Integrand size = 15, antiderivative size = 72 \[ \int \frac {\cos ^5(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=-\frac {(3 a-b) (a+b) \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2}}+\frac {\sin (x)}{b^2}+\frac {(a+b)^2 \sin (x)}{2 a b^2 \left (a+b \sin ^2(x)\right )} \] Output:
-1/2*(3*a-b)*(a+b)*arctan(b^(1/2)*sin(x)/a^(1/2))/a^(3/2)/b^(5/2)+sin(x)/b ^2+1/2*(a+b)^2*sin(x)/a/b^2/(a+b*sin(x)^2)
Time = 0.45 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.64 \[ \int \frac {\cos ^5(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {\frac {\left (3 a^2+2 a b-b^2\right ) \arctan \left (\frac {\sqrt {a} \csc (x)}{\sqrt {b}}\right )}{a^{3/2}}+\frac {\left (-3 a^2-2 a b+b^2\right ) \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{a^{3/2}}+4 \sqrt {b} \sin (x)+\frac {4 \sqrt {b} (a+b)^2 \sin (x)}{a (2 a+b-b \cos (2 x))}}{4 b^{5/2}} \] Input:
Integrate[Cos[x]^5/(a + b*Sin[x]^2)^2,x]
Output:
(((3*a^2 + 2*a*b - b^2)*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]])/a^(3/2) + ((-3*a ^2 - 2*a*b + b^2)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/a^(3/2) + 4*Sqrt[b]*Si n[x] + (4*Sqrt[b]*(a + b)^2*Sin[x])/(a*(2*a + b - b*Cos[2*x])))/(4*b^(5/2) )
Time = 0.29 (sec) , antiderivative size = 72, 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, 3669, 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.
\(\displaystyle \int \frac {\cos ^5(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (x)^5}{\left (a+b \sin (x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \int \frac {\left (1-\sin ^2(x)\right )^2}{\left (a+b \sin ^2(x)\right )^2}d\sin (x)\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \int \left (\frac {1}{b^2}-\frac {a^2+2 b (a+b) \sin ^2(x)-b^2}{b^2 \left (a+b \sin ^2(x)\right )^2}\right )d\sin (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {(3 a-b) (a+b) \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2}}+\frac {(a+b)^2 \sin (x)}{2 a b^2 \left (a+b \sin ^2(x)\right )}+\frac {\sin (x)}{b^2}\) |
Input:
Int[Cos[x]^5/(a + b*Sin[x]^2)^2,x]
Output:
-1/2*((3*a - b)*(a + b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(a^(3/2)*b^(5/2) ) + Sin[x]/b^2 + ((a + b)^2*Sin[x])/(2*a*b^2*(a + b*Sin[x]^2))
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[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 1.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07
method | result | size |
derivativedivides | \(\frac {\sin \left (x \right )}{b^{2}}-\frac {-\frac {\left (a^{2}+2 a b +b^{2}\right ) \sin \left (x \right )}{2 a \left (a +b \sin \left (x \right )^{2}\right )}+\frac {\left (3 a^{2}+2 a b -b^{2}\right ) \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}}{b^{2}}\) | \(77\) |
default | \(\frac {\sin \left (x \right )}{b^{2}}-\frac {-\frac {\left (a^{2}+2 a b +b^{2}\right ) \sin \left (x \right )}{2 a \left (a +b \sin \left (x \right )^{2}\right )}+\frac {\left (3 a^{2}+2 a b -b^{2}\right ) \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}}{b^{2}}\) | \(77\) |
risch | \(-\frac {i {\mathrm e}^{i x}}{2 b^{2}}+\frac {i {\mathrm e}^{-i x}}{2 b^{2}}-\frac {i \left (a^{2}+2 a b +b^{2}\right ) \left ({\mathrm e}^{3 i x}-{\mathrm e}^{i x}\right )}{b^{2} a \left (-{\mathrm e}^{4 i x} b +4 \,{\mathrm e}^{2 i x} a +2 \,{\mathrm e}^{2 i x} b -b \right )}-\frac {3 a \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, b^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{2 \sqrt {-a b}\, b}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, a}+\frac {3 a \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, b^{2}}+\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{2 \sqrt {-a b}\, b}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, a}\) | \(293\) |
Input:
int(cos(x)^5/(a+b*sin(x)^2)^2,x,method=_RETURNVERBOSE)
Output:
sin(x)/b^2-1/b^2*(-1/2*(a^2+2*a*b+b^2)/a*sin(x)/(a+b*sin(x)^2)+1/2*(3*a^2+ 2*a*b-b^2)/a/(a*b)^(1/2)*arctan(b*sin(x)/(a*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (60) = 120\).
Time = 0.12 (sec) , antiderivative size = 296, normalized size of antiderivative = 4.11 \[ \int \frac {\cos ^5(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\left [-\frac {{\left (3 \, a^{3} + 5 \, a^{2} b + a b^{2} - b^{3} - {\left (3 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} \cos \left (x\right )^{2}\right )} \sqrt {-a b} \log \left (-\frac {b \cos \left (x\right )^{2} + 2 \, \sqrt {-a b} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) - 2 \, {\left (2 \, a^{2} b^{2} \cos \left (x\right )^{2} - 3 \, a^{3} b - 4 \, a^{2} b^{2} - a b^{3}\right )} \sin \left (x\right )}{4 \, {\left (a^{2} b^{4} \cos \left (x\right )^{2} - a^{3} b^{3} - a^{2} b^{4}\right )}}, \frac {{\left (3 \, a^{3} + 5 \, a^{2} b + a b^{2} - b^{3} - {\left (3 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} \cos \left (x\right )^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} \sin \left (x\right )}{a}\right ) + {\left (2 \, a^{2} b^{2} \cos \left (x\right )^{2} - 3 \, a^{3} b - 4 \, a^{2} b^{2} - a b^{3}\right )} \sin \left (x\right )}{2 \, {\left (a^{2} b^{4} \cos \left (x\right )^{2} - a^{3} b^{3} - a^{2} b^{4}\right )}}\right ] \] Input:
integrate(cos(x)^5/(a+b*sin(x)^2)^2,x, algorithm="fricas")
Output:
[-1/4*((3*a^3 + 5*a^2*b + a*b^2 - b^3 - (3*a^2*b + 2*a*b^2 - b^3)*cos(x)^2 )*sqrt(-a*b)*log(-(b*cos(x)^2 + 2*sqrt(-a*b)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) - 2*(2*a^2*b^2*cos(x)^2 - 3*a^3*b - 4*a^2*b^2 - a*b^3)*sin(x))/(a ^2*b^4*cos(x)^2 - a^3*b^3 - a^2*b^4), 1/2*((3*a^3 + 5*a^2*b + a*b^2 - b^3 - (3*a^2*b + 2*a*b^2 - b^3)*cos(x)^2)*sqrt(a*b)*arctan(sqrt(a*b)*sin(x)/a) + (2*a^2*b^2*cos(x)^2 - 3*a^3*b - 4*a^2*b^2 - a*b^3)*sin(x))/(a^2*b^4*cos (x)^2 - a^3*b^3 - a^2*b^4)]
Timed out. \[ \int \frac {\cos ^5(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\text {Timed out} \] Input:
integrate(cos(x)**5/(a+b*sin(x)**2)**2,x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^5(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (x\right )}{2 \, {\left (a b^{3} \sin \left (x\right )^{2} + a^{2} b^{2}\right )}} + \frac {\sin \left (x\right )}{b^{2}} - \frac {{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a b^{2}} \] Input:
integrate(cos(x)^5/(a+b*sin(x)^2)^2,x, algorithm="maxima")
Output:
1/2*(a^2 + 2*a*b + b^2)*sin(x)/(a*b^3*sin(x)^2 + a^2*b^2) + sin(x)/b^2 - 1 /2*(3*a^2 + 2*a*b - b^2)*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*a*b^2)
Time = 0.38 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^5(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {\sin \left (x\right )}{b^{2}} - \frac {{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a b^{2}} + \frac {a^{2} \sin \left (x\right ) + 2 \, a b \sin \left (x\right ) + b^{2} \sin \left (x\right )}{2 \, {\left (b \sin \left (x\right )^{2} + a\right )} a b^{2}} \] Input:
integrate(cos(x)^5/(a+b*sin(x)^2)^2,x, algorithm="giac")
Output:
sin(x)/b^2 - 1/2*(3*a^2 + 2*a*b - b^2)*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a* b)*a*b^2) + 1/2*(a^2*sin(x) + 2*a*b*sin(x) + b^2*sin(x))/((b*sin(x)^2 + a) *a*b^2)
Time = 34.57 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.33 \[ \int \frac {\cos ^5(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {\sin \left (x\right )}{b^2}+\frac {\sin \left (x\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,a\,\left (b^3\,{\sin \left (x\right )}^2+a\,b^2\right )}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sin \left (x\right )\,\left (a+b\right )\,\left (3\,a-b\right )}{\sqrt {a}\,\left (3\,a^2+2\,a\,b-b^2\right )}\right )\,\left (a+b\right )\,\left (3\,a-b\right )}{2\,a^{3/2}\,b^{5/2}} \] Input:
int(cos(x)^5/(a + b*sin(x)^2)^2,x)
Output:
sin(x)/b^2 + (sin(x)*(2*a*b + a^2 + b^2))/(2*a*(b^3*sin(x)^2 + a*b^2)) - ( atan((b^(1/2)*sin(x)*(a + b)*(3*a - b))/(a^(1/2)*(2*a*b + 3*a^2 - b^2)))*( a + b)*(3*a - b))/(2*a^(3/2)*b^(5/2))
Time = 0.26 (sec) , antiderivative size = 2026, normalized size of antiderivative = 28.14 \[ \int \frac {\cos ^5(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(cos(x)^5/(a+b*sin(x)^2)^2,x)
Output:
(6*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan( (tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*sin(x)**2*a* *2*b + 4*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b) *atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*sin(x) **2*a*b**2 - 2*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))* sin(x)**2*b**3 + 6*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b )))*a**3 + 4*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a* *2*b - 2*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b) *atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a*b**2 - 6*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt (a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*sin(x)**2*a**2*b**2 - 4*sqrt(a )*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2* sqrt(b)*sqrt(a + b) + a + 2*b)))*sin(x)**2*a*b**3 + 2*sqrt(a)*sqrt(2*sqrt( b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*sin(x)**2*b**4 - 6*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b) ))*a**3*b - 4*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan(x...