Integrand size = 17, antiderivative size = 36 \[ \int \cos ^3(x) \left (a+b \cos ^2(x)\right )^3 \sin (x) \, dx=\frac {a \left (a+b \cos ^2(x)\right )^4}{8 b^2}-\frac {\left (a+b \cos ^2(x)\right )^5}{10 b^2} \] Output:
1/8*a*(a+b*cos(x)^2)^4/b^2-1/10*(a+b*cos(x)^2)^5/b^2
Leaf count is larger than twice the leaf count of optimal. \(137\) vs. \(2(36)=72\).
Time = 0.27 (sec) , antiderivative size = 137, normalized size of antiderivative = 3.81 \[ \int \cos ^3(x) \left (a+b \cos ^2(x)\right )^3 \sin (x) \, dx=\frac {1}{32} \left (-12 a^2 b \cos ^4(x)-8 a b^2 \cos ^6(x)-2 b^3 \cos ^8(x)-4 a^3 \cos (2 x)-4 a^2 b \cos ^3(x) \cos (3 x)-a^3 \cos (4 x)-\frac {1}{32} a b^2 (48 \cos (2 x)+36 \cos (4 x)+16 \cos (6 x)+3 \cos (8 x))-\frac {1}{320} b^3 (140 \cos (2 x)+100 \cos (4 x)+50 \cos (6 x)+15 \cos (8 x)+2 \cos (10 x))\right ) \] Input:
Integrate[Cos[x]^3*(a + b*Cos[x]^2)^3*Sin[x],x]
Output:
(-12*a^2*b*Cos[x]^4 - 8*a*b^2*Cos[x]^6 - 2*b^3*Cos[x]^8 - 4*a^3*Cos[2*x] - 4*a^2*b*Cos[x]^3*Cos[3*x] - a^3*Cos[4*x] - (a*b^2*(48*Cos[2*x] + 36*Cos[4 *x] + 16*Cos[6*x] + 3*Cos[8*x]))/32 - (b^3*(140*Cos[2*x] + 100*Cos[4*x] + 50*Cos[6*x] + 15*Cos[8*x] + 2*Cos[10*x]))/320)/32
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3042, 4835, 243, 49, 2009}
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.
\(\displaystyle \int \sin (x) \cos ^3(x) \left (a+b \cos ^2(x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (x) \cos (x)^3 \left (a+b \cos (x)^2\right )^3dx\) |
\(\Big \downarrow \) 4835 |
\(\displaystyle -\int \cos ^3(x) \left (b \cos ^2(x)+a\right )^3d\cos (x)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {1}{2} \int \cos ^2(x) \left (b \cos ^2(x)+a\right )^3d\cos ^2(x)\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {\left (b \cos ^2(x)+a\right )^4}{b}-\frac {a \left (b \cos ^2(x)+a\right )^3}{b}\right )d\cos ^2(x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {a \left (a+b \cos ^2(x)\right )^4}{4 b^2}-\frac {\left (a+b \cos ^2(x)\right )^5}{5 b^2}\right )\) |
Input:
Int[Cos[x]^3*(a + b*Cos[x]^2)^3*Sin[x],x]
Output:
((a*(a + b*Cos[x]^2)^4)/(4*b^2) - (a + b*Cos[x]^2)^5/(5*b^2))/2
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto rs[Cos[c*(a + b*x)], x]}, Simp[-d/(b*c) Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a + b* x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])
Time = 229.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(-\frac {b^{3} \cos \left (x \right )^{10}}{10}-\frac {3 a \,b^{2} \cos \left (x \right )^{8}}{8}-\frac {a^{2} b \cos \left (x \right )^{6}}{2}-\frac {a^{3} \cos \left (x \right )^{4}}{4}\) | \(40\) |
default | \(-\frac {b^{3} \cos \left (x \right )^{10}}{10}-\frac {3 a \,b^{2} \cos \left (x \right )^{8}}{8}-\frac {a^{2} b \cos \left (x \right )^{6}}{2}-\frac {a^{3} \cos \left (x \right )^{4}}{4}\) | \(40\) |
parts | \(-\frac {b^{3} \cos \left (x \right )^{10}}{10}-\frac {3 a \,b^{2} \cos \left (x \right )^{8}}{8}-\frac {a^{2} b \cos \left (x \right )^{6}}{2}-\frac {a^{3} \cos \left (x \right )^{4}}{4}\) | \(40\) |
parallelrisch | \(\frac {\left (-64 a^{3}-120 a^{2} b -84 a \,b^{2}-21 b^{3}\right ) \cos \left (2 x \right )}{512}+\frac {\left (-8 a^{3}-24 a^{2} b -21 a \,b^{2}-6 b^{3}\right ) \cos \left (4 x \right )}{256}-\frac {\left (a +\frac {3 b}{4}\right )^{2} b \cos \left (6 x \right )}{64}+\frac {\left (-3 a \,b^{2}-2 b^{3}\right ) \cos \left (8 x \right )}{1024}-\frac {b^{3} \cos \left (10 x \right )}{5120}+\frac {5 a^{3}}{32}+\frac {11 a^{2} b}{32}+\frac {279 a \,b^{2}}{1024}+\frac {193 b^{3}}{2560}\) | \(123\) |
risch | \(-\frac {b^{3} \cos \left (10 x \right )}{5120}-\frac {3 \cos \left (8 x \right ) a \,b^{2}}{1024}-\frac {\cos \left (8 x \right ) b^{3}}{512}-\frac {\cos \left (6 x \right ) a^{2} b}{64}-\frac {3 \cos \left (6 x \right ) a \,b^{2}}{128}-\frac {9 \cos \left (6 x \right ) b^{3}}{1024}-\frac {\cos \left (4 x \right ) a^{3}}{32}-\frac {3 \cos \left (4 x \right ) a^{2} b}{32}-\frac {21 \cos \left (4 x \right ) a \,b^{2}}{256}-\frac {3 \cos \left (4 x \right ) b^{3}}{128}-\frac {\cos \left (2 x \right ) a^{3}}{8}-\frac {15 \cos \left (2 x \right ) a^{2} b}{64}-\frac {21 \cos \left (2 x \right ) a \,b^{2}}{128}-\frac {21 \cos \left (2 x \right ) b^{3}}{512}\) | \(135\) |
orering | \(\frac {3 \sin \left (x \right )^{4} \left (a +b \cos \left (x \right )^{2}\right )^{3}}{32}-\frac {3 \cos \left (x \right )^{8} \sin \left (x \right )^{2} b^{3}}{512}+\frac {3 \cos \left (x \right )^{6} \sin \left (x \right )^{4} b^{3}}{512}-\frac {3 \cos \left (x \right )^{4} \left (a +b \cos \left (x \right )^{2}\right )^{2} \sin \left (x \right )^{2} b}{32}-\frac {5 \cos \left (x \right )^{4} \left (a +b \cos \left (x \right )^{2}\right )^{3}}{32}+\frac {3 \cos \left (x \right )^{2} \left (a +b \cos \left (x \right )^{2}\right )^{3} \sin \left (x \right )^{2}}{16}-\frac {\cos \left (x \right )^{4} \sin \left (x \right )^{6} b^{3}}{128}+\frac {49 b^{3} \cos \left (x \right )^{10}}{5120}+\frac {21 \cos \left (x \right )^{2} \sin \left (x \right )^{8} b^{3}}{1024}+\frac {9 \cos \left (x \right )^{6} \left (a +b \cos \left (x \right )^{2}\right ) \sin \left (x \right )^{2} b^{2}}{256}+\frac {63 \sin \left (x \right )^{10} b^{3}}{2560}+\frac {3 \cos \left (x \right )^{2} \left (a +b \cos \left (x \right )^{2}\right )^{2} \sin \left (x \right )^{4} b}{16}-\frac {21 \cos \left (x \right )^{4} \left (a +b \cos \left (x \right )^{2}\right ) \sin \left (x \right )^{4} b^{2}}{512}+\frac {\cos \left (x \right )^{6} \left (a +b \cos \left (x \right )^{2}\right )^{2} b}{8}+\frac {25 \sin \left (x \right )^{6} \left (a +b \cos \left (x \right )^{2}\right ) b^{2} \cos \left (x \right )^{2}}{256}-\frac {55 \cos \left (x \right )^{8} \left (a +b \cos \left (x \right )^{2}\right ) b^{2}}{1024}+\frac {5 \sin \left (x \right )^{6} \left (a +b \cos \left (x \right )^{2}\right )^{2} b}{32}+\frac {105 \sin \left (x \right )^{8} \left (a +b \cos \left (x \right )^{2}\right ) b^{2}}{1024}\) | \(297\) |
Input:
int(cos(x)^3*(a+b*cos(x)^2)^3*sin(x),x,method=_RETURNVERBOSE)
Output:
-1/10*b^3*cos(x)^10-3/8*a*b^2*cos(x)^8-1/2*a^2*b*cos(x)^6-1/4*a^3*cos(x)^4
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08 \[ \int \cos ^3(x) \left (a+b \cos ^2(x)\right )^3 \sin (x) \, dx=-\frac {1}{10} \, b^{3} \cos \left (x\right )^{10} - \frac {3}{8} \, a b^{2} \cos \left (x\right )^{8} - \frac {1}{2} \, a^{2} b \cos \left (x\right )^{6} - \frac {1}{4} \, a^{3} \cos \left (x\right )^{4} \] Input:
integrate(cos(x)^3*(a+b*cos(x)^2)^3*sin(x),x, algorithm="fricas")
Output:
-1/10*b^3*cos(x)^10 - 3/8*a*b^2*cos(x)^8 - 1/2*a^2*b*cos(x)^6 - 1/4*a^3*co s(x)^4
Time = 1.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \cos ^3(x) \left (a+b \cos ^2(x)\right )^3 \sin (x) \, dx=- \frac {a^{3} \cos ^{4}{\left (x \right )}}{4} - \frac {a^{2} b \cos ^{6}{\left (x \right )}}{2} - \frac {3 a b^{2} \cos ^{8}{\left (x \right )}}{8} - \frac {b^{3} \cos ^{10}{\left (x \right )}}{10} \] Input:
integrate(cos(x)**3*(a+b*cos(x)**2)**3*sin(x),x)
Output:
-a**3*cos(x)**4/4 - a**2*b*cos(x)**6/2 - 3*a*b**2*cos(x)**8/8 - b**3*cos(x )**10/10
Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (32) = 64\).
Time = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.86 \[ \int \cos ^3(x) \left (a+b \cos ^2(x)\right )^3 \sin (x) \, dx=\frac {1}{10} \, b^{3} \sin \left (x\right )^{10} - \frac {1}{8} \, {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \sin \left (x\right )^{8} + \frac {1}{2} \, {\left (a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} \sin \left (x\right )^{6} - \frac {1}{4} \, {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \sin \left (x\right )^{4} + \frac {1}{2} \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sin \left (x\right )^{2} \] Input:
integrate(cos(x)^3*(a+b*cos(x)^2)^3*sin(x),x, algorithm="maxima")
Output:
1/10*b^3*sin(x)^10 - 1/8*(3*a*b^2 + 4*b^3)*sin(x)^8 + 1/2*(a^2*b + 3*a*b^2 + 2*b^3)*sin(x)^6 - 1/4*(a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*sin(x)^4 + 1/2* (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sin(x)^2
Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08 \[ \int \cos ^3(x) \left (a+b \cos ^2(x)\right )^3 \sin (x) \, dx=-\frac {1}{10} \, b^{3} \cos \left (x\right )^{10} - \frac {3}{8} \, a b^{2} \cos \left (x\right )^{8} - \frac {1}{2} \, a^{2} b \cos \left (x\right )^{6} - \frac {1}{4} \, a^{3} \cos \left (x\right )^{4} \] Input:
integrate(cos(x)^3*(a+b*cos(x)^2)^3*sin(x),x, algorithm="giac")
Output:
-1/10*b^3*cos(x)^10 - 3/8*a*b^2*cos(x)^8 - 1/2*a^2*b*cos(x)^6 - 1/4*a^3*co s(x)^4
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08 \[ \int \cos ^3(x) \left (a+b \cos ^2(x)\right )^3 \sin (x) \, dx=-\frac {a^3\,{\cos \left (x\right )}^4}{4}-\frac {a^2\,b\,{\cos \left (x\right )}^6}{2}-\frac {3\,a\,b^2\,{\cos \left (x\right )}^8}{8}-\frac {b^3\,{\cos \left (x\right )}^{10}}{10} \] Input:
int(cos(x)^3*sin(x)*(a + b*cos(x)^2)^3,x)
Output:
- (a^3*cos(x)^4)/4 - (b^3*cos(x)^10)/10 - (a^2*b*cos(x)^6)/2 - (3*a*b^2*co s(x)^8)/8
Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14 \[ \int \cos ^3(x) \left (a+b \cos ^2(x)\right )^3 \sin (x) \, dx=\frac {\cos \left (x \right )^{4} \left (-4 \cos \left (x \right )^{6} b^{3}-15 \cos \left (x \right )^{4} a \,b^{2}-20 \cos \left (x \right )^{2} a^{2} b -10 a^{3}\right )}{40} \] Input:
int(cos(x)^3*(a+b*cos(x)^2)^3*sin(x),x)
Output:
(cos(x)**4*( - 4*cos(x)**6*b**3 - 15*cos(x)**4*a*b**2 - 20*cos(x)**2*a**2* b - 10*a**3))/40