Integrand size = 13, antiderivative size = 49 \[ \int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx=2 \sqrt {2} \arcsin \left (\sqrt {2} \sin (x)\right )-\frac {5}{2} \arctan \left (\frac {\sin (x)}{\sqrt {\cos (2 x)}}\right )-\frac {1}{2} \sqrt {\cos (2 x)} \sec (x) \tan (x) \] Output:
-5/2*arctan(sin(x)/cos(2*x)^(1/2))+2*arcsin(sin(x)*2^(1/2))*2^(1/2)-1/2*se c(x)*cos(2*x)^(1/2)*tan(x)
Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx=\frac {1}{2} \left (4 \sqrt {2} \arcsin \left (\sqrt {2} \sin (x)\right )-5 \arctan \left (\frac {\sin (x)}{\sqrt {\cos (2 x)}}\right )-\sqrt {\cos (2 x)} \sec (x) \tan (x)\right ) \] Input:
Integrate[Cos[2*x]^(3/2)*Sec[x]^3,x]
Output:
(4*Sqrt[2]*ArcSin[Sqrt[2]*Sin[x]] - 5*ArcTan[Sin[x]/Sqrt[Cos[2*x]]] - Sqrt [Cos[2*x]]*Sec[x]*Tan[x])/2
Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.39, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 4864, 315, 25, 398, 223, 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.
\(\displaystyle \int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (2 x)^{3/2}}{\cos (x)^3}dx\) |
\(\Big \downarrow \) 4864 |
\(\displaystyle \int \frac {\left (1-2 \sin ^2(x)\right )^{3/2}}{\left (1-\sin ^2(x)\right )^2}d\sin (x)\) |
\(\Big \downarrow \) 315 |
\(\displaystyle -\frac {1}{2} \int -\frac {3-8 \sin ^2(x)}{\sqrt {1-2 \sin ^2(x)} \left (1-\sin ^2(x)\right )}d\sin (x)-\frac {\sqrt {1-2 \sin ^2(x)} \sin (x)}{2 \left (1-\sin ^2(x)\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \int \frac {3-8 \sin ^2(x)}{\sqrt {1-2 \sin ^2(x)} \left (1-\sin ^2(x)\right )}d\sin (x)-\frac {\sin (x) \sqrt {1-2 \sin ^2(x)}}{2 \left (1-\sin ^2(x)\right )}\) |
\(\Big \downarrow \) 398 |
\(\displaystyle \frac {1}{2} \left (8 \int \frac {1}{\sqrt {1-2 \sin ^2(x)}}d\sin (x)-5 \int \frac {1}{\sqrt {1-2 \sin ^2(x)} \left (1-\sin ^2(x)\right )}d\sin (x)\right )-\frac {\sin (x) \sqrt {1-2 \sin ^2(x)}}{2 \left (1-\sin ^2(x)\right )}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{2} \left (4 \sqrt {2} \arcsin \left (\sqrt {2} \sin (x)\right )-5 \int \frac {1}{\sqrt {1-2 \sin ^2(x)} \left (1-\sin ^2(x)\right )}d\sin (x)\right )-\frac {\sin (x) \sqrt {1-2 \sin ^2(x)}}{2 \left (1-\sin ^2(x)\right )}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {1}{2} \left (4 \sqrt {2} \arcsin \left (\sqrt {2} \sin (x)\right )-5 \int \frac {1}{\frac {\sin ^2(x)}{1-2 \sin ^2(x)}+1}d\frac {\sin (x)}{\sqrt {1-2 \sin ^2(x)}}\right )-\frac {\sin (x) \sqrt {1-2 \sin ^2(x)}}{2 \left (1-\sin ^2(x)\right )}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (4 \sqrt {2} \arcsin \left (\sqrt {2} \sin (x)\right )-5 \arctan \left (\frac {\sin (x)}{\sqrt {1-2 \sin ^2(x)}}\right )\right )-\frac {\sin (x) \sqrt {1-2 \sin ^2(x)}}{2 \left (1-\sin ^2(x)\right )}\) |
Input:
Int[Cos[2*x]^(3/2)*Sec[x]^3,x]
Output:
(4*Sqrt[2]*ArcSin[Sqrt[2]*Sin[x]] - 5*ArcTan[Sin[x]/Sqrt[1 - 2*Sin[x]^2]]) /2 - (Sin[x]*Sqrt[1 - 2*Sin[x]^2])/(2*(1 - Sin[x]^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] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
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[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 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[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = Free Factors[Sin[c*(a + b*x)], x]}, Simp[d/(b*c) Subst[Int[SubstFor[(1 - d^2*x ^2)^((n - 1)/2), Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && Integer Q[(n - 1)/2] && NonsumQ[u] && (EqQ[F, Cos] || EqQ[F, cos])
Leaf count of result is larger than twice the leaf count of optimal. \(99\) vs. \(2(37)=74\).
Time = 0.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.04
method | result | size |
default | \(-\frac {\sqrt {\left (2 \cos \left (x \right )^{2}-1\right ) \sin \left (x \right )^{2}}\, \left (4 \sqrt {2}\, \arcsin \left (4 \cos \left (x \right )^{2}-3\right ) \cos \left (x \right )^{2}-5 \arctan \left (\frac {3 \cos \left (x \right )^{2}-2}{2 \sqrt {-2 \sin \left (x \right )^{4}+\sin \left (x \right )^{2}}}\right ) \cos \left (x \right )^{2}+2 \sqrt {-2 \sin \left (x \right )^{4}+\sin \left (x \right )^{2}}\right )}{4 \cos \left (x \right )^{2} \sin \left (x \right ) \sqrt {2 \cos \left (x \right )^{2}-1}}\) | \(100\) |
Input:
int(cos(2*x)^(3/2)/cos(x)^3,x,method=_RETURNVERBOSE)
Output:
-1/4*((2*cos(x)^2-1)*sin(x)^2)^(1/2)*(4*2^(1/2)*arcsin(4*cos(x)^2-3)*cos(x )^2-5*arctan(1/2*(3*cos(x)^2-2)/(-2*sin(x)^4+sin(x)^2)^(1/2))*cos(x)^2+2*( -2*sin(x)^4+sin(x)^2)^(1/2))/cos(x)^2/sin(x)/(2*cos(x)^2-1)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (37) = 74\).
Time = 0.10 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.41 \[ \int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx=-\frac {2 \, \sqrt {2} \arctan \left (\frac {{\left (32 \, \sqrt {2} \cos \left (x\right )^{4} - 48 \, \sqrt {2} \cos \left (x\right )^{2} + 17 \, \sqrt {2}\right )} \sqrt {2 \, \cos \left (x\right )^{2} - 1}}{8 \, {\left (8 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{2} + 3\right )} \sin \left (x\right )}\right ) \cos \left (x\right )^{2} - 5 \, \arctan \left (\frac {3 \, \cos \left (x\right )^{2} - 2}{2 \, \sqrt {2 \, \cos \left (x\right )^{2} - 1} \sin \left (x\right )}\right ) \cos \left (x\right )^{2} + 2 \, \sqrt {2 \, \cos \left (x\right )^{2} - 1} \sin \left (x\right )}{4 \, \cos \left (x\right )^{2}} \] Input:
integrate(cos(2*x)^(3/2)/cos(x)^3,x, algorithm="fricas")
Output:
-1/4*(2*sqrt(2)*arctan(1/8*(32*sqrt(2)*cos(x)^4 - 48*sqrt(2)*cos(x)^2 + 17 *sqrt(2))*sqrt(2*cos(x)^2 - 1)/((8*cos(x)^4 - 10*cos(x)^2 + 3)*sin(x)))*co s(x)^2 - 5*arctan(1/2*(3*cos(x)^2 - 2)/(sqrt(2*cos(x)^2 - 1)*sin(x)))*cos( x)^2 + 2*sqrt(2*cos(x)^2 - 1)*sin(x))/cos(x)^2
Timed out. \[ \int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx=\text {Timed out} \] Input:
integrate(cos(2*x)**(3/2)/cos(x)**3,x)
Output:
Timed out
\[ \int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx=\int { \frac {\cos \left (2 \, x\right )^{\frac {3}{2}}}{\cos \left (x\right )^{3}} \,d x } \] Input:
integrate(cos(2*x)^(3/2)/cos(x)^3,x, algorithm="maxima")
Output:
integrate(cos(2*x)^(3/2)/cos(x)^3, x)
\[ \int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx=\int { \frac {\cos \left (2 \, x\right )^{\frac {3}{2}}}{\cos \left (x\right )^{3}} \,d x } \] Input:
integrate(cos(2*x)^(3/2)/cos(x)^3,x, algorithm="giac")
Output:
integrate(cos(2*x)^(3/2)/cos(x)^3, x)
Timed out. \[ \int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx=\int \frac {{\cos \left (2\,x\right )}^{3/2}}{{\cos \left (x\right )}^3} \,d x \] Input:
int(cos(2*x)^(3/2)/cos(x)^3,x)
Output:
int(cos(2*x)^(3/2)/cos(x)^3, x)
\[ \int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx=\int \frac {\sqrt {\cos \left (2 x \right )}\, \cos \left (2 x \right )}{\cos \left (x \right )^{3}}d x \] Input:
int(cos(2*x)^(3/2)/cos(x)^3,x)
Output:
int((sqrt(cos(2*x))*cos(2*x))/cos(x)**3,x)