Integrand size = 12, antiderivative size = 68 \[ \int \left (4-5 \sec ^2(x)\right )^{3/2} \, dx=8 \arctan \left (\frac {2 \tan (x)}{\sqrt {-1-5 \tan ^2(x)}}\right )-\frac {7}{2} \sqrt {5} \arctan \left (\frac {\sqrt {5} \tan (x)}{\sqrt {-1-5 \tan ^2(x)}}\right )-\frac {5}{2} \tan (x) \sqrt {-1-5 \tan ^2(x)} \] Output:
8*arctan(2*tan(x)/(-1-5*tan(x)^2)^(1/2))-7/2*arctan(5^(1/2)*tan(x)/(-1-5*t an(x)^2)^(1/2))*5^(1/2)-5/2*(-1-5*tan(x)^2)^(1/2)*tan(x)
Result contains complex when optimal does not.
Time = 0.15 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.69 \[ \int \left (4-5 \sec ^2(x)\right )^{3/2} \, dx=-\frac {\left (-5+4 \cos ^2(x)\right ) \sec (x) \sqrt {4-5 \sec ^2(x)} \left (7 \sqrt {5} \arctan \left (\frac {\sqrt {5} \sin (x)}{\sqrt {-3+2 \cos (2 x)}}\right ) \cos ^2(x)+16 i \cos ^2(x) \log \left (\sqrt {-3+2 \cos (2 x)}+2 i \sin (x)\right )+5 \sqrt {-3+2 \cos (2 x)} \sin (x)\right )}{2 (-3+2 \cos (2 x))^{3/2}} \] Input:
Integrate[(4 - 5*Sec[x]^2)^(3/2),x]
Output:
-1/2*((-5 + 4*Cos[x]^2)*Sec[x]*Sqrt[4 - 5*Sec[x]^2]*(7*Sqrt[5]*ArcTan[(Sqr t[5]*Sin[x])/Sqrt[-3 + 2*Cos[2*x]]]*Cos[x]^2 + (16*I)*Cos[x]^2*Log[Sqrt[-3 + 2*Cos[2*x]] + (2*I)*Sin[x]] + 5*Sqrt[-3 + 2*Cos[2*x]]*Sin[x]))/(-3 + 2* Cos[2*x])^(3/2)
Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 4616, 318, 25, 398, 224, 216, 291, 216}
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.
| \(\displaystyle \int \left (4-5 \sec ^2(x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (4-5 \sec (x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4616 |
| \(\displaystyle \int \frac {\left (-5 \tan ^2(x)-1\right )^{3/2}}{\tan ^2(x)+1}d\tan (x)\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle \frac {1}{2} \int -\frac {35 \tan ^2(x)+3}{\sqrt {-5 \tan ^2(x)-1} \left (\tan ^2(x)+1\right )}d\tan (x)-\frac {5}{2} \tan (x) \sqrt {-5 \tan ^2(x)-1}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int \frac {35 \tan ^2(x)+3}{\sqrt {-5 \tan ^2(x)-1} \left (\tan ^2(x)+1\right )}d\tan (x)-\frac {5}{2} \sqrt {-5 \tan ^2(x)-1} \tan (x)\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {1}{2} \left (32 \int \frac {1}{\sqrt {-5 \tan ^2(x)-1} \left (\tan ^2(x)+1\right )}d\tan (x)-35 \int \frac {1}{\sqrt {-5 \tan ^2(x)-1}}d\tan (x)\right )-\frac {5}{2} \tan (x) \sqrt {-5 \tan ^2(x)-1}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {1}{2} \left (32 \int \frac {1}{\sqrt {-5 \tan ^2(x)-1} \left (\tan ^2(x)+1\right )}d\tan (x)-35 \int \frac {1}{\frac {5 \tan ^2(x)}{-5 \tan ^2(x)-1}+1}d\frac {\tan (x)}{\sqrt {-5 \tan ^2(x)-1}}\right )-\frac {5}{2} \tan (x) \sqrt {-5 \tan ^2(x)-1}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{2} \left (32 \int \frac {1}{\sqrt {-5 \tan ^2(x)-1} \left (\tan ^2(x)+1\right )}d\tan (x)-7 \sqrt {5} \arctan \left (\frac {\sqrt {5} \tan (x)}{\sqrt {-5 \tan ^2(x)-1}}\right )\right )-\frac {5}{2} \tan (x) \sqrt {-5 \tan ^2(x)-1}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {1}{2} \left (32 \int \frac {1}{\frac {4 \tan ^2(x)}{-5 \tan ^2(x)-1}+1}d\frac {\tan (x)}{\sqrt {-5 \tan ^2(x)-1}}-7 \sqrt {5} \arctan \left (\frac {\sqrt {5} \tan (x)}{\sqrt {-5 \tan ^2(x)-1}}\right )\right )-\frac {5}{2} \tan (x) \sqrt {-5 \tan ^2(x)-1}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{2} \left (16 \arctan \left (\frac {2 \tan (x)}{\sqrt {-5 \tan ^2(x)-1}}\right )-7 \sqrt {5} \arctan \left (\frac {\sqrt {5} \tan (x)}{\sqrt {-5 \tan ^2(x)-1}}\right )\right )-\frac {5}{2} \tan (x) \sqrt {-5 \tan ^2(x)-1}\) |
Input:
Int[(4 - 5*Sec[x]^2)^(3/2),x]
Output:
(16*ArcTan[(2*Tan[x])/Sqrt[-1 - 5*Tan[x]^2]] - 7*Sqrt[5]*ArcTan[(Sqrt[5]*T an[x])/Sqrt[-1 - 5*Tan[x]^2]])/2 - (5*Tan[x]*Sqrt[-1 - 5*Tan[x]^2])/2
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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]
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]
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]
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]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b + b*ff^2*x^2)^p /(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && NeQ[a + b, 0] && NeQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(199\) vs. \(2(54)=108\).
Time = 7.18 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.94
| method | result | size |
| default | \(\frac {\left (4-5 \sec \left (x \right )^{2}\right )^{\frac {3}{2}} \left (-7 \cos \left (x \right )^{3} \sqrt {5}\, \arctan \left (\frac {\left (4 \sin \left (x \right )-1\right ) \sqrt {5}}{5 \left (1+\cos \left (x \right )\right ) \sqrt {\frac {4 \cos \left (x \right )^{2}-5}{\left (1+\cos \left (x \right )\right )^{2}}}}\right )-7 \cos \left (x \right )^{3} \sqrt {5}\, \arctan \left (\frac {\left (4 \sin \left (x \right )+1\right ) \sqrt {5}}{5 \left (1+\cos \left (x \right )\right ) \sqrt {\frac {4 \cos \left (x \right )^{2}-5}{\left (1+\cos \left (x \right )\right )^{2}}}}\right )+32 \cos \left (x \right )^{3} \arctan \left (\frac {2 \csc \left (x \right )-2 \cot \left (x \right )}{\sqrt {\frac {4 \cos \left (x \right )^{2}-5}{\left (1+\cos \left (x \right )\right )^{2}}}}\right )+\cos \left (x \right ) \left (-10 \cos \left (x \right )-10\right ) \sin \left (x \right ) \sqrt {\frac {4 \cos \left (x \right )^{2}-5}{\left (1+\cos \left (x \right )\right )^{2}}}\right )}{4 \sqrt {\frac {4 \cos \left (x \right )^{2}-5}{\left (1+\cos \left (x \right )\right )^{2}}}\, \left (4 \cos \left (x \right )^{3}+4 \cos \left (x \right )^{2}-5 \cos \left (x \right )-5\right )}\) | \(200\) |
Input:
int((4-5*sec(x)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/4*(4-5*sec(x)^2)^(3/2)/((4*cos(x)^2-5)/(1+cos(x))^2)^(1/2)/(4*cos(x)^3+4 *cos(x)^2-5*cos(x)-5)*(-7*cos(x)^3*5^(1/2)*arctan(1/5*(4*sin(x)-1)/(1+cos( x))/((4*cos(x)^2-5)/(1+cos(x))^2)^(1/2)*5^(1/2))-7*cos(x)^3*5^(1/2)*arctan (1/5*(4*sin(x)+1)/(1+cos(x))/((4*cos(x)^2-5)/(1+cos(x))^2)^(1/2)*5^(1/2))+ 32*cos(x)^3*arctan(2/((4*cos(x)^2-5)/(1+cos(x))^2)^(1/2)*(csc(x)-cot(x)))+ cos(x)*(-10*cos(x)-10)*sin(x)*((4*cos(x)^2-5)/(1+cos(x))^2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (54) = 108\).
Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.91 \[ \int \left (4-5 \sec ^2(x)\right )^{3/2} \, dx=\frac {7 \, \sqrt {5} \arctan \left (\frac {\sqrt {5} \sqrt {\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \cos \left (x\right )}{5 \, \sin \left (x\right )}\right ) \cos \left (x\right ) + 8 \, \arctan \left (\frac {4 \, {\left (8 \, \cos \left (x\right )^{3} - 9 \, \cos \left (x\right )\right )} \sqrt {\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \sin \left (x\right ) + \cos \left (x\right ) \sin \left (x\right )}{64 \, \cos \left (x\right )^{4} - 143 \, \cos \left (x\right )^{2} + 80}\right ) \cos \left (x\right ) - 8 \, \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right ) \cos \left (x\right ) - 5 \, \sqrt {\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \sin \left (x\right )}{2 \, \cos \left (x\right )} \] Input:
integrate((4-5*sec(x)^2)^(3/2),x, algorithm="fricas")
Output:
1/2*(7*sqrt(5)*arctan(1/5*sqrt(5)*sqrt((4*cos(x)^2 - 5)/cos(x)^2)*cos(x)/s in(x))*cos(x) + 8*arctan((4*(8*cos(x)^3 - 9*cos(x))*sqrt((4*cos(x)^2 - 5)/ cos(x)^2)*sin(x) + cos(x)*sin(x))/(64*cos(x)^4 - 143*cos(x)^2 + 80))*cos(x ) - 8*arctan(sin(x)/cos(x))*cos(x) - 5*sqrt((4*cos(x)^2 - 5)/cos(x)^2)*sin (x))/cos(x)
\[ \int \left (4-5 \sec ^2(x)\right )^{3/2} \, dx=\int \left (4 - 5 \sec ^{2}{\left (x \right )}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((4-5*sec(x)**2)**(3/2),x)
Output:
Integral((4 - 5*sec(x)**2)**(3/2), x)
\[ \int \left (4-5 \sec ^2(x)\right )^{3/2} \, dx=\int { {\left (-5 \, \sec \left (x\right )^{2} + 4\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((4-5*sec(x)^2)^(3/2),x, algorithm="maxima")
Output:
integrate((-5*sec(x)^2 + 4)^(3/2), x)
\[ \int \left (4-5 \sec ^2(x)\right )^{3/2} \, dx=\int { {\left (-5 \, \sec \left (x\right )^{2} + 4\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((4-5*sec(x)^2)^(3/2),x, algorithm="giac")
Output:
integrate((-5*sec(x)^2 + 4)^(3/2), x)
Timed out. \[ \int \left (4-5 \sec ^2(x)\right )^{3/2} \, dx=\int {\left (4-\frac {5}{{\cos \left (x\right )}^2}\right )}^{3/2} \,d x \] Input:
int((4 - 5/cos(x)^2)^(3/2),x)
Output:
int((4 - 5/cos(x)^2)^(3/2), x)
\[ \int \left (4-5 \sec ^2(x)\right )^{3/2} \, dx=4 \left (\int \sqrt {-5 \sec \left (x \right )^{2}+4}d x \right )-5 \left (\int \sqrt {-5 \sec \left (x \right )^{2}+4}\, \sec \left (x \right )^{2}d x \right ) \] Input:
int((4-5*sec(x)^2)^(3/2),x)
Output:
4*int(sqrt( - 5*sec(x)**2 + 4),x) - 5*int(sqrt( - 5*sec(x)**2 + 4)*sec(x)* *2,x)