Integrand size = 25, antiderivative size = 165 \[ \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {(a-b)^{3/2} \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}+\frac {(5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 f}-\frac {a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f} \]
-(a-b)^(3/2)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f-1/1 5*(15*a^2-20*a*b+3*b^2)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/a/f+1/15*(5*a- 6*b)*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/f-1/5*a*cot(f*x+e)^5*(a+b*tan(f *x+e)^2)^(1/2)/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 9.79 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.85 \[ \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {\cos (e+f x) \left (b+a \cot ^2(e+f x)\right )^2 \left (a \left (-2 b+3 a \cot ^2(e+f x)\right ) \operatorname {Hypergeometric2F1}\left (1,1,-\frac {1}{2},\frac {(a-b) \sin ^2(e+f x)}{a}\right )+2 (a-b) (a+b+(a-b) \cos (2 (e+f x))) \operatorname {Hypergeometric2F1}\left (2,2,\frac {1}{2},\frac {(a-b) \sin ^2(e+f x)}{a}\right )\right ) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f} \]
-1/15*(Cos[e + f*x]*(b + a*Cot[e + f*x]^2)^2*(a*(-2*b + 3*a*Cot[e + f*x]^2 )*Hypergeometric2F1[1, 1, -1/2, ((a - b)*Sin[e + f*x]^2)/a] + 2*(a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])*Hypergeometric2F1[2, 2, 1/2, ((a - b)*Sin [e + f*x]^2)/a])*Sin[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(a^3*f)
Time = 0.42 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4153, 376, 25, 445, 27, 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.
| \(\displaystyle \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \tan (e+f x)^2\right )^{3/2}}{\tan (e+f x)^6}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (b \tan ^2(e+f x)+a\right )^{3/2}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 376 |
| \(\displaystyle \frac {\frac {1}{5} \int -\frac {\cot ^4(e+f x) \left ((4 a-5 b) b \tan ^2(e+f x)+a (5 a-6 b)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {1}{5} \int \frac {\cot ^4(e+f x) \left ((4 a-5 b) b \tan ^2(e+f x)+a (5 a-6 b)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {\int \frac {a \cot ^2(e+f x) \left (15 a^2-20 b a+3 b^2+2 (5 a-6 b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{3 a}+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\cot ^2(e+f x) \left (15 a^2-20 b a+3 b^2+2 (5 a-6 b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-\frac {\int \frac {15 a (a-b)^2}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a}-\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}\right )+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-15 (a-b)^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}\right )+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-15 (a-b)^2 \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}\right )+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}-15 (a-b)^{3/2} \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )\right )+\frac {1}{3} (5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )-\frac {1}{5} a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\) |
(-1/5*(a*Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2]) + (((5*a - 6*b)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/3 + (-15*(a - b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]] - ((15*a^2 - 20*a*b + 3*b^ 2)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/a)/3)/5)/f
3.4.18.3.1 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[c*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1 )/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^ 2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b*c - a*d)*(m + 1) + 2*c*(b*c*(p + 1) + a* d*(q - 1)) + d*((b*c - a*d)*(m + 1) + 2*b*c*(p + q))*x^2, x], x], x] /; Fre eQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && 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. \(970\) vs. \(2(147)=294\).
Time = 5.23 (sec) , antiderivative size = 971, normalized size of antiderivative = 5.88
1/480/f/a/(a-b)^(1/2)*((a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+ 1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)/((-cos(f*x+e)+1)^2 *csc(f*x+e)^2-1)^2)^(3/2)*((-cos(f*x+e)+1)^2*csc(f*x+e)^2-1)^3/(a*(-cos(f* x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e) +1)^2*csc(f*x+e)^2+a)^(3/2)/(-cos(f*x+e)+1)^5*(-480*arctan(1/2*(a*(-cos(f* x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e) +1)^2*csc(f*x+e)^2+a)^(1/2)/(-cos(f*x+e)+1)*sin(f*x+e)/(a-b)^(1/2))*a^3*(- cos(f*x+e)+1)^5+960*arctan(1/2*(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos (f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)/(-co s(f*x+e)+1)*sin(f*x+e)/(a-b)^(1/2))*a^2*b*(-cos(f*x+e)+1)^5-480*arctan(1/2 *(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b* (-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/2)/(-cos(f*x+e)+1)*sin(f*x+e)/(a-b)^( 1/2))*a*b^2*(-cos(f*x+e)+1)^5+240*a^2*(a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2* a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+a)^(1/ 2)*(-cos(f*x+e)+1)^4*(a-b)^(1/2)*sin(f*x+e)-240*a*b*(a*(-cos(f*x+e)+1)^4*c sc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e)+1)^2*csc(f *x+e)^2+a)^(1/2)*(-cos(f*x+e)+1)^4*(a-b)^(1/2)*sin(f*x+e)-20*a*(a*(-cos(f* x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b*(-cos(f*x+e) +1)^2*csc(f*x+e)^2+a)^(3/2)*(-cos(f*x+e)+1)^2*(a-b)^(1/2)*sin(f*x+e)^3+3*( a*(-cos(f*x+e)+1)^4*csc(f*x+e)^4-2*a*(-cos(f*x+e)+1)^2*csc(f*x+e)^2+4*b...
Time = 0.36 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.33 \[ \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (a^{2} - a b\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{5} + 4 \, {\left ({\left (15 \, a^{2} - 20 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{2} - 6 \, a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, a f \tan \left (f x + e\right )^{5}}, -\frac {15 \, {\left (a^{2} - a b\right )} \sqrt {a - b} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{5} + 2 \, {\left ({\left (15 \, a^{2} - 20 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{2} - 6 \, a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, a f \tan \left (f x + e\right )^{5}}\right ] \]
[-1/60*(15*(a^2 - a*b)*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*tan(f*x + e)^2 + 1))*tan(f*x + e)^5 + 4*((15*a^2 - 20*a*b + 3*b^2)* tan(f*x + e)^4 - (5*a^2 - 6*a*b)*tan(f*x + e)^2 + 3*a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a*f*tan(f*x + e)^5), -1/30*(15*(a^2 - a*b)*sqrt(a - b)*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)^5 + 2*((15*a^2 - 20*a*b + 3*b^2)*tan(f*x + e)^4 - (5*a^2 - 6*a*b)*tan(f*x + e)^2 + 3*a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a*f *tan(f*x + e)^5)]
Timed out. \[ \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
\[ \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \cot ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^6\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]