\(\int \frac {\csc ^5(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [145]

Optimal result

Integrand size = 25, antiderivative size = 237 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left (3 a^2-30 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{8 a^{9/2} f}-\frac {(5 a-7 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(23 a-35 b) b \sec (e+f x)}{24 a^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {5 (11 a-21 b) b \sec (e+f x)}{24 a^4 f \sqrt {a-b+b \sec ^2(e+f x)}} \] Output:

-1/8*(3*a^2-30*a*b+35*b^2)*arctanh(a^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2) 
^(1/2))/a^(9/2)/f-1/8*(5*a-7*b)*cot(f*x+e)*csc(f*x+e)/a^2/f/(a-b+b*sec(f*x 
+e)^2)^(3/2)-1/4*cot(f*x+e)^3*csc(f*x+e)/a/f/(a-b+b*sec(f*x+e)^2)^(3/2)-1/ 
24*(23*a-35*b)*b*sec(f*x+e)/a^3/f/(a-b+b*sec(f*x+e)^2)^(3/2)-5/24*(11*a-21 
*b)*b*sec(f*x+e)/a^4/f/(a-b+b*sec(f*x+e)^2)^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1132\) vs. \(2(237)=474\).

Time = 7.42 (sec) , antiderivative size = 1132, normalized size of antiderivative = 4.78 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

Integrate[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(5/2),x]
 

Output:

(Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f* 
x)])]*((4*b^2*Cos[e + f*x])/(3*a^3*(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*( 
e + f*x)])^2) - (2*(2*a*b*Cos[e + f*x] - 3*b^2*Cos[e + f*x]))/(a^4*(a + b 
+ a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])) + ((-3*a*Cos[e + f*x] + 11*b*C 
os[e + f*x])*Csc[e + f*x]^2)/(8*a^4) - (Cot[e + f*x]*Csc[e + f*x]^3)/(4*a^ 
3)))/f + ((3*a^2 - 30*a*b + 35*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*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*S 
qrt[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 + T 
an[(e + f*x)/2]^2)^2]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/ 
2]^2)^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 
*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*...
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4147, 25, 372, 402, 25, 402, 25, 27, 402, 27, 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.

\(\displaystyle \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^5 \left (a+b \tan (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4147

\(\displaystyle \frac {\int -\frac {\sec ^4(e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {\sec ^4(e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 372

\(\displaystyle \frac {\frac {\int \frac {2 (2 a-3 b) \sec ^2(e+f x)+a-b}{\left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\int -\frac {(3 a-7 b) (a-b)-4 (5 a-7 b) b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{2 a}+\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\int \frac {(3 a-7 b) (a-b)-4 (5 a-7 b) b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\int -\frac {(a-b) \left ((9 a-35 b) (a-b)-2 (23 a-35 b) b \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a (a-b)}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\int \frac {(a-b) \left ((9 a-35 b) (a-b)-2 (23 a-35 b) b \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a (a-b)}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\int \frac {(9 a-35 b) (a-b)-2 (23 a-35 b) b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {5 b (11 a-21 b) \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\int -\frac {3 (a-b) \left (3 a^2-30 b a+35 b^2\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a (a-b)}}{3 a}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {3 \left (3 a^2-30 a b+35 b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a}+\frac {5 b (11 a-21 b) \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {3 \left (3 a^2-30 a b+35 b^2\right ) \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}}{a}+\frac {5 b (11 a-21 b) \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {3 \left (3 a^2-30 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{a^{3/2}}+\frac {5 b (11 a-21 b) \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\)

Input:

Int[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(5/2),x]
 

Output:

(-1/4*Sec[e + f*x]/(a*(1 - Sec[e + f*x]^2)^2*(a - b + b*Sec[e + f*x]^2)^(3 
/2)) + (((5*a - 7*b)*Sec[e + f*x])/(2*a*(1 - Sec[e + f*x]^2)*(a - b + b*Se 
c[e + f*x]^2)^(3/2)) - (((23*a - 35*b)*b*Sec[e + f*x])/(3*a*(a - b + b*Sec 
[e + f*x]^2)^(3/2)) + ((3*(3*a^2 - 30*a*b + 35*b^2)*ArcTanh[(Sqrt[a]*Sec[e 
 + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/a^(3/2) + (5*(11*a - 21*b)*b*Sec 
[e + f*x])/(a*Sqrt[a - b + b*Sec[e + f*x]^2]))/(3*a))/(2*a))/(4*a))/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 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 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 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 402

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

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. \(4401\) vs. \(2(213)=426\).

Time = 162.72 (sec) , antiderivative size = 4402, normalized size of antiderivative = 18.57

method	result	size
default	\(\text {Expression too large to display}\)	\(4402\)

Input:

int(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/48/f/a^(21/2)/(a^2-2*a*b+b^2)/(2*(-b*(a-b))^(1/2)-a+2*b)^4/(2*(-b*(a-b) 
)^(1/2)+a-2*b)^4*((3510*cos(f*x+e)^9+3510*cos(f*x+e)^8-6060*cos(f*x+e)^7-6 
060*cos(f*x+e)^6+3114*cos(f*x+e)^5+3114*cos(f*x+e)^4-468*cos(f*x+e)^3-468* 
cos(f*x+e)^2+9*cos(f*x+e)+9)*sin(f*x+e)^4*((a*cos(f*x+e)^2+b*sin(f*x+e)^2) 
/(cos(f*x+e)+1)^2)^(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^18*b^4+cos(f*x+e)^6*(-1560*cos(f*x+e)^6+3770*cos(f*x+e)^4-2866*cos(f* 
x+e)^2+668)*a^(41/2)*b^2+(-7008*cos(f*x+e)^6+7946*cos(f*x+e)^4-2212*cos(f* 
x+e)^2+110)*sin(f*x+e)^6*a^(35/2)*b^5+(-4248*cos(f*x+e)^4+3166*cos(f*x+e)^ 
2-430)*sin(f*x+e)^8*a^(33/2)*b^6+(-1440*cos(f*x+e)^2+530)*sin(f*x+e)^10*a^ 
(31/2)*b^7+cos(f*x+e)^8*(288*cos(f*x+e)^4-558*cos(f*x+e)^2+246)*a^(43/2)*b 
+(3504*cos(f*x+e)^7+3504*cos(f*x+e)^6-4476*cos(f*x+e)^5-4476*cos(f*x+e)^4+ 
1500*cos(f*x+e)^3+1500*cos(f*x+e)^2-108*cos(f*x+e)-108)*sin(f*x+e)^6*((a*c 
os(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(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*co 
s(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)+co 
s(f*x+e)*b+b)/(cos(f*x+e)+1))*a^17*b^5+(2124*cos(f*x+e)^5+2124*cos(f*x+e)^ 
4-1788*cos(f*x+e)^3-1788*cos(f*x+e)^2+294*cos(f*x+e)+294)*sin(f*x+e)^8*((a 
*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*ln(2/a^(1/2)*(a^(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 522 vs. \(2 (213) = 426\).

Time = 0.36 (sec) , antiderivative size = 1054, normalized size of antiderivative = 4.45 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

[1/48*(3*((3*a^4 - 36*a^3*b + 98*a^2*b^2 - 100*a*b^3 + 35*b^4)*cos(f*x + e 
)^8 - 2*(3*a^4 - 39*a^3*b + 131*a^2*b^2 - 165*a*b^3 + 70*b^4)*cos(f*x + e) 
^6 + (3*a^4 - 48*a^3*b + 233*a^2*b^2 - 390*a*b^3 + 210*b^4)*cos(f*x + e)^4 
 + 3*a^2*b^2 - 30*a*b^3 + 35*b^4 + 2*(3*a^3*b - 36*a^2*b^2 + 95*a*b^3 - 70 
*b^4)*cos(f*x + e)^2)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*s 
qrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(co 
s(f*x + e)^2 - 1)) + 2*(3*(3*a^4 - 33*a^3*b + 65*a^2*b^2 - 35*a*b^3)*cos(f 
*x + e)^7 - (15*a^4 - 177*a^3*b + 445*a^2*b^2 - 315*a*b^3)*cos(f*x + e)^5 
- (78*a^3*b - 305*a^2*b^2 + 315*a*b^3)*cos(f*x + e)^3 - 5*(11*a^2*b^2 - 21 
*a*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/( 
(a^7 - 2*a^6*b + a^5*b^2)*f*cos(f*x + e)^8 + a^5*b^2*f - 2*(a^7 - 3*a^6*b 
+ 2*a^5*b^2)*f*cos(f*x + e)^6 + (a^7 - 6*a^6*b + 6*a^5*b^2)*f*cos(f*x + e) 
^4 + 2*(a^6*b - 2*a^5*b^2)*f*cos(f*x + e)^2), -1/24*(3*((3*a^4 - 36*a^3*b 
+ 98*a^2*b^2 - 100*a*b^3 + 35*b^4)*cos(f*x + e)^8 - 2*(3*a^4 - 39*a^3*b + 
131*a^2*b^2 - 165*a*b^3 + 70*b^4)*cos(f*x + e)^6 + (3*a^4 - 48*a^3*b + 233 
*a^2*b^2 - 390*a*b^3 + 210*b^4)*cos(f*x + e)^4 + 3*a^2*b^2 - 30*a*b^3 + 35 
*b^4 + 2*(3*a^3*b - 36*a^2*b^2 + 95*a*b^3 - 70*b^4)*cos(f*x + e)^2)*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*(3*a^4 - 33*a^3*b + 65*a^2*b^2 
 - 35*a*b^3)*cos(f*x + e)^7 - (15*a^4 - 177*a^3*b + 445*a^2*b^2 - 315*a...
 

Sympy [F]

\[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\csc ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(csc(f*x+e)**5/(a+b*tan(f*x+e)**2)**(5/2),x)
 

Output:

Integral(csc(e + f*x)**5/(a + b*tan(e + f*x)**2)**(5/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1648 vs. \(2 (213) = 426\).

Time = 2.49 (sec) , antiderivative size = 1648, normalized size of antiderivative = 6.95 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

1/192*(((((3*((a^17*b^2 - 2*a^16*b^3 + a^15*b^4)*tan(1/2*f*x + 1/2*e)^2/(a 
^18*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 2*a^17*b^3*sgn(tan(1/2*f*x + 1/2 
*e)^2 - 1) + a^16*b^4*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)) + (5*a^17*b^2 - 24* 
a^16*b^3 + 33*a^15*b^4 - 14*a^14*b^5)/(a^18*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 
 - 1) - 2*a^17*b^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + a^16*b^4*sgn(tan(1/2* 
f*x + 1/2*e)^2 - 1)))*tan(1/2*f*x + 1/2*e)^2 - 2*(45*a^17*b^2 - 498*a^16*b 
^3 + 1421*a^15*b^4 - 1528*a^14*b^5 + 560*a^13*b^6)/(a^18*b^2*sgn(tan(1/2*f 
*x + 1/2*e)^2 - 1) - 2*a^17*b^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + a^16*b^4 
*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)))*tan(1/2*f*x + 1/2*e)^2 + 6*(25*a^17*b^2 
 - 248*a^16*b^3 + 989*a^15*b^4 - 1894*a^14*b^5 + 1688*a^13*b^6 - 560*a^12* 
b^7)/(a^18*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 2*a^17*b^3*sgn(tan(1/2*f* 
x + 1/2*e)^2 - 1) + a^16*b^4*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)))*tan(1/2*f*x 
 + 1/2*e)^2 - 3*(35*a^17*b^2 - 102*a^16*b^3 - 365*a^15*b^4 + 1664*a^14*b^5 
 - 2000*a^13*b^6 + 768*a^12*b^7)/(a^18*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) 
 - 2*a^17*b^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + a^16*b^4*sgn(tan(1/2*f*x + 
 1/2*e)^2 - 1)))*tan(1/2*f*x + 1/2*e)^2 + (27*a^17*b^2 + 264*a^16*b^3 - 12 
49*a^15*b^4 + 1598*a^14*b^5 - 640*a^13*b^6)/(a^18*b^2*sgn(tan(1/2*f*x + 1/ 
2*e)^2 - 1) - 2*a^17*b^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + a^16*b^4*sgn(ta 
n(1/2*f*x + 1/2*e)^2 - 1)))/(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)^(3/2) - 24*(3*a^2 - 30*a*b +...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Hanged} \] Input:

int(1/(sin(e + f*x)^5*(a + b*tan(e + f*x)^2)^(5/2)),x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )^{5}}{\tan \left (f x +e \right )^{6} b^{3}+3 \tan \left (f x +e \right )^{4} a \,b^{2}+3 \tan \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^(5/2),x)
 

Output:

int((sqrt(tan(e + f*x)**2*b + a)*csc(e + f*x)**5)/(tan(e + f*x)**6*b**3 + 
3*tan(e + f*x)**4*a*b**2 + 3*tan(e + f*x)**2*a**2*b + a**3),x)