Integrand size = 10, antiderivative size = 60 \[ \int \left (4-5 \sin ^2(x)\right )^4 \, dx=\frac {7923 x}{128}+\frac {15245}{128} \cos (x) \sin (x)-\frac {3825}{64} \cos (x) \sin ^3(x)+\frac {35}{16} \cos (x) \sin (x) \left (4-5 \sin ^2(x)\right )^2+\frac {5}{8} \cos (x) \sin (x) \left (4-5 \sin ^2(x)\right )^3 \] Output:
7923/128*x+15245/128*cos(x)*sin(x)-3825/64*cos(x)*sin(x)^3+35/16*cos(x)*si n(x)*(4-5*sin(x)^2)^2+5/8*cos(x)*sin(x)*(4-5*sin(x)^2)^3
Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.63 \[ \int \left (4-5 \sin ^2(x)\right )^4 \, dx=\frac {7923 x}{128}+\frac {1665}{32} \sin (2 x)+\frac {1975}{128} \sin (4 x)+\frac {125}{32} \sin (6 x)+\frac {625 \sin (8 x)}{1024} \] Input:
Integrate[(4 - 5*Sin[x]^2)^4,x]
Output:
(7923*x)/128 + (1665*Sin[2*x])/32 + (1975*Sin[4*x])/128 + (125*Sin[6*x])/3 2 + (625*Sin[8*x])/1024
Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {3042, 3659, 27, 3042, 3649, 3042, 3648}
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.
\(\displaystyle \int \left (4-5 \sin ^2(x)\right )^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (4-5 \sin (x)^2\right )^4dx\) |
\(\Big \downarrow \) 3659 |
\(\displaystyle \frac {1}{8} \int 3 \left (36-35 \sin ^2(x)\right ) \left (4-5 \sin ^2(x)\right )^2dx+\frac {5}{8} \sin (x) \left (4-5 \sin ^2(x)\right )^3 \cos (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{8} \int \left (36-35 \sin ^2(x)\right ) \left (4-5 \sin ^2(x)\right )^2dx+\frac {5}{8} \sin (x) \left (4-5 \sin ^2(x)\right )^3 \cos (x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{8} \int \left (36-35 \sin (x)^2\right ) \left (4-5 \sin (x)^2\right )^2dx+\frac {5}{8} \sin (x) \left (4-5 \sin ^2(x)\right )^3 \cos (x)\) |
\(\Big \downarrow \) 3649 |
\(\displaystyle \frac {3}{8} \left (\frac {1}{6} \int \left (724-765 \sin ^2(x)\right ) \left (4-5 \sin ^2(x)\right )dx+\frac {35}{6} \sin (x) \left (4-5 \sin ^2(x)\right )^2 \cos (x)\right )+\frac {5}{8} \sin (x) \left (4-5 \sin ^2(x)\right )^3 \cos (x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{8} \left (\frac {1}{6} \int \left (724-765 \sin (x)^2\right ) \left (4-5 \sin (x)^2\right )dx+\frac {35}{6} \sin (x) \left (4-5 \sin ^2(x)\right )^2 \cos (x)\right )+\frac {5}{8} \sin (x) \left (4-5 \sin ^2(x)\right )^3 \cos (x)\) |
\(\Big \downarrow \) 3648 |
\(\displaystyle \frac {5}{8} \sin (x) \left (4-5 \sin ^2(x)\right )^3 \cos (x)+\frac {3}{8} \left (\frac {1}{6} \left (\frac {7923 x}{8}-\frac {3825}{4} \sin ^3(x) \cos (x)+\frac {15245}{8} \sin (x) \cos (x)\right )+\frac {35}{6} \sin (x) \left (4-5 \sin ^2(x)\right )^2 \cos (x)\right )\) |
Input:
Int[(4 - 5*Sin[x]^2)^4,x]
Output:
(5*Cos[x]*Sin[x]*(4 - 5*Sin[x]^2)^3)/8 + (3*((35*Cos[x]*Sin[x]*(4 - 5*Sin[ x]^2)^2)/6 + ((7923*x)/8 + (15245*Cos[x]*Sin[x])/8 - (3825*Cos[x]*Sin[x]^3 )/4)/6))/8
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)*((A_.) + (B_.)*sin[(e_.) + (f_ .)*(x_)]^2), x_Symbol] :> Simp[(4*A*(2*a + b) + B*(4*a + 3*b))*(x/8), x] + (-Simp[b*B*Cos[e + f*x]*(Sin[e + f*x]^3/(4*f)), x] - Simp[(4*A*b + B*(4*a + 3*b))*Cos[e + f*x]*(Sin[e + f*x]/(8*f)), x]) /; FreeQ[{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[(-B)*Cos[e + f*x]*Sin[e + f*x]*((a + b* Sin[e + f*x]^2)^p/(2*f*(p + 1))), x] + Simp[1/(2*(p + 1)) Int[(a + b*Sin[ e + f*x]^2)^(p - 1)*Simp[a*B + 2*a*A*(p + 1) + (2*A*b*(p + 1) + B*(b + 2*a* p + 2*b*p))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && G tQ[p, 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*f*p)), x] + Sim p[1/(2*p) Int[(a + b*Sin[e + f*x]^2)^(p - 2)*Simp[a*(b + 2*a*p) + b*(2*a + b)*(2*p - 1)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[ a + b, 0] && GtQ[p, 1]
Time = 11.89 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.48
method | result | size |
risch | \(\frac {7923 x}{128}+\frac {625 \sin \left (8 x \right )}{1024}+\frac {125 \sin \left (6 x \right )}{32}+\frac {1975 \sin \left (4 x \right )}{128}+\frac {1665 \sin \left (2 x \right )}{32}\) | \(29\) |
parallelrisch | \(\frac {7923 x}{128}+\frac {625 \sin \left (8 x \right )}{1024}+\frac {125 \sin \left (6 x \right )}{32}+\frac {1975 \sin \left (4 x \right )}{128}+\frac {1665 \sin \left (2 x \right )}{32}\) | \(29\) |
default | \(-\frac {625 \left (\sin \left (x \right )^{7}+\frac {7 \sin \left (x \right )^{5}}{6}+\frac {35 \sin \left (x \right )^{3}}{24}+\frac {35 \sin \left (x \right )}{16}\right ) \cos \left (x \right )}{8}+\frac {7923 x}{128}+\frac {1000 \left (\sin \left (x \right )^{5}+\frac {5 \sin \left (x \right )^{3}}{4}+\frac {15 \sin \left (x \right )}{8}\right ) \cos \left (x \right )}{3}-600 \left (\sin \left (x \right )^{3}+\frac {3 \sin \left (x \right )}{2}\right ) \cos \left (x \right )+640 \cos \left (x \right ) \sin \left (x \right )\) | \(68\) |
parts | \(-\frac {625 \left (\sin \left (x \right )^{7}+\frac {7 \sin \left (x \right )^{5}}{6}+\frac {35 \sin \left (x \right )^{3}}{24}+\frac {35 \sin \left (x \right )}{16}\right ) \cos \left (x \right )}{8}+\frac {7923 x}{128}+\frac {1000 \left (\sin \left (x \right )^{5}+\frac {5 \sin \left (x \right )^{3}}{4}+\frac {15 \sin \left (x \right )}{8}\right ) \cos \left (x \right )}{3}-600 \left (\sin \left (x \right )^{3}+\frac {3 \sin \left (x \right )}{2}\right ) \cos \left (x \right )+640 \cos \left (x \right ) \sin \left (x \right )\) | \(68\) |
norman | \(\frac {\frac {7923 x}{128}-\frac {27975 \tan \left (\frac {x}{2}\right )^{3}}{64}+\frac {263005 \tan \left (\frac {x}{2}\right )^{5}}{64}-\frac {324175 \tan \left (\frac {x}{2}\right )^{7}}{64}+\frac {324175 \tan \left (\frac {x}{2}\right )^{9}}{64}-\frac {263005 \tan \left (\frac {x}{2}\right )^{11}}{64}+\frac {27975 \tan \left (\frac {x}{2}\right )^{13}}{64}-\frac {24845 \tan \left (\frac {x}{2}\right )^{15}}{64}+\frac {7923 x \tan \left (\frac {x}{2}\right )^{2}}{16}+\frac {55461 x \tan \left (\frac {x}{2}\right )^{4}}{32}+\frac {55461 x \tan \left (\frac {x}{2}\right )^{6}}{16}+\frac {277305 x \tan \left (\frac {x}{2}\right )^{8}}{64}+\frac {55461 x \tan \left (\frac {x}{2}\right )^{10}}{16}+\frac {55461 x \tan \left (\frac {x}{2}\right )^{12}}{32}+\frac {7923 x \tan \left (\frac {x}{2}\right )^{14}}{16}+\frac {7923 x \tan \left (\frac {x}{2}\right )^{16}}{128}+\frac {24845 \tan \left (\frac {x}{2}\right )}{64}}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{8}}\) | \(150\) |
Input:
int((4-5*sin(x)^2)^4,x,method=_RETURNVERBOSE)
Output:
7923/128*x+625/1024*sin(8*x)+125/32*sin(6*x)+1975/128*sin(4*x)+1665/32*sin (2*x)
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.52 \[ \int \left (4-5 \sin ^2(x)\right )^4 \, dx=\frac {5}{128} \, {\left (2000 \, \cos \left (x\right )^{7} + 200 \, \cos \left (x\right )^{5} + 1210 \, \cos \left (x\right )^{3} + 1559 \, \cos \left (x\right )\right )} \sin \left (x\right ) + \frac {7923}{128} \, x \] Input:
integrate((4-5*sin(x)^2)^4,x, algorithm="fricas")
Output:
5/128*(2000*cos(x)^7 + 200*cos(x)^5 + 1210*cos(x)^3 + 1559*cos(x))*sin(x) + 7923/128*x
Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (66) = 132\).
Time = 0.47 (sec) , antiderivative size = 274, normalized size of antiderivative = 4.57 \[ \int \left (4-5 \sin ^2(x)\right )^4 \, dx=\frac {21875 x \sin ^{8}{\left (x \right )}}{128} + \frac {21875 x \sin ^{6}{\left (x \right )} \cos ^{2}{\left (x \right )}}{32} - 625 x \sin ^{6}{\left (x \right )} + \frac {65625 x \sin ^{4}{\left (x \right )} \cos ^{4}{\left (x \right )}}{64} - 1875 x \sin ^{4}{\left (x \right )} \cos ^{2}{\left (x \right )} + 900 x \sin ^{4}{\left (x \right )} + \frac {21875 x \sin ^{2}{\left (x \right )} \cos ^{6}{\left (x \right )}}{32} - 1875 x \sin ^{2}{\left (x \right )} \cos ^{4}{\left (x \right )} + 1800 x \sin ^{2}{\left (x \right )} \cos ^{2}{\left (x \right )} - 640 x \sin ^{2}{\left (x \right )} + \frac {21875 x \cos ^{8}{\left (x \right )}}{128} - 625 x \cos ^{6}{\left (x \right )} + 900 x \cos ^{4}{\left (x \right )} - 640 x \cos ^{2}{\left (x \right )} + 256 x - \frac {58125 \sin ^{7}{\left (x \right )} \cos {\left (x \right )}}{128} - \frac {319375 \sin ^{5}{\left (x \right )} \cos ^{3}{\left (x \right )}}{384} + 1375 \sin ^{5}{\left (x \right )} \cos {\left (x \right )} - \frac {240625 \sin ^{3}{\left (x \right )} \cos ^{5}{\left (x \right )}}{384} + \frac {5000 \sin ^{3}{\left (x \right )} \cos ^{3}{\left (x \right )}}{3} - 1500 \sin ^{3}{\left (x \right )} \cos {\left (x \right )} - \frac {21875 \sin {\left (x \right )} \cos ^{7}{\left (x \right )}}{128} + 625 \sin {\left (x \right )} \cos ^{5}{\left (x \right )} - 900 \sin {\left (x \right )} \cos ^{3}{\left (x \right )} + 640 \sin {\left (x \right )} \cos {\left (x \right )} \] Input:
integrate((4-5*sin(x)**2)**4,x)
Output:
21875*x*sin(x)**8/128 + 21875*x*sin(x)**6*cos(x)**2/32 - 625*x*sin(x)**6 + 65625*x*sin(x)**4*cos(x)**4/64 - 1875*x*sin(x)**4*cos(x)**2 + 900*x*sin(x )**4 + 21875*x*sin(x)**2*cos(x)**6/32 - 1875*x*sin(x)**2*cos(x)**4 + 1800* x*sin(x)**2*cos(x)**2 - 640*x*sin(x)**2 + 21875*x*cos(x)**8/128 - 625*x*co s(x)**6 + 900*x*cos(x)**4 - 640*x*cos(x)**2 + 256*x - 58125*sin(x)**7*cos( x)/128 - 319375*sin(x)**5*cos(x)**3/384 + 1375*sin(x)**5*cos(x) - 240625*s in(x)**3*cos(x)**5/384 + 5000*sin(x)**3*cos(x)**3/3 - 1500*sin(x)**3*cos(x ) - 21875*sin(x)*cos(x)**7/128 + 625*sin(x)*cos(x)**5 - 900*sin(x)*cos(x)* *3 + 640*sin(x)*cos(x)
Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.50 \[ \int \left (4-5 \sin ^2(x)\right )^4 \, dx=-\frac {125}{8} \, \sin \left (2 \, x\right )^{3} + \frac {7923}{128} \, x + \frac {625}{1024} \, \sin \left (8 \, x\right ) + \frac {1975}{128} \, \sin \left (4 \, x\right ) + \frac {255}{4} \, \sin \left (2 \, x\right ) \] Input:
integrate((4-5*sin(x)^2)^4,x, algorithm="maxima")
Output:
-125/8*sin(2*x)^3 + 7923/128*x + 625/1024*sin(8*x) + 1975/128*sin(4*x) + 2 55/4*sin(2*x)
Time = 0.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.47 \[ \int \left (4-5 \sin ^2(x)\right )^4 \, dx=\frac {7923}{128} \, x + \frac {625}{1024} \, \sin \left (8 \, x\right ) + \frac {125}{32} \, \sin \left (6 \, x\right ) + \frac {1975}{128} \, \sin \left (4 \, x\right ) + \frac {1665}{32} \, \sin \left (2 \, x\right ) \] Input:
integrate((4-5*sin(x)^2)^4,x, algorithm="giac")
Output:
7923/128*x + 625/1024*sin(8*x) + 125/32*sin(6*x) + 1975/128*sin(4*x) + 166 5/32*sin(2*x)
Time = 37.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.60 \[ \int \left (4-5 \sin ^2(x)\right )^4 \, dx=\frac {7923\,x}{128}+\frac {\frac {7795\,{\mathrm {tan}\left (x\right )}^7}{128}+\frac {29435\,{\mathrm {tan}\left (x\right )}^5}{128}+\frac {36485\,{\mathrm {tan}\left (x\right )}^3}{128}+\frac {24845\,\mathrm {tan}\left (x\right )}{128}}{{\left ({\mathrm {tan}\left (x\right )}^2+1\right )}^4} \] Input:
int((5*sin(x)^2 - 4)^4,x)
Output:
(7923*x)/128 + ((24845*tan(x))/128 + (36485*tan(x)^3)/128 + (29435*tan(x)^ 5)/128 + (7795*tan(x)^7)/128)/(tan(x)^2 + 1)^4
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.57 \[ \int \left (4-5 \sin ^2(x)\right )^4 \, dx=-\frac {625 \cos \left (x \right ) \sin \left (x \right )^{7}}{8}+\frac {3875 \cos \left (x \right ) \sin \left (x \right )^{5}}{16}-\frac {19025 \cos \left (x \right ) \sin \left (x \right )^{3}}{64}+\frac {24845 \cos \left (x \right ) \sin \left (x \right )}{128}+\frac {7923 x}{128} \] Input:
int((4-5*sin(x)^2)^4,x)
Output:
( - 10000*cos(x)*sin(x)**7 + 31000*cos(x)*sin(x)**5 - 38050*cos(x)*sin(x)* *3 + 24845*cos(x)*sin(x) + 7923*x)/128