Integrand size = 12, antiderivative size = 90 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\frac {5 \arctan \left (\sinh \left (a+b x^2\right )\right )}{32 b}+\frac {5 \text {sech}\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{32 b}+\frac {5 \text {sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac {\text {sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b} \]
5/32*arctan(sinh(b*x^2+a))/b+5/32*sech(b*x^2+a)*tanh(b*x^2+a)/b+5/48*sech( b*x^2+a)^3*tanh(b*x^2+a)/b+1/12*sech(b*x^2+a)^5*tanh(b*x^2+a)/b
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\frac {5 \arctan \left (\sinh \left (a+b x^2\right )\right )}{32 b}+\frac {5 \text {sech}\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{32 b}+\frac {5 \text {sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac {\text {sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b} \]
(5*ArcTan[Sinh[a + b*x^2]])/(32*b) + (5*Sech[a + b*x^2]*Tanh[a + b*x^2])/( 32*b) + (5*Sech[a + b*x^2]^3*Tanh[a + b*x^2])/(48*b) + (Sech[a + b*x^2]^5* Tanh[a + b*x^2])/(12*b)
Time = 0.50 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.16, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5959, 3042, 4255, 3042, 4255, 3042, 4255, 3042, 4257}
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.
| \(\displaystyle \int x \text {sech}^7\left (a+b x^2\right ) \, dx\) |
| \(\Big \downarrow \) 5959 |
| \(\displaystyle \frac {1}{2} \int \text {sech}^7\left (b x^2+a\right )dx^2\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \csc \left (i b x^2+i a+\frac {\pi }{2}\right )^7dx^2\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \int \text {sech}^5\left (b x^2+a\right )dx^2+\frac {\tanh \left (a+b x^2\right ) \text {sech}^5\left (a+b x^2\right )}{6 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {\tanh \left (a+b x^2\right ) \text {sech}^5\left (a+b x^2\right )}{6 b}+\frac {5}{6} \int \csc \left (i b x^2+i a+\frac {\pi }{2}\right )^5dx^2\right )\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {3}{4} \int \text {sech}^3\left (b x^2+a\right )dx^2+\frac {\tanh \left (a+b x^2\right ) \text {sech}^3\left (a+b x^2\right )}{4 b}\right )+\frac {\tanh \left (a+b x^2\right ) \text {sech}^5\left (a+b x^2\right )}{6 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {\tanh \left (a+b x^2\right ) \text {sech}^5\left (a+b x^2\right )}{6 b}+\frac {5}{6} \left (\frac {\tanh \left (a+b x^2\right ) \text {sech}^3\left (a+b x^2\right )}{4 b}+\frac {3}{4} \int \csc \left (i b x^2+i a+\frac {\pi }{2}\right )^3dx^2\right )\right )\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \text {sech}\left (b x^2+a\right )dx^2+\frac {\tanh \left (a+b x^2\right ) \text {sech}\left (a+b x^2\right )}{2 b}\right )+\frac {\tanh \left (a+b x^2\right ) \text {sech}^3\left (a+b x^2\right )}{4 b}\right )+\frac {\tanh \left (a+b x^2\right ) \text {sech}^5\left (a+b x^2\right )}{6 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {\tanh \left (a+b x^2\right ) \text {sech}^5\left (a+b x^2\right )}{6 b}+\frac {5}{6} \left (\frac {\tanh \left (a+b x^2\right ) \text {sech}^3\left (a+b x^2\right )}{4 b}+\frac {3}{4} \left (\frac {\tanh \left (a+b x^2\right ) \text {sech}\left (a+b x^2\right )}{2 b}+\frac {1}{2} \int \csc \left (i b x^2+i a+\frac {\pi }{2}\right )dx^2\right )\right )\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\arctan \left (\sinh \left (a+b x^2\right )\right )}{2 b}+\frac {\tanh \left (a+b x^2\right ) \text {sech}\left (a+b x^2\right )}{2 b}\right )+\frac {\tanh \left (a+b x^2\right ) \text {sech}^3\left (a+b x^2\right )}{4 b}\right )+\frac {\tanh \left (a+b x^2\right ) \text {sech}^5\left (a+b x^2\right )}{6 b}\right )\) |
((Sech[a + b*x^2]^5*Tanh[a + b*x^2])/(6*b) + (5*((Sech[a + b*x^2]^3*Tanh[a + b*x^2])/(4*b) + (3*(ArcTan[Sinh[a + b*x^2]]/(2*b) + (Sech[a + b*x^2]*Ta nh[a + b*x^2])/(2*b)))/4))/6)/2
3.1.15.3.1 Defintions of rubi rules used
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo l] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sech[c + d*x] )^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 1)/n], 0] && IntegerQ[p]
Time = 1.38 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.69
| method | result | size |
| derivativedivides | \(\frac {\left (\frac {\operatorname {sech}\left (b \,x^{2}+a \right )^{5}}{6}+\frac {5 \operatorname {sech}\left (b \,x^{2}+a \right )^{3}}{24}+\frac {5 \,\operatorname {sech}\left (b \,x^{2}+a \right )}{16}\right ) \tanh \left (b \,x^{2}+a \right )+\frac {5 \arctan \left ({\mathrm e}^{b \,x^{2}+a}\right )}{8}}{2 b}\) | \(62\) |
| default | \(\frac {\left (\frac {\operatorname {sech}\left (b \,x^{2}+a \right )^{5}}{6}+\frac {5 \operatorname {sech}\left (b \,x^{2}+a \right )^{3}}{24}+\frac {5 \,\operatorname {sech}\left (b \,x^{2}+a \right )}{16}\right ) \tanh \left (b \,x^{2}+a \right )+\frac {5 \arctan \left ({\mathrm e}^{b \,x^{2}+a}\right )}{8}}{2 b}\) | \(62\) |
| risch | \(\frac {{\mathrm e}^{b \,x^{2}+a} \left (15 \,{\mathrm e}^{10 b \,x^{2}+10 a}+85 \,{\mathrm e}^{8 b \,x^{2}+8 a}+198 \,{\mathrm e}^{6 b \,x^{2}+6 a}-198 \,{\mathrm e}^{4 b \,x^{2}+4 a}-85 \,{\mathrm e}^{2 b \,x^{2}+2 a}-15\right )}{48 b \left ({\mathrm e}^{2 b \,x^{2}+2 a}+1\right )^{6}}+\frac {5 i \ln \left ({\mathrm e}^{b \,x^{2}+a}+i\right )}{32 b}-\frac {5 i \ln \left ({\mathrm e}^{b \,x^{2}+a}-i\right )}{32 b}\) | \(133\) |
| parallelrisch | \(\frac {15 i \left (-10-\cosh \left (6 b \,x^{2}+6 a \right )-6 \cosh \left (4 b \,x^{2}+4 a \right )-15 \cosh \left (2 b \,x^{2}+2 a \right )\right ) \ln \left (\tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )-i\right )+15 i \left (10+\cosh \left (6 b \,x^{2}+6 a \right )+6 \cosh \left (4 b \,x^{2}+4 a \right )+15 \cosh \left (2 b \,x^{2}+2 a \right )\right ) \ln \left (\tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )+i\right )+396 \sinh \left (b \,x^{2}+a \right )+170 \sinh \left (3 b \,x^{2}+3 a \right )+30 \sinh \left (5 b \,x^{2}+5 a \right )}{96 b \left (10+\cosh \left (6 b \,x^{2}+6 a \right )+6 \cosh \left (4 b \,x^{2}+4 a \right )+15 \cosh \left (2 b \,x^{2}+2 a \right )\right )}\) | \(200\) |
1/2/b*((1/6*sech(b*x^2+a)^5+5/24*sech(b*x^2+a)^3+5/16*sech(b*x^2+a))*tanh( b*x^2+a)+5/8*arctan(exp(b*x^2+a)))
Leaf count of result is larger than twice the leaf count of optimal. 1918 vs. \(2 (82) = 164\).
Time = 0.26 (sec) , antiderivative size = 1918, normalized size of antiderivative = 21.31 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\text {Too large to display} \]
1/48*(15*cosh(b*x^2 + a)^11 + 165*cosh(b*x^2 + a)*sinh(b*x^2 + a)^10 + 15* sinh(b*x^2 + a)^11 + 5*(165*cosh(b*x^2 + a)^2 + 17)*sinh(b*x^2 + a)^9 + 85 *cosh(b*x^2 + a)^9 + 45*(55*cosh(b*x^2 + a)^3 + 17*cosh(b*x^2 + a))*sinh(b *x^2 + a)^8 + 18*(275*cosh(b*x^2 + a)^4 + 170*cosh(b*x^2 + a)^2 + 11)*sinh (b*x^2 + a)^7 + 198*cosh(b*x^2 + a)^7 + 42*(165*cosh(b*x^2 + a)^5 + 170*co sh(b*x^2 + a)^3 + 33*cosh(b*x^2 + a))*sinh(b*x^2 + a)^6 + 18*(385*cosh(b*x ^2 + a)^6 + 595*cosh(b*x^2 + a)^4 + 231*cosh(b*x^2 + a)^2 - 11)*sinh(b*x^2 + a)^5 - 198*cosh(b*x^2 + a)^5 + 90*(55*cosh(b*x^2 + a)^7 + 119*cosh(b*x^ 2 + a)^5 + 77*cosh(b*x^2 + a)^3 - 11*cosh(b*x^2 + a))*sinh(b*x^2 + a)^4 + 5*(495*cosh(b*x^2 + a)^8 + 1428*cosh(b*x^2 + a)^6 + 1386*cosh(b*x^2 + a)^4 - 396*cosh(b*x^2 + a)^2 - 17)*sinh(b*x^2 + a)^3 - 85*cosh(b*x^2 + a)^3 + 3*(275*cosh(b*x^2 + a)^9 + 1020*cosh(b*x^2 + a)^7 + 1386*cosh(b*x^2 + a)^5 - 660*cosh(b*x^2 + a)^3 - 85*cosh(b*x^2 + a))*sinh(b*x^2 + a)^2 + 15*(cos h(b*x^2 + a)^12 + 12*cosh(b*x^2 + a)*sinh(b*x^2 + a)^11 + sinh(b*x^2 + a)^ 12 + 6*(11*cosh(b*x^2 + a)^2 + 1)*sinh(b*x^2 + a)^10 + 6*cosh(b*x^2 + a)^1 0 + 20*(11*cosh(b*x^2 + a)^3 + 3*cosh(b*x^2 + a))*sinh(b*x^2 + a)^9 + 15*( 33*cosh(b*x^2 + a)^4 + 18*cosh(b*x^2 + a)^2 + 1)*sinh(b*x^2 + a)^8 + 15*co sh(b*x^2 + a)^8 + 24*(33*cosh(b*x^2 + a)^5 + 30*cosh(b*x^2 + a)^3 + 5*cosh (b*x^2 + a))*sinh(b*x^2 + a)^7 + 4*(231*cosh(b*x^2 + a)^6 + 315*cosh(b*x^2 + a)^4 + 105*cosh(b*x^2 + a)^2 + 5)*sinh(b*x^2 + a)^6 + 20*cosh(b*x^2 ...
\[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\int x \operatorname {sech}^{7}{\left (a + b x^{2} \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (82) = 164\).
Time = 0.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.02 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=-\frac {5 \, \arctan \left (e^{\left (-b x^{2} - a\right )}\right )}{16 \, b} + \frac {15 \, e^{\left (-b x^{2} - a\right )} + 85 \, e^{\left (-3 \, b x^{2} - 3 \, a\right )} + 198 \, e^{\left (-5 \, b x^{2} - 5 \, a\right )} - 198 \, e^{\left (-7 \, b x^{2} - 7 \, a\right )} - 85 \, e^{\left (-9 \, b x^{2} - 9 \, a\right )} - 15 \, e^{\left (-11 \, b x^{2} - 11 \, a\right )}}{48 \, b {\left (6 \, e^{\left (-2 \, b x^{2} - 2 \, a\right )} + 15 \, e^{\left (-4 \, b x^{2} - 4 \, a\right )} + 20 \, e^{\left (-6 \, b x^{2} - 6 \, a\right )} + 15 \, e^{\left (-8 \, b x^{2} - 8 \, a\right )} + 6 \, e^{\left (-10 \, b x^{2} - 10 \, a\right )} + e^{\left (-12 \, b x^{2} - 12 \, a\right )} + 1\right )}} \]
-5/16*arctan(e^(-b*x^2 - a))/b + 1/48*(15*e^(-b*x^2 - a) + 85*e^(-3*b*x^2 - 3*a) + 198*e^(-5*b*x^2 - 5*a) - 198*e^(-7*b*x^2 - 7*a) - 85*e^(-9*b*x^2 - 9*a) - 15*e^(-11*b*x^2 - 11*a))/(b*(6*e^(-2*b*x^2 - 2*a) + 15*e^(-4*b*x^ 2 - 4*a) + 20*e^(-6*b*x^2 - 6*a) + 15*e^(-8*b*x^2 - 8*a) + 6*e^(-10*b*x^2 - 10*a) + e^(-12*b*x^2 - 12*a) + 1))
Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.62 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\frac {5 \, {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x^{2} + 2 \, a\right )} - 1\right )} e^{\left (-b x^{2} - a\right )}\right )\right )}}{64 \, b} + \frac {15 \, {\left (e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}\right )}^{5} + 160 \, {\left (e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}\right )}^{3} + 528 \, e^{\left (b x^{2} + a\right )} - 528 \, e^{\left (-b x^{2} - a\right )}}{48 \, {\left ({\left (e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}\right )}^{2} + 4\right )}^{3} b} \]
5/64*(pi + 2*arctan(1/2*(e^(2*b*x^2 + 2*a) - 1)*e^(-b*x^2 - a)))/b + 1/48* (15*(e^(b*x^2 + a) - e^(-b*x^2 - a))^5 + 160*(e^(b*x^2 + a) - e^(-b*x^2 - a))^3 + 528*e^(b*x^2 + a) - 528*e^(-b*x^2 - a))/(((e^(b*x^2 + a) - e^(-b*x ^2 - a))^2 + 4)^3*b)
Time = 0.11 (sec) , antiderivative size = 395, normalized size of antiderivative = 4.39 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\frac {5\,\mathrm {atan}\left (\frac {{\mathrm {e}}^a\,{\mathrm {e}}^{b\,x^2}\,\sqrt {b^2}}{b}\right )}{16\,\sqrt {b^2}}-\frac {8\,{\mathrm {e}}^{3\,b\,x^2+3\,a}}{3\,b\,\left (5\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+10\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+10\,{\mathrm {e}}^{6\,b\,x^2+6\,a}+5\,{\mathrm {e}}^{8\,b\,x^2+8\,a}+{\mathrm {e}}^{10\,b\,x^2+10\,a}+1\right )}-\frac {{\mathrm {e}}^{b\,x^2+a}}{b\,\left (4\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+6\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+4\,{\mathrm {e}}^{6\,b\,x^2+6\,a}+{\mathrm {e}}^{8\,b\,x^2+8\,a}+1\right )}+\frac {5\,{\mathrm {e}}^{b\,x^2+a}}{24\,b\,\left (2\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+{\mathrm {e}}^{4\,b\,x^2+4\,a}+1\right )}-\frac {16\,{\mathrm {e}}^{5\,b\,x^2+5\,a}}{3\,b\,\left (6\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+15\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+20\,{\mathrm {e}}^{6\,b\,x^2+6\,a}+15\,{\mathrm {e}}^{8\,b\,x^2+8\,a}+6\,{\mathrm {e}}^{10\,b\,x^2+10\,a}+{\mathrm {e}}^{12\,b\,x^2+12\,a}+1\right )}+\frac {{\mathrm {e}}^{b\,x^2+a}}{6\,b\,\left (3\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+3\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+{\mathrm {e}}^{6\,b\,x^2+6\,a}+1\right )}+\frac {5\,{\mathrm {e}}^{b\,x^2+a}}{16\,b\,\left ({\mathrm {e}}^{2\,b\,x^2+2\,a}+1\right )} \]
(5*atan((exp(a)*exp(b*x^2)*(b^2)^(1/2))/b))/(16*(b^2)^(1/2)) - (8*exp(3*a + 3*b*x^2))/(3*b*(5*exp(2*a + 2*b*x^2) + 10*exp(4*a + 4*b*x^2) + 10*exp(6* a + 6*b*x^2) + 5*exp(8*a + 8*b*x^2) + exp(10*a + 10*b*x^2) + 1)) - exp(a + b*x^2)/(b*(4*exp(2*a + 2*b*x^2) + 6*exp(4*a + 4*b*x^2) + 4*exp(6*a + 6*b* x^2) + exp(8*a + 8*b*x^2) + 1)) + (5*exp(a + b*x^2))/(24*b*(2*exp(2*a + 2* b*x^2) + exp(4*a + 4*b*x^2) + 1)) - (16*exp(5*a + 5*b*x^2))/(3*b*(6*exp(2* a + 2*b*x^2) + 15*exp(4*a + 4*b*x^2) + 20*exp(6*a + 6*b*x^2) + 15*exp(8*a + 8*b*x^2) + 6*exp(10*a + 10*b*x^2) + exp(12*a + 12*b*x^2) + 1)) + exp(a + b*x^2)/(6*b*(3*exp(2*a + 2*b*x^2) + 3*exp(4*a + 4*b*x^2) + exp(6*a + 6*b* x^2) + 1)) + (5*exp(a + b*x^2))/(16*b*(exp(2*a + 2*b*x^2) + 1))