Integrand size = 15, antiderivative size = 56 \[ \int \frac {\cos ^5(x)}{a+b \cos ^2(x)} \, dx=\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{b^{5/2} \sqrt {a+b}}-\frac {(a-b) \sin (x)}{b^2}-\frac {\sin ^3(x)}{3 b} \] Output:
a^2*arctanh(b^(1/2)*sin(x)/(a+b)^(1/2))/b^(5/2)/(a+b)^(1/2)-(a-b)*sin(x)/b ^2-1/3*sin(x)^3/b
Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.54 \[ \int \frac {\cos ^5(x)}{a+b \cos ^2(x)} \, dx=\frac {\frac {6 a^2 \left (-\log \left (\sqrt {a+b}-\sqrt {b} \sin (x)\right )+\log \left (\sqrt {a+b}+\sqrt {b} \sin (x)\right )\right )}{\sqrt {a+b}}+3 \sqrt {b} (-4 a+3 b) \sin (x)+b^{3/2} \sin (3 x)}{12 b^{5/2}} \] Input:
Integrate[Cos[x]^5/(a + b*Cos[x]^2),x]
Output:
((6*a^2*(-Log[Sqrt[a + b] - Sqrt[b]*Sin[x]] + Log[Sqrt[a + b] + Sqrt[b]*Si n[x]]))/Sqrt[a + b] + 3*Sqrt[b]*(-4*a + 3*b)*Sin[x] + b^(3/2)*Sin[3*x])/(1 2*b^(5/2))
Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3665, 300, 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.
\(\displaystyle \int \frac {\cos ^5(x)}{a+b \cos ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (x+\frac {\pi }{2}\right )^5}{a+b \sin \left (x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle \int \frac {\left (1-\sin ^2(x)\right )^2}{a-b \sin ^2(x)+b}d\sin (x)\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \int \left (\frac {a^2}{b^2 \left (a-b \sin ^2(x)+b\right )}-\frac {a-b}{b^2}-\frac {\sin ^2(x)}{b}\right )d\sin (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{b^{5/2} \sqrt {a+b}}-\frac {(a-b) \sin (x)}{b^2}-\frac {\sin ^3(x)}{3 b}\) |
Input:
Int[Cos[x]^5/(a + b*Cos[x]^2),x]
Output:
(a^2*ArcTanh[(Sqrt[b]*Sin[x])/Sqrt[a + b]])/(b^(5/2)*Sqrt[a + b]) - ((a - b)*Sin[x])/b^2 - Sin[x]^3/(3*b)
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.68 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {\frac {b \sin \left (x \right )^{3}}{3}+\sin \left (x \right ) a -b \sin \left (x \right )}{b^{2}}+\frac {a^{2} \operatorname {arctanh}\left (\frac {b \sin \left (x \right )}{\sqrt {\left (a +b \right ) b}}\right )}{b^{2} \sqrt {\left (a +b \right ) b}}\) | \(50\) |
risch | \(\frac {i {\mathrm e}^{i x} a}{2 b^{2}}-\frac {3 i {\mathrm e}^{i x}}{8 b}-\frac {i {\mathrm e}^{-i x} a}{2 b^{2}}+\frac {3 i {\mathrm e}^{-i x}}{8 b}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i x}}{\sqrt {a b +b^{2}}}-1\right )}{2 \sqrt {a b +b^{2}}\, b^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i x}}{\sqrt {a b +b^{2}}}-1\right )}{2 \sqrt {a b +b^{2}}\, b^{2}}+\frac {\sin \left (3 x \right )}{12 b}\) | \(147\) |
Input:
int(cos(x)^5/(a+b*cos(x)^2),x,method=_RETURNVERBOSE)
Output:
-1/b^2*(1/3*b*sin(x)^3+sin(x)*a-b*sin(x))+a^2/b^2/((a+b)*b)^(1/2)*arctanh( b*sin(x)/((a+b)*b)^(1/2))
Time = 0.17 (sec) , antiderivative size = 191, normalized size of antiderivative = 3.41 \[ \int \frac {\cos ^5(x)}{a+b \cos ^2(x)} \, dx=\left [\frac {3 \, \sqrt {a b + b^{2}} a^{2} \log \left (-\frac {b \cos \left (x\right )^{2} - 2 \, \sqrt {a b + b^{2}} \sin \left (x\right ) - a - 2 \, b}{b \cos \left (x\right )^{2} + a}\right ) - 2 \, {\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3} - {\left (a b^{2} + b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \, {\left (a b^{3} + b^{4}\right )}}, -\frac {3 \, \sqrt {-a b - b^{2}} a^{2} \arctan \left (\frac {\sqrt {-a b - b^{2}} \sin \left (x\right )}{a + b}\right ) + {\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3} - {\left (a b^{2} + b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{3 \, {\left (a b^{3} + b^{4}\right )}}\right ] \] Input:
integrate(cos(x)^5/(a+b*cos(x)^2),x, algorithm="fricas")
Output:
[1/6*(3*sqrt(a*b + b^2)*a^2*log(-(b*cos(x)^2 - 2*sqrt(a*b + b^2)*sin(x) - a - 2*b)/(b*cos(x)^2 + a)) - 2*(3*a^2*b + a*b^2 - 2*b^3 - (a*b^2 + b^3)*co s(x)^2)*sin(x))/(a*b^3 + b^4), -1/3*(3*sqrt(-a*b - b^2)*a^2*arctan(sqrt(-a *b - b^2)*sin(x)/(a + b)) + (3*a^2*b + a*b^2 - 2*b^3 - (a*b^2 + b^3)*cos(x )^2)*sin(x))/(a*b^3 + b^4)]
Timed out. \[ \int \frac {\cos ^5(x)}{a+b \cos ^2(x)} \, dx=\text {Timed out} \] Input:
integrate(cos(x)**5/(a+b*cos(x)**2),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^5(x)}{a+b \cos ^2(x)} \, dx=-\frac {a^{2} \log \left (\frac {b \sin \left (x\right ) - \sqrt {{\left (a + b\right )} b}}{b \sin \left (x\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} b^{2}} - \frac {b \sin \left (x\right )^{3} + 3 \, {\left (a - b\right )} \sin \left (x\right )}{3 \, b^{2}} \] Input:
integrate(cos(x)^5/(a+b*cos(x)^2),x, algorithm="maxima")
Output:
-1/2*a^2*log((b*sin(x) - sqrt((a + b)*b))/(b*sin(x) + sqrt((a + b)*b)))/(s qrt((a + b)*b)*b^2) - 1/3*(b*sin(x)^3 + 3*(a - b)*sin(x))/b^2
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.16 \[ \int \frac {\cos ^5(x)}{a+b \cos ^2(x)} \, dx=-\frac {a^{2} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} b^{2}} - \frac {b^{2} \sin \left (x\right )^{3} + 3 \, a b \sin \left (x\right ) - 3 \, b^{2} \sin \left (x\right )}{3 \, b^{3}} \] Input:
integrate(cos(x)^5/(a+b*cos(x)^2),x, algorithm="giac")
Output:
-a^2*arctan(b*sin(x)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*b^2) - 1/3*(b^2*s in(x)^3 + 3*a*b*sin(x) - 3*b^2*sin(x))/b^3
Time = 0.06 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^5(x)}{a+b \cos ^2(x)} \, dx=\frac {a^2\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sin \left (x\right )}{\sqrt {a+b}}\right )}{b^{5/2}\,\sqrt {a+b}}-\frac {{\sin \left (x\right )}^3}{3\,b}-\sin \left (x\right )\,\left (\frac {a+b}{b^2}-\frac {2}{b}\right ) \] Input:
int(cos(x)^5/(a + b*cos(x)^2),x)
Output:
(a^2*atanh((b^(1/2)*sin(x))/(a + b)^(1/2)))/(b^(5/2)*(a + b)^(1/2)) - sin( x)^3/(3*b) - sin(x)*((a + b)/b^2 - 2/b)
Time = 0.17 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.09 \[ \int \frac {\cos ^5(x)}{a+b \cos ^2(x)} \, dx=\frac {-3 \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {x}{2}\right )^{2}+\sqrt {a +b}-2 \sqrt {b}\, \tan \left (\frac {x}{2}\right )\right ) a^{2}+3 \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {x}{2}\right )^{2}+\sqrt {a +b}+2 \sqrt {b}\, \tan \left (\frac {x}{2}\right )\right ) a^{2}-2 \sin \left (x \right )^{3} a \,b^{2}-2 \sin \left (x \right )^{3} b^{3}-6 \sin \left (x \right ) a^{2} b +6 \sin \left (x \right ) b^{3}}{6 b^{3} \left (a +b \right )} \] Input:
int(cos(x)^5/(a+b*cos(x)^2),x)
Output:
( - 3*sqrt(b)*sqrt(a + b)*log(sqrt(a + b)*tan(x/2)**2 + sqrt(a + b) - 2*sq rt(b)*tan(x/2))*a**2 + 3*sqrt(b)*sqrt(a + b)*log(sqrt(a + b)*tan(x/2)**2 + sqrt(a + b) + 2*sqrt(b)*tan(x/2))*a**2 - 2*sin(x)**3*a*b**2 - 2*sin(x)**3 *b**3 - 6*sin(x)*a**2*b + 6*sin(x)*b**3)/(6*b**3*(a + b))