Integrand size = 25, antiderivative size = 120 \[ \int \frac {\cot ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}+\frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a^2 f}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a f} \]
arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f/(a-b)^(1/2)+1/3* (3*a+2*b)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/a^2/f-1/3*cot(f*x+e)^3*(a+b* tan(f*x+e)^2)^(1/2)/a/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 11.48 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.18 \[ \int \frac {\cot ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\cos ^2(e+f x) \cot ^3(e+f x) \left (1+\frac {b \tan ^2(e+f x)}{a}\right ) \left (-24 (a-b) b \operatorname {Hypergeometric2F1}\left (2,2,\frac {5}{2},\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )-8 (a-b) \, _3F_2\left (2,2,2;1,\frac {5}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^2+\frac {6 a \arcsin \left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \left (a^2-4 a b \tan ^2(e+f x)-8 b^2 \tan ^4(e+f x)\right )}{\sqrt {\frac {(a-b) \sin ^2(2 (e+f x)) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}\right )}{9 a^3 f \sqrt {a+b \tan ^2(e+f x)}} \]
-1/9*(Cos[e + f*x]^2*Cot[e + f*x]^3*(1 + (b*Tan[e + f*x]^2)/a)*(-24*(a - b )*b*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^ 2*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2) - 8*(a - b)*HypergeometricPFQ[{2, 2, 2}, {1, 5/2}, ((a - b)*Sin[e + f*x]^2)/a]*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2)^2 + (6*a*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*(a^2 - 4*a*b*Ta n[e + f*x]^2 - 8*b^2*Tan[e + f*x]^4))/Sqrt[((a - b)*Sin[2*(e + f*x)]^2*(a + b*Tan[e + f*x]^2))/a^2]))/(a^3*f*Sqrt[a + b*Tan[e + f*x]^2])
Time = 0.33 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4153, 382, 25, 445, 27, 291, 216}
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.
| \(\displaystyle \int \frac {\cot ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^4 \sqrt {a+b \tan (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {\int \frac {\cot ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 382 |
| \(\displaystyle \frac {\frac {\int -\frac {\cot ^2(e+f x) \left (2 b \tan ^2(e+f x)+3 a+2 b\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{3 a}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\cot ^2(e+f x) \left (2 b \tan ^2(e+f x)+3 a+2 b\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{3 a}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {3 a^2}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a}-\frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-3 a \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {-\frac {-3 a \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-\frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {-\frac {-\frac {3 a \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b}}-\frac {(3 a+2 b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}-\frac {\cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}}{f}\) |
(-1/3*(Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/a - ((-3*a*ArcTan[(Sqrt[ a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/Sqrt[a - b] - ((3*a + 2* b)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/a)/(3*a))/f
3.4.31.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ (a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b* x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m + 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Leaf count of result is larger than twice the leaf count of optimal. \(362\) vs. \(2(106)=212\).
Time = 5.34 (sec) , antiderivative size = 363, normalized size of antiderivative = 3.02
| method | result | size |
| default | \(-\frac {-3 \sin \left (f x +e \right )^{2} \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {a -b}}\right ) a^{2}+2 \sqrt {a -b}\, b^{2} \sin \left (f x +e \right )^{2} \tan \left (f x +e \right )-3 \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {a -b}}\right ) a^{2} \sin \left (f x +e \right ) \tan \left (f x +e \right )-2 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {a -b}\, a b -4 \sqrt {a -b}\, a^{2} \cos \left (f x +e \right )^{2} \cot \left (f x +e \right )+3 \sqrt {a -b}\, a b \tan \left (f x +e \right )+3 \sqrt {a -b}\, a^{2} \cot \left (f x +e \right )}{3 f \,a^{2} \sqrt {a -b}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}\, \left (\cos \left (f x +e \right )^{2}-1\right )}\) | \(363\) |
-1/3/f/a^2/(a-b)^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)/(cos(f*x+e)^2-1)*(-3*sin(f *x+e)^2*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*arctan(1/ (a-b)^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*(cot( f*x+e)+csc(f*x+e)))*a^2+2*(a-b)^(1/2)*b^2*sin(f*x+e)^2*tan(f*x+e)-3*((a*co s(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*arctan(1/(a-b)^(1/2)*(( a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*(cot(f*x+e)+csc(f*x +e)))*a^2*sin(f*x+e)*tan(f*x+e)-2*cos(f*x+e)*sin(f*x+e)*(a-b)^(1/2)*a*b-4* (a-b)^(1/2)*a^2*cos(f*x+e)^2*cot(f*x+e)+3*(a-b)^(1/2)*a*b*tan(f*x+e)+3*(a- b)^(1/2)*a^2*cot(f*x+e))
Time = 0.34 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.99 \[ \int \frac {\cot ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [-\frac {3 \, a^{2} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} - 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{3} - 4 \, {\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + a b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left (a^{3} - a^{2} b\right )} f \tan \left (f x + e\right )^{3}}, \frac {3 \, \sqrt {a - b} a^{2} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{3} + 2 \, {\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + a b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left (a^{3} - a^{2} b\right )} f \tan \left (f x + e\right )^{3}}\right ] \]
[-1/12*(3*a^2*sqrt(-a + b)*log(-((a^2 - 8*a*b + 8*b^2)*tan(f*x + e)^4 - 2* (3*a^2 - 4*a*b)*tan(f*x + e)^2 + a^2 - 4*((a - 2*b)*tan(f*x + e)^3 - a*tan (f*x + e))*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b))/(tan(f*x + e)^4 + 2*ta n(f*x + e)^2 + 1))*tan(f*x + e)^3 - 4*((3*a^2 - a*b - 2*b^2)*tan(f*x + e)^ 2 - a^2 + a*b)*sqrt(b*tan(f*x + e)^2 + a))/((a^3 - a^2*b)*f*tan(f*x + e)^3 ), 1/6*(3*sqrt(a - b)*a^2*arctan(-2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) *tan(f*x + e)/((a - 2*b)*tan(f*x + e)^2 - a))*tan(f*x + e)^3 + 2*((3*a^2 - a*b - 2*b^2)*tan(f*x + e)^2 - a^2 + a*b)*sqrt(b*tan(f*x + e)^2 + a))/((a^ 3 - a^2*b)*f*tan(f*x + e)^3)]
\[ \int \frac {\cot ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\cot ^{4}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\cot ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )^{4}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\cot ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )^{4}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\cot ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^4}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]