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

Optimal. Leaf size=57 \[ -\frac {3}{4} \log (\tan (x))+\frac {3}{8} \log \left (4+9 \tan ^2(x)\right )-\frac {\cot (x)}{4 \sqrt {4+9 \tan ^2(x)}}-\frac {7 \tan (x)}{8 \sqrt {4+9 \tan ^2(x)}} \]

[Out]

-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*tan(x)/(4+9*tan(x)^2)^(1/2)

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Rubi [A]
time = 0.75, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {6874, 197, 277, 272, 36, 29, 31} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

(-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*Ta
n[x]^2])

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 6874

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

Rubi steps

\begin {align*} \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 &=\text {Subst}\left (\int \frac {1+x^2-3 x \sqrt {4+9 x^2}}{x^2 \left (4+9 x^2\right )^{3/2}} \, dx,x,\tan (x)\right )\\ &=\text {Subst}\left (\int \left (\frac {1}{\left (4+9 x^2\right )^{3/2}}+\frac {1}{x^2 \left (4+9 x^2\right )^{3/2}}-\frac {3}{x \left (4+9 x^2\right )}\right ) \, dx,x,\tan (x)\right )\\ &=-\left (3 \text {Subst}\left (\int \frac {1}{x \left (4+9 x^2\right )} \, dx,x,\tan (x)\right )\right )+\text {Subst}\left (\int \frac {1}{\left (4+9 x^2\right )^{3/2}} \, dx,x,\tan (x)\right )+\text {Subst}\left (\int \frac {1}{x^2 \left (4+9 x^2\right )^{3/2}} \, dx,x,\tan (x)\right )\\ &=-\frac {\cot (x)}{4 \sqrt {4+9 \tan ^2(x)}}+\frac {\tan (x)}{4 \sqrt {4+9 \tan ^2(x)}}-\frac {3}{2} \text {Subst}\left (\int \frac {1}{x (4+9 x)} \, dx,x,\tan ^2(x)\right )-\frac {9}{2} \text {Subst}\left (\int \frac {1}{\left (4+9 x^2\right )^{3/2}} \, dx,x,\tan (x)\right )\\ &=-\frac {\cot (x)}{4 \sqrt {4+9 \tan ^2(x)}}-\frac {7 \tan (x)}{8 \sqrt {4+9 \tan ^2(x)}}-\frac {3}{8} \text {Subst}\left (\int \frac {1}{x} \, dx,x,\tan ^2(x)\right )+\frac {27}{8} \text {Subst}\left (\int \frac {1}{4+9 x} \, dx,x,\tan ^2(x)\right )\\ &=-\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)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(116\) vs. \(2(57)=114\).
time = 0.70, size = 116, normalized size = 2.04 \begin {gather*} \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)}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

(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]
)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(116\) vs. \(2(45)=90\).
time = 0.67, size = 117, normalized size = 2.05

method	result	size
default	\(-\frac {-3 \left (\cos ^{3}\left (x \right )\right ) \sin \left (x \right ) \left (-\frac {5 \left (\cos ^{2}\left (x \right )\right )-9}{\cos \left (x \right )^{2}}\right )^{\frac {3}{2}} \ln \left (-\frac {5 \left (\cos ^{2}\left (x \right )\right )-9}{\left (1+\cos \left (x \right )\right )^{2}}\right )+6 \left (\cos ^{3}\left (x \right )\right ) \sin \left (x \right ) \left (-\frac {5 \left (\cos ^{2}\left (x \right )\right )-9}{\cos \left (x \right )^{2}}\right )^{\frac {3}{2}} \ln \left (-\frac {\cos \left (x \right )-1}{\sin \left (x \right )}\right )+25 \left (\cos ^{4}\left (x \right )\right )-80 \left (\cos ^{2}\left (x \right )\right )+63}{8 \cos \left (x \right )^{3} \sin \left (x \right ) \left (-\frac {5 \left (\cos ^{2}\left (x \right )\right )-9}{\cos \left (x \right )^{2}}\right )^{\frac {3}{2}}}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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=_RETURNV
ERBOSE)

[Out]

-1/8*(-3*cos(x)^3*sin(x)*(-(5*cos(x)^2-9)/cos(x)^2)^(3/2)*ln(-(5*cos(x)^2-9)/(1+cos(x))^2)+6*cos(x)^3*sin(x)*(
-(5*cos(x)^2-9)/cos(x)^2)^(3/2)*ln(-(cos(x)-1)/sin(x))+25*cos(x)^4-80*cos(x)^2+63)/cos(x)^3/sin(x)/(-(5*cos(x)
^2-9)/cos(x)^2)^(3/2)

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Maxima [A]
time = 2.71, size = 47, normalized size = 0.82 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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, algorith
m="maxima")

[Out]

-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
))

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Fricas [A]
time = 1.56, size = 84, normalized size = 1.47 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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, algorith
m="fricas")

[Out]

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))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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, algorith
m="giac")

[Out]

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)

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Mupad [B]
time = 1.47, size = 113, normalized size = 1.98 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*log((cos(2*x) + sin(2*x)*1i)*(5*cos(2*x) - 13)))/8 - (3*log(cos(2*x)*852930i - 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)

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