-i\right )}{8 a}+\frac {i b \ln \left ({\mathrm e}^{x}-i\right )}{2 a^{2}}-\frac {i \ln \left ({\mathrm e}^{x}-i\right ) b^{2}}{a^{3}}+\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 x}-\frac {2 \sqrt {-\left (a +b \right ) b}\, {\mathrm e}^{x}}{b}-1\right )}{2 \left (a +b \right ) a^{3}}-\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 x}+\frac {2 \sqrt {-\left (a +b \right ) b}\, {\mathrm e}^{x}}{b}-1\right )}{2 \left (a +b \right ) a^{3}}\) | \(232\) |
Input:
int(sech(x)^5/(a+b*cosh(x)^2),x,method=_RETURNVERBOSE)
Output:
2/a^3*(((-5/8*a^2+1/2*a*b)*tanh(1/2*x)^7+(3/8*a^2+1/2*a*b)*tanh(1/2*x)^5+( -3/8*a^2-1/2*a*b)*tanh(1/2*x)^3+(5/8*a^2-1/2*a*b)*tanh(1/2*x))/(tanh(1/2*x )^2+1)^4+1/8*(3*a^2-4*a*b+8*b^2)*arctan(tanh(1/2*x)))-2*b^3/a^3*(1/2/(a+b) ^(1/2)/b^(1/2)*arctan(1/2*(2*(a+b)^(1/2)*tanh(1/2*x)+2*a^(1/2))/b^(1/2))+1 /2/(a+b)^(1/2)/b^(1/2)*arctan(1/2*(2*(a+b)^(1/2)*tanh(1/2*x)-2*a^(1/2))/b^ (1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 1673 vs. \(2 (76) = 152\).
Time = 0.14 (sec) , antiderivative size = 3239, normalized size of antiderivative = 35.99 \[ \int \frac {\text {sech}^5(x)}{a+b \cosh ^2(x)} \, dx=\text {Too large to display} \] Input:
integrate(sech(x)^5/(a+b*cosh(x)^2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\text {sech}^5(x)}{a+b \cosh ^2(x)} \, dx=\int \frac {\operatorname {sech}^{5}{\left (x \right )}}{a + b \cosh ^{2}{\left (x \right )}}\, dx \] Input:
integrate(sech(x)**5/(a+b*cosh(x)**2),x)
Output:
Integral(sech(x)**5/(a + b*cosh(x)**2), x)
\[ \int \frac {\text {sech}^5(x)}{a+b \cosh ^2(x)} \, dx=\int { \frac {\operatorname {sech}\left (x\right )^{5}}{b \cosh \left (x\right )^{2} + a} \,d x } \] Input:
integrate(sech(x)^5/(a+b*cosh(x)^2),x, algorithm="maxima")
Output:
1/4*((3*a - 4*b)*e^(7*x) + (11*a - 4*b)*e^(5*x) - (11*a - 4*b)*e^(3*x) - ( 3*a - 4*b)*e^x)/(a^2*e^(8*x) + 4*a^2*e^(6*x) + 6*a^2*e^(4*x) + 4*a^2*e^(2* x) + a^2) + 1/4*(3*a^2 - 4*a*b + 8*b^2)*arctan(e^x)/a^3 - 32*integrate(1/1 6*(b^3*e^(3*x) + b^3*e^x)/(a^3*b*e^(4*x) + a^3*b + 2*(2*a^4 + a^3*b)*e^(2* x)), x)
Exception generated. \[ \int \frac {\text {sech}^5(x)}{a+b \cosh ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sech(x)^5/(a+b*cosh(x)^2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Time = 45.50 (sec) , antiderivative size = 1305, normalized size of antiderivative = 14.50 \[ \int \frac {\text {sech}^5(x)}{a+b \cosh ^2(x)} \, dx=\text {Too large to display} \] Input:
int(1/(cosh(x)^5*(a + b*cosh(x)^2)),x)
Output:
(atan((exp(x)*(243*a^12*(a^6)^(1/2) + 5024*b^6*(a^6)^(3/2) + 18432*b^12*(a ^6)^(1/2) + 6912*a^2*b^10*(a^6)^(1/2) + 30720*a^3*b^9*(a^6)^(1/2) - 26880* a^4*b^8*(a^6)^(1/2) + 24192*a^5*b^7*(a^6)^(1/2) - 13408*a^7*b^5*(a^6)^(1/2 ) + 17160*a^8*b^4*(a^6)^(1/2) - 9540*a^9*b^3*(a^6)^(1/2) + 4563*a^10*b^2*( a^6)^(1/2) - 9216*a*b^11*(a^6)^(1/2) - 1134*a^11*b*(a^6)^(1/2)))/(81*a^13* (9*a^4 - 24*a^3*b - 64*a*b^3 + 64*b^4 + 64*a^2*b^2)^(1/2) - 270*a^12*b*(9* a^4 - 24*a^3*b - 64*a*b^3 + 64*b^4 + 64*a^2*b^2)^(1/2) + 2304*a^3*b^10*(9* a^4 - 24*a^3*b - 64*a*b^3 + 64*b^4 + 64*a^2*b^2)^(1/2) + 3840*a^6*b^7*(9*a ^4 - 24*a^3*b - 64*a*b^3 + 64*b^4 + 64*a^2*b^2)^(1/2) - 1440*a^7*b^6*(9*a^ 4 - 24*a^3*b - 64*a*b^3 + 64*b^4 + 64*a^2*b^2)^(1/2) + 864*a^8*b^5*(9*a^4 - 24*a^3*b - 64*a*b^3 + 64*b^4 + 64*a^2*b^2)^(1/2) + 1600*a^9*b^4*(9*a^4 - 24*a^3*b - 64*a*b^3 + 64*b^4 + 64*a^2*b^2)^(1/2) - 1200*a^10*b^3*(9*a^4 - 24*a^3*b - 64*a*b^3 + 64*b^4 + 64*a^2*b^2)^(1/2) + 945*a^11*b^2*(9*a^4 - 24*a^3*b - 64*a*b^3 + 64*b^4 + 64*a^2*b^2)^(1/2)))*(9*a^4 - 24*a^3*b - 64* a*b^3 + 64*b^4 + 64*a^2*b^2)^(1/2))/(4*(a^6)^(1/2)) - (6*exp(x))/(a*(3*exp (2*x) + 3*exp(4*x) + exp(6*x) + 1)) + ((b^5)^(1/2)*(2*atan((exp(x)*((2*(48 *b^8*(a^6*b + a^7)^(1/2) + 40*a^3*b^5*(a^6*b + a^7)^(1/2) - 15*a^4*b^4*(a^ 6*b + a^7)^(1/2) + 9*a^5*b^3*(a^6*b + a^7)^(1/2)))/(a^11*b*(a + b)*(a*b + a^2)*(a^6*b + a^7)^(1/2)*(b^5)^(1/2)*(48*a*b^5 - 6*a^5*b + 9*a^6 + 48*b^6 + 40*a^3*b^3 + 25*a^4*b^2)) - (4*(96*a^4*(b^5)^(3/2) + 18*a^9*(b^5)^(1/...
Time = 0.38 (sec) , antiderivative size = 2834, normalized size of antiderivative = 31.49 \[ \int \frac {\text {sech}^5(x)}{a+b \cosh ^2(x)} \, dx =\text {Too large to display} \] Input:
int(sech(x)^5/(a+b*cosh(x)^2),x)
Output:
(3*e**(8*x)*atan(e**x)*a**3 - e**(8*x)*atan(e**x)*a**2*b + 4*e**(8*x)*atan (e**x)*a*b**2 + 8*e**(8*x)*atan(e**x)*b**3 + 12*e**(6*x)*atan(e**x)*a**3 - 4*e**(6*x)*atan(e**x)*a**2*b + 16*e**(6*x)*atan(e**x)*a*b**2 + 32*e**(6*x )*atan(e**x)*b**3 + 18*e**(4*x)*atan(e**x)*a**3 - 6*e**(4*x)*atan(e**x)*a* *2*b + 24*e**(4*x)*atan(e**x)*a*b**2 + 48*e**(4*x)*atan(e**x)*b**3 + 12*e* *(2*x)*atan(e**x)*a**3 - 4*e**(2*x)*atan(e**x)*a**2*b + 16*e**(2*x)*atan(e **x)*a*b**2 + 32*e**(2*x)*atan(e**x)*b**3 + 3*atan(e**x)*a**3 - atan(e**x) *a**2*b + 4*atan(e**x)*a*b**2 + 8*atan(e**x)*b**3 + 4*e**(8*x)*sqrt(b)*sqr t(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt (b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*b + 16*e**(6*x)*sqrt(b)*sqrt(a )*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b) *sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*b + 24*e**(4*x)*sqrt(b)*sqrt(a)*s qrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sq rt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*b + 16*e**(2*x)*sqrt(b)*sqrt(a)*sqrt (a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt( 2*sqrt(a)*sqrt(a + b) + 2*a + b)))*b + 4*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt( 2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqr t(a + b) + 2*a + b)))*b - 4*e**(8*x)*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*a* b - 4*e**(8*x)*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x...