\(\int \csc (e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [107]

Optimal result
Mathematica [B] (verified)
Rubi [A] (verified)
Maple [B] (warning: unable to verify)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 127 \[ \int \csc (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}+\frac {(3 a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {b \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 f} \] Output:

-a^(3/2)*arctanh(a^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2)^(1/2))/f+1/2*(3*a 
-b)*b^(1/2)*arctanh(b^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2)^(1/2))/f+1/2*b 
*sec(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/f
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(492\) vs. \(2(127)=254\).

Time = 3.31 (sec) , antiderivative size = 492, normalized size of antiderivative = 3.87 \[ \int \csc (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right ) \left (-4 \sqrt {b} (-3 a+b) \text {arctanh}\left (\frac {-\sqrt {a} \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{2 \sqrt {b}}\right ) \cos ^2(e+f x)+4 a^{3/2} \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right ) \cos ^2(e+f x)+a^{3/2} \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )+a^{3/2} \cos (2 (e+f x)) \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )+\frac {b \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}{\sqrt {2}}+\frac {b \cos (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}{\sqrt {2}}\right ) \sec (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{4 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}} \] Input:

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

Output:

(Sec[(e + f*x)/2]^2*(-4*Sqrt[b]*(-3*a + b)*ArcTanh[(-(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])]*Cos[e + f*x]^2 + 4*a^(3/2)*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]]*Cos[ 
e + f*x]^2 + a^(3/2)*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]] + a^(3/2)*Cos[2*(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*Sqrt[(a + b + (a - b)*Cos[2*( 
e + f*x)])*Sec[(e + f*x)/2]^4])/Sqrt[2] + (b*Cos[e + f*x]*Sqrt[(a + b + (a 
 - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])/Sqrt[2])*Sec[e + f*x]*Sqrt[(a 
 + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(4*f*Sqrt[(a + b + (a - 
b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4147, 25, 318, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{2} \left (\sqrt {b} (3 a-b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )-2 a^2 \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)\right )+\frac {1}{2} b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{2} \left (\sqrt {b} (3 a-b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )-2 a^2 \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}\right )+\frac {1}{2} b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{2} \left (\sqrt {b} (3 a-b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )\right )+\frac {1}{2} b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}\)

Input:

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

Output:

((-2*a^(3/2)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2] 
] + (3*a - b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e 
+ f*x]^2]])/2 + (b*Sec[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/2)/f
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 318
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 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. \(665\) vs. \(2(109)=218\).

Time = 8.36 (sec) , antiderivative size = 666, normalized size of antiderivative = 5.24

method result size
default \(-\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (a^{\frac {3}{2}} \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 ) b \cos \left (f x +e \right )^{3}+a^{\frac {3}{2}} \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 ) b \cos \left (f x +e \right )^{3}+b^{\frac {5}{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 ) \cos \left (f x +e \right )^{3}-3 b^{\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 ) a \cos \left (f x +e \right )^{3}+b^{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}}}\, \left (-\cos \left (f x +e \right )^{2}-\cos \left (f x +e \right )\right )\right )}{2 f 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}}}\, \left (\cos \left (f x +e \right )^{2} \left (\cos \left (f x +e \right )+1\right ) a +\left (\cos \left (f x +e \right )+1\right ) \sin \left (f x +e \right )^{2} b \right )}\) \(666\)

Input:

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

Output:

-1/2/f/b*(a+b*tan(f*x+e)^2)^(3/2)/((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f* 
x+e)+1)^2)^(1/2)/(cos(f*x+e)^2*(cos(f*x+e)+1)*a+(cos(f*x+e)+1)*sin(f*x+e)^ 
2*b)*(a^(3/2)*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*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)*b*cos(f*x+e)^3+a^(3/ 
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))*b*cos(f*x+e)^3+b^(5 
/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))*cos(f*x+e)^3-3*b^(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))*a*cos(f*x+e)^3+b^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)^2-cos(f*x+e)))
 

Fricas [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 809, normalized size of antiderivative = 6.37 \[ \int \csc (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/4*(2*a^(3/2)*cos(f*x + e)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sq 
rt(((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)) - (3*a - b)*sqrt(b)*cos(f*x + e)*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) + 2*b*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos( 
f*x + e)^2))/(f*cos(f*x + e)), 1/2*((3*a - b)*sqrt(-b)*arctan(-sqrt(-b)*sq 
rt(((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))*cos(f*x + e) + a^(3/2)*cos(f*x + e)*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)*co 
s(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) + b*sqrt(((a - b)*cos(f*x + e)^2 
 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), -1/4*(4*sqrt(-a)*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))*cos(f*x + e) + (3*a - b)*sqrt(b)*cos(f*x + e)*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) - 2*b*sqrt(((a - b)*cos(f*x 
 + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), -1/2*(2*sqrt(-a)*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))*cos(f*x + e) - (3*a - b)*sqrt(-b)*arctan(-sqr 
t(-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))*cos(f*x + e) - b*sqrt(((a - b)*cos(f*x + e)^2 ...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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