\(\int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [305]

Optimal result

Integrand size = 25, antiderivative size = 167 \[ \int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {a-b} \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f} \] Output:

-(a-b)^(1/2)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f-1/1 
5*(15*a^2-5*a*b-2*b^2)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/a^2/f+1/15*(5*a 
-b)*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/a/f-1/5*cot(f*x+e)^5*(a+b*tan(f* 
x+e)^2)^(1/2)/f
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 8.44 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.92 \[ \int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\cos ^4(e+f x) \cot ^5(e+f x) \left (1+\frac {b \tan ^2(e+f x)}{a}\right ) \left (8 (a-b) \, _3F_2\left (2,2,2;1,\frac {3}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^3+8 (a-b) \operatorname {Hypergeometric2F1}\left (2,2,\frac {3}{2},\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \left (-2 a+3 b \tan ^2(e+f x)\right )+\frac {a^2 \sec ^4(e+f x) \left (\arcsin \left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}+\sqrt {\cos ^2(e+f x)+\frac {b \sin ^2(e+f x)}{a}}\right ) \left (3 a^2-4 a b \tan ^2(e+f x)+8 b^2 \tan ^4(e+f x)\right )}{\sqrt {\cos ^2(e+f x)+\frac {b \sin ^2(e+f x)}{a}}}\right )}{15 a^3 f \sqrt {a+b \tan ^2(e+f x)}} \] Input:

Integrate[Cot[e + f*x]^6*Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

-1/15*(Cos[e + f*x]^4*Cot[e + f*x]^5*(1 + (b*Tan[e + f*x]^2)/a)*(8*(a - b) 
*HypergeometricPFQ[{2, 2, 2}, {1, 3/2}, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e 
+ f*x]^2*(a + b*Tan[e + f*x]^2)^3 + 8*(a - b)*Hypergeometric2F1[2, 2, 3/2, 
 ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^2*(-2*a 
 + 3*b*Tan[e + f*x]^2) + (a^2*Sec[e + f*x]^4*(ArcSin[Sqrt[((a - b)*Sin[e + 
 f*x]^2)/a]]*Sqrt[((a - b)*Sin[e + f*x]^2)/a] + Sqrt[Cos[e + f*x]^2 + (b*S 
in[e + f*x]^2)/a])*(3*a^2 - 4*a*b*Tan[e + f*x]^2 + 8*b^2*Tan[e + f*x]^4))/ 
Sqrt[Cos[e + f*x]^2 + (b*Sin[e + f*x]^2)/a]))/(a^3*f*Sqrt[a + b*Tan[e + f* 
x]^2])
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4153, 377, 25, 445, 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.

\(\displaystyle \int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \tan (e+f x)^2}}{\tan (e+f x)^6}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \sqrt {b \tan ^2(e+f x)+a}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 377

\(\displaystyle \frac {\frac {1}{5} \int -\frac {\cot ^4(e+f x) \left (4 b \tan ^2(e+f x)+5 a-b\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {1}{5} \int \frac {\cot ^4(e+f x) \left (4 b \tan ^2(e+f x)+5 a-b\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {1}{5} \left (\frac {\int \frac {\cot ^2(e+f x) \left (15 a^2-5 b a-2 b^2+2 (5 a-b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{3 a}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {1}{5} \left (\frac {-\frac {\int \frac {15 a^2 (a-b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a}-\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {-15 a (a-b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{5} \left (\frac {-15 a (a-b) \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{3 a}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{5} \left (\frac {-\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}-15 a \sqrt {a-b} \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{3 a}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{3 a}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}\)

Input:

Int[Cot[e + f*x]^6*Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

(-1/5*(Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2]) + (((5*a - b)*Cot[e + f* 
x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(3*a) + (-15*a*Sqrt[a - b]*ArcTan[(Sqrt[a 
 - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]] - ((15*a^2 - 5*a*b - 2*b^2 
)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/a)/(3*a))/5)/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

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])
 

rule 291

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]
 

rule 377

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/(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 - 1)*Simp[b*c*(m + 1) + 2*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) 
+ 2*b*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b 
*c - a*d, 0] && LtQ[0, q, 1] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m 
, 2, p, q, x]
 

rule 445

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4153

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]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(323\) vs. \(2(149)=298\).

Time = 25.00 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.94

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right )^{3} \left (-15 \cos \left (f x +e \right )+15\right ) \sqrt {a -b}\, \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )}{\sqrt {a -b}\, \left (\cos \left (f x +e \right )-1\right )}\right ) a^{2}+\left (23 \cos \left (f x +e \right )^{4}-35 \cos \left (f x +e \right )^{2}+15\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a^{2}+\left (6 \cos \left (f x +e \right )^{2}-5\right ) \sin \left (f x +e \right )^{2} \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a b -2 \sin \left (f x +e \right )^{4} \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b^{2}\right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}\, \cot \left (f x +e \right ) \csc \left (f x +e \right )^{4}}{15 f \,a^{2} \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\)	\(324\)

Input:

int(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/15/f/a^2*(sin(f*x+e)^3*(-15*cos(f*x+e)+15)*(a-b)^(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)*sin(f*x+e)/( 
cos(f*x+e)-1))*a^2+(23*cos(f*x+e)^4-35*cos(f*x+e)^2+15)*((a*cos(f*x+e)^2+b 
*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^2+(6*cos(f*x+e)^2-5)*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)*a*b-2*sin(f*x+e) 
^4*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*b^2)*(a+b*tan( 
f*x+e)^2)^(1/2)/((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*c 
ot(f*x+e)*csc(f*x+e)^4
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.25 \[ \int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {15 \, a^{2} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} - 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{5} - 4 \, {\left ({\left (15 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{2} - a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, a^{2} f \tan \left (f x + e\right )^{5}}, -\frac {15 \, \sqrt {a - b} a^{2} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{5} + 2 \, {\left ({\left (15 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{2} - a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, a^{2} f \tan \left (f x + e\right )^{5}}\right ] \] Input:

integrate(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/60*(15*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)^5 - 4*((15*a^2 - 5*a*b - 2*b^2)*tan(f*x + 
e)^4 - (5*a^2 - a*b)*tan(f*x + e)^2 + 3*a^2)*sqrt(b*tan(f*x + e)^2 + a))/( 
a^2*f*tan(f*x + e)^5), -1/30*(15*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)^5 + 2*((15*a^2 - 5*a*b - 2*b^2)*tan(f*x + e)^4 - (5*a^2 - a*b)*tan 
(f*x + e)^2 + 3*a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a^2*f*tan(f*x + e)^5)]
 

Sympy [F]

\[ \int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \cot ^{6}{\left (e + f x \right )}\, dx \] Input:

integrate(cot(f*x+e)**6*(a+b*tan(f*x+e)**2)**(1/2),x)
 

Output:

Integral(sqrt(a + b*tan(e + f*x)**2)*cot(e + f*x)**6, x)
 

Maxima [F]

\[ \int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int { \sqrt {b \tan \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{6} \,d x } \] Input:

integrate(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*tan(f*x + e)^2 + a)*cot(f*x + e)^6, x)
 

Giac [F]

\[ \int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int { \sqrt {b \tan \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{6} \,d x } \] Input:

integrate(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*tan(f*x + e)^2 + a)*cot(f*x + e)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^6\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \] Input:

int(cot(e + f*x)^6*(a + b*tan(e + f*x)^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(cot(e + f*x)^6*(a + b*tan(e + f*x)^2)^(1/2), x)
 

Reduce [F]

\[ \int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \cot \left (f x +e \right )^{6} \sqrt {\tan \left (f x +e \right )^{2} b +a}d x \] Input:

int(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(1/2),x)
 

Output:

int(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(1/2),x)