Integrand size = 15, antiderivative size = 61 \[ \int \frac {\csc ^4(x)}{a+b \cos ^2(x)} \, dx=-\frac {b^2 \arctan \left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{5/2}}-\frac {(a+2 b) \cot (x)}{(a+b)^2}-\frac {\cot ^3(x)}{3 (a+b)} \] Output:
-b^2*arctan((a+b)^(1/2)*cot(x)/a^(1/2))/a^(1/2)/(a+b)^(5/2)-(a+2*b)*cot(x) /(a+b)^2-cot(x)^3/(3*a+3*b)
Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97 \[ \int \frac {\csc ^4(x)}{a+b \cos ^2(x)} \, dx=\frac {b^2 \arctan \left (\frac {\sqrt {a} \tan (x)}{\sqrt {a+b}}\right )}{\sqrt {a} (a+b)^{5/2}}-\frac {\cot (x) \left (2 a+5 b+(a+b) \csc ^2(x)\right )}{3 (a+b)^2} \] Input:
Integrate[Csc[x]^4/(a + b*Cos[x]^2),x]
Output:
(b^2*ArcTan[(Sqrt[a]*Tan[x])/Sqrt[a + b]])/(Sqrt[a]*(a + b)^(5/2)) - (Cot[ x]*(2*a + 5*b + (a + b)*Csc[x]^2))/(3*(a + b)^2)
Time = 0.29 (sec) , antiderivative size = 61, 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, 3670, 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.
\(\displaystyle \int \frac {\csc ^4(x)}{a+b \cos ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos \left (x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (x+\frac {\pi }{2}\right )^2\right )}dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle -\int \frac {\left (\cot ^2(x)+1\right )^2}{(a+b) \cot ^2(x)+a}d\cot (x)\) |
\(\Big \downarrow \) 300 |
\(\displaystyle -\int \left (\frac {b^2}{(a+b)^2 \left ((a+b) \cot ^2(x)+a\right )}+\frac {\cot ^2(x)}{a+b}+\frac {a+2 b}{(a+b)^2}\right )d\cot (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b^2 \arctan \left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{5/2}}-\frac {\cot ^3(x)}{3 (a+b)}-\frac {(a+2 b) \cot (x)}{(a+b)^2}\) |
Input:
Int[Csc[x]^4/(a + b*Cos[x]^2),x]
Output:
-((b^2*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^(5/2))) - (( a + 2*b)*Cot[x])/(a + b)^2 - Cot[x]^3/(3*(a + 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[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Time = 0.43 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {b^{2} \arctan \left (\frac {a \tan \left (x \right )}{\sqrt {\left (a +b \right ) a}}\right )}{\left (a +b \right )^{2} \sqrt {\left (a +b \right ) a}}-\frac {1}{3 \left (a +b \right ) \tan \left (x \right )^{3}}-\frac {a +2 b}{\left (a +b \right )^{2} \tan \left (x \right )}\) | \(57\) |
risch | \(-\frac {2 i \left (3 \,{\mathrm e}^{4 i x} b -6 a \,{\mathrm e}^{2 i x}-12 \,{\mathrm e}^{2 i x} b +2 a +5 b \right )}{3 \left ({\mathrm e}^{2 i x}-1\right )^{3} \left (a +b \right )^{2}}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2}}+\frac {b^{2} \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2}}\) | \(224\) |
Input:
int(csc(x)^4/(a+b*cos(x)^2),x,method=_RETURNVERBOSE)
Output:
1/(a+b)^2*b^2/((a+b)*a)^(1/2)*arctan(a*tan(x)/((a+b)*a)^(1/2))-1/3/(a+b)/t an(x)^3-(a+2*b)/(a+b)^2/tan(x)
Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (51) = 102\).
Time = 0.20 (sec) , antiderivative size = 396, normalized size of antiderivative = 6.49 \[ \int \frac {\csc ^4(x)}{a+b \cos ^2(x)} \, dx=\left [\frac {4 \, {\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (x\right )^{3} + 3 \, {\left (b^{2} \cos \left (x\right )^{2} - b^{2}\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (x\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{3} - a \cos \left (x\right )\right )} \sqrt {-a^{2} - a b} \sin \left (x\right ) + a^{2}}{b^{2} \cos \left (x\right )^{4} + 2 \, a b \cos \left (x\right )^{2} + a^{2}}\right ) \sin \left (x\right ) - 12 \, {\left (a^{3} + 3 \, a^{2} b + 2 \, a b^{2}\right )} \cos \left (x\right )}{12 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} - {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}, \frac {2 \, {\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (x\right )^{3} + 3 \, {\left (b^{2} \cos \left (x\right )^{2} - b^{2}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a}{2 \, \sqrt {a^{2} + a b} \cos \left (x\right ) \sin \left (x\right )}\right ) \sin \left (x\right ) - 6 \, {\left (a^{3} + 3 \, a^{2} b + 2 \, a b^{2}\right )} \cos \left (x\right )}{6 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} - {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}\right ] \] Input:
integrate(csc(x)^4/(a+b*cos(x)^2),x, algorithm="fricas")
Output:
[1/12*(4*(2*a^3 + 7*a^2*b + 5*a*b^2)*cos(x)^3 + 3*(b^2*cos(x)^2 - b^2)*sqr t(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*cos(x)^4 - 2*(4*a^2 + 3*a*b)*cos( x)^2 + 4*((2*a + b)*cos(x)^3 - a*cos(x))*sqrt(-a^2 - a*b)*sin(x) + a^2)/(b ^2*cos(x)^4 + 2*a*b*cos(x)^2 + a^2))*sin(x) - 12*(a^3 + 3*a^2*b + 2*a*b^2) *cos(x))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 - (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(x)^2)*sin(x)), 1/6*(2*(2*a^3 + 7*a^2*b + 5*a*b^2)*cos(x)^3 + 3*(b^2*cos(x)^2 - b^2)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(x)^2 - a) /(sqrt(a^2 + a*b)*cos(x)*sin(x)))*sin(x) - 6*(a^3 + 3*a^2*b + 2*a*b^2)*cos (x))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 - (a^4 + 3*a^3*b + 3*a^2*b^2 + a* b^3)*cos(x)^2)*sin(x))]
\[ \int \frac {\csc ^4(x)}{a+b \cos ^2(x)} \, dx=\int \frac {\csc ^{4}{\left (x \right )}}{a + b \cos ^{2}{\left (x \right )}}\, dx \] Input:
integrate(csc(x)**4/(a+b*cos(x)**2),x)
Output:
Integral(csc(x)**4/(a + b*cos(x)**2), x)
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15 \[ \int \frac {\csc ^4(x)}{a+b \cos ^2(x)} \, dx=\frac {b^{2} \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {3 \, {\left (a + 2 \, b\right )} \tan \left (x\right )^{2} + a + b}{3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (x\right )^{3}} \] Input:
integrate(csc(x)^4/(a+b*cos(x)^2),x, algorithm="maxima")
Output:
b^2*arctan(a*tan(x)/sqrt((a + b)*a))/(sqrt((a + b)*a)*(a^2 + 2*a*b + b^2)) - 1/3*(3*(a + 2*b)*tan(x)^2 + a + b)/((a^2 + 2*a*b + b^2)*tan(x)^3)
Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.48 \[ \int \frac {\csc ^4(x)}{a+b \cos ^2(x)} \, dx=\frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )} b^{2}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a^{2} + a b}} - \frac {3 \, a \tan \left (x\right )^{2} + 6 \, b \tan \left (x\right )^{2} + a + b}{3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (x\right )^{3}} \] Input:
integrate(csc(x)^4/(a+b*cos(x)^2),x, algorithm="giac")
Output:
(pi*floor(x/pi + 1/2)*sgn(a) + arctan(a*tan(x)/sqrt(a^2 + a*b)))*b^2/((a^2 + 2*a*b + b^2)*sqrt(a^2 + a*b)) - 1/3*(3*a*tan(x)^2 + 6*b*tan(x)^2 + a + b)/((a^2 + 2*a*b + b^2)*tan(x)^3)
Time = 1.00 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10 \[ \int \frac {\csc ^4(x)}{a+b \cos ^2(x)} \, dx=\frac {b^2\,\mathrm {atan}\left (\frac {\sqrt {a}\,\mathrm {tan}\left (x\right )\,\left (a^2+2\,a\,b+b^2\right )}{{\left (a+b\right )}^{5/2}}\right )}{\sqrt {a}\,{\left (a+b\right )}^{5/2}}-\frac {\frac {1}{3\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (x\right )}^2\,\left (a+2\,b\right )}{{\left (a+b\right )}^2}}{{\mathrm {tan}\left (x\right )}^3} \] Input:
int(1/(sin(x)^4*(a + b*cos(x)^2)),x)
Output:
(b^2*atan((a^(1/2)*tan(x)*(2*a*b + a^2 + b^2))/(a + b)^(5/2)))/(a^(1/2)*(a + b)^(5/2)) - (1/(3*(a + b)) + (tan(x)^2*(a + 2*b))/(a + b)^2)/tan(x)^3
Time = 0.19 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.57 \[ \int \frac {\csc ^4(x)}{a+b \cos ^2(x)} \, dx=\frac {3 \sqrt {a}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {x}{2}\right )-\sqrt {b}}{\sqrt {a}}\right ) \sin \left (x \right )^{3} b^{2}+3 \sqrt {a}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {x}{2}\right )+\sqrt {b}}{\sqrt {a}}\right ) \sin \left (x \right )^{3} b^{2}-2 \cos \left (x \right ) \sin \left (x \right )^{2} a^{3}-7 \cos \left (x \right ) \sin \left (x \right )^{2} a^{2} b -5 \cos \left (x \right ) \sin \left (x \right )^{2} a \,b^{2}-\cos \left (x \right ) a^{3}-2 \cos \left (x \right ) a^{2} b -\cos \left (x \right ) a \,b^{2}}{3 \sin \left (x \right )^{3} a \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )} \] Input:
int(csc(x)^4/(a+b*cos(x)^2),x)
Output:
(3*sqrt(a)*sqrt(a + b)*atan((sqrt(a + b)*tan(x/2) - sqrt(b))/sqrt(a))*sin( x)**3*b**2 + 3*sqrt(a)*sqrt(a + b)*atan((sqrt(a + b)*tan(x/2) + sqrt(b))/s qrt(a))*sin(x)**3*b**2 - 2*cos(x)*sin(x)**2*a**3 - 7*cos(x)*sin(x)**2*a**2 *b - 5*cos(x)*sin(x)**2*a*b**2 - cos(x)*a**3 - 2*cos(x)*a**2*b - cos(x)*a* b**2)/(3*sin(x)**3*a*(a**3 + 3*a**2*b + 3*a*b**2 + b**3))