Integrand size = 12, antiderivative size = 59 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^{5/2}} \, dx=\frac {5 E\left (x\left |\frac {3}{4}\right .\right )}{12}-\frac {\operatorname {EllipticF}\left (x,\frac {3}{4}\right )}{24}-\frac {\cos (x) \sin (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}-\frac {5 \cos (x) \sin (x)}{8 \sqrt {4-3 \sin ^2(x)}} \] Output:
5/12*EllipticE(sin(x),1/2*3^(1/2))-1/24*InverseJacobiAM(x,1/2*3^(1/2))-1/4 *cos(x)*sin(x)/(4-3*sin(x)^2)^(3/2)-5/8*cos(x)*sin(x)/(4-3*sin(x)^2)^(1/2)
Time = 0.42 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^{5/2}} \, dx=\frac {1}{96} \left (40 E\left (x\left |\frac {3}{4}\right .\right )-4 \operatorname {EllipticF}\left (x,\frac {3}{4}\right )-\frac {3 \sqrt {2} (58 \sin (2 x)+15 \sin (4 x))}{(5+3 \cos (2 x))^{3/2}}\right ) \] Input:
Integrate[(4 - 3*Sin[x]^2)^(-5/2),x]
Output:
(40*EllipticE[x, 3/4] - 4*EllipticF[x, 3/4] - (3*Sqrt[2]*(58*Sin[2*x] + 15 *Sin[4*x]))/(5 + 3*Cos[2*x])^(3/2))/96
Time = 0.50 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.17, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 3663, 27, 3042, 3652, 27, 3042, 3651, 3042, 3656, 3661}
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.
| \(\displaystyle \int \frac {1}{\left (4-3 \sin ^2(x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (4-3 \sin (x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3663 |
| \(\displaystyle -\frac {1}{12} \int -\frac {3 \left (\sin ^2(x)+2\right )}{\left (4-3 \sin ^2(x)\right )^{3/2}}dx-\frac {\sin (x) \cos (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \frac {\sin ^2(x)+2}{\left (4-3 \sin ^2(x)\right )^{3/2}}dx-\frac {\sin (x) \cos (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\sin (x)^2+2}{\left (4-3 \sin (x)^2\right )^{3/2}}dx-\frac {\sin (x) \cos (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 3652 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{4} \int \frac {2 \left (6-5 \sin ^2(x)\right )}{\sqrt {4-3 \sin ^2(x)}}dx-\frac {5 \sin (x) \cos (x)}{2 \sqrt {4-3 \sin ^2(x)}}\right )-\frac {\sin (x) \cos (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {6-5 \sin ^2(x)}{\sqrt {4-3 \sin ^2(x)}}dx-\frac {5 \sin (x) \cos (x)}{2 \sqrt {4-3 \sin ^2(x)}}\right )-\frac {\sin (x) \cos (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {6-5 \sin (x)^2}{\sqrt {4-3 \sin (x)^2}}dx-\frac {5 \sin (x) \cos (x)}{2 \sqrt {4-3 \sin ^2(x)}}\right )-\frac {\sin (x) \cos (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 3651 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {5}{3} \int \sqrt {4-3 \sin ^2(x)}dx-\frac {2}{3} \int \frac {1}{\sqrt {4-3 \sin ^2(x)}}dx\right )-\frac {5 \sin (x) \cos (x)}{2 \sqrt {4-3 \sin ^2(x)}}\right )-\frac {\sin (x) \cos (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {5}{3} \int \sqrt {4-3 \sin (x)^2}dx-\frac {2}{3} \int \frac {1}{\sqrt {4-3 \sin (x)^2}}dx\right )-\frac {5 \sin (x) \cos (x)}{2 \sqrt {4-3 \sin ^2(x)}}\right )-\frac {\sin (x) \cos (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 3656 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {10 E\left (x\left |\frac {3}{4}\right .\right )}{3}-\frac {2}{3} \int \frac {1}{\sqrt {4-3 \sin (x)^2}}dx\right )-\frac {5 \sin (x) \cos (x)}{2 \sqrt {4-3 \sin ^2(x)}}\right )-\frac {\sin (x) \cos (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 3661 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {10 E\left (x\left |\frac {3}{4}\right .\right )}{3}-\frac {\operatorname {EllipticF}\left (x,\frac {3}{4}\right )}{3}\right )-\frac {5 \sin (x) \cos (x)}{2 \sqrt {4-3 \sin ^2(x)}}\right )-\frac {\sin (x) \cos (x)}{4 \left (4-3 \sin ^2(x)\right )^{3/2}}\) |
Input:
Int[(4 - 3*Sin[x]^2)^(-5/2),x]
Output:
-1/4*(Cos[x]*Sin[x])/(4 - 3*Sin[x]^2)^(3/2) + (((10*EllipticE[x, 3/4])/3 - EllipticF[x, 3/4]/3)/2 - (5*Cos[x]*Sin[x])/(2*Sqrt[4 - 3*Sin[x]^2]))/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[B/b Int[Sqrt[a + b*Sin[e + f*x]^2], x] , x] + Simp[(A*b - a*B)/b Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /; Fre eQ[{a, b, e, f, A, B}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b - a*B))*Cos[e + f*x]*Sin[e + f*x ]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(a + b)*(p + 1))), x] - Simp[1/(2* a*(a + b)*(p + 1)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*( p + 1) + b*(2*p + 3)) + 2*(A*b - a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a ]/f)*EllipticE[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(S qrt[a]*f))*EllipticF[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*C os[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a + b))), x] + Simp[1/(2*a*(p + 1)*(a + b)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0] && LtQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(189\) vs. \(2(52)=104\).
Time = 0.75 (sec) , antiderivative size = 190, normalized size of antiderivative = 3.22
| method | result | size |
| default | \(-\frac {\sqrt {-\left (-4+3 \sin \left (x \right )^{2}\right ) \cos \left (x \right )^{2}}\, \left (3 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {3 \cos \left (x \right )^{2}+1}\, \operatorname {EllipticF}\left (\sin \left (x \right ), \frac {\sqrt {3}}{2}\right ) \cos \left (x \right )^{2}-30 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {3 \cos \left (x \right )^{2}+1}\, \operatorname {EllipticE}\left (\sin \left (x \right ), \frac {\sqrt {3}}{2}\right ) \cos \left (x \right )^{2}+45 \cos \left (x \right )^{4} \sin \left (x \right )+\sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {3 \cos \left (x \right )^{2}+1}\, \operatorname {EllipticF}\left (\sin \left (x \right ), \frac {\sqrt {3}}{2}\right )-10 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {3 \cos \left (x \right )^{2}+1}\, \operatorname {EllipticE}\left (\sin \left (x \right ), \frac {\sqrt {3}}{2}\right )+21 \cos \left (x \right )^{2} \sin \left (x \right )\right ) \sqrt {3 \cos \left (x \right )^{4}+\cos \left (x \right )^{2}}}{24 \left (9 \cos \left (x \right )^{4}+6 \cos \left (x \right )^{2}+1\right ) \cos \left (x \right )^{3} \sqrt {4-3 \sin \left (x \right )^{2}}}\) | \(190\) |
Input:
int(1/(4-3*sin(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/24*(-(-4+3*sin(x)^2)*cos(x)^2)^(1/2)/(9*cos(x)^4+6*cos(x)^2+1)/cos(x)^3 *(3*(cos(x)^2)^(1/2)*(3*cos(x)^2+1)^(1/2)*EllipticF(sin(x),1/2*3^(1/2))*co s(x)^2-30*(cos(x)^2)^(1/2)*(3*cos(x)^2+1)^(1/2)*EllipticE(sin(x),1/2*3^(1/ 2))*cos(x)^2+45*cos(x)^4*sin(x)+(cos(x)^2)^(1/2)*(3*cos(x)^2+1)^(1/2)*Elli pticF(sin(x),1/2*3^(1/2))-10*(cos(x)^2)^(1/2)*(3*cos(x)^2+1)^(1/2)*Ellipti cE(sin(x),1/2*3^(1/2))+21*cos(x)^2*sin(x))*(3*cos(x)^4+cos(x)^2)^(1/2)/(4- 3*sin(x)^2)^(1/2)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 3.05 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^{5/2}} \, dx=-\frac {18 \, {\left (15 \, \cos \left (x\right )^{3} + 7 \, \cos \left (x\right )\right )} \sqrt {3 \, \cos \left (x\right )^{2} + 1} \sin \left (x\right ) - 5 \, {\left (9 \, \cos \left (x\right )^{4} + 6 \, \cos \left (x\right )^{2} + 1\right )} E(\arcsin \left (\frac {1}{3} i \, \sqrt {3} \cos \left (x\right ) - \frac {1}{3} \, \sqrt {3} \sin \left (x\right )\right )\,|\,9) - 5 \, {\left (9 \, \cos \left (x\right )^{4} + 6 \, \cos \left (x\right )^{2} + 1\right )} E(\arcsin \left (-\frac {1}{3} i \, \sqrt {3} \cos \left (x\right ) - \frac {1}{3} \, \sqrt {3} \sin \left (x\right )\right )\,|\,9) + 68 \, {\left (9 \, \cos \left (x\right )^{4} + 6 \, \cos \left (x\right )^{2} + 1\right )} F(\arcsin \left (\frac {1}{3} i \, \sqrt {3} \cos \left (x\right ) - \frac {1}{3} \, \sqrt {3} \sin \left (x\right )\right )\,|\,9) + 68 \, {\left (9 \, \cos \left (x\right )^{4} + 6 \, \cos \left (x\right )^{2} + 1\right )} F(\arcsin \left (-\frac {1}{3} i \, \sqrt {3} \cos \left (x\right ) - \frac {1}{3} \, \sqrt {3} \sin \left (x\right )\right )\,|\,9)}{144 \, {\left (9 \, \cos \left (x\right )^{4} + 6 \, \cos \left (x\right )^{2} + 1\right )}} \] Input:
integrate(1/(4-3*sin(x)^2)^(5/2),x, algorithm="fricas")
Output:
-1/144*(18*(15*cos(x)^3 + 7*cos(x))*sqrt(3*cos(x)^2 + 1)*sin(x) - 5*(9*cos (x)^4 + 6*cos(x)^2 + 1)*elliptic_e(arcsin(1/3*I*sqrt(3)*cos(x) - 1/3*sqrt( 3)*sin(x)), 9) - 5*(9*cos(x)^4 + 6*cos(x)^2 + 1)*elliptic_e(arcsin(-1/3*I* sqrt(3)*cos(x) - 1/3*sqrt(3)*sin(x)), 9) + 68*(9*cos(x)^4 + 6*cos(x)^2 + 1 )*elliptic_f(arcsin(1/3*I*sqrt(3)*cos(x) - 1/3*sqrt(3)*sin(x)), 9) + 68*(9 *cos(x)^4 + 6*cos(x)^2 + 1)*elliptic_f(arcsin(-1/3*I*sqrt(3)*cos(x) - 1/3* sqrt(3)*sin(x)), 9))/(9*cos(x)^4 + 6*cos(x)^2 + 1)
\[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (4 - 3 \sin ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(4-3*sin(x)**2)**(5/2),x)
Output:
Integral((4 - 3*sin(x)**2)**(-5/2), x)
\[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-3 \, \sin \left (x\right )^{2} + 4\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(4-3*sin(x)^2)^(5/2),x, algorithm="maxima")
Output:
integrate((-3*sin(x)^2 + 4)^(-5/2), x)
\[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-3 \, \sin \left (x\right )^{2} + 4\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(4-3*sin(x)^2)^(5/2),x, algorithm="giac")
Output:
integrate((-3*sin(x)^2 + 4)^(-5/2), x)
Timed out. \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (4-3\,{\sin \left (x\right )}^2\right )}^{5/2}} \,d x \] Input:
int(1/(4 - 3*sin(x)^2)^(5/2),x)
Output:
int(1/(4 - 3*sin(x)^2)^(5/2), x)
\[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^{5/2}} \, dx=-\left (\int \frac {\sqrt {-3 \sin \left (x \right )^{2}+4}}{27 \sin \left (x \right )^{6}-108 \sin \left (x \right )^{4}+144 \sin \left (x \right )^{2}-64}d x \right ) \] Input:
int(1/(4-3*sin(x)^2)^(5/2),x)
Output:
- int(sqrt( - 3*sin(x)**2 + 4)/(27*sin(x)**6 - 108*sin(x)**4 + 144*sin(x) **2 - 64),x)