Integrand size = 12, antiderivative size = 90 \[ \int x \sec ^7\left (a+b x^2\right ) \, dx=\frac {5 \text {arctanh}\left (\sin \left (a+b x^2\right )\right )}{32 b}+\frac {5 \sec \left (a+b x^2\right ) \tan \left (a+b x^2\right )}{32 b}+\frac {5 \sec ^3\left (a+b x^2\right ) \tan \left (a+b x^2\right )}{48 b}+\frac {\sec ^5\left (a+b x^2\right ) \tan \left (a+b x^2\right )}{12 b} \] Output:
5/32*arctanh(sin(b*x^2+a))/b+5/32*sec(b*x^2+a)*tan(b*x^2+a)/b+5/48*sec(b*x ^2+a)^3*tan(b*x^2+a)/b+1/12*sec(b*x^2+a)^5*tan(b*x^2+a)/b
Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00 \[ \int x \sec ^7\left (a+b x^2\right ) \, dx=\frac {5 \text {arctanh}\left (\sin \left (a+b x^2\right )\right )}{32 b}+\frac {5 \sec \left (a+b x^2\right ) \tan \left (a+b x^2\right )}{32 b}+\frac {5 \sec ^3\left (a+b x^2\right ) \tan \left (a+b x^2\right )}{48 b}+\frac {\sec ^5\left (a+b x^2\right ) \tan \left (a+b x^2\right )}{12 b} \] Input:
Integrate[x*Sec[a + b*x^2]^7,x]
Output:
(5*ArcTanh[Sin[a + b*x^2]])/(32*b) + (5*Sec[a + b*x^2]*Tan[a + b*x^2])/(32 *b) + (5*Sec[a + b*x^2]^3*Tan[a + b*x^2])/(48*b) + (Sec[a + b*x^2]^5*Tan[a + b*x^2])/(12*b)
Time = 0.47 (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 = {4692, 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.
| \(\displaystyle \int x \sec ^7\left (a+b x^2\right ) \, dx\) |
| \(\Big \downarrow \) 4692 |
| \(\displaystyle \frac {1}{2} \int \sec ^7\left (b x^2+a\right )dx^2\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \csc \left (b x^2+a+\frac {\pi }{2}\right )^7dx^2\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \int \sec ^5\left (b x^2+a\right )dx^2+\frac {\tan \left (a+b x^2\right ) \sec ^5\left (a+b x^2\right )}{6 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \int \csc \left (b x^2+a+\frac {\pi }{2}\right )^5dx^2+\frac {\tan \left (a+b x^2\right ) \sec ^5\left (a+b x^2\right )}{6 b}\right )\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {3}{4} \int \sec ^3\left (b x^2+a\right )dx^2+\frac {\tan \left (a+b x^2\right ) \sec ^3\left (a+b x^2\right )}{4 b}\right )+\frac {\tan \left (a+b x^2\right ) \sec ^5\left (a+b x^2\right )}{6 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {3}{4} \int \csc \left (b x^2+a+\frac {\pi }{2}\right )^3dx^2+\frac {\tan \left (a+b x^2\right ) \sec ^3\left (a+b x^2\right )}{4 b}\right )+\frac {\tan \left (a+b x^2\right ) \sec ^5\left (a+b x^2\right )}{6 b}\right )\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \sec \left (b x^2+a\right )dx^2+\frac {\tan \left (a+b x^2\right ) \sec \left (a+b x^2\right )}{2 b}\right )+\frac {\tan \left (a+b x^2\right ) \sec ^3\left (a+b x^2\right )}{4 b}\right )+\frac {\tan \left (a+b x^2\right ) \sec ^5\left (a+b x^2\right )}{6 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (b x^2+a+\frac {\pi }{2}\right )dx^2+\frac {\tan \left (a+b x^2\right ) \sec \left (a+b x^2\right )}{2 b}\right )+\frac {\tan \left (a+b x^2\right ) \sec ^3\left (a+b x^2\right )}{4 b}\right )+\frac {\tan \left (a+b x^2\right ) \sec ^5\left (a+b x^2\right )}{6 b}\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\text {arctanh}\left (\sin \left (a+b x^2\right )\right )}{2 b}+\frac {\tan \left (a+b x^2\right ) \sec \left (a+b x^2\right )}{2 b}\right )+\frac {\tan \left (a+b x^2\right ) \sec ^3\left (a+b x^2\right )}{4 b}\right )+\frac {\tan \left (a+b x^2\right ) \sec ^5\left (a+b x^2\right )}{6 b}\right )\) |
Input:
Int[x*Sec[a + b*x^2]^7,x]
Output:
((Sec[a + b*x^2]^5*Tan[a + b*x^2])/(6*b) + (5*((Sec[a + b*x^2]^3*Tan[a + b *x^2])/(4*b) + (3*(ArcTanh[Sin[a + b*x^2]]/(2*b) + (Sec[a + b*x^2]*Tan[a + b*x^2])/(2*b)))/4))/6)/2
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_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[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 = 0.50 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.80
| method | result | size |
| derivativedivides | \(\frac {-\left (-\frac {\sec \left (b \,x^{2}+a \right )^{5}}{6}-\frac {5 \sec \left (b \,x^{2}+a \right )^{3}}{24}-\frac {5 \sec \left (b \,x^{2}+a \right )}{16}\right ) \tan \left (b \,x^{2}+a \right )+\frac {5 \ln \left (\sec \left (b \,x^{2}+a \right )+\tan \left (b \,x^{2}+a \right )\right )}{16}}{2 b}\) | \(72\) |
| default | \(\frac {-\left (-\frac {\sec \left (b \,x^{2}+a \right )^{5}}{6}-\frac {5 \sec \left (b \,x^{2}+a \right )^{3}}{24}-\frac {5 \sec \left (b \,x^{2}+a \right )}{16}\right ) \tan \left (b \,x^{2}+a \right )+\frac {5 \ln \left (\sec \left (b \,x^{2}+a \right )+\tan \left (b \,x^{2}+a \right )\right )}{16}}{2 b}\) | \(72\) |
| risch | \(-\frac {i \left (15 \,{\mathrm e}^{11 i \left (b \,x^{2}+a \right )}+85 \,{\mathrm e}^{9 i \left (b \,x^{2}+a \right )}+198 \,{\mathrm e}^{7 i \left (b \,x^{2}+a \right )}-198 \,{\mathrm e}^{5 i \left (b \,x^{2}+a \right )}-85 \,{\mathrm e}^{3 i \left (b \,x^{2}+a \right )}-15 \,{\mathrm e}^{i \left (b \,x^{2}+a \right )}\right )}{48 b \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}+1\right )^{6}}+\frac {5 \ln \left ({\mathrm e}^{i \left (b \,x^{2}+a \right )}+i\right )}{32 b}-\frac {5 \ln \left ({\mathrm e}^{i \left (b \,x^{2}+a \right )}-i\right )}{32 b}\) | \(142\) |
| parallelrisch | \(\frac {\left (-225 \cos \left (2 b \,x^{2}+2 a \right )-90 \cos \left (4 b \,x^{2}+4 a \right )-15 \cos \left (6 b \,x^{2}+6 a \right )-150\right ) \ln \left (\tan \left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )-1\right )+\left (225 \cos \left (2 b \,x^{2}+2 a \right )+90 \cos \left (4 b \,x^{2}+4 a \right )+15 \cos \left (6 b \,x^{2}+6 a \right )+150\right ) \ln \left (\tan \left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )+1\right )+396 \sin \left (b \,x^{2}+a \right )+170 \sin \left (3 b \,x^{2}+3 a \right )+30 \sin \left (5 b \,x^{2}+5 a \right )}{96 b \left (10+\cos \left (6 b \,x^{2}+6 a \right )+6 \cos \left (4 b \,x^{2}+4 a \right )+15 \cos \left (2 b \,x^{2}+2 a \right )\right )}\) | \(196\) |
Input:
int(x*sec(b*x^2+a)^7,x,method=_RETURNVERBOSE)
Output:
1/2/b*(-(-1/6*sec(b*x^2+a)^5-5/24*sec(b*x^2+a)^3-5/16*sec(b*x^2+a))*tan(b* x^2+a)+5/16*ln(sec(b*x^2+a)+tan(b*x^2+a)))
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11 \[ \int x \sec ^7\left (a+b x^2\right ) \, dx=\frac {15 \, \cos \left (b x^{2} + a\right )^{6} \log \left (\sin \left (b x^{2} + a\right ) + 1\right ) - 15 \, \cos \left (b x^{2} + a\right )^{6} \log \left (-\sin \left (b x^{2} + a\right ) + 1\right ) + 2 \, {\left (15 \, \cos \left (b x^{2} + a\right )^{4} + 10 \, \cos \left (b x^{2} + a\right )^{2} + 8\right )} \sin \left (b x^{2} + a\right )}{192 \, b \cos \left (b x^{2} + a\right )^{6}} \] Input:
integrate(x*sec(b*x^2+a)^7,x, algorithm="fricas")
Output:
1/192*(15*cos(b*x^2 + a)^6*log(sin(b*x^2 + a) + 1) - 15*cos(b*x^2 + a)^6*l og(-sin(b*x^2 + a) + 1) + 2*(15*cos(b*x^2 + a)^4 + 10*cos(b*x^2 + a)^2 + 8 )*sin(b*x^2 + a))/(b*cos(b*x^2 + a)^6)
\[ \int x \sec ^7\left (a+b x^2\right ) \, dx=\int x \sec ^{7}{\left (a + b x^{2} \right )}\, dx \] Input:
integrate(x*sec(b*x**2+a)**7,x)
Output:
Integral(x*sec(a + b*x**2)**7, x)
Leaf count of result is larger than twice the leaf count of optimal. 2838 vs. \(2 (82) = 164\).
Time = 0.21 (sec) , antiderivative size = 2838, normalized size of antiderivative = 31.53 \[ \int x \sec ^7\left (a+b x^2\right ) \, dx=\text {Too large to display} \] Input:
integrate(x*sec(b*x^2+a)^7,x, algorithm="maxima")
Output:
1/192*(4*(15*sin(11*b*x^2 + 11*a) + 85*sin(9*b*x^2 + 9*a) + 198*sin(7*b*x^ 2 + 7*a) - 198*sin(5*b*x^2 + 5*a) - 85*sin(3*b*x^2 + 3*a) - 15*sin(b*x^2 + a))*cos(12*b*x^2 + 12*a) - 60*(6*sin(10*b*x^2 + 10*a) + 15*sin(8*b*x^2 + 8*a) + 20*sin(6*b*x^2 + 6*a) + 15*sin(4*b*x^2 + 4*a) + 6*sin(2*b*x^2 + 2*a ))*cos(11*b*x^2 + 11*a) + 24*(85*sin(9*b*x^2 + 9*a) + 198*sin(7*b*x^2 + 7* a) - 198*sin(5*b*x^2 + 5*a) - 85*sin(3*b*x^2 + 3*a) - 15*sin(b*x^2 + a))*c os(10*b*x^2 + 10*a) - 340*(15*sin(8*b*x^2 + 8*a) + 20*sin(6*b*x^2 + 6*a) + 15*sin(4*b*x^2 + 4*a) + 6*sin(2*b*x^2 + 2*a))*cos(9*b*x^2 + 9*a) + 60*(19 8*sin(7*b*x^2 + 7*a) - 198*sin(5*b*x^2 + 5*a) - 85*sin(3*b*x^2 + 3*a) - 15 *sin(b*x^2 + a))*cos(8*b*x^2 + 8*a) - 792*(20*sin(6*b*x^2 + 6*a) + 15*sin( 4*b*x^2 + 4*a) + 6*sin(2*b*x^2 + 2*a))*cos(7*b*x^2 + 7*a) - 80*(198*sin(5* b*x^2 + 5*a) + 85*sin(3*b*x^2 + 3*a) + 15*sin(b*x^2 + a))*cos(6*b*x^2 + 6* a) + 2376*(5*sin(4*b*x^2 + 4*a) + 2*sin(2*b*x^2 + 2*a))*cos(5*b*x^2 + 5*a) - 300*(17*sin(3*b*x^2 + 3*a) + 3*sin(b*x^2 + a))*cos(4*b*x^2 + 4*a) - 15* (2*(6*cos(10*b*x^2 + 10*a) + 15*cos(8*b*x^2 + 8*a) + 20*cos(6*b*x^2 + 6*a) + 15*cos(4*b*x^2 + 4*a) + 6*cos(2*b*x^2 + 2*a) + 1)*cos(12*b*x^2 + 12*a) + cos(12*b*x^2 + 12*a)^2 + 12*(15*cos(8*b*x^2 + 8*a) + 20*cos(6*b*x^2 + 6* a) + 15*cos(4*b*x^2 + 4*a) + 6*cos(2*b*x^2 + 2*a) + 1)*cos(10*b*x^2 + 10*a ) + 36*cos(10*b*x^2 + 10*a)^2 + 30*(20*cos(6*b*x^2 + 6*a) + 15*cos(4*b*x^2 + 4*a) + 6*cos(2*b*x^2 + 2*a) + 1)*cos(8*b*x^2 + 8*a) + 225*cos(8*b*x^...
Time = 0.49 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94 \[ \int x \sec ^7\left (a+b x^2\right ) \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (b x^{2} + a\right )^{5} - 40 \, \sin \left (b x^{2} + a\right )^{3} + 33 \, \sin \left (b x^{2} + a\right )\right )}}{{\left (\sin \left (b x^{2} + a\right )^{2} - 1\right )}^{3}} - 15 \, \log \left (\sin \left (b x^{2} + a\right ) + 1\right ) + 15 \, \log \left (-\sin \left (b x^{2} + a\right ) + 1\right )}{192 \, b} \] Input:
integrate(x*sec(b*x^2+a)^7,x, algorithm="giac")
Output:
-1/192*(2*(15*sin(b*x^2 + a)^5 - 40*sin(b*x^2 + a)^3 + 33*sin(b*x^2 + a))/ (sin(b*x^2 + a)^2 - 1)^3 - 15*log(sin(b*x^2 + a) + 1) + 15*log(-sin(b*x^2 + a) + 1))/b
Time = 27.86 (sec) , antiderivative size = 496, normalized size of antiderivative = 5.51 \[ \int x \sec ^7\left (a+b x^2\right ) \, dx =\text {Too large to display} \] Input:
int(x/cos(a + b*x^2)^7,x)
Output:
(5*log(- (x*5i)/8 - (5*x*exp(a*1i)*exp(b*x^2*1i))/8))/(32*b) - (5*log((x*5 i)/8 - (5*x*exp(a*1i)*exp(b*x^2*1i))/8))/(32*b) + (exp(a*3i + b*x^2*3i)*8i )/(3*b*(5*exp(a*2i + b*x^2*2i) + 10*exp(a*4i + b*x^2*4i) + 10*exp(a*6i + b *x^2*6i) + 5*exp(a*8i + b*x^2*8i) + exp(a*10i + b*x^2*10i) + 1)) - (exp(a* 1i + b*x^2*1i)*1i)/(6*b*(3*exp(a*2i + b*x^2*2i) + 3*exp(a*4i + b*x^2*4i) + exp(a*6i + b*x^2*6i) + 1)) - (exp(a*1i + b*x^2*1i)*5i)/(16*b*(exp(a*2i + b*x^2*2i) + 1)) + (exp(a*5i + b*x^2*5i)*16i)/(3*b*(6*exp(a*2i + b*x^2*2i) + 15*exp(a*4i + b*x^2*4i) + 20*exp(a*6i + b*x^2*6i) + 15*exp(a*8i + b*x^2* 8i) + 6*exp(a*10i + b*x^2*10i) + exp(a*12i + b*x^2*12i) + 1)) + (exp(a*1i + b*x^2*1i)*1i)/(b*(4*exp(a*2i + b*x^2*2i) + 6*exp(a*4i + b*x^2*4i) + 4*ex p(a*6i + b*x^2*6i) + exp(a*8i + b*x^2*8i) + 1)) - (exp(a*1i + b*x^2*1i)*5i )/(24*b*(2*exp(a*2i + b*x^2*2i) + exp(a*4i + b*x^2*4i) + 1))
Time = 0.16 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.96 \[ \int x \sec ^7\left (a+b x^2\right ) \, dx=\frac {-15 \,\mathrm {log}\left (\tan \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b \,x^{2}+a \right )^{6}+45 \,\mathrm {log}\left (\tan \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b \,x^{2}+a \right )^{4}-45 \,\mathrm {log}\left (\tan \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b \,x^{2}+a \right )^{2}+15 \,\mathrm {log}\left (\tan \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )-1\right )+15 \,\mathrm {log}\left (\tan \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b \,x^{2}+a \right )^{6}-45 \,\mathrm {log}\left (\tan \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b \,x^{2}+a \right )^{4}+45 \,\mathrm {log}\left (\tan \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b \,x^{2}+a \right )^{2}-15 \,\mathrm {log}\left (\tan \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )+1\right )-15 \sin \left (b \,x^{2}+a \right )^{5}+40 \sin \left (b \,x^{2}+a \right )^{3}-33 \sin \left (b \,x^{2}+a \right )}{96 b \left (\sin \left (b \,x^{2}+a \right )^{6}-3 \sin \left (b \,x^{2}+a \right )^{4}+3 \sin \left (b \,x^{2}+a \right )^{2}-1\right )} \] Input:
int(x*sec(b*x^2+a)^7,x)
Output:
( - 15*log(tan((a + b*x**2)/2) - 1)*sin(a + b*x**2)**6 + 45*log(tan((a + b *x**2)/2) - 1)*sin(a + b*x**2)**4 - 45*log(tan((a + b*x**2)/2) - 1)*sin(a + b*x**2)**2 + 15*log(tan((a + b*x**2)/2) - 1) + 15*log(tan((a + b*x**2)/2 ) + 1)*sin(a + b*x**2)**6 - 45*log(tan((a + b*x**2)/2) + 1)*sin(a + b*x**2 )**4 + 45*log(tan((a + b*x**2)/2) + 1)*sin(a + b*x**2)**2 - 15*log(tan((a + b*x**2)/2) + 1) - 15*sin(a + b*x**2)**5 + 40*sin(a + b*x**2)**3 - 33*sin (a + b*x**2))/(96*b*(sin(a + b*x**2)**6 - 3*sin(a + b*x**2)**4 + 3*sin(a + b*x**2)**2 - 1))