\(\int \cot ^6(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [401]

Optimal result

Integrand size = 25, antiderivative size = 165 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 f}-\frac {(a+b) \cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f} \] Output:

-a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f-1/15*(15* 
a^2+10*a*b-2*b^2)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)/f+1/15*(5*a- 
b)*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/f-1/5*(a+b)*cot(f*x+e)^5*(a+b+b 
*tan(f*x+e)^2)^(1/2)/f
 

Mathematica [C] (verified)

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

Time = 0.97 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.84 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {2 \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (-\frac {3}{4} (a+2 b+a \cos (2 (e+f x)))^2 \csc ^2(e+f x)+\frac {5 (a+b)^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {a \sin ^2(e+f x)}{a+b}\right )}{\sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}}\right )}{15 (a+b) f (a+2 b+a \cos (2 (e+f x)))} \] Input:

Integrate[Cot[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(2*Cot[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2)*((-3*(a + 2*b + a*Cos[2*(e 
+ f*x)])^2*Csc[e + f*x]^2)/4 + (5*(a + b)^2*Hypergeometric2F1[-3/2, -3/2, 
-1/2, (a*Sin[e + f*x]^2)/(a + b)])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b) 
]))/(15*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)]))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4629, 2075, 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.

\(\displaystyle \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^{3/2}}{\tan (e+f x)^6}dx\)

\(\Big \downarrow \) 4629

\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^{3/2}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2075

\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 376

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {1}{5} \left (\frac {\int \frac {(a+b) \cot ^2(e+f x) \left (15 a^2+10 b a-2 b^2+2 (5 a-b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{3 (a+b)}+\frac {1}{3} (5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {1}{5} (a+b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\cot ^2(e+f x) \left (15 a^2+10 b a-2 b^2+2 (5 a-b) b \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{3} (5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {1}{5} (a+b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-\frac {\int \frac {15 a^2 (a+b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a+b}-\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}\right )+\frac {1}{3} (5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {1}{5} (a+b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-15 a^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}\right )+\frac {1}{3} (5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {1}{5} (a+b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-15 a^2 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}\right )+\frac {1}{3} (5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {1}{5} (a+b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-15 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )-\frac {\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}\right )+\frac {1}{3} (5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {1}{5} (a+b) \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

Input:

Int[Cot[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(-1/5*((a + b)*Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2]) + (((5*a - b 
)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/3 + (-15*a^(3/2)*ArcTan[( 
Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]] - ((15*a^2 + 10*a*b 
- 2*b^2)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b))/3)/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 376

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]
 

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 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

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

rule 4629

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(449\) vs. \(2(147)=294\).

Time = 4.70 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.73

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right )^{3} \left (-15 \cos \left (f x +e \right )+15\right ) a^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right )+\sin \left (f x +e \right )^{3} \left (-15 \cos \left (f x +e \right )+15\right ) \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) a^{2} b +\left (23 \cos \left (f x +e \right )^{4}-35 \cos \left (f x +e \right )^{2}+15\right ) \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2}+\left (20 \cos \left (f x +e \right )^{4}-24 \cos \left (f x +e \right )^{2}+10\right ) \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b +\left (5 \cos \left (f x +e \right )^{2}-2\right ) \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \cot \left (f x +e \right )^{3} \csc \left (f x +e \right )^{2}}{15 f \left (a +b \right ) \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (b +a \cos \left (f x +e \right )^{2}\right )}\)	\(450\)

Input:

int(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/15/f/(a+b)/(-a)^(1/2)*(sin(f*x+e)^3*(-15*cos(f*x+e)+15)*a^3*ln(4*(-a)^( 
1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*( 
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)+sin(f*x+e)^3*(- 
15*cos(f*x+e)+15)*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 
/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4* 
sin(f*x+e)*a)*a^2*b+(23*cos(f*x+e)^4-35*cos(f*x+e)^2+15)*(-a)^(1/2)*((b+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+(20*cos(f*x+e)^4-24*cos(f*x+e)^2 
+10)*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+(5*cos(f*x 
+e)^2-2)*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2)*(a+b* 
sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)/(b+a*cos(f 
*x+e)^2)*cot(f*x+e)^3*csc(f*x+e)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (147) = 294\).

Time = 2.66 (sec) , antiderivative size = 767, normalized size of antiderivative = 4.65 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx =\text {Too large to display} \] Input:

integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/120*(15*((a^2 + a*b)*cos(f*x + e)^4 - 2*(a^2 + a*b)*cos(f*x + e)^2 + a^ 
2 + a*b)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + 
 e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b 
+ 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos 
(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 
 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 
- b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)* 
sin(f*x + e))*sin(f*x + e) - 8*((23*a^2 + 20*a*b)*cos(f*x + e)^5 - (35*a^2 
 + 24*a*b - 5*b^2)*cos(f*x + e)^3 + (15*a^2 + 10*a*b - 2*b^2)*cos(f*x + e) 
)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a + b)*f*cos(f*x + e)^4 
- 2*(a + b)*f*cos(f*x + e)^2 + (a + b)*f)*sin(f*x + e)), 1/60*(15*((a^2 + 
a*b)*cos(f*x + e)^4 - 2*(a^2 + a*b)*cos(f*x + e)^2 + a^2 + a*b)*sqrt(a)*ar 
ctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a 
*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2 
)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2) 
*sin(f*x + e)))*sin(f*x + e) - 4*((23*a^2 + 20*a*b)*cos(f*x + e)^5 - (35*a 
^2 + 24*a*b - 5*b^2)*cos(f*x + e)^3 + (15*a^2 + 10*a*b - 2*b^2)*cos(f*x + 
e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a + b)*f*cos(f*x + e)^ 
4 - 2*(a + b)*f*cos(f*x + e)^2 + (a + b)*f)*sin(f*x + e))]
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(cot(f*x+e)**6*(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{6} \,d x } \] Input:

integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^6, x)
 

Giac [F]

\[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{6} \,d x } \] Input:

integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \] Input:

int(cot(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2),x)
 

Output:

int(cot(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cot \left (f x +e \right )^{6} \sec \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cot \left (f x +e \right )^{6}d x \right ) a \] Input:

int(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x)
 

Output:

int(sqrt(sec(e + f*x)**2*b + a)*cot(e + f*x)**6*sec(e + f*x)**2,x)*b + int 
(sqrt(sec(e + f*x)**2*b + a)*cot(e + f*x)**6,x)*a