Integrand size = 23, antiderivative size = 393 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\frac {b^{4/3} \left (a^2-3 a^{2/3} b^{4/3}+2 b^2\right ) \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{a} \left (a^2-b^2\right )^2 f}-\frac {1}{4 (a+b) f (1-\cos (e+f x))}-\frac {1}{4 (a-b) f (1+\cos (e+f x))}-\frac {(2 a+5 b) \log (1-\cos (e+f x))}{4 (a+b)^2 f}-\frac {(2 a-5 b) \log (1+\cos (e+f x))}{4 (a-b)^2 f}-\frac {b^{4/3} \left (a^2+3 a^{2/3} b^{4/3}+2 b^2\right ) \log \left (\sqrt [3]{b}+\sqrt [3]{a} \cos (e+f x)\right )}{3 \sqrt [3]{a} \left (a^2-b^2\right )^2 f}+\frac {b^{4/3} \left (a^2+3 a^{2/3} b^{4/3}+2 b^2\right ) \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+a^{2/3} \cos ^2(e+f x)\right )}{6 \sqrt [3]{a} \left (a^2-b^2\right )^2 f}-\frac {b^2 \left (2 a^2+b^2\right ) \log \left (b+a \cos ^3(e+f x)\right )}{3 a \left (a^2-b^2\right )^2 f} \] Output:
1/3*b^(4/3)*(a^2-3*a^(2/3)*b^(4/3)+2*b^2)*arctan(1/3*(b^(1/3)-2*a^(1/3)*co s(f*x+e))*3^(1/2)/b^(1/3))*3^(1/2)/a^(1/3)/(a^2-b^2)^2/f-1/4/(a+b)/f/(1-co s(f*x+e))-1/4/(a-b)/f/(1+cos(f*x+e))-1/4*(2*a+5*b)*ln(1-cos(f*x+e))/(a+b)^ 2/f-1/4*(2*a-5*b)*ln(1+cos(f*x+e))/(a-b)^2/f-1/3*b^(4/3)*(a^2+3*a^(2/3)*b^ (4/3)+2*b^2)*ln(b^(1/3)+a^(1/3)*cos(f*x+e))/a^(1/3)/(a^2-b^2)^2/f+1/6*b^(4 /3)*(a^2+3*a^(2/3)*b^(4/3)+2*b^2)*ln(b^(2/3)-a^(1/3)*b^(1/3)*cos(f*x+e)+a^ (2/3)*cos(f*x+e)^2)/a^(1/3)/(a^2-b^2)^2/f-1/3*b^2*(2*a^2+b^2)*ln(b+a*cos(f *x+e)^3)/a/(a^2-b^2)^2/f
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 1.65 (sec) , antiderivative size = 336, normalized size of antiderivative = 0.85 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\frac {-\frac {3 \csc ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}+\frac {12 (-2 a+5 b) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{(a-b)^2}-\frac {12 (2 a+5 b) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(a+b)^2}+\frac {8 b^2 \left (3 \left (2 a^2+b^2\right ) \log \left (\sec ^2\left (\frac {1}{2} (e+f x)\right )\right )+(-a+b) \text {RootSum}\left [-8 a+12 a \text {$\#$1}-6 a \text {$\#$1}^2+a \text {$\#$1}^3-b \text {$\#$1}^3\&,\frac {8 a^2 \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 a b \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-6 a^2 \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}+2 a^2 \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}^2+b^2 \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}^2}{4 a-4 a \text {$\#$1}+a \text {$\#$1}^2-b \text {$\#$1}^2}\&\right ]\right )}{a \left (a^2-b^2\right )^2}-\frac {3 \sec ^2\left (\frac {1}{2} (e+f x)\right )}{a-b}}{24 f} \] Input:
Integrate[Cot[e + f*x]^3/(a + b*Sec[e + f*x]^3),x]
Output:
((-3*Csc[(e + f*x)/2]^2)/(a + b) + (12*(-2*a + 5*b)*Log[Cos[(e + f*x)/2]]) /(a - b)^2 - (12*(2*a + 5*b)*Log[Sin[(e + f*x)/2]])/(a + b)^2 + (8*b^2*(3* (2*a^2 + b^2)*Log[Sec[(e + f*x)/2]^2] + (-a + b)*RootSum[-8*a + 12*a*#1 - 6*a*#1^2 + a*#1^3 - b*#1^3 & , (8*a^2*Log[1 - #1 + Tan[(e + f*x)/2]^2] - 4 *a*b*Log[1 - #1 + Tan[(e + f*x)/2]^2] - 6*a^2*Log[1 - #1 + Tan[(e + f*x)/2 ]^2]*#1 + 2*a^2*Log[1 - #1 + Tan[(e + f*x)/2]^2]*#1^2 + b^2*Log[1 - #1 + T an[(e + f*x)/2]^2]*#1^2)/(4*a - 4*a*#1 + a*#1^2 - b*#1^2) & ]))/(a*(a^2 - b^2)^2) - (3*Sec[(e + f*x)/2]^2)/(a - b))/(24*f)
Time = 0.84 (sec) , antiderivative size = 375, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4626, 7276, 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.
| \(\displaystyle \int \frac {\cot ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^3 \left (a+b \sec (e+f x)^3\right )}dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle -\frac {\int \frac {\cos ^6(e+f x)}{\left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^3(e+f x)+b\right )}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle -\frac {\int \left (\frac {\left (a^2-3 b \cos (e+f x) a+2 b^2+\left (2 a^2+b^2\right ) \cos ^2(e+f x)\right ) b^2}{\left (a^2-b^2\right )^2 \left (a \cos ^3(e+f x)+b\right )}+\frac {2 a+5 b}{4 (a+b)^2 (\cos (e+f x)-1)}+\frac {2 a-5 b}{4 (a-b)^2 (\cos (e+f x)+1)}+\frac {1}{4 (a+b) (\cos (e+f x)-1)^2}-\frac {1}{4 (a-b) (\cos (e+f x)+1)^2}\right )d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {b^2 \left (2 a^2+b^2\right ) \log \left (a \cos ^3(e+f x)+b\right )}{3 a \left (a^2-b^2\right )^2}-\frac {b^{4/3} \left (-3 a^{2/3} b^{4/3}+a^2+2 b^2\right ) \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{a} \left (a^2-b^2\right )^2}-\frac {b^{4/3} \left (3 a^{2/3} b^{4/3}+a^2+2 b^2\right ) \log \left (a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}\right )}{6 \sqrt [3]{a} \left (a^2-b^2\right )^2}+\frac {b^{4/3} \left (3 a^{2/3} b^{4/3}+a^2+2 b^2\right ) \log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} \left (a^2-b^2\right )^2}+\frac {1}{4 (a+b) (1-\cos (e+f x))}+\frac {1}{4 (a-b) (\cos (e+f x)+1)}+\frac {(2 a+5 b) \log (1-\cos (e+f x))}{4 (a+b)^2}+\frac {(2 a-5 b) \log (\cos (e+f x)+1)}{4 (a-b)^2}}{f}\) |
Input:
Int[Cot[e + f*x]^3/(a + b*Sec[e + f*x]^3),x]
Output:
-((-((b^(4/3)*(a^2 - 3*a^(2/3)*b^(4/3) + 2*b^2)*ArcTan[(b^(1/3) - 2*a^(1/3 )*Cos[e + f*x])/(Sqrt[3]*b^(1/3))])/(Sqrt[3]*a^(1/3)*(a^2 - b^2)^2)) + 1/( 4*(a + b)*(1 - Cos[e + f*x])) + 1/(4*(a - b)*(1 + Cos[e + f*x])) + ((2*a + 5*b)*Log[1 - Cos[e + f*x]])/(4*(a + b)^2) + ((2*a - 5*b)*Log[1 + Cos[e + f*x]])/(4*(a - b)^2) + (b^(4/3)*(a^2 + 3*a^(2/3)*b^(4/3) + 2*b^2)*Log[b^(1 /3) + a^(1/3)*Cos[e + f*x]])/(3*a^(1/3)*(a^2 - b^2)^2) - (b^(4/3)*(a^2 + 3 *a^(2/3)*b^(4/3) + 2*b^2)*Log[b^(2/3) - a^(1/3)*b^(1/3)*Cos[e + f*x] + a^( 2/3)*Cos[e + f*x]^2])/(6*a^(1/3)*(a^2 - b^2)^2) + (b^2*(2*a^2 + b^2)*Log[b + a*Cos[e + f*x]^3])/(3*a*(a^2 - b^2)^2))/f)
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]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 5.51 (sec) , antiderivative size = 374, normalized size of antiderivative = 0.95
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{\left (4 a -4 b \right ) \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-2 a +5 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{4 \left (a -b \right )^{2}}+\frac {1}{\left (4 a +4 b \right ) \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (-2 a -5 b \right ) \ln \left (\cos \left (f x +e \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {\left (\left (a^{2}+2 b^{2}\right ) \left (\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {\ln \left (\cos \left (f x +e \right )^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}\right )-3 a b \left (-\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\ln \left (\cos \left (f x +e \right )^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )+\frac {\left (2 a^{2}+b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{3}\right )}{3 a}\right ) b^{2}}{\left (a -b \right )^{2} \left (a +b \right )^{2}}}{f}\) | \(374\) |
| default | \(\frac {-\frac {1}{\left (4 a -4 b \right ) \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-2 a +5 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{4 \left (a -b \right )^{2}}+\frac {1}{\left (4 a +4 b \right ) \left (\cos \left (f x +e \right )-1\right )}+\frac {\left (-2 a -5 b \right ) \ln \left (\cos \left (f x +e \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {\left (\left (a^{2}+2 b^{2}\right ) \left (\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {\ln \left (\cos \left (f x +e \right )^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}\right )-3 a b \left (-\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\ln \left (\cos \left (f x +e \right )^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )+\frac {\left (2 a^{2}+b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{3}\right )}{3 a}\right ) b^{2}}{\left (a -b \right )^{2} \left (a +b \right )^{2}}}{f}\) | \(374\) |
| risch | \(-\frac {5 i b x}{2 \left (a^{2}-2 a b +b^{2}\right )}-i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (27 a^{7} f^{3}-54 a^{5} b^{2} f^{3}+27 a^{3} b^{4} f^{3}\right ) \textit {\_Z}^{3}+\left (54 i a^{4} b^{2} f^{2}+27 i a^{2} b^{4} f^{2}\right ) \textit {\_Z}^{2}-9 \textit {\_Z} a \,b^{4} f -i b^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\left (\left (\frac {54 a^{6} f^{2}}{a^{3} b +8 a \,b^{3}}-\frac {108 a^{4} b^{2} f^{2}}{a^{3} b +8 a \,b^{3}}+\frac {54 a^{2} b^{4} f^{2}}{a^{3} b +8 a \,b^{3}}\right ) \textit {\_R}^{2}+\left (\frac {6 i a^{5} f}{a^{3} b +8 a \,b^{3}}+\frac {96 i f \,a^{3} b^{2}}{a^{3} b +8 a \,b^{3}}+\frac {60 i f \,b^{4} a}{a^{3} b +8 a \,b^{3}}\right ) \textit {\_R} -\frac {4 a^{2} b^{2}}{a^{3} b +8 a \,b^{3}}-\frac {14 b^{4}}{a^{3} b +8 a \,b^{3}}\right ) {\mathrm e}^{i \left (f x +e \right )}+1\right )\right )+\frac {i a e}{f \left (a^{2}+2 a b +b^{2}\right )}-\frac {5 i b e}{2 f \left (a^{2}-2 a b +b^{2}\right )}+\frac {i a e}{f \left (a^{2}-2 a b +b^{2}\right )}+\frac {5 i b e}{2 f \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i a^{2} b^{4} f^{2} e}{a^{7} f^{3}-2 a^{5} b^{2} f^{3}+a^{3} b^{4} f^{3}}+\frac {4 i a^{4} b^{2} f^{3} x}{a^{7} f^{3}-2 a^{5} b^{2} f^{3}+a^{3} b^{4} f^{3}}-\frac {i x}{a}+\frac {4 i a^{4} b^{2} f^{2} e}{a^{7} f^{3}-2 a^{5} b^{2} f^{3}+a^{3} b^{4} f^{3}}+\frac {5 i b x}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {i a x}{a^{2}+2 a b +b^{2}}+\frac {2 i a^{2} b^{4} f^{3} x}{a^{7} f^{3}-2 a^{5} b^{2} f^{3}+a^{3} b^{4} f^{3}}-\frac {b \,{\mathrm e}^{3 i \left (f x +e \right )}-2 \,{\mathrm e}^{2 i \left (f x +e \right )} a +b \,{\mathrm e}^{i \left (f x +e \right )}}{f \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) a}{f \left (a^{2}+2 a b +b^{2}\right )}-\frac {5 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{2 f \left (a^{2}+2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) a}{f \left (a^{2}-2 a b +b^{2}\right )}+\frac {5 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{2 f \left (a^{2}-2 a b +b^{2}\right )}+\frac {i a x}{a^{2}-2 a b +b^{2}}\) | \(828\) |
Input:
int(cot(f*x+e)^3/(a+b*sec(f*x+e)^3),x,method=_RETURNVERBOSE)
Output:
1/f*(-1/(4*a-4*b)/(1+cos(f*x+e))+1/4/(a-b)^2*(-2*a+5*b)*ln(1+cos(f*x+e))+1 /(4*a+4*b)/(cos(f*x+e)-1)+1/4/(a+b)^2*(-2*a-5*b)*ln(cos(f*x+e)-1)-((a^2+2* b^2)*(1/3/a/(b/a)^(2/3)*ln(cos(f*x+e)+(b/a)^(1/3))-1/6/a/(b/a)^(2/3)*ln(co s(f*x+e)^2-(b/a)^(1/3)*cos(f*x+e)+(b/a)^(2/3))+1/3/a/(b/a)^(2/3)*3^(1/2)*a rctan(1/3*3^(1/2)*(2/(b/a)^(1/3)*cos(f*x+e)-1)))-3*a*b*(-1/3/a/(b/a)^(1/3) *ln(cos(f*x+e)+(b/a)^(1/3))+1/6/a/(b/a)^(1/3)*ln(cos(f*x+e)^2-(b/a)^(1/3)* cos(f*x+e)+(b/a)^(2/3))+1/3*3^(1/2)/a/(b/a)^(1/3)*arctan(1/3*3^(1/2)*(2/(b /a)^(1/3)*cos(f*x+e)-1)))+1/3*(2*a^2+b^2)/a*ln(b+a*cos(f*x+e)^3))*b^2/(a-b )^2/(a+b)^2)
Result contains complex when optimal does not.
Time = 2.09 (sec) , antiderivative size = 10746, normalized size of antiderivative = 27.34 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\text {Too large to display} \] Input:
integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\int \frac {\cot ^{3}{\left (e + f x \right )}}{a + b \sec ^{3}{\left (e + f x \right )}}\, dx \] Input:
integrate(cot(f*x+e)**3/(a+b*sec(f*x+e)**3),x)
Output:
Integral(cot(e + f*x)**3/(a + b*sec(e + f*x)**3), x)
Time = 0.13 (sec) , antiderivative size = 488, normalized size of antiderivative = 1.24 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx =\text {Too large to display} \] Input:
integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^3),x, algorithm="maxima")
Output:
1/36*(4*sqrt(3)*(a^2*b^3*(9*(b/a)^(2/3) + 4) - a^3*b^2*(3*(b/a)^(1/3) + 4* b/a) - 2*a*b^4*(3*(b/a)^(1/3) + b/a) + 2*b^5)*arctan(-1/3*sqrt(3)*((b/a)^( 1/3) - 2*cos(f*x + e))/(b/a)^(1/3))/((a^6*(b/a)^(2/3) - 2*a^4*b^2*(b/a)^(2 /3) + a^2*b^4*(b/a)^(2/3))*(b/a)^(1/3)) - 6*(a^2*b^2*(4*(b/a)^(2/3) - 1) + 2*b^4*((b/a)^(2/3) - 1) - 3*a*b^3*(b/a)^(1/3))*log(cos(f*x + e)^2 - (b/a) ^(1/3)*cos(f*x + e) + (b/a)^(2/3))/(a^5*(b/a)^(2/3) - 2*a^3*b^2*(b/a)^(2/3 ) + a*b^4*(b/a)^(2/3)) - 12*(a^2*b^2*(2*(b/a)^(2/3) + 1) + b^4*((b/a)^(2/3 ) + 2) + 3*a*b^3*(b/a)^(1/3))*log((b/a)^(1/3) + cos(f*x + e))/(a^5*(b/a)^( 2/3) - 2*a^3*b^2*(b/a)^(2/3) + a*b^4*(b/a)^(2/3)) - 9*(2*a - 5*b)*log(cos( f*x + e) + 1)/(a^2 - 2*a*b + b^2) - 9*(2*a + 5*b)*log(cos(f*x + e) - 1)/(a ^2 + 2*a*b + b^2) - 18*(b*cos(f*x + e) - a)/((a^2 - b^2)*cos(f*x + e)^2 - a^2 + b^2))/f
Time = 0.21 (sec) , antiderivative size = 545, normalized size of antiderivative = 1.39 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=-\frac {{\left (3 \, a^{6} b^{3} f \left (-\frac {b}{a}\right )^{\frac {1}{3}} - 6 \, a^{4} b^{5} f \left (-\frac {b}{a}\right )^{\frac {1}{3}} + 3 \, a^{2} b^{7} f \left (-\frac {b}{a}\right )^{\frac {1}{3}} - a^{7} b^{2} f + 3 \, a^{3} b^{6} f - 2 \, a b^{8} f\right )} \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {b}{a}\right )^{\frac {1}{3}} + \cos \left (f x + e\right ) \right |}\right )}{3 \, {\left (a^{9} b f^{2} - 4 \, a^{7} b^{3} f^{2} + 6 \, a^{5} b^{5} f^{2} - 4 \, a^{3} b^{7} f^{2} + a b^{9} f^{2}\right )}} - \frac {{\left (2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | a \cos \left (f x + e\right )^{3} + b \right |}\right )}{3 \, {\left (a^{5} f - 2 \, a^{3} b^{2} f + a b^{4} f\right )}} - \frac {{\left (2 \, a + 5 \, b\right )} \log \left ({\left | -\cos \left (f x + e\right ) + 1 \right |}\right )}{4 \, {\left (a^{2} f + 2 \, a b f + b^{2} f\right )}} - \frac {{\left (2 \, a - 5 \, b\right )} \log \left ({\left | -\cos \left (f x + e\right ) - 1 \right |}\right )}{4 \, {\left (a^{2} f - 2 \, a b f + b^{2} f\right )}} - \frac {{\left (3 \, \left (-a^{2} b\right )^{\frac {2}{3}} b^{2} + {\left (a^{2} b + 2 \, b^{3}\right )} \left (-a^{2} b\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {b}{a}\right )^{\frac {1}{3}} + 2 \, \cos \left (f x + e\right )\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{{\left (\sqrt {3} a^{5} - 2 \, \sqrt {3} a^{3} b^{2} + \sqrt {3} a b^{4}\right )} f} + \frac {{\left (3 \, \left (-a^{2} b\right )^{\frac {2}{3}} b^{2} - {\left (a^{2} b + 2 \, b^{3}\right )} \left (-a^{2} b\right )^{\frac {1}{3}}\right )} \log \left (\cos \left (f x + e\right )^{2} + \left (-\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x + e\right ) + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f} + \frac {a^{3} - a b^{2} - {\left (a^{2} b - b^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} f {\left (\cos \left (f x + e\right ) + 1\right )} {\left (\cos \left (f x + e\right ) - 1\right )}} \] Input:
integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^3),x, algorithm="giac")
Output:
-1/3*(3*a^6*b^3*f*(-b/a)^(1/3) - 6*a^4*b^5*f*(-b/a)^(1/3) + 3*a^2*b^7*f*(- b/a)^(1/3) - a^7*b^2*f + 3*a^3*b^6*f - 2*a*b^8*f)*(-b/a)^(1/3)*log(abs(-(- b/a)^(1/3) + cos(f*x + e)))/(a^9*b*f^2 - 4*a^7*b^3*f^2 + 6*a^5*b^5*f^2 - 4 *a^3*b^7*f^2 + a*b^9*f^2) - 1/3*(2*a^2*b^2 + b^4)*log(abs(a*cos(f*x + e)^3 + b))/(a^5*f - 2*a^3*b^2*f + a*b^4*f) - 1/4*(2*a + 5*b)*log(abs(-cos(f*x + e) + 1))/(a^2*f + 2*a*b*f + b^2*f) - 1/4*(2*a - 5*b)*log(abs(-cos(f*x + e) - 1))/(a^2*f - 2*a*b*f + b^2*f) - (3*(-a^2*b)^(2/3)*b^2 + (a^2*b + 2*b^ 3)*(-a^2*b)^(1/3))*arctan(1/3*sqrt(3)*((-b/a)^(1/3) + 2*cos(f*x + e))/(-b/ a)^(1/3))/((sqrt(3)*a^5 - 2*sqrt(3)*a^3*b^2 + sqrt(3)*a*b^4)*f) + 1/6*(3*( -a^2*b)^(2/3)*b^2 - (a^2*b + 2*b^3)*(-a^2*b)^(1/3))*log(cos(f*x + e)^2 + ( -b/a)^(1/3)*cos(f*x + e) + (-b/a)^(2/3))/((a^5 - 2*a^3*b^2 + a*b^4)*f) + 1 /2*(a^3 - a*b^2 - (a^2*b - b^3)*cos(f*x + e))/((a + b)^2*(a - b)^2*f*(cos( f*x + e) + 1)*(cos(f*x + e) - 1))
Time = 30.03 (sec) , antiderivative size = 58699, normalized size of antiderivative = 149.36 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\text {Too large to display} \] Input:
int(cot(e + f*x)^3/(a + b/cos(e + f*x)^3),x)
Output:
-(a^3*cos(e/2 + (f*x)/2)^4 + a^3*sin(e/2 + (f*x)/2)^4 - a*b^2*cos(e/2 + (f *x)/2)^4 + a*b^2*sin(e/2 + (f*x)/2)^4 + 2*a^2*b*sin(e/2 + (f*x)/2)^4 - 8*a ^3*log((cos(e/2 + (f*x)/2)^2 + sin(e/2 + (f*x)/2)^2)/cos(e/2 + (f*x)/2)^2) *cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2)^2 + 8*b^3*log((cos(e/2 + (f*x)/2) ^2 + sin(e/2 + (f*x)/2)^2)/cos(e/2 + (f*x)/2)^2)*cos(e/2 + (f*x)/2)^2*sin( e/2 + (f*x)/2)^2 + 8*a^3*cos(e/2 + (f*x)/2)^2*log(sin(e/2 + (f*x)/2)/cos(e /2 + (f*x)/2))*sin(e/2 + (f*x)/2)^2 - 8*a^4*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2)^2*symsum(log((131072*(980*b^11*cos(e/2 + (f*x)/2)^2 + 336*b^11*s in(e/2 + (f*x)/2)^2 + 1764*a^2*b^9*cos(e/2 + (f*x)/2)^2 + 392*a^3*b^8*cos( e/2 + (f*x)/2)^2 + 640*root(54*a^5*b^2*z^3 - 27*a^3*b^4*z^3 - 27*a^7*z^3 - 54*a^4*b^2*z^2 - 27*a^2*b^4*z^2 - 9*a*b^4*z - b^4, z, k)^2*b^13*sin(e/2 + (f*x)/2)^2 + 32*root(54*a^5*b^2*z^3 - 27*a^3*b^4*z^3 - 27*a^7*z^3 - 54*a^ 4*b^2*z^2 - 27*a^2*b^4*z^2 - 9*a*b^4*z - b^4, z, k)^3*b^14*sin(e/2 + (f*x) /2)^2 - 1176*a^2*b^9*sin(e/2 + (f*x)/2)^2 - 784*a^3*b^8*sin(e/2 + (f*x)/2) ^2 + 952*root(54*a^5*b^2*z^3 - 27*a^3*b^4*z^3 - 27*a^7*z^3 - 54*a^4*b^2*z^ 2 - 27*a^2*b^4*z^2 - 9*a*b^4*z - b^4, z, k)*b^12*cos(e/2 + (f*x)/2)^2 + 23 52*a*b^10*cos(e/2 + (f*x)/2)^2 + 1944*root(54*a^5*b^2*z^3 - 27*a^3*b^4*z^3 - 27*a^7*z^3 - 54*a^4*b^2*z^2 - 27*a^2*b^4*z^2 - 9*a*b^4*z - b^4, z, k)*b ^12*sin(e/2 + (f*x)/2)^2 - 56*a*b^10*sin(e/2 + (f*x)/2)^2 + 304*root(54*a^ 5*b^2*z^3 - 27*a^3*b^4*z^3 - 27*a^7*z^3 - 54*a^4*b^2*z^2 - 27*a^2*b^4*z...
\[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\text {too large to display} \] Input:
int(cot(f*x+e)^3/(a+b*sec(f*x+e)^3),x)
Output:
(128*int(1/(tan((e + f*x)/2)**9*a**5 + 19*tan((e + f*x)/2)**9*a**4*b - 50* tan((e + f*x)/2)**9*a**3*b**2 + 26*tan((e + f*x)/2)**9*a**2*b**3 + 17*tan( (e + f*x)/2)**9*a*b**4 - 13*tan((e + f*x)/2)**9*b**5 - 3*tan((e + f*x)/2)* *7*a**5 - 63*tan((e + f*x)/2)**7*a**4*b + 30*tan((e + f*x)/2)**7*a**3*b**2 + 102*tan((e + f*x)/2)**7*a**2*b**3 - 27*tan((e + f*x)/2)**7*a*b**4 - 39* tan((e + f*x)/2)**7*b**5 + 3*tan((e + f*x)/2)**5*a**5 + 57*tan((e + f*x)/2 )**5*a**4*b - 150*tan((e + f*x)/2)**5*a**3*b**2 + 78*tan((e + f*x)/2)**5*a **2*b**3 + 51*tan((e + f*x)/2)**5*a*b**4 - 39*tan((e + f*x)/2)**5*b**5 - t an((e + f*x)/2)**3*a**5 - 21*tan((e + f*x)/2)**3*a**4*b + 10*tan((e + f*x) /2)**3*a**3*b**2 + 34*tan((e + f*x)/2)**3*a**2*b**3 - 9*tan((e + f*x)/2)** 3*a*b**4 - 13*tan((e + f*x)/2)**3*b**5),x)*tan((e + f*x)/2)**2*a**7*b**2*f + 2400*int(1/(tan((e + f*x)/2)**9*a**5 + 19*tan((e + f*x)/2)**9*a**4*b - 50*tan((e + f*x)/2)**9*a**3*b**2 + 26*tan((e + f*x)/2)**9*a**2*b**3 + 17*t an((e + f*x)/2)**9*a*b**4 - 13*tan((e + f*x)/2)**9*b**5 - 3*tan((e + f*x)/ 2)**7*a**5 - 63*tan((e + f*x)/2)**7*a**4*b + 30*tan((e + f*x)/2)**7*a**3*b **2 + 102*tan((e + f*x)/2)**7*a**2*b**3 - 27*tan((e + f*x)/2)**7*a*b**4 - 39*tan((e + f*x)/2)**7*b**5 + 3*tan((e + f*x)/2)**5*a**5 + 57*tan((e + f*x )/2)**5*a**4*b - 150*tan((e + f*x)/2)**5*a**3*b**2 + 78*tan((e + f*x)/2)** 5*a**2*b**3 + 51*tan((e + f*x)/2)**5*a*b**4 - 39*tan((e + f*x)/2)**5*b**5 - tan((e + f*x)/2)**3*a**5 - 21*tan((e + f*x)/2)**3*a**4*b + 10*tan((e ...