\(\int \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [297]

Optimal result

Integrand size = 25, antiderivative size = 115 \[ \int \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {(2 a-b) \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{2 \sqrt {a} f}-\frac {\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}-\frac {\cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f} \] Output:

1/2*(2*a-b)*arctanh((a+b*tan(f*x+e)^2)^(1/2)/a^(1/2))/a^(1/2)/f-(a-b)^(1/2 
)*arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))/f-1/2*cot(f*x+e)^2*(a+b*ta 
n(f*x+e)^2)^(1/2)/f
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00 \[ \int \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {(2 a-b) \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )-\sqrt {a} \left (2 \sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+\cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}\right )}{2 \sqrt {a} f} \] Input:

Integrate[Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

((2*a - b)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]] - Sqrt[a]*(2*Sqrt[a 
 - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]] + Cot[e + f*x]^2*Sqr 
t[a + b*Tan[e + f*x]^2]))/(2*Sqrt[a]*f)
 

Rubi [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4153, 354, 110, 27, 174, 73, 221}

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.

\(\displaystyle \int \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \tan (e+f x)^2}}{\tan (e+f x)^3}dx\)

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 110

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

(((2*(2*a - b)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/Sqrt[a] - 4*Sq 
rt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/2 - Cot[e + f*x 
]*Sqrt[a + b*Tan[e + f*x]^2])/(2*f)
 

Defintions of rubi rules used

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 110

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[1/((m + 1)*(b*e - a*f))   Int[(a + b*x)^(m + 1) 
*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + 
p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt 
Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, 
 m + n])
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(933\) vs. \(2(97)=194\).

Time = 6.86 (sec) , antiderivative size = 934, normalized size of antiderivative = 8.12

method	result	size
default	\(\text {Expression too large to display}\)	\(934\)

Input:

int(cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4/f/a^(5/2)/(a-b)^(1/2)*((cos(f*x+e)-1)*(a-b)^(1/2)*ln(2/a^(1/2)*(a^(1/ 
2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+((a 
*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-a*cos(f*x+e) 
+cos(f*x+e)*b+b)/(cos(f*x+e)+1))*a^2*b+(4*cos(f*x+e)-4)*a^(5/2)*ln(4*(a-b) 
^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e) 
+4*(a-b)^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)+4* 
a*cos(f*x+e)-4*cos(f*x+e)*b)*b+(1-cos(f*x+e))*(a-b)^(1/2)*ln(2*(2*((a*cos( 
f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)*sin(f*x+e)^2+a*si 
n(f*x+e)^2-a*cos(f*x+e)^2+2*b*cos(f*x+e)^2+2*a*cos(f*x+e)-4*cos(f*x+e)*b-a 
+2*b)/(cos(f*x+e)-1)^2)*a^2*b+(2-2*cos(f*x+e))*(a-b)^(1/2)*ln(2/a^(1/2)*(a 
^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e) 
+((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-a*cos(f* 
x+e)+cos(f*x+e)*b+b)/(cos(f*x+e)+1))*a^3+(-4*cos(f*x+e)+4)*a^(7/2)*ln(4*(a 
-b)^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x 
+e)+4*(a-b)^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2) 
+4*a*cos(f*x+e)-4*cos(f*x+e)*b)+(2*cos(f*x+e)-2)*(a-b)^(1/2)*ln(2*(2*((a*c 
os(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)*sin(f*x+e)^2+a 
*sin(f*x+e)^2-a*cos(f*x+e)^2+2*b*cos(f*x+e)^2+2*a*cos(f*x+e)-4*cos(f*x+e)* 
b-a+2*b)/(cos(f*x+e)-1)^2)*a^3+2*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x 
+e)+1)^2)^(1/2)*(a-b)^(1/2)*a^(5/2)*cos(f*x+e))*(a+b*tan(f*x+e)^2)^(1/2...
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 571, normalized size of antiderivative = 4.97 \[ \int \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {2 \, \sqrt {a - b} a \log \left (\frac {b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 2 \, a - b}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} - {\left (2 \, a - b\right )} \sqrt {a} \log \left (\frac {b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a} + 2 \, a}{\tan \left (f x + e\right )^{2}}\right ) \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{4 \, a f \tan \left (f x + e\right )^{2}}, \frac {4 \, a \sqrt {-a + b} \arctan \left (\frac {\sqrt {-a + b}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\right ) \tan \left (f x + e\right )^{2} - {\left (2 \, a - b\right )} \sqrt {a} \log \left (\frac {b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a} + 2 \, a}{\tan \left (f x + e\right )^{2}}\right ) \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{4 \, a f \tan \left (f x + e\right )^{2}}, -\frac {\sqrt {-a} {\left (2 \, a - b\right )} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\right ) \tan \left (f x + e\right )^{2} - \sqrt {a - b} a \log \left (\frac {b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 2 \, a - b}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} + \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{2 \, a f \tan \left (f x + e\right )^{2}}, -\frac {\sqrt {-a} {\left (2 \, a - b\right )} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\right ) \tan \left (f x + e\right )^{2} - 2 \, a \sqrt {-a + b} \arctan \left (\frac {\sqrt {-a + b}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\right ) \tan \left (f x + e\right )^{2} + \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{2 \, a f \tan \left (f x + e\right )^{2}}\right ] \] Input:

integrate(cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/4*(2*sqrt(a - b)*a*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a) 
*sqrt(a - b) + 2*a - b)/(tan(f*x + e)^2 + 1))*tan(f*x + e)^2 - (2*a - b)*s 
qrt(a)*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a) + 2*a) 
/tan(f*x + e)^2)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*a)/(a*f*tan 
(f*x + e)^2), 1/4*(4*a*sqrt(-a + b)*arctan(sqrt(-a + b)/sqrt(b*tan(f*x + e 
)^2 + a))*tan(f*x + e)^2 - (2*a - b)*sqrt(a)*log((b*tan(f*x + e)^2 - 2*sqr 
t(b*tan(f*x + e)^2 + a)*sqrt(a) + 2*a)/tan(f*x + e)^2)*tan(f*x + e)^2 - 2* 
sqrt(b*tan(f*x + e)^2 + a)*a)/(a*f*tan(f*x + e)^2), -1/2*(sqrt(-a)*(2*a - 
b)*arctan(sqrt(-a)/sqrt(b*tan(f*x + e)^2 + a))*tan(f*x + e)^2 - sqrt(a - b 
)*a*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 2*a 
 - b)/(tan(f*x + e)^2 + 1))*tan(f*x + e)^2 + sqrt(b*tan(f*x + e)^2 + a)*a) 
/(a*f*tan(f*x + e)^2), -1/2*(sqrt(-a)*(2*a - b)*arctan(sqrt(-a)/sqrt(b*tan 
(f*x + e)^2 + a))*tan(f*x + e)^2 - 2*a*sqrt(-a + b)*arctan(sqrt(-a + b)/sq 
rt(b*tan(f*x + e)^2 + a))*tan(f*x + e)^2 + sqrt(b*tan(f*x + e)^2 + a)*a)/( 
a*f*tan(f*x + e)^2)]
 

Sympy [F]

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

integrate(cot(f*x+e)**3*(a+b*tan(f*x+e)**2)**(1/2),x)
 

Output:

Integral(sqrt(a + b*tan(e + f*x)**2)*cot(e + f*x)**3, x)
 

Maxima [F]

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

integrate(cot(f*x+e)^3*(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)^3, x)
 

Giac [F(-2)]

Exception generated. \[ \int \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [B] (verification not implemented)

Time = 7.47 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.07 \[ \int \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {a}\,b^4\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{2\,\left (\frac {a\,b^4}{2}-\frac {3\,b^5}{4}+\frac {b^6}{4\,a}\right )}-\frac {3\,b^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{4\,\sqrt {a}\,\left (\frac {a\,b^4}{2}-\frac {3\,b^5}{4}+\frac {b^6}{4\,a}\right )}+\frac {b^6\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{4\,a^{3/2}\,\left (\frac {a\,b^4}{2}-\frac {3\,b^5}{4}+\frac {b^6}{4\,a}\right )}\right )\,\left (2\,a-b\right )}{2\,\sqrt {a}\,f}-\frac {\mathrm {atanh}\left (\frac {b^4\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a-b}}{2\,\left (\frac {a\,b^4}{2}-\frac {b^5}{2}\right )}\right )\,\sqrt {a-b}}{f}-\frac {b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{2\,\left (f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )-a\,f\right )} \] Input:

int(cot(e + f*x)^3*(a + b*tan(e + f*x)^2)^(1/2),x)
 

Output:

(atanh((a^(1/2)*b^4*(a + b*tan(e + f*x)^2)^(1/2))/(2*((a*b^4)/2 - (3*b^5)/ 
4 + b^6/(4*a))) - (3*b^5*(a + b*tan(e + f*x)^2)^(1/2))/(4*a^(1/2)*((a*b^4) 
/2 - (3*b^5)/4 + b^6/(4*a))) + (b^6*(a + b*tan(e + f*x)^2)^(1/2))/(4*a^(3/ 
2)*((a*b^4)/2 - (3*b^5)/4 + b^6/(4*a))))*(2*a - b))/(2*a^(1/2)*f) - (atanh 
((b^4*(a + b*tan(e + f*x)^2)^(1/2)*(a - b)^(1/2))/(2*((a*b^4)/2 - b^5/2))) 
*(a - b)^(1/2))/f - (b*(a + b*tan(e + f*x)^2)^(1/2))/(2*(f*(a + b*tan(e + 
f*x)^2) - a*f))
 

Reduce [F]

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

int(cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x)
 

Output:

int(sqrt(tan(e + f*x)**2*b + a)*cot(e + f*x)**3,x)