\(\int \frac {\csc ^2(x) (\sec ^2(x)-3 \tan (x) \sqrt {4 \sec ^2(x)+5 \tan ^2(x)})}{(4 \sec ^2(x)+5 \tan ^2(x))^{3/2}} \, dx\) [439]

Optimal result

Integrand size = 48, antiderivative size = 57 \[ \int \frac {\csc ^2(x) \left (\sec ^2(x)-3 \tan (x) \sqrt {4 \sec ^2(x)+5 \tan ^2(x)}\right )}{\left (4 \sec ^2(x)+5 \tan ^2(x)\right )^{3/2}} \, dx=-\frac {3}{4} \log (\tan (x))+\frac {3}{8} \log \left (4+9 \tan ^2(x)\right )-\frac {\cot (x)}{4 \sqrt {4+9 \tan ^2(x)}}-\frac {7 \tan (x)}{8 \sqrt {4+9 \tan ^2(x)}} \] Output:

-3/4*ln(tan(x))+3/8*ln(4+9*tan(x)^2)-1/4*cot(x)/(4+9*tan(x)^2)^(1/2)-7/8*t 
an(x)/(4+9*tan(x)^2)^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(116\) vs. \(2(57)=114\).

Time = 4.47 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.04 \[ \int \frac {\csc ^2(x) \left (\sec ^2(x)-3 \tan (x) \sqrt {4 \sec ^2(x)+5 \tan ^2(x)}\right )}{\left (4 \sec ^2(x)+5 \tan ^2(x)\right )^{3/2}} \, dx=\frac {5 \cot (x)+6 \sqrt {\frac {13-5 \cos (2 x)}{1+\cos (2 x)}} \log \left (1+7 \tan ^2\left (\frac {x}{2}\right )+\tan ^4\left (\frac {x}{2}\right )\right )-9 \csc (x) \sec (x)-5 \tan (x)-6 \sqrt {2} \log \left (\tan \left (\frac {x}{2}\right )\right ) \sqrt {-5+13 \sec ^2(x)+5 \tan ^2(x)}}{16 \sqrt {\frac {13-5 \cos (2 x)}{1+\cos (2 x)}}} \] Input:

Integrate[(Csc[x]^2*(Sec[x]^2 - 3*Tan[x]*Sqrt[4*Sec[x]^2 + 5*Tan[x]^2]))/( 
4*Sec[x]^2 + 5*Tan[x]^2)^(3/2),x]
 

Output:

(5*Cot[x] + 6*Sqrt[(13 - 5*Cos[2*x])/(1 + Cos[2*x])]*Log[1 + 7*Tan[x/2]^2 
+ Tan[x/2]^4] - 9*Csc[x]*Sec[x] - 5*Tan[x] - 6*Sqrt[2]*Log[Tan[x/2]]*Sqrt[ 
-5 + 13*Sec[x]^2 + 5*Tan[x]^2])/(16*Sqrt[(13 - 5*Cos[2*x])/(1 + Cos[2*x])] 
)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3042, 4889, 7293, 2009}

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.

\(\displaystyle \int \frac {\csc ^2(x) \left (\sec ^2(x)-3 \tan (x) \sqrt {5 \tan ^2(x)+4 \sec ^2(x)}\right )}{\left (5 \tan ^2(x)+4 \sec ^2(x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc (x)^2 \left (\sec (x)^2-3 \tan (x) \sqrt {5 \tan (x)^2+4 \sec (x)^2}\right )}{\left (5 \tan (x)^2+4 \sec (x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4889

\(\displaystyle \int \frac {\left (\tan ^2(x)-3 \sqrt {9 \tan ^2(x)+4} \tan (x)+1\right ) \cot ^2(x)}{\left (9 \tan ^2(x)+4\right )^{3/2}}d\tan (x)\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {1}{\left (9 \tan ^2(x)+4\right )^{3/2}}+\frac {\cot ^2(x)}{\left (9 \tan ^2(x)+4\right )^{3/2}}-\frac {3 \cot (x)}{9 \tan ^2(x)+4}\right )d\tan (x)\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {7 \tan (x)}{8 \sqrt {9 \tan ^2(x)+4}}+\frac {3}{8} \log \left (9 \tan ^2(x)+4\right )-\frac {3}{4} \log (\tan (x))-\frac {\cot (x)}{4 \sqrt {9 \tan ^2(x)+4}}\)

Input:

Int[(Csc[x]^2*(Sec[x]^2 - 3*Tan[x]*Sqrt[4*Sec[x]^2 + 5*Tan[x]^2]))/(4*Sec[ 
x]^2 + 5*Tan[x]^2)^(3/2),x]
 

Output:

(-3*Log[Tan[x]])/4 + (3*Log[4 + 9*Tan[x]^2])/8 - Cot[x]/(4*Sqrt[4 + 9*Tan[ 
x]^2]) - (7*Tan[x])/(8*Sqrt[4 + 9*Tan[x]^2])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4889

Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors 
[Tan[v], x]}, Simp[d/Coefficient[v, x, 1]   Subst[Int[SubstFor[1/(1 + d^2*x 
^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /;  !FalseQ[v] && FunctionOfQ[N 
onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] &&  !MatchQ[ 
u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I 
ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 4.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.09

method	result	size
parts	\(-\frac {\sec \left (x \right )^{3} \csc \left (x \right ) \left (25 \cos \left (x \right )^{4}-80 \cos \left (x \right )^{2}+63\right ) \sqrt {4}}{16 \left (-5+9 \sec \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {3 \ln \left (-1+\cos \left (x \right )\right )}{8}+\frac {3 \ln \left (5 \cos \left (x \right )^{2}-9\right )}{8}-\frac {3 \ln \left (1+\cos \left (x \right )\right )}{8}\)	\(62\)
default	\(-\frac {-3 \left (-5+9 \sec \left (x \right )^{2}\right )^{\frac {3}{2}} \ln \left (-\frac {5 \cos \left (x \right )^{2}-9}{\left (1+\cos \left (x \right )\right )^{2}}\right )+6 \left (-5+9 \sec \left (x \right )^{2}\right )^{\frac {3}{2}} \ln \left (\csc \left (x \right )-\cot \left (x \right )\right )+25 \cot \left (x \right )-80 \sec \left (x \right ) \csc \left (x \right )+63 \sec \left (x \right )^{3} \csc \left (x \right )}{8 \left (-5+9 \sec \left (x \right )^{2}\right )^{\frac {3}{2}}}\)	\(81\)

Input:

int((sec(x)^2-3*(4*sec(x)^2+5*tan(x)^2)^(1/2)*tan(x))/sin(x)^2/(4*sec(x)^2 
+5*tan(x)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/16*sec(x)^3*csc(x)*(25*cos(x)^4-80*cos(x)^2+63)/(-5+9*sec(x)^2)^(3/2)*4 
^(1/2)-3/8*ln(-1+cos(x))+3/8*ln(5*cos(x)^2-9)-3/8*ln(1+cos(x))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.47 \[ \int \frac {\csc ^2(x) \left (\sec ^2(x)-3 \tan (x) \sqrt {4 \sec ^2(x)+5 \tan ^2(x)}\right )}{\left (4 \sec ^2(x)+5 \tan ^2(x)\right )^{3/2}} \, dx=\frac {3 \, {\left (5 \, \cos \left (x\right )^{2} - 9\right )} \log \left (-\frac {5}{4} \, \cos \left (x\right )^{2} + \frac {9}{4}\right ) \sin \left (x\right ) - 6 \, {\left (5 \, \cos \left (x\right )^{2} - 9\right )} \log \left (\frac {1}{2} \, \sin \left (x\right )\right ) \sin \left (x\right ) - {\left (5 \, \cos \left (x\right )^{3} - 7 \, \cos \left (x\right )\right )} \sqrt {-\frac {5 \, \cos \left (x\right )^{2} - 9}{\cos \left (x\right )^{2}}}}{8 \, {\left (5 \, \cos \left (x\right )^{2} - 9\right )} \sin \left (x\right )} \] Input:

integrate((sec(x)^2-3*(4*sec(x)^2+5*tan(x)^2)^(1/2)*tan(x))/sin(x)^2/(4*se 
c(x)^2+5*tan(x)^2)^(3/2),x, algorithm="fricas")
 

Output:

1/8*(3*(5*cos(x)^2 - 9)*log(-5/4*cos(x)^2 + 9/4)*sin(x) - 6*(5*cos(x)^2 - 
9)*log(1/2*sin(x))*sin(x) - (5*cos(x)^3 - 7*cos(x))*sqrt(-(5*cos(x)^2 - 9) 
/cos(x)^2))/((5*cos(x)^2 - 9)*sin(x))
 

Sympy [F]

\[ \int \frac {\csc ^2(x) \left (\sec ^2(x)-3 \tan (x) \sqrt {4 \sec ^2(x)+5 \tan ^2(x)}\right )}{\left (4 \sec ^2(x)+5 \tan ^2(x)\right )^{3/2}} \, dx=\int \frac {- 3 \sqrt {5 \tan ^{2}{\left (x \right )} + 4 \sec ^{2}{\left (x \right )}} \tan {\left (x \right )} + \sec ^{2}{\left (x \right )}}{\left (5 \tan ^{2}{\left (x \right )} + 4 \sec ^{2}{\left (x \right )}\right )^{\frac {3}{2}} \sin ^{2}{\left (x \right )}}\, dx \] Input:

integrate((sec(x)**2-3*(4*sec(x)**2+5*tan(x)**2)**(1/2)*tan(x))/sin(x)**2/ 
(4*sec(x)**2+5*tan(x)**2)**(3/2),x)
 

Output:

Integral((-3*sqrt(5*tan(x)**2 + 4*sec(x)**2)*tan(x) + sec(x)**2)/((5*tan(x 
)**2 + 4*sec(x)**2)**(3/2)*sin(x)**2), x)
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {\csc ^2(x) \left (\sec ^2(x)-3 \tan (x) \sqrt {4 \sec ^2(x)+5 \tan ^2(x)}\right )}{\left (4 \sec ^2(x)+5 \tan ^2(x)\right )^{3/2}} \, dx=-\frac {7 \, \tan \left (x\right )}{8 \, \sqrt {9 \, \tan \left (x\right )^{2} + 4}} - \frac {1}{4 \, \sqrt {9 \, \tan \left (x\right )^{2} + 4} \tan \left (x\right )} + \frac {3}{8} \, \log \left (9 \, \tan \left (x\right )^{2} + 4\right ) - \frac {3}{4} \, \log \left (\tan \left (x\right )\right ) \] Input:

integrate((sec(x)^2-3*(4*sec(x)^2+5*tan(x)^2)^(1/2)*tan(x))/sin(x)^2/(4*se 
c(x)^2+5*tan(x)^2)^(3/2),x, algorithm="maxima")
 

Output:

-7/8*tan(x)/sqrt(9*tan(x)^2 + 4) - 1/4/(sqrt(9*tan(x)^2 + 4)*tan(x)) + 3/8 
*log(9*tan(x)^2 + 4) - 3/4*log(tan(x))
 

Giac [F]

\[ \int \frac {\csc ^2(x) \left (\sec ^2(x)-3 \tan (x) \sqrt {4 \sec ^2(x)+5 \tan ^2(x)}\right )}{\left (4 \sec ^2(x)+5 \tan ^2(x)\right )^{3/2}} \, dx=\int { \frac {\sec \left (x\right )^{2} - 3 \, \sqrt {4 \, \sec \left (x\right )^{2} + 5 \, \tan \left (x\right )^{2}} \tan \left (x\right )}{{\left (4 \, \sec \left (x\right )^{2} + 5 \, \tan \left (x\right )^{2}\right )}^{\frac {3}{2}} \sin \left (x\right )^{2}} \,d x } \] Input:

integrate((sec(x)^2-3*(4*sec(x)^2+5*tan(x)^2)^(1/2)*tan(x))/sin(x)^2/(4*se 
c(x)^2+5*tan(x)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate((sec(x)^2 - 3*sqrt(4*sec(x)^2 + 5*tan(x)^2)*tan(x))/((4*sec(x)^2 
 + 5*tan(x)^2)^(3/2)*sin(x)^2), x)
 

Mupad [B] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.98 \[ \int \frac {\csc ^2(x) \left (\sec ^2(x)-3 \tan (x) \sqrt {4 \sec ^2(x)+5 \tan ^2(x)}\right )}{\left (4 \sec ^2(x)+5 \tan ^2(x)\right )^{3/2}} \, dx=\frac {3\,\ln \left (\left (\cos \left (2\,x\right )+\sin \left (2\,x\right )\,1{}\mathrm {i}\right )\,\left (5\,\cos \left (2\,x\right )-13\right )\right )}{8}-\frac {3\,\ln \left (\cos \left (2\,x\right )\,852930{}\mathrm {i}-852930\,\sin \left (2\,x\right )-852930{}\mathrm {i}\right )}{4}-\frac {\frac {18\,\sin \left (2\,x\right )\,\sqrt {13-5\,\cos \left (2\,x\right )}}{\sqrt {\cos \left (2\,x\right )+1}}-\frac {5\,\sin \left (4\,x\right )\,\sqrt {13-5\,\cos \left (2\,x\right )}}{\sqrt {\cos \left (2\,x\right )+1}}}{80\,{\cos \left (2\,x\right )}^2-288\,\cos \left (2\,x\right )+208} \] Input:

int((1/cos(x)^2 - 3*tan(x)*(4/cos(x)^2 + 5*tan(x)^2)^(1/2))/(sin(x)^2*(4/c 
os(x)^2 + 5*tan(x)^2)^(3/2)),x)
 

Output:

(3*log((cos(2*x) + sin(2*x)*1i)*(5*cos(2*x) - 13)))/8 - (3*log(cos(2*x)*85 
2930i - 852930*sin(2*x) - 852930i))/4 - ((18*sin(2*x)*(13 - 5*cos(2*x))^(1 
/2))/(cos(2*x) + 1)^(1/2) - (5*sin(4*x)*(13 - 5*cos(2*x))^(1/2))/(cos(2*x) 
 + 1)^(1/2))/(80*cos(2*x)^2 - 288*cos(2*x) + 208)
 

Reduce [F]

\[ \int \frac {\csc ^2(x) \left (\sec ^2(x)-3 \tan (x) \sqrt {4 \sec ^2(x)+5 \tan ^2(x)}\right )}{\left (4 \sec ^2(x)+5 \tan ^2(x)\right )^{3/2}} \, dx=\int \frac {\sec \left (x \right )^{2}}{4 \sqrt {4 \sec \left (x \right )^{2}+5 \tan \left (x \right )^{2}}\, \sec \left (x \right )^{2} \sin \left (x \right )^{2}+5 \sqrt {4 \sec \left (x \right )^{2}+5 \tan \left (x \right )^{2}}\, \sin \left (x \right )^{2} \tan \left (x \right )^{2}}d x -3 \left (\int \frac {\tan \left (x \right )}{4 \sec \left (x \right )^{2} \sin \left (x \right )^{2}+5 \sin \left (x \right )^{2} \tan \left (x \right )^{2}}d x \right ) \] Input:

int((sec(x)^2-3*(4*sec(x)^2+5*tan(x)^2)^(1/2)*tan(x))/sin(x)^2/(4*sec(x)^2 
+5*tan(x)^2)^(3/2),x)
 

Output:

int(sec(x)**2/(4*sqrt(4*sec(x)**2 + 5*tan(x)**2)*sec(x)**2*sin(x)**2 + 5*s 
qrt(4*sec(x)**2 + 5*tan(x)**2)*sin(x)**2*tan(x)**2),x) - 3*int(tan(x)/(4*s 
ec(x)**2*sin(x)**2 + 5*sin(x)**2*tan(x)**2),x)