\(\int \csc ^3(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [108]

Optimal result

Integrand size = 25, antiderivative size = 167 \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {\sqrt {a} (a+3 b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {\sqrt {b} (3 a+b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {b \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}-\frac {\cot (e+f x) \csc (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{2 f} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1012\) vs. \(2(167)=334\).

Time = 6.80 (sec) , antiderivative size = 1012, normalized size of antiderivative = 6.06 \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

(Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f* 
x)])]*(-1/2*(a*Cot[e + f*x]*Csc[e + f*x]) + (b*Sec[e + f*x])/2))/f + (((a^ 
2 - b^2)*(1 + Cos[e + f*x])*Sqrt[(1 + Cos[2*(e + f*x)])/(1 + Cos[e + f*x]) 
^2]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(4*Sqr 
t[a]*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])] - Sqrt[b]*(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]] + 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]]))*(-1 + Tan[(e + f*x) 
/2]^2)*(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)/(1 + Tan[(e + f*x)/2]^2)^2])/(4*Sqrt[a]*Sqrt[b]*Sqrt[a 
 + b + (a - b)*Cos[2*(e + f*x)]]*Sqrt[(-1 + Tan[(e + f*x)/2]^2)^2]*Sqrt[4* 
b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]) - ((a^2 + 6*a*b + b 
^2)*(1 + Cos[e + f*x])*Sqrt[(1 + Cos[2*(e + f*x)])/(1 + Cos[e + f*x])^2]*S 
qrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(4*Sqrt[a]* 
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])] + Sqrt[b]*(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]] + 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]]))*(-1 + Tan[(e + f*x)/2...
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4147, 369, 403, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 369

\(\displaystyle \frac {\frac {\sec (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 \left (1-\sec ^2(e+f x)\right )}-\frac {1}{2} \int \frac {\sqrt {b \sec ^2(e+f x)+a-b} \left (4 b \sec ^2(e+f x)+a-b\right )}{1-\sec ^2(e+f x)}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 403

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

((Sec[e + f*x]*(a - b + b*Sec[e + f*x]^2)^(3/2))/(2*(1 - Sec[e + f*x]^2)) 
+ (-(Sqrt[a]*(a + 3*b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e 
 + f*x]^2]]) + Sqrt[b]*(3*a + 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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 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 403

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

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. \(1031\) vs. \(2(145)=290\).

Time = 8.48 (sec) , antiderivative size = 1032, normalized size of antiderivative = 6.18

method	result	size
default	\(\text {Expression too large to display}\)	\(1032\)

Input:

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

Output:

1/4/f/b/a*(cos(f*x+e)^2*(cos(f*x+e)-1)*a^(5/2)*ln(2/a^(1/2)*(a^(1/2)*((a*c 
os(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)^2*(cos(f*x+e)-1)*a^(5/2)*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)^2*(3*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^2+cos(f*x+e)^2*(3*cos(f*x 
+e)-3)*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^2+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))*a*(-2*cos(f*x+e)^3+2*cos(f*x+e)^2)+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^2*(-6*cos(f*x+e)^3+6*cos(f*x+e)^ 
2)-2*cos(f*x+e)^2*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1...
 

Fricas [A] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 1059, normalized size of antiderivative = 6.34 \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*(((a + 3*b)*cos(f*x + e)^3 - (a + 3*b)*cos(f*x + e))*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)) + ((3*a + b)*cos(f* 
x + e)^3 - (3*a + b)*cos(f*x + e))*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) + 2*((a + b)*cos(f*x + e)^2 - b)*sqrt(((a - b)*cos(f 
*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), 1/4*( 
2*((3*a + b)*cos(f*x + e)^3 - (3*a + b)*cos(f*x + e))*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)) + ((a + 3*b)*cos(f*x + e)^3 - (a + 3*b)*cos(f*x + 
 e))*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 + b)*cos(f*x + e)^2 - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f 
*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), -1/4*(2*((a + 3*b)*cos(f* 
x + e)^3 - (a + 3*b)*cos(f*x + e))*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)) - ((3*a + b)*cos(f*x + e)^3 - (3*a + b)*cos(f*x + e))*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) - 2*((a + b)*cos(f*x + e)^2 
 - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1467 vs. \(2 (145) = 290\).

Time = 2.59 (sec) , antiderivative size = 1467, normalized size of antiderivative = 8.78 \[ \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*(sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan( 
1/2*f*x + 1/2*e)^2 + a)*a*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 4*(a^2*sgn(tan 
(1/2*f*x + 1/2*e)^2 - 1) + 3*a*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))*arctan(- 
(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan( 
1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))/sqrt(-a))/sqrt(-a) + 
 8*(3*a*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + b^2*sgn(tan(1/2*f*x + 1/2*e)^2 
 - 1))*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 
1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - 
sqrt(a))/sqrt(-b))/sqrt(-b) - 2*(a^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 3*a 
*b*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))*log(abs((sqrt(a)*tan(1/2*f*x + 1/2*e)^ 
2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1 
/2*f*x + 1/2*e)^2 + a))*sqrt(a) - a + 2*b))/sqrt(a) + 2*((sqrt(a)*tan(1/2* 
f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^ 
2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 
 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*t 
an(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a*b*sgn(tan(1/2*f 
*x + 1/2*e)^2 - 1) - a^(5/2)*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))/((sqrt(a)*ta 
n(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1 
/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2 - a) + 16*((sqrt(a)*tan(1/2*f 
*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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