Integrand size = 13, antiderivative size = 1 \[ \int \frac {\sec ^6(x)}{a+b \cot (x)} \, dx=0 \] Output:
0
Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.
Time = 0.45 (sec) , antiderivative size = 95, normalized size of antiderivative = 95.00 \[ \int \frac {\sec ^6(x)}{a+b \cot (x)} \, dx=\frac {-60 b \left (a^2+b^2\right )^2 \log (b+a \tan (x))-15 a^4 b \sec ^4(x)+60 a \left (a^2+b^2\right )^2 \tan (x)-30 a^2 b \left (a^2+b^2\right ) \tan ^2(x)+20 a^3 \left (2 a^2+b^2\right ) \tan ^3(x)+12 a^5 \tan ^5(x)}{60 a^6} \] Input:
Integrate[Sec[x]^6/(a + b*Cot[x]),x]
Output:
(-60*b*(a^2 + b^2)^2*Log[b + a*Tan[x]] - 15*a^4*b*Sec[x]^4 + 60*a*(a^2 + b ^2)^2*Tan[x] - 30*a^2*b*(a^2 + b^2)*Tan[x]^2 + 20*a^3*(2*a^2 + b^2)*Tan[x] ^3 + 12*a^5*Tan[x]^5)/(60*a^6)
Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.
Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 130.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3999, 522, 2009}
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.
\(\displaystyle \int \frac {\sec ^6(x)}{a+b \cot (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (x+\frac {\pi }{2}\right )^6 \left (a-b \tan \left (x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3999 |
\(\displaystyle -b \int \frac {\left (\cot ^2(x) b^2+b^2\right )^2 \tan ^6(x)}{b^6 (a+b \cot (x))}d(b \cot (x))\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -b \int \left (\frac {\tan ^6(x)}{a b^2}-\frac {\tan ^5(x)}{a^2 b}+\frac {\left (b^4+2 a^2 b^2\right ) \tan ^4(x)}{a^3 b^4}+\frac {\left (-2 a^2-b^2\right ) \tan ^3(x)}{a^4 b}+\frac {\left (a^2+b^2\right )^2 \tan ^2(x)}{a^5 b^2}-\frac {\left (a^2+b^2\right )^2 \tan (x)}{a^6 b}+\frac {\left (a^2+b^2\right )^2}{a^6 (a+b \cot (x))}\right )d(b \cot (x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -b \left (\frac {\tan ^4(x)}{4 a^2}-\frac {\left (a^2+b^2\right )^2 \log (b \cot (x))}{a^6}+\frac {\left (a^2+b^2\right )^2 \log (a+b \cot (x))}{a^6}-\frac {\left (a^2+b^2\right )^2 \tan (x)}{a^5 b}+\frac {\left (2 a^2+b^2\right ) \tan ^2(x)}{2 a^4}-\frac {\left (2 a^2+b^2\right ) \tan ^3(x)}{3 a^3 b}-\frac {\tan ^5(x)}{5 a b}\right )\) |
Input:
Int[Sec[x]^6/(a + b*Cot[x]),x]
Output:
-(b*(-(((a^2 + b^2)^2*Log[b*Cot[x]])/a^6) + ((a^2 + b^2)^2*Log[a + b*Cot[x ]])/a^6 - ((a^2 + b^2)^2*Tan[x])/(a^5*b) + ((2*a^2 + b^2)*Tan[x]^2)/(2*a^4 ) - ((2*a^2 + b^2)*Tan[x]^3)/(3*a^3*b) + Tan[x]^4/(4*a^2) - Tan[x]^5/(5*a* b)))
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[b/f Subst[Int[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/2]
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 32.54 (sec) , antiderivative size = 117, normalized size of antiderivative = 117.00
method | result | size |
default | \(\frac {\frac {\tan \left (x \right )^{5} a^{4}}{5}-\frac {b \tan \left (x \right )^{4} a^{3}}{4}+\frac {2 a^{4} \tan \left (x \right )^{3}}{3}+\frac {a^{2} b^{2} \tan \left (x \right )^{3}}{3}-a^{3} b \tan \left (x \right )^{2}-\frac {a \,b^{3} \tan \left (x \right )^{2}}{2}+a^{4} \tan \left (x \right )+2 a^{2} b^{2} \tan \left (x \right )+b^{4} \tan \left (x \right )}{a^{5}}-\frac {b \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \ln \left (\tan \left (x \right ) a +b \right )}{a^{6}}\) | \(117\) |
risch | \(\frac {2 i \left (45 i a \,b^{3} {\mathrm e}^{4 i x}+45 i a \,b^{3} {\mathrm e}^{6 i x}+15 a^{2} b^{2} {\mathrm e}^{8 i x}+15 b^{4} {\mathrm e}^{8 i x}+15 i a^{3} b \,{\mathrm e}^{2 i x}+75 i a^{3} b \,{\mathrm e}^{6 i x}+90 a^{2} b^{2} {\mathrm e}^{6 i x}+60 b^{4} {\mathrm e}^{6 i x}+75 i a^{3} b \,{\mathrm e}^{4 i x}+15 i a \,b^{3} {\mathrm e}^{8 i x}+80 a^{4} {\mathrm e}^{4 i x}+160 a^{2} b^{2} {\mathrm e}^{4 i x}+90 b^{4} {\mathrm e}^{4 i x}+15 i a \,b^{3} {\mathrm e}^{2 i x}+15 i a^{3} b \,{\mathrm e}^{8 i x}+40 a^{4} {\mathrm e}^{2 i x}+110 a^{2} b^{2} {\mathrm e}^{2 i x}+60 b^{4} {\mathrm e}^{2 i x}+8 a^{4}+25 a^{2} b^{2}+15 b^{4}\right )}{15 a^{5} \left ({\mathrm e}^{2 i x}+1\right )^{5}}+\frac {b \ln \left ({\mathrm e}^{2 i x}+1\right )}{a^{2}}+\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i x}+1\right )}{a^{4}}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i x}+1\right )}{a^{6}}-\frac {b \ln \left ({\mathrm e}^{2 i x}+\frac {i b -a}{i b +a}\right )}{a^{2}}-\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i x}+\frac {i b -a}{i b +a}\right )}{a^{4}}-\frac {b^{5} \ln \left ({\mathrm e}^{2 i x}+\frac {i b -a}{i b +a}\right )}{a^{6}}\) | \(382\) |
Input:
int(sec(x)^6/(a+b*cot(x)),x,method=_RETURNVERBOSE)
Output:
1/a^5*(1/5*tan(x)^5*a^4-1/4*b*tan(x)^4*a^3+2/3*a^4*tan(x)^3+1/3*a^2*b^2*ta n(x)^3-a^3*b*tan(x)^2-1/2*a*b^3*tan(x)^2+a^4*tan(x)+2*a^2*b^2*tan(x)+b^4*t an(x))-b*(a^4+2*a^2*b^2+b^4)/a^6*ln(tan(x)*a+b)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.10 (sec) , antiderivative size = 170, normalized size of antiderivative = 170.00 \[ \int \frac {\sec ^6(x)}{a+b \cot (x)} \, dx=-\frac {30 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{5} \log \left (2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}\right ) - 30 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{5} \log \left (\cos \left (x\right )^{2}\right ) + 15 \, a^{4} b \cos \left (x\right ) + 30 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (x\right )^{3} - 4 \, {\left (3 \, a^{5} + {\left (8 \, a^{5} + 25 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (x\right )^{4} + {\left (4 \, a^{5} + 5 \, a^{3} b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{60 \, a^{6} \cos \left (x\right )^{5}} \] Input:
integrate(sec(x)^6/(a+b*cot(x)),x, algorithm="fricas")
Output:
-1/60*(30*(a^4*b + 2*a^2*b^3 + b^5)*cos(x)^5*log(2*a*b*cos(x)*sin(x) - (a^ 2 - b^2)*cos(x)^2 + a^2) - 30*(a^4*b + 2*a^2*b^3 + b^5)*cos(x)^5*log(cos(x )^2) + 15*a^4*b*cos(x) + 30*(a^4*b + a^2*b^3)*cos(x)^3 - 4*(3*a^5 + (8*a^5 + 25*a^3*b^2 + 15*a*b^4)*cos(x)^4 + (4*a^5 + 5*a^3*b^2)*cos(x)^2)*sin(x)) /(a^6*cos(x)^5)
\[ \int \frac {\sec ^6(x)}{a+b \cot (x)} \, dx=\int \frac {\sec ^{6}{\left (x \right )}}{a + b \cot {\left (x \right )}}\, dx \] Input:
integrate(sec(x)**6/(a+b*cot(x)),x)
Output:
Integral(sec(x)**6/(a + b*cot(x)), x)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.03 (sec) , antiderivative size = 111, normalized size of antiderivative = 111.00 \[ \int \frac {\sec ^6(x)}{a+b \cot (x)} \, dx=\frac {12 \, a^{4} \tan \left (x\right )^{5} - 15 \, a^{3} b \tan \left (x\right )^{4} + 20 \, {\left (2 \, a^{4} + a^{2} b^{2}\right )} \tan \left (x\right )^{3} - 30 \, {\left (2 \, a^{3} b + a b^{3}\right )} \tan \left (x\right )^{2} + 60 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (x\right )}{60 \, a^{5}} - \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (a \tan \left (x\right ) + b\right )}{a^{6}} \] Input:
integrate(sec(x)^6/(a+b*cot(x)),x, algorithm="maxima")
Output:
1/60*(12*a^4*tan(x)^5 - 15*a^3*b*tan(x)^4 + 20*(2*a^4 + a^2*b^2)*tan(x)^3 - 30*(2*a^3*b + a*b^3)*tan(x)^2 + 60*(a^4 + 2*a^2*b^2 + b^4)*tan(x))/a^5 - (a^4*b + 2*a^2*b^3 + b^5)*log(a*tan(x) + b)/a^6
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 121.00 \[ \int \frac {\sec ^6(x)}{a+b \cot (x)} \, dx=\frac {12 \, a^{4} \tan \left (x\right )^{5} - 15 \, a^{3} b \tan \left (x\right )^{4} + 40 \, a^{4} \tan \left (x\right )^{3} + 20 \, a^{2} b^{2} \tan \left (x\right )^{3} - 60 \, a^{3} b \tan \left (x\right )^{2} - 30 \, a b^{3} \tan \left (x\right )^{2} + 60 \, a^{4} \tan \left (x\right ) + 120 \, a^{2} b^{2} \tan \left (x\right ) + 60 \, b^{4} \tan \left (x\right )}{60 \, a^{5}} - \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | a \tan \left (x\right ) + b \right |}\right )}{a^{6}} \] Input:
integrate(sec(x)^6/(a+b*cot(x)),x, algorithm="giac")
Output:
1/60*(12*a^4*tan(x)^5 - 15*a^3*b*tan(x)^4 + 40*a^4*tan(x)^3 + 20*a^2*b^2*t an(x)^3 - 60*a^3*b*tan(x)^2 - 30*a*b^3*tan(x)^2 + 60*a^4*tan(x) + 120*a^2* b^2*tan(x) + 60*b^4*tan(x))/a^5 - (a^4*b + 2*a^2*b^3 + b^5)*log(abs(a*tan( x) + b))/a^6
Time = 8.87 (sec) , antiderivative size = 118, normalized size of antiderivative = 118.00 \[ \int \frac {\sec ^6(x)}{a+b \cot (x)} \, dx={\mathrm {tan}\left (x\right )}^3\,\left (\frac {2}{3\,a}+\frac {b^2}{3\,a^3}\right )+\frac {{\mathrm {tan}\left (x\right )}^5}{5\,a}+\mathrm {tan}\left (x\right )\,\left (\frac {1}{a}+\frac {b^2\,\left (\frac {2}{a}+\frac {b^2}{a^3}\right )}{a^2}\right )-\frac {\ln \left (b+a\,\mathrm {tan}\left (x\right )\right )\,\left (a^4\,b+2\,a^2\,b^3+b^5\right )}{a^6}-\frac {b\,{\mathrm {tan}\left (x\right )}^4}{4\,a^2}-\frac {b\,{\mathrm {tan}\left (x\right )}^2\,\left (\frac {2}{a}+\frac {b^2}{a^3}\right )}{2\,a} \] Input:
int(1/(cos(x)^6*(a + b*cot(x))),x)
Output:
tan(x)^3*(2/(3*a) + b^2/(3*a^3)) + tan(x)^5/(5*a) + tan(x)*(1/a + (b^2*(2/ a + b^2/a^3))/a^2) - (log(b + a*tan(x))*(a^4*b + b^5 + 2*a^2*b^3))/a^6 - ( b*tan(x)^4)/(4*a^2) - (b*tan(x)^2*(2/a + b^2/a^3))/(2*a)
Time = 0.18 (sec) , antiderivative size = 784, normalized size of antiderivative = 784.00 \[ \int \frac {\sec ^6(x)}{a+b \cot (x)} \, dx =\text {Too large to display} \] Input:
int(sec(x)^6/(a+b*cot(x)),x)
Output:
(60*cos(x)*log(tan(x/2) - 1)*sin(x)**4*a**4*b + 120*cos(x)*log(tan(x/2) - 1)*sin(x)**4*a**2*b**3 + 60*cos(x)*log(tan(x/2) - 1)*sin(x)**4*b**5 - 120* cos(x)*log(tan(x/2) - 1)*sin(x)**2*a**4*b - 240*cos(x)*log(tan(x/2) - 1)*s in(x)**2*a**2*b**3 - 120*cos(x)*log(tan(x/2) - 1)*sin(x)**2*b**5 + 60*cos( x)*log(tan(x/2) - 1)*a**4*b + 120*cos(x)*log(tan(x/2) - 1)*a**2*b**3 + 60* cos(x)*log(tan(x/2) - 1)*b**5 + 60*cos(x)*log(tan(x/2) + 1)*sin(x)**4*a**4 *b + 120*cos(x)*log(tan(x/2) + 1)*sin(x)**4*a**2*b**3 + 60*cos(x)*log(tan( x/2) + 1)*sin(x)**4*b**5 - 120*cos(x)*log(tan(x/2) + 1)*sin(x)**2*a**4*b - 240*cos(x)*log(tan(x/2) + 1)*sin(x)**2*a**2*b**3 - 120*cos(x)*log(tan(x/2 ) + 1)*sin(x)**2*b**5 + 60*cos(x)*log(tan(x/2) + 1)*a**4*b + 120*cos(x)*lo g(tan(x/2) + 1)*a**2*b**3 + 60*cos(x)*log(tan(x/2) + 1)*b**5 - 60*cos(x)*l og(tan(x/2)**2*b - 2*tan(x/2)*a - b)*sin(x)**4*a**4*b - 120*cos(x)*log(tan (x/2)**2*b - 2*tan(x/2)*a - b)*sin(x)**4*a**2*b**3 - 60*cos(x)*log(tan(x/2 )**2*b - 2*tan(x/2)*a - b)*sin(x)**4*b**5 + 120*cos(x)*log(tan(x/2)**2*b - 2*tan(x/2)*a - b)*sin(x)**2*a**4*b + 240*cos(x)*log(tan(x/2)**2*b - 2*tan (x/2)*a - b)*sin(x)**2*a**2*b**3 + 120*cos(x)*log(tan(x/2)**2*b - 2*tan(x/ 2)*a - b)*sin(x)**2*b**5 - 60*cos(x)*log(tan(x/2)**2*b - 2*tan(x/2)*a - b) *a**4*b - 120*cos(x)*log(tan(x/2)**2*b - 2*tan(x/2)*a - b)*a**2*b**3 - 60* cos(x)*log(tan(x/2)**2*b - 2*tan(x/2)*a - b)*b**5 - 3*cos(x)*sin(x)**4*a** 4*b + 6*cos(x)*sin(x)**4*a**2*b**3 + 36*cos(x)*sin(x)**2*a**4*b + 18*co...