Integrand size = 22, antiderivative size = 110 \[ \int x^5 \sec ^7\left (a+b x^3\right ) \tan \left (a+b x^3\right ) \, dx=-\frac {5 \text {arctanh}\left (\sin \left (a+b x^3\right )\right )}{336 b^2}+\frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {5 \sec \left (a+b x^3\right ) \tan \left (a+b x^3\right )}{336 b^2}-\frac {5 \sec ^3\left (a+b x^3\right ) \tan \left (a+b x^3\right )}{504 b^2}-\frac {\sec ^5\left (a+b x^3\right ) \tan \left (a+b x^3\right )}{126 b^2} \] Output:
-5/336*arctanh(sin(b*x^3+a))/b^2+1/21*x^3*sec(b*x^3+a)^7/b-5/336*sec(b*x^3 +a)*tan(b*x^3+a)/b^2-5/504*sec(b*x^3+a)^3*tan(b*x^3+a)/b^2-1/126*sec(b*x^3 +a)^5*tan(b*x^3+a)/b^2
Leaf count is larger than twice the leaf count of optimal. \(352\) vs. \(2(110)=220\).
Time = 0.76 (sec) , antiderivative size = 352, normalized size of antiderivative = 3.20 \[ \int x^5 \sec ^7\left (a+b x^3\right ) \tan \left (a+b x^3\right ) \, dx=\frac {\sec ^7\left (a+b x^3\right ) \left (3072 b x^3+105 \cos \left (5 \left (a+b x^3\right )\right ) \log \left (\cos \left (\frac {1}{2} \left (a+b x^3\right )\right )-\sin \left (\frac {1}{2} \left (a+b x^3\right )\right )\right )+15 \cos \left (7 \left (a+b x^3\right )\right ) \log \left (\cos \left (\frac {1}{2} \left (a+b x^3\right )\right )-\sin \left (\frac {1}{2} \left (a+b x^3\right )\right )\right )+525 \cos \left (a+b x^3\right ) \left (\log \left (\cos \left (\frac {1}{2} \left (a+b x^3\right )\right )-\sin \left (\frac {1}{2} \left (a+b x^3\right )\right )\right )-\log \left (\cos \left (\frac {1}{2} \left (a+b x^3\right )\right )+\sin \left (\frac {1}{2} \left (a+b x^3\right )\right )\right )\right )+315 \cos \left (3 \left (a+b x^3\right )\right ) \left (\log \left (\cos \left (\frac {1}{2} \left (a+b x^3\right )\right )-\sin \left (\frac {1}{2} \left (a+b x^3\right )\right )\right )-\log \left (\cos \left (\frac {1}{2} \left (a+b x^3\right )\right )+\sin \left (\frac {1}{2} \left (a+b x^3\right )\right )\right )\right )-105 \cos \left (5 \left (a+b x^3\right )\right ) \log \left (\cos \left (\frac {1}{2} \left (a+b x^3\right )\right )+\sin \left (\frac {1}{2} \left (a+b x^3\right )\right )\right )-15 \cos \left (7 \left (a+b x^3\right )\right ) \log \left (\cos \left (\frac {1}{2} \left (a+b x^3\right )\right )+\sin \left (\frac {1}{2} \left (a+b x^3\right )\right )\right )-566 \sin \left (2 \left (a+b x^3\right )\right )-200 \sin \left (4 \left (a+b x^3\right )\right )-30 \sin \left (6 \left (a+b x^3\right )\right )\right )}{64512 b^2} \] Input:
Integrate[x^5*Sec[a + b*x^3]^7*Tan[a + b*x^3],x]
Output:
(Sec[a + b*x^3]^7*(3072*b*x^3 + 105*Cos[5*(a + b*x^3)]*Log[Cos[(a + b*x^3) /2] - Sin[(a + b*x^3)/2]] + 15*Cos[7*(a + b*x^3)]*Log[Cos[(a + b*x^3)/2] - Sin[(a + b*x^3)/2]] + 525*Cos[a + b*x^3]*(Log[Cos[(a + b*x^3)/2] - Sin[(a + b*x^3)/2]] - Log[Cos[(a + b*x^3)/2] + Sin[(a + b*x^3)/2]]) + 315*Cos[3* (a + b*x^3)]*(Log[Cos[(a + b*x^3)/2] - Sin[(a + b*x^3)/2]] - Log[Cos[(a + b*x^3)/2] + Sin[(a + b*x^3)/2]]) - 105*Cos[5*(a + b*x^3)]*Log[Cos[(a + b*x ^3)/2] + Sin[(a + b*x^3)/2]] - 15*Cos[7*(a + b*x^3)]*Log[Cos[(a + b*x^3)/2 ] + Sin[(a + b*x^3)/2]] - 566*Sin[2*(a + b*x^3)] - 200*Sin[4*(a + b*x^3)] - 30*Sin[6*(a + b*x^3)]))/(64512*b^2)
Time = 0.58 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {4244, 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.
| \(\displaystyle \int x^5 \tan \left (a+b x^3\right ) \sec ^7\left (a+b x^3\right ) \, dx\) |
| \(\Big \downarrow \) 4244 |
| \(\displaystyle \frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {\int x^2 \sec ^7\left (b x^3+a\right )dx}{7 b}\) |
| \(\Big \downarrow \) 4692 |
| \(\displaystyle \frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {\int \sec ^7\left (b x^3+a\right )dx^3}{21 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {\int \csc \left (b x^3+a+\frac {\pi }{2}\right )^7dx^3}{21 b}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {\frac {5}{6} \int \sec ^5\left (b x^3+a\right )dx^3+\frac {\tan \left (a+b x^3\right ) \sec ^5\left (a+b x^3\right )}{6 b}}{21 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {\frac {5}{6} \int \csc \left (b x^3+a+\frac {\pi }{2}\right )^5dx^3+\frac {\tan \left (a+b x^3\right ) \sec ^5\left (a+b x^3\right )}{6 b}}{21 b}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {\frac {5}{6} \left (\frac {3}{4} \int \sec ^3\left (b x^3+a\right )dx^3+\frac {\tan \left (a+b x^3\right ) \sec ^3\left (a+b x^3\right )}{4 b}\right )+\frac {\tan \left (a+b x^3\right ) \sec ^5\left (a+b x^3\right )}{6 b}}{21 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {\frac {5}{6} \left (\frac {3}{4} \int \csc \left (b x^3+a+\frac {\pi }{2}\right )^3dx^3+\frac {\tan \left (a+b x^3\right ) \sec ^3\left (a+b x^3\right )}{4 b}\right )+\frac {\tan \left (a+b x^3\right ) \sec ^5\left (a+b x^3\right )}{6 b}}{21 b}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \sec \left (b x^3+a\right )dx^3+\frac {\tan \left (a+b x^3\right ) \sec \left (a+b x^3\right )}{2 b}\right )+\frac {\tan \left (a+b x^3\right ) \sec ^3\left (a+b x^3\right )}{4 b}\right )+\frac {\tan \left (a+b x^3\right ) \sec ^5\left (a+b x^3\right )}{6 b}}{21 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (b x^3+a+\frac {\pi }{2}\right )dx^3+\frac {\tan \left (a+b x^3\right ) \sec \left (a+b x^3\right )}{2 b}\right )+\frac {\tan \left (a+b x^3\right ) \sec ^3\left (a+b x^3\right )}{4 b}\right )+\frac {\tan \left (a+b x^3\right ) \sec ^5\left (a+b x^3\right )}{6 b}}{21 b}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {x^3 \sec ^7\left (a+b x^3\right )}{21 b}-\frac {\frac {5}{6} \left (\frac {3}{4} \left (\frac {\text {arctanh}\left (\sin \left (a+b x^3\right )\right )}{2 b}+\frac {\tan \left (a+b x^3\right ) \sec \left (a+b x^3\right )}{2 b}\right )+\frac {\tan \left (a+b x^3\right ) \sec ^3\left (a+b x^3\right )}{4 b}\right )+\frac {\tan \left (a+b x^3\right ) \sec ^5\left (a+b x^3\right )}{6 b}}{21 b}\) |
Input:
Int[x^5*Sec[a + b*x^3]^7*Tan[a + b*x^3],x]
Output:
(x^3*Sec[a + b*x^3]^7)/(21*b) - ((Sec[a + b*x^3]^5*Tan[a + b*x^3])/(6*b) + (5*((Sec[a + b*x^3]^3*Tan[a + b*x^3])/(4*b) + (3*(ArcTanh[Sin[a + b*x^3]] /(2*b) + (Sec[a + b*x^3]*Tan[a + b*x^3])/(2*b)))/4))/6)/(21*b)
Int[(x_)^(m_.)*Sec[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tan[(a_.) + (b_.)*(x_)^( n_.)]^(q_.), x_Symbol] :> Simp[x^(m - n + 1)*(Sec[a + b*x^n]^p/(b*n*p)), x] - Simp[(m - n + 1)/(b*n*p) Int[x^(m - n)*Sec[a + b*x^n]^p, x], x] /; Fre eQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m, n] && EqQ[q, 1]
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]
Result contains complex when optimal does not.
Time = 69.18 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.45
| method | result | size |
| risch | \(\frac {i \left (15 \,{\mathrm e}^{13 i \left (b \,x^{3}+a \right )}-3072 i b \,x^{3} {\mathrm e}^{7 i \left (b \,x^{3}+a \right )}+100 \,{\mathrm e}^{11 i \left (b \,x^{3}+a \right )}+283 \,{\mathrm e}^{9 i \left (b \,x^{3}+a \right )}-283 \,{\mathrm e}^{5 i \left (b \,x^{3}+a \right )}-100 \,{\mathrm e}^{3 i \left (b \,x^{3}+a \right )}-15 \,{\mathrm e}^{i \left (b \,x^{3}+a \right )}\right )}{504 b^{2} \left ({\mathrm e}^{2 i \left (b \,x^{3}+a \right )}+1\right )^{7}}-\frac {5 \ln \left ({\mathrm e}^{i \left (b \,x^{3}+a \right )}+i\right )}{336 b^{2}}+\frac {5 \ln \left ({\mathrm e}^{i \left (b \,x^{3}+a \right )}-i\right )}{336 b^{2}}\) | \(160\) |
Input:
int(x^5*sec(b*x^3+a)^7*tan(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
1/504*I/b^2/(exp(2*I*(b*x^3+a))+1)^7*(15*exp(13*I*(b*x^3+a))-3072*I*b*x^3* exp(7*I*(b*x^3+a))+100*exp(11*I*(b*x^3+a))+283*exp(9*I*(b*x^3+a))-283*exp( 5*I*(b*x^3+a))-100*exp(3*I*(b*x^3+a))-15*exp(I*(b*x^3+a)))-5/336/b^2*ln(ex p(I*(b*x^3+a))+I)+5/336/b^2*ln(exp(I*(b*x^3+a))-I)
Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05 \[ \int x^5 \sec ^7\left (a+b x^3\right ) \tan \left (a+b x^3\right ) \, dx=-\frac {15 \, \cos \left (b x^{3} + a\right )^{7} \log \left (\sin \left (b x^{3} + a\right ) + 1\right ) - 15 \, \cos \left (b x^{3} + a\right )^{7} \log \left (-\sin \left (b x^{3} + a\right ) + 1\right ) - 96 \, b x^{3} + 2 \, {\left (15 \, \cos \left (b x^{3} + a\right )^{5} + 10 \, \cos \left (b x^{3} + a\right )^{3} + 8 \, \cos \left (b x^{3} + a\right )\right )} \sin \left (b x^{3} + a\right )}{2016 \, b^{2} \cos \left (b x^{3} + a\right )^{7}} \] Input:
integrate(x^5*sec(b*x^3+a)^7*tan(b*x^3+a),x, algorithm="fricas")
Output:
-1/2016*(15*cos(b*x^3 + a)^7*log(sin(b*x^3 + a) + 1) - 15*cos(b*x^3 + a)^7 *log(-sin(b*x^3 + a) + 1) - 96*b*x^3 + 2*(15*cos(b*x^3 + a)^5 + 10*cos(b*x ^3 + a)^3 + 8*cos(b*x^3 + a))*sin(b*x^3 + a))/(b^2*cos(b*x^3 + a)^7)
\[ \int x^5 \sec ^7\left (a+b x^3\right ) \tan \left (a+b x^3\right ) \, dx=\int x^{5} \tan {\left (a + b x^{3} \right )} \sec ^{7}{\left (a + b x^{3} \right )}\, dx \] Input:
integrate(x**5*sec(b*x**3+a)**7*tan(b*x**3+a),x)
Output:
Integral(x**5*tan(a + b*x**3)*sec(a + b*x**3)**7, x)
Leaf count of result is larger than twice the leaf count of optimal. 3830 vs. \(2 (100) = 200\).
Time = 0.26 (sec) , antiderivative size = 3830, normalized size of antiderivative = 34.82 \[ \int x^5 \sec ^7\left (a+b x^3\right ) \tan \left (a+b x^3\right ) \, dx=\text {Too large to display} \] Input:
integrate(x^5*sec(b*x^3+a)^7*tan(b*x^3+a),x, algorithm="maxima")
Output:
1/2016*(4*(3072*b*x^3*cos(7*b*x^3 + 7*a) - 15*sin(13*b*x^3 + 13*a) - 100*s in(11*b*x^3 + 11*a) - 283*sin(9*b*x^3 + 9*a) + 283*sin(5*b*x^3 + 5*a) + 10 0*sin(3*b*x^3 + 3*a) + 15*sin(b*x^3 + a))*cos(14*b*x^3 + 14*a) + 420*(sin( 12*b*x^3 + 12*a) + 3*sin(10*b*x^3 + 10*a) + 5*sin(8*b*x^3 + 8*a) + 5*sin(6 *b*x^3 + 6*a) + 3*sin(4*b*x^3 + 4*a) + sin(2*b*x^3 + 2*a))*cos(13*b*x^3 + 13*a) + 28*(3072*b*x^3*cos(7*b*x^3 + 7*a) - 100*sin(11*b*x^3 + 11*a) - 283 *sin(9*b*x^3 + 9*a) + 283*sin(5*b*x^3 + 5*a) + 100*sin(3*b*x^3 + 3*a) + 15 *sin(b*x^3 + a))*cos(12*b*x^3 + 12*a) + 2800*(3*sin(10*b*x^3 + 10*a) + 5*s in(8*b*x^3 + 8*a) + 5*sin(6*b*x^3 + 6*a) + 3*sin(4*b*x^3 + 4*a) + sin(2*b* x^3 + 2*a))*cos(11*b*x^3 + 11*a) + 84*(3072*b*x^3*cos(7*b*x^3 + 7*a) - 283 *sin(9*b*x^3 + 9*a) + 283*sin(5*b*x^3 + 5*a) + 100*sin(3*b*x^3 + 3*a) + 15 *sin(b*x^3 + a))*cos(10*b*x^3 + 10*a) + 7924*(5*sin(8*b*x^3 + 8*a) + 5*sin (6*b*x^3 + 6*a) + 3*sin(4*b*x^3 + 4*a) + sin(2*b*x^3 + 2*a))*cos(9*b*x^3 + 9*a) + 140*(3072*b*x^3*cos(7*b*x^3 + 7*a) + 283*sin(5*b*x^3 + 5*a) + 100* sin(3*b*x^3 + 3*a) + 15*sin(b*x^3 + a))*cos(8*b*x^3 + 8*a) + 12288*(35*b*x ^3*cos(6*b*x^3 + 6*a) + 21*b*x^3*cos(4*b*x^3 + 4*a) + 7*b*x^3*cos(2*b*x^3 + 2*a) + b*x^3)*cos(7*b*x^3 + 7*a) + 140*(283*sin(5*b*x^3 + 5*a) + 100*sin (3*b*x^3 + 3*a) + 15*sin(b*x^3 + a))*cos(6*b*x^3 + 6*a) - 7924*(3*sin(4*b* x^3 + 4*a) + sin(2*b*x^3 + 2*a))*cos(5*b*x^3 + 5*a) + 420*(20*sin(3*b*x^3 + 3*a) + 3*sin(b*x^3 + a))*cos(4*b*x^3 + 4*a) + 15*(2*(7*cos(12*b*x^3 +...
Leaf count of result is larger than twice the leaf count of optimal. 1363 vs. \(2 (100) = 200\).
Time = 0.67 (sec) , antiderivative size = 1363, normalized size of antiderivative = 12.39 \[ \int x^5 \sec ^7\left (a+b x^3\right ) \tan \left (a+b x^3\right ) \, dx=\text {Too large to display} \] Input:
integrate(x^5*sec(b*x^3+a)^7*tan(b*x^3+a),x, algorithm="giac")
Output:
-1/2016*(96*(b*x^3 + a)*tan(1/2*b*x^3 + 1/2*a)^14 + 15*log(2*(tan(1/2*b*x^ 3 + 1/2*a)^2 + 2*tan(1/2*b*x^3 + 1/2*a) + 1)/(tan(1/2*b*x^3 + 1/2*a)^2 + 1 ))*tan(1/2*b*x^3 + 1/2*a)^14 - 15*log(2*(tan(1/2*b*x^3 + 1/2*a)^2 - 2*tan( 1/2*b*x^3 + 1/2*a) + 1)/(tan(1/2*b*x^3 + 1/2*a)^2 + 1))*tan(1/2*b*x^3 + 1/ 2*a)^14 + 672*(b*x^3 + a)*tan(1/2*b*x^3 + 1/2*a)^12 - 105*log(2*(tan(1/2*b *x^3 + 1/2*a)^2 + 2*tan(1/2*b*x^3 + 1/2*a) + 1)/(tan(1/2*b*x^3 + 1/2*a)^2 + 1))*tan(1/2*b*x^3 + 1/2*a)^12 + 105*log(2*(tan(1/2*b*x^3 + 1/2*a)^2 - 2* tan(1/2*b*x^3 + 1/2*a) + 1)/(tan(1/2*b*x^3 + 1/2*a)^2 + 1))*tan(1/2*b*x^3 + 1/2*a)^12 + 132*tan(1/2*b*x^3 + 1/2*a)^13 + 2016*(b*x^3 + a)*tan(1/2*b*x ^3 + 1/2*a)^10 + 315*log(2*(tan(1/2*b*x^3 + 1/2*a)^2 + 2*tan(1/2*b*x^3 + 1 /2*a) + 1)/(tan(1/2*b*x^3 + 1/2*a)^2 + 1))*tan(1/2*b*x^3 + 1/2*a)^10 - 315 *log(2*(tan(1/2*b*x^3 + 1/2*a)^2 - 2*tan(1/2*b*x^3 + 1/2*a) + 1)/(tan(1/2* b*x^3 + 1/2*a)^2 + 1))*tan(1/2*b*x^3 + 1/2*a)^10 - 112*tan(1/2*b*x^3 + 1/2 *a)^11 + 3360*(b*x^3 + a)*tan(1/2*b*x^3 + 1/2*a)^8 - 525*log(2*(tan(1/2*b* x^3 + 1/2*a)^2 + 2*tan(1/2*b*x^3 + 1/2*a) + 1)/(tan(1/2*b*x^3 + 1/2*a)^2 + 1))*tan(1/2*b*x^3 + 1/2*a)^8 + 525*log(2*(tan(1/2*b*x^3 + 1/2*a)^2 - 2*ta n(1/2*b*x^3 + 1/2*a) + 1)/(tan(1/2*b*x^3 + 1/2*a)^2 + 1))*tan(1/2*b*x^3 + 1/2*a)^8 + 340*tan(1/2*b*x^3 + 1/2*a)^9 + 3360*(b*x^3 + a)*tan(1/2*b*x^3 + 1/2*a)^6 + 525*log(2*(tan(1/2*b*x^3 + 1/2*a)^2 + 2*tan(1/2*b*x^3 + 1/2*a) + 1)/(tan(1/2*b*x^3 + 1/2*a)^2 + 1))*tan(1/2*b*x^3 + 1/2*a)^6 - 525*lo...
Time = 31.18 (sec) , antiderivative size = 730, normalized size of antiderivative = 6.64 \[ \int x^5 \sec ^7\left (a+b x^3\right ) \tan \left (a+b x^3\right ) \, dx =\text {Too large to display} \] Input:
int((x^5*tan(a + b*x^3))/cos(a + b*x^3)^7,x)
Output:
(5*log(x^2*(exp(a*1i + b*x^3*1i) - 1i)))/(336*b^2) - ((8*exp(a*1i + b*x^3* 1i)*(15*b*x^3 - 8i))/(315*b^2) - (8*exp(a*3i + b*x^3*3i)*(35*b*x^3 - 12i)) /(315*b^2))/(5*exp(a*2i + b*x^3*2i) + 10*exp(a*4i + b*x^3*4i) + 10*exp(a*6 i + b*x^3*6i) + 5*exp(a*8i + b*x^3*8i) + exp(a*10i + b*x^3*10i) + 1) - (5* log(x^2*(exp(a*1i + b*x^3*1i) + 1i)))/(336*b^2) - ((16*exp(a*3i + b*x^3*3i )*(5*b*x^3 - 1i))/(63*b^2) - (16*exp(a*5i + b*x^3*5i)*(7*b*x^3 - 1i))/(63* b^2))/(6*exp(a*2i + b*x^3*2i) + 15*exp(a*4i + b*x^3*4i) + 20*exp(a*6i + b* x^3*6i) + 15*exp(a*8i + b*x^3*8i) + 6*exp(a*10i + b*x^3*10i) + exp(a*12i + b*x^3*12i) + 1) - ((64*x^3*exp(a*5i + b*x^3*5i))/(21*b) - (64*x^3*exp(a*7 i + b*x^3*7i))/(21*b))/(7*exp(a*2i + b*x^3*2i) + 21*exp(a*4i + b*x^3*4i) + 35*exp(a*6i + b*x^3*6i) + 35*exp(a*8i + b*x^3*8i) + 21*exp(a*10i + b*x^3* 10i) + 7*exp(a*12i + b*x^3*12i) + exp(a*14i + b*x^3*14i) + 1) + (exp(a*1i + b*x^3*1i)*1i)/(63*b^2*(3*exp(a*2i + b*x^3*2i) + 3*exp(a*4i + b*x^3*4i) + exp(a*6i + b*x^3*6i) + 1)) + (exp(a*1i + b*x^3*1i)*5i)/(168*b^2*(exp(a*2i + b*x^3*2i) + 1)) + (exp(a*1i + b*x^3*1i)*5i)/(252*b^2*(2*exp(a*2i + b*x^ 3*2i) + exp(a*4i + b*x^3*4i) + 1)) + (2*exp(a*1i + b*x^3*1i)*(60*b*x^3 - 4 7i))/(315*b^2*(4*exp(a*2i + b*x^3*2i) + 6*exp(a*4i + b*x^3*4i) + 4*exp(a*6 i + b*x^3*6i) + exp(a*8i + b*x^3*8i) + 1))
Time = 0.19 (sec) , antiderivative size = 370, normalized size of antiderivative = 3.36 \[ \int x^5 \sec ^7\left (a+b x^3\right ) \tan \left (a+b x^3\right ) \, dx=\frac {15 \cos \left (b \,x^{3}+a \right ) \mathrm {log}\left (\tan \left (\frac {b \,x^{3}}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b \,x^{3}+a \right )^{6}-45 \cos \left (b \,x^{3}+a \right ) \mathrm {log}\left (\tan \left (\frac {b \,x^{3}}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b \,x^{3}+a \right )^{4}+45 \cos \left (b \,x^{3}+a \right ) \mathrm {log}\left (\tan \left (\frac {b \,x^{3}}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b \,x^{3}+a \right )^{2}-15 \cos \left (b \,x^{3}+a \right ) \mathrm {log}\left (\tan \left (\frac {b \,x^{3}}{2}+\frac {a}{2}\right )-1\right )-15 \cos \left (b \,x^{3}+a \right ) \mathrm {log}\left (\tan \left (\frac {b \,x^{3}}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b \,x^{3}+a \right )^{6}+45 \cos \left (b \,x^{3}+a \right ) \mathrm {log}\left (\tan \left (\frac {b \,x^{3}}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b \,x^{3}+a \right )^{4}-45 \cos \left (b \,x^{3}+a \right ) \mathrm {log}\left (\tan \left (\frac {b \,x^{3}}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b \,x^{3}+a \right )^{2}+15 \cos \left (b \,x^{3}+a \right ) \mathrm {log}\left (\tan \left (\frac {b \,x^{3}}{2}+\frac {a}{2}\right )+1\right )+15 \cos \left (b \,x^{3}+a \right ) \sin \left (b \,x^{3}+a \right )^{5}-40 \cos \left (b \,x^{3}+a \right ) \sin \left (b \,x^{3}+a \right )^{3}+33 \cos \left (b \,x^{3}+a \right ) \sin \left (b \,x^{3}+a \right )-48 b \,x^{3}}{1008 \cos \left (b \,x^{3}+a \right ) b^{2} \left (\sin \left (b \,x^{3}+a \right )^{6}-3 \sin \left (b \,x^{3}+a \right )^{4}+3 \sin \left (b \,x^{3}+a \right )^{2}-1\right )} \] Input:
int(x^5*sec(b*x^3+a)^7*tan(b*x^3+a),x)
Output:
(15*cos(a + b*x**3)*log(tan((a + b*x**3)/2) - 1)*sin(a + b*x**3)**6 - 45*c os(a + b*x**3)*log(tan((a + b*x**3)/2) - 1)*sin(a + b*x**3)**4 + 45*cos(a + b*x**3)*log(tan((a + b*x**3)/2) - 1)*sin(a + b*x**3)**2 - 15*cos(a + b*x **3)*log(tan((a + b*x**3)/2) - 1) - 15*cos(a + b*x**3)*log(tan((a + b*x**3 )/2) + 1)*sin(a + b*x**3)**6 + 45*cos(a + b*x**3)*log(tan((a + b*x**3)/2) + 1)*sin(a + b*x**3)**4 - 45*cos(a + b*x**3)*log(tan((a + b*x**3)/2) + 1)* sin(a + b*x**3)**2 + 15*cos(a + b*x**3)*log(tan((a + b*x**3)/2) + 1) + 15* cos(a + b*x**3)*sin(a + b*x**3)**5 - 40*cos(a + b*x**3)*sin(a + b*x**3)**3 + 33*cos(a + b*x**3)*sin(a + b*x**3) - 48*b*x**3)/(1008*cos(a + b*x**3)*b **2*(sin(a + b*x**3)**6 - 3*sin(a + b*x**3)**4 + 3*sin(a + b*x**3)**2 - 1) )