\(\int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [96]

Optimal result

Integrand size = 25, antiderivative size = 127 \[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 \sqrt {a} f}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 f} \] Output:

-1/2*(a+b)*arctanh(a^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(1/2)/ 
f+b^(1/2)*arctanh(b^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2)^(1/2))/f-1/2*cot 
(f*x+e)*csc(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/f
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(586\) vs. \(2(127)=254\).

Time = 3.77 (sec) , antiderivative size = 586, normalized size of antiderivative = 4.61 \[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\cot (e+f x) \csc (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (-a \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )-b \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )+a \cos (e+f x) \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )+b \cos (e+f x) \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )+\frac {\sqrt {a} \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}{\sqrt {2}}-16 \sqrt {a} \sqrt {b} \text {arctanh}\left (\frac {-\sqrt {a} \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{2 \sqrt {b}}\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )-4 (a+b) \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )}{2 \sqrt {a} f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}} \] Input:

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

Output:

-1/2*(Cot[e + f*x]*Csc[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Se 
c[e + f*x]^2]*(-(a*Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*T 
an[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]]) - b*Log[a - 2*b - a*T 
an[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + 
 f*x)/2]^2)^2]] + a*Cos[e + f*x]*Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt 
[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]] + b*Cos[ 
e + f*x]*Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f* 
x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]] + (Sqrt[a]*Sqrt[(a + b + (a - b) 
*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])/Sqrt[2] - 16*Sqrt[a]*Sqrt[b]*ArcTa 
nh[(-(Sqrt[a]*(-1 + Tan[(e + f*x)/2]^2)) + Sqrt[4*b*Tan[(e + f*x)/2]^2 + a 
*(-1 + Tan[(e + f*x)/2]^2)^2])/(2*Sqrt[b])]*Sin[(e + f*x)/2]^2 - 4*(a + b) 
*ArcTanh[Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e 
 + f*x)/2]^2)^2]/Sqrt[a]]*Sin[(e + f*x)/2]^2))/(Sqrt[a]*f*Sqrt[(a + b + (a 
 - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4147, 369, 398, 224, 219, 291, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 369

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

((-(((a + b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2] 
])/Sqrt[a]) + 2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[ 
e + f*x]^2]])/2 + (Sec[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(2*(1 - Se 
c[e + f*x]^2)))/f
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 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 369

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

rule 398

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

rule 3042

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

rule 4147

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ 
m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 
)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( 
m - 1)/2]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(760\) vs. \(2(109)=218\).

Time = 6.49 (sec) , antiderivative size = 761, normalized size of antiderivative = 5.99

method	result	size
default	\(-\frac {\left (\left (1-\cos \left (f x +e \right )\right ) \ln \left (\frac {2 \sqrt {a}\, \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}}}\, \cos \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}}}\, \sqrt {a}-2 a \cos \left (f x +e \right )+2 \cos \left (f x +e \right ) b +2 b}{\sqrt {a}\, \left (\cos \left (f x +e \right )+1\right )}\right ) a b +\left (1-\cos \left (f x +e \right )\right ) \ln \left (\frac {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}}}\, \sqrt {a}\, \sin \left (f x +e \right )^{2}+2 a \sin \left (f x +e \right )^{2}-2 a \cos \left (f x +e \right )^{2}+4 b \cos \left (f x +e \right )^{2}+4 a \cos \left (f x +e \right )-8 \cos \left (f x +e \right ) b -2 a +4 b}{\left (\cos \left (f x +e \right )-1\right )^{2}}\right ) a b +\left (1-\cos \left (f x +e \right )\right ) \ln \left (\frac {2 \sqrt {a}\, \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}}}\, \cos \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}}}\, \sqrt {a}-2 a \cos \left (f x +e \right )+2 \cos \left (f x +e \right ) b +2 b}{\sqrt {a}\, \left (\cos \left (f x +e \right )+1\right )}\right ) a^{2}+\left (4 \cos \left (f x +e \right )-4\right ) \sqrt {b}\, a^{\frac {3}{2}} \ln \left (\frac {4 b \cot \left (f x +e \right )^{2}-8 b \cot \left (f x +e \right ) \csc \left (f x +e \right )+4 b \csc \left (f x +e \right )^{2}+8 \sqrt {b}\, \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}}}+4 b}{\cot \left (f x +e \right )^{2}-2 \csc \left (f x +e \right ) \cot \left (f x +e \right )+\csc \left (f x +e \right )^{2}-1}\right )+\left (1-\cos \left (f x +e \right )\right ) \ln \left (\frac {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}}}\, \sqrt {a}\, \sin \left (f x +e \right )^{2}+2 a \sin \left (f x +e \right )^{2}-2 a \cos \left (f x +e \right )^{2}+4 b \cos \left (f x +e \right )^{2}+4 a \cos \left (f x +e \right )-8 \cos \left (f x +e \right ) b -2 a +4 b}{\left (\cos \left (f x +e \right )-1\right )^{2}}\right ) a^{2}+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^{\frac {3}{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 f \,a^{\frac {3}{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}}}}\)	\(761\)

Input:

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

Output:

-1/4/f/a^(3/2)*((1-cos(f*x+e))*ln(2/a^(1/2)*(a^(1/2)*((a*cos(f*x+e)^2+b*si 
n(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*b+(1-cos(f*x+e))*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*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*b+(1 
-cos(f*x+e))*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+(4*co 
s(f*x+e)-4)*b^(1/2)*a^(3/2)*ln(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e) 
+b*csc(f*x+e)^2+2*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^ 
2)^(1/2)+b)/(cot(f*x+e)^2-2*csc(f*x+e)*cot(f*x+e)+csc(f*x+e)^2-1))+(1-cos( 
f*x+e))*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*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^2+2*((a*cos(f*x+e)^2+b*s 
in(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(3/2))*(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)*cot(f*x+e)*csc(f*x+e 
)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (109) = 218\).

Time = 0.58 (sec) , antiderivative size = 915, normalized size of antiderivative = 7.20 \[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*(2*a*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 
 ((a + b)*cos(f*x + e)^2 - a - b)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 
 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) 
+ a + b)/(cos(f*x + e)^2 - 1)) + 2*(a*cos(f*x + e)^2 - a)*sqrt(b)*log(-((a 
 - b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x 
 + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2))/(a*f*cos(f*x + e)^2 - a*f), 
-1/2*(((a + b)*cos(f*x + e)^2 - a - b)*sqrt(-a)*arctan(-sqrt(-a)*sqrt(((a 
- b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/((a - b)*cos(f*x + e 
)^2 + b)) - a*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + 
e) - (a*cos(f*x + e)^2 - a)*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt( 
b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/c 
os(f*x + e)^2))/(a*f*cos(f*x + e)^2 - a*f), 1/4*(4*(a*cos(f*x + e)^2 - a)* 
sqrt(-b)*arctan(-sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2 
)*cos(f*x + e)/((a - b)*cos(f*x + e)^2 + b)) + 2*a*sqrt(((a - b)*cos(f*x + 
 e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + ((a + b)*cos(f*x + e)^2 - a - b) 
*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x 
+ e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)))/( 
a*f*cos(f*x + e)^2 - a*f), -1/2*(((a + b)*cos(f*x + e)^2 - a - b)*sqrt(-a) 
*arctan(-sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f* 
x + e)/((a - b)*cos(f*x + e)^2 + b)) - 2*(a*cos(f*x + e)^2 - a)*sqrt(-b...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F(-2)]

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

integrate(csc(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 [F(-1)]

Timed out. \[ \int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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