Integrand size = 23, antiderivative size = 154 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {b^4}{4 a^3 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {b^3 (2 a+b)}{a^3 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac {\csc ^2(e+f x)}{2 (a+b)^3 f}-\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^3 (a+b)^4 f}-\frac {(a+4 b) \log (\sin (e+f x))}{(a+b)^4 f} \] Output:
1/4*b^4/a^3/(a+b)^2/f/(b+a*cos(f*x+e)^2)^2-b^3*(2*a+b)/a^3/(a+b)^3/f/(b+a* cos(f*x+e)^2)-1/2*csc(f*x+e)^2/(a+b)^3/f-1/2*b^2*(6*a^2+4*a*b+b^2)*ln(b+a* cos(f*x+e)^2)/a^3/(a+b)^4/f-(a+4*b)*ln(sin(f*x+e))/(a+b)^4/f
Time = 1.28 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.14 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x)))^3 \sec ^6(e+f x) \left (2 (a+b) \csc ^2(e+f x)+4 (a+4 b) \log (\sin (e+f x))+\frac {2 b^2 \left (6 a^2+4 a b+b^2\right ) \log \left (a+b-a \sin ^2(e+f x)\right )}{a^3}-\frac {b^4 (a+b)^2}{a^3 \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {4 b^3 (a+b) (2 a+b)}{a^3 \left (a+b-a \sin ^2(e+f x)\right )}\right )}{32 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^3} \] Input:
Integrate[Cot[e + f*x]^3/(a + b*Sec[e + f*x]^2)^3,x]
Output:
-1/32*((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*(2*(a + b)*Csc[e + f*x]^2 + 4*(a + 4*b)*Log[Sin[e + f*x]] + (2*b^2*(6*a^2 + 4*a*b + b^2)*Log[ a + b - a*Sin[e + f*x]^2])/a^3 - (b^4*(a + b)^2)/(a^3*(a + b - a*Sin[e + f *x]^2)^2) + (4*b^3*(a + b)*(2*a + b))/(a^3*(a + b - a*Sin[e + f*x]^2))))/( (a + b)^4*f*(a + b*Sec[e + f*x]^2)^3)
Time = 0.40 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4626, 354, 99, 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.
| \(\displaystyle \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^3 \left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle -\frac {\int \frac {\cos ^9(e+f x)}{\left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right )^3}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle -\frac {\int \frac {\cos ^8(e+f x)}{\left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right )^3}d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {\int \left (\frac {b^4}{a^2 (a+b)^2 \left (a \cos ^2(e+f x)+b\right )^3}-\frac {2 (2 a+b) b^3}{a^2 (a+b)^3 \left (a \cos ^2(e+f x)+b\right )^2}+\frac {\left (6 a^2+4 b a+b^2\right ) b^2}{a^2 (a+b)^4 \left (a \cos ^2(e+f x)+b\right )}+\frac {a+4 b}{(a+b)^4 \left (\cos ^2(e+f x)-1\right )}+\frac {1}{(a+b)^3 \left (\cos ^2(e+f x)-1\right )^2}\right )d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\frac {b^4}{2 a^3 (a+b)^2 \left (a \cos ^2(e+f x)+b\right )^2}+\frac {2 b^3 (2 a+b)}{a^3 (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac {b^2 \left (6 a^2+4 a b+b^2\right ) \log \left (a \cos ^2(e+f x)+b\right )}{a^3 (a+b)^4}+\frac {1}{(a+b)^3 \left (1-\cos ^2(e+f x)\right )}+\frac {(a+4 b) \log \left (1-\cos ^2(e+f x)\right )}{(a+b)^4}}{2 f}\) |
Input:
Int[Cot[e + f*x]^3/(a + b*Sec[e + f*x]^2)^3,x]
Output:
-1/2*(1/((a + b)^3*(1 - Cos[e + f*x]^2)) - b^4/(2*a^3*(a + b)^2*(b + a*Cos [e + f*x]^2)^2) + (2*b^3*(2*a + b))/(a^3*(a + b)^3*(b + a*Cos[e + f*x]^2)) + ((a + 4*b)*Log[1 - Cos[e + f*x]^2])/(a + b)^4 + (b^2*(6*a^2 + 4*a*b + b ^2)*Log[b + a*Cos[e + f*x]^2])/(a^3*(a + b)^4))/f
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Time = 17.36 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.25
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -4 b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{4}}-\frac {b^{2} \left (-\frac {b^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (6 a^{2}+4 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{3}}+\frac {2 b \left (2 a^{2}+3 a b +b^{2}\right )}{a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}\right )}{2 \left (a +b \right )^{4}}-\frac {1}{4 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -4 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{4}}}{f}\) | \(193\) |
| default | \(\frac {\frac {1}{4 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -4 b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{4}}-\frac {b^{2} \left (-\frac {b^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (6 a^{2}+4 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{3}}+\frac {2 b \left (2 a^{2}+3 a b +b^{2}\right )}{a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}\right )}{2 \left (a +b \right )^{4}}-\frac {1}{4 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -4 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{4}}}{f}\) | \(193\) |
| risch | \(\text {Expression too large to display}\) | \(1067\) |
Input:
int(cot(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
Output:
1/f*(1/4/(a+b)^3/(-1+cos(f*x+e))+1/2*(-a-4*b)/(a+b)^4*ln(-1+cos(f*x+e))-1/ 2*b^2/(a+b)^4*(-1/2*b^2*(a^2+2*a*b+b^2)/a^3/(b+a*cos(f*x+e)^2)^2+(6*a^2+4* a*b+b^2)/a^3*ln(b+a*cos(f*x+e)^2)+2/a^3*b*(2*a^2+3*a*b+b^2)/(b+a*cos(f*x+e )^2))-1/4/(a+b)^3/(1+cos(f*x+e))+1/2*(-a-4*b)/(a+b)^4*ln(1+cos(f*x+e)))
Leaf count of result is larger than twice the leaf count of optimal. 584 vs. \(2 (148) = 296\).
Time = 0.76 (sec) , antiderivative size = 584, normalized size of antiderivative = 3.79 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {2 \, a^{4} b^{2} + 2 \, a^{3} b^{3} + 7 \, a^{2} b^{4} + 10 \, a b^{5} + 3 \, b^{6} + 2 \, {\left (a^{6} + a^{5} b - 4 \, a^{3} b^{3} - 6 \, a^{2} b^{4} - 2 \, a b^{5}\right )} \cos \left (f x + e\right )^{4} + {\left (4 \, a^{5} b + 4 \, a^{4} b^{2} + 8 \, a^{3} b^{3} + 5 \, a^{2} b^{4} - 6 \, a b^{5} - 3 \, b^{6}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left ({\left (6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \cos \left (f x + e\right )^{6} - 6 \, a^{2} b^{4} - 4 \, a b^{5} - b^{6} - {\left (6 \, a^{4} b^{2} - 8 \, a^{3} b^{3} - 7 \, a^{2} b^{4} - 2 \, a b^{5}\right )} \cos \left (f x + e\right )^{4} - {\left (12 \, a^{3} b^{3} + 2 \, a^{2} b^{4} - 2 \, a b^{5} - b^{6}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 4 \, {\left ({\left (a^{6} + 4 \, a^{5} b\right )} \cos \left (f x + e\right )^{6} - a^{4} b^{2} - 4 \, a^{3} b^{3} - {\left (a^{6} + 2 \, a^{5} b - 8 \, a^{4} b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (2 \, a^{5} b + 7 \, a^{4} b^{2} - 4 \, a^{3} b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{4 \, {\left ({\left (a^{9} + 4 \, a^{8} b + 6 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + a^{5} b^{4}\right )} f \cos \left (f x + e\right )^{6} - {\left (a^{9} + 2 \, a^{8} b - 2 \, a^{7} b^{2} - 8 \, a^{6} b^{3} - 7 \, a^{5} b^{4} - 2 \, a^{4} b^{5}\right )} f \cos \left (f x + e\right )^{4} - {\left (2 \, a^{8} b + 7 \, a^{7} b^{2} + 8 \, a^{6} b^{3} + 2 \, a^{5} b^{4} - 2 \, a^{4} b^{5} - a^{3} b^{6}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{7} b^{2} + 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} + 4 \, a^{4} b^{5} + a^{3} b^{6}\right )} f\right )}} \] Input:
integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
Output:
1/4*(2*a^4*b^2 + 2*a^3*b^3 + 7*a^2*b^4 + 10*a*b^5 + 3*b^6 + 2*(a^6 + a^5*b - 4*a^3*b^3 - 6*a^2*b^4 - 2*a*b^5)*cos(f*x + e)^4 + (4*a^5*b + 4*a^4*b^2 + 8*a^3*b^3 + 5*a^2*b^4 - 6*a*b^5 - 3*b^6)*cos(f*x + e)^2 - 2*((6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*cos(f*x + e)^6 - 6*a^2*b^4 - 4*a*b^5 - b^6 - (6*a^4 *b^2 - 8*a^3*b^3 - 7*a^2*b^4 - 2*a*b^5)*cos(f*x + e)^4 - (12*a^3*b^3 + 2*a ^2*b^4 - 2*a*b^5 - b^6)*cos(f*x + e)^2)*log(a*cos(f*x + e)^2 + b) - 4*((a^ 6 + 4*a^5*b)*cos(f*x + e)^6 - a^4*b^2 - 4*a^3*b^3 - (a^6 + 2*a^5*b - 8*a^4 *b^2)*cos(f*x + e)^4 - (2*a^5*b + 7*a^4*b^2 - 4*a^3*b^3)*cos(f*x + e)^2)*l og(1/2*sin(f*x + e)))/((a^9 + 4*a^8*b + 6*a^7*b^2 + 4*a^6*b^3 + a^5*b^4)*f *cos(f*x + e)^6 - (a^9 + 2*a^8*b - 2*a^7*b^2 - 8*a^6*b^3 - 7*a^5*b^4 - 2*a ^4*b^5)*f*cos(f*x + e)^4 - (2*a^8*b + 7*a^7*b^2 + 8*a^6*b^3 + 2*a^5*b^4 - 2*a^4*b^5 - a^3*b^6)*f*cos(f*x + e)^2 - (a^7*b^2 + 4*a^6*b^3 + 6*a^5*b^4 + 4*a^4*b^5 + a^3*b^6)*f)
Timed out. \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(cot(f*x+e)**3/(a+b*sec(f*x+e)**2)**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (148) = 296\).
Time = 0.04 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.23 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {2 \, {\left (6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4}} + \frac {2 \, {\left (a + 4 \, b\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {2 \, a^{5} + 4 \, a^{4} b + 2 \, a^{3} b^{2} + 2 \, {\left (a^{5} - 4 \, a^{2} b^{3} - 2 \, a b^{4}\right )} \sin \left (f x + e\right )^{4} - {\left (4 \, a^{5} + 4 \, a^{4} b - 8 \, a^{2} b^{3} - 11 \, a b^{4} - 3 \, b^{5}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3}\right )} \sin \left (f x + e\right )^{6} - 2 \, {\left (a^{8} + 4 \, a^{7} b + 6 \, a^{6} b^{2} + 4 \, a^{5} b^{3} + a^{4} b^{4}\right )} \sin \left (f x + e\right )^{4} + {\left (a^{8} + 5 \, a^{7} b + 10 \, a^{6} b^{2} + 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} + a^{3} b^{5}\right )} \sin \left (f x + e\right )^{2}}}{4 \, f} \] Input:
integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
Output:
-1/4*(2*(6*a^2*b^2 + 4*a*b^3 + b^4)*log(a*sin(f*x + e)^2 - a - b)/(a^7 + 4 *a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4) + 2*(a + 4*b)*log(sin(f*x + e)^2 )/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + (2*a^5 + 4*a^4*b + 2*a^3*b ^2 + 2*(a^5 - 4*a^2*b^3 - 2*a*b^4)*sin(f*x + e)^4 - (4*a^5 + 4*a^4*b - 8*a ^2*b^3 - 11*a*b^4 - 3*b^5)*sin(f*x + e)^2)/((a^8 + 3*a^7*b + 3*a^6*b^2 + a ^5*b^3)*sin(f*x + e)^6 - 2*(a^8 + 4*a^7*b + 6*a^6*b^2 + 4*a^5*b^3 + a^4*b^ 4)*sin(f*x + e)^4 + (a^8 + 5*a^7*b + 10*a^6*b^2 + 10*a^5*b^3 + 5*a^4*b^4 + a^3*b^5)*sin(f*x + e)^2))/f
Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (148) = 296\).
Time = 0.17 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.98 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {{\left (6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \log \left ({\left | a \cos \left (f x + e\right )^{2} + b \right |}\right )}{2 \, {\left (a^{7} f + 4 \, a^{6} b f + 6 \, a^{5} b^{2} f + 4 \, a^{4} b^{3} f + a^{3} b^{4} f\right )}} - \frac {{\left (a + 4 \, b\right )} \log \left ({\left | -\cos \left (f x + e\right )^{2} + 1 \right |}\right )}{2 \, {\left (a^{4} f + 4 \, a^{3} b f + 6 \, a^{2} b^{2} f + 4 \, a b^{3} f + b^{4} f\right )}} + \frac {2 \, {\left (a^{5} + a^{4} b - 4 \, a^{2} b^{3} - 6 \, a b^{4} - 2 \, b^{5}\right )} \cos \left (f x + e\right )^{4} + \frac {{\left (4 \, a^{5} b + 4 \, a^{4} b^{2} + 8 \, a^{3} b^{3} + 5 \, a^{2} b^{4} - 6 \, a b^{5} - 3 \, b^{6}\right )} \cos \left (f x + e\right )^{2}}{a} + \frac {2 \, a^{4} b^{2} + 2 \, a^{3} b^{3} + 7 \, a^{2} b^{4} + 10 \, a b^{5} + 3 \, b^{6}}{a}}{4 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )}^{2} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\left (a + b\right )}^{4} a^{2} f} \] Input:
integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
Output:
-1/2*(6*a^2*b^2 + 4*a*b^3 + b^4)*log(abs(a*cos(f*x + e)^2 + b))/(a^7*f + 4 *a^6*b*f + 6*a^5*b^2*f + 4*a^4*b^3*f + a^3*b^4*f) - 1/2*(a + 4*b)*log(abs( -cos(f*x + e)^2 + 1))/(a^4*f + 4*a^3*b*f + 6*a^2*b^2*f + 4*a*b^3*f + b^4*f ) + 1/4*(2*(a^5 + a^4*b - 4*a^2*b^3 - 6*a*b^4 - 2*b^5)*cos(f*x + e)^4 + (4 *a^5*b + 4*a^4*b^2 + 8*a^3*b^3 + 5*a^2*b^4 - 6*a*b^5 - 3*b^6)*cos(f*x + e) ^2/a + (2*a^4*b^2 + 2*a^3*b^3 + 7*a^2*b^4 + 10*a*b^5 + 3*b^6)/a)/((a*cos(f *x + e)^2 + b)^2*(cos(f*x + e)^2 - 1)*(a + b)^4*a^2*f)
Time = 19.69 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.77 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (-4\,a^2\,b+7\,a\,b^2+2\,b^3\right )}{4\,a^2\,\left (a^2+2\,a\,b+b^2\right )}-\frac {1}{2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (-a^2\,b^2+3\,a\,b^3+b^4\right )}{2\,a^2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a^2+2\,a\,b+b^2\right )+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^6\right )}+\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a^3\,f}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a+4\,b\right )}{f\,\left (a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4\right )}-\frac {b^2\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )\,\left (6\,a^2+4\,a\,b+b^2\right )}{2\,a^3\,f\,{\left (a+b\right )}^4} \] Input:
int(cot(e + f*x)^3/(a + b/cos(e + f*x)^2)^3,x)
Output:
((tan(e + f*x)^2*(7*a*b^2 - 4*a^2*b + 2*b^3))/(4*a^2*(2*a*b + a^2 + b^2)) - 1/(2*(a + b)) + (tan(e + f*x)^4*(3*a*b^3 + b^4 - a^2*b^2))/(2*a^2*(a + b )*(2*a*b + a^2 + b^2)))/(f*(tan(e + f*x)^2*(2*a*b + a^2 + b^2) + tan(e + f *x)^4*(2*a*b + 2*b^2) + b^2*tan(e + f*x)^6)) + log(tan(e + f*x)^2 + 1)/(2* a^3*f) - (log(tan(e + f*x))*(a + 4*b))/(f*(4*a*b^3 + 4*a^3*b + a^4 + b^4 + 6*a^2*b^2)) - (b^2*log(a + b + b*tan(e + f*x)^2)*(4*a*b + 6*a^2 + b^2))/( 2*a^3*f*(a + b)^4)
Time = 0.29 (sec) , antiderivative size = 2392, normalized size of antiderivative = 15.53 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:
int(cot(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x)
Output:
(4*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**6*a**6 + 16*log(tan((e + f*x )/2)**2 + 1)*sin(e + f*x)**6*a**5*b + 24*log(tan((e + f*x)/2)**2 + 1)*sin( e + f*x)**6*a**4*b**2 + 16*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**6*a* *3*b**3 + 4*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**6*a**2*b**4 - 8*log (tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a**6 - 40*log(tan((e + f*x)/2)** 2 + 1)*sin(e + f*x)**4*a**5*b - 80*log(tan((e + f*x)/2)**2 + 1)*sin(e + f* x)**4*a**4*b**2 - 80*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a**3*b** 3 - 40*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a**2*b**4 - 8*log(tan( (e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a*b**5 + 4*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a**6 + 24*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2* a**5*b + 60*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a**4*b**2 + 80*lo g(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a**3*b**3 + 60*log(tan((e + f*x )/2)**2 + 1)*sin(e + f*x)**2*a**2*b**4 + 24*log(tan((e + f*x)/2)**2 + 1)*s in(e + f*x)**2*a*b**5 + 4*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*b** 6 - 12*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan(( e + f*x)/2))*sin(e + f*x)**6*a**4*b**2 - 8*log(sqrt(a + b)*tan((e + f*x)/2 )**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**6*a**3*b**3 - 2*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**6*a**2*b**4 + 24*log(sqrt(a + b)*tan((e + f*x)/2) **2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**4*a**4*b*...