Integrand size = 10, antiderivative size = 65 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=-\frac {3}{8} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cot (x)}{\sqrt {a \csc ^2(x)}}\right )-\frac {3}{8} a^2 \cot (x) \sqrt {a \csc ^2(x)}-\frac {1}{4} a \cot (x) \left (a \csc ^2(x)\right )^{3/2} \] Output:
-3/8*a^(5/2)*arctanh(a^(1/2)*cot(x)/(a*csc(x)^2)^(1/2))-3/8*a^2*cot(x)*(a* csc(x)^2)^(1/2)-1/4*a*cot(x)*(a*csc(x)^2)^(3/2)
Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\frac {1}{64} \left (a \csc ^2(x)\right )^{5/2} \sin (x) \left (-22 \cos (x)+6 \left (\cos (3 x)+4 \left (-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )\right ) \sin ^4(x)\right )\right ) \] Input:
Integrate[(a*Csc[x]^2)^(5/2),x]
Output:
((a*Csc[x]^2)^(5/2)*Sin[x]*(-22*Cos[x] + 6*(Cos[3*x] + 4*(-Log[Cos[x/2]] + Log[Sin[x/2]])*Sin[x]^4)))/64
Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4610, 211, 211, 224, 219}
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.
\(\displaystyle \int \left (a \csc ^2(x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \sec \left (x+\frac {\pi }{2}\right )^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 4610 |
\(\displaystyle -a \int \left (a \cot ^2(x)+a\right )^{3/2}d\cot (x)\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -a \left (\frac {3}{4} a \int \sqrt {a \cot ^2(x)+a}d\cot (x)+\frac {1}{4} \cot (x) \left (a \cot ^2(x)+a\right )^{3/2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {a \cot ^2(x)+a}}d\cot (x)+\frac {1}{2} \cot (x) \sqrt {a \cot ^2(x)+a}\right )+\frac {1}{4} \cot (x) \left (a \cot ^2(x)+a\right )^{3/2}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {a \cot ^2(x)}{a \cot ^2(x)+a}}d\frac {\cot (x)}{\sqrt {a \cot ^2(x)+a}}+\frac {1}{2} \cot (x) \sqrt {a \cot ^2(x)+a}\right )+\frac {1}{4} \cot (x) \left (a \cot ^2(x)+a\right )^{3/2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -a \left (\frac {3}{4} a \left (\frac {1}{2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cot (x)}{\sqrt {a \cot ^2(x)+a}}\right )+\frac {1}{2} \cot (x) \sqrt {a \cot ^2(x)+a}\right )+\frac {1}{4} \cot (x) \left (a \cot ^2(x)+a\right )^{3/2}\right )\) |
Input:
Int[(a*Csc[x]^2)^(5/2),x]
Output:
-(a*((Cot[x]*(a + a*Cot[x]^2)^(3/2))/4 + (3*a*((Sqrt[a]*ArcTanh[(Sqrt[a]*C ot[x])/Sqrt[a + a*Cot[x]^2]])/2 + (Cot[x]*Sqrt[a + a*Cot[x]^2])/2))/4))
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFac tors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] && !IntegerQ[p]
Time = 0.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.62
method | result | size |
default | \(\left (\frac {3 \ln \left (-\cot \left (x \right )+\csc \left (x \right )\right ) \sin \left (x \right )}{8}+\frac {3 \cot \left (x \right )^{3}}{8}-\frac {5 \csc \left (x \right )^{2} \cot \left (x \right )}{8}\right ) \sqrt {a \csc \left (x \right )^{2}}\, a^{2}\) | \(40\) |
risch | \(-\frac {i a^{2} \sqrt {-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}\, \left (3 \,{\mathrm e}^{6 i x}-11 \,{\mathrm e}^{4 i x}-11 \,{\mathrm e}^{2 i x}+3\right )}{4 \left ({\mathrm e}^{2 i x}-1\right )^{3}}-\frac {3 a^{2} \sqrt {-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{i x}+1\right ) \sin \left (x \right )}{4}+\frac {3 a^{2} \sqrt {-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{i x}-1\right ) \sin \left (x \right )}{4}\) | \(127\) |
Input:
int((a*csc(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
(3/8*ln(-cot(x)+csc(x))*sin(x)+3/8*cot(x)^3-5/8*csc(x)^2*cot(x))*(a*csc(x) ^2)^(1/2)*a^2
Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.23 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=-\frac {{\left (6 \, a^{2} \cos \left (x\right )^{3} - 10 \, a^{2} \cos \left (x\right ) + 3 \, {\left (a^{2} \cos \left (x\right )^{4} - 2 \, a^{2} \cos \left (x\right )^{2} + a^{2}\right )} \log \left (-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1}\right )\right )} \sqrt {-\frac {a}{\cos \left (x\right )^{2} - 1}}}{16 \, {\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )} \] Input:
integrate((a*csc(x)^2)^(5/2),x, algorithm="fricas")
Output:
-1/16*(6*a^2*cos(x)^3 - 10*a^2*cos(x) + 3*(a^2*cos(x)^4 - 2*a^2*cos(x)^2 + a^2)*log(-(cos(x) - 1)/(cos(x) + 1)))*sqrt(-a/(cos(x)^2 - 1))/((cos(x)^2 - 1)*sin(x))
\[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\int \left (a \csc ^{2}{\left (x \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((a*csc(x)**2)**(5/2),x)
Output:
Integral((a*csc(x)**2)**(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 1113 vs. \(2 (49) = 98\).
Time = 0.30 (sec) , antiderivative size = 1113, normalized size of antiderivative = 17.12 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\text {Too large to display} \] Input:
integrate((a*csc(x)^2)^(5/2),x, algorithm="maxima")
Output:
1/8*(88*a^2*cos(3*x)*sin(2*x) - 24*a^2*cos(x)*sin(2*x) + 24*a^2*cos(2*x)*s in(x) - 6*a^2*sin(x) + 3*(a^2*cos(8*x)^2 + 16*a^2*cos(6*x)^2 + 36*a^2*cos( 4*x)^2 + 16*a^2*cos(2*x)^2 + a^2*sin(8*x)^2 + 16*a^2*sin(6*x)^2 + 36*a^2*s in(4*x)^2 - 48*a^2*sin(4*x)*sin(2*x) + 16*a^2*sin(2*x)^2 - 8*a^2*cos(2*x) + a^2 - 2*(4*a^2*cos(6*x) - 6*a^2*cos(4*x) + 4*a^2*cos(2*x) - a^2)*cos(8*x ) - 8*(6*a^2*cos(4*x) - 4*a^2*cos(2*x) + a^2)*cos(6*x) - 12*(4*a^2*cos(2*x ) - a^2)*cos(4*x) - 4*(2*a^2*sin(6*x) - 3*a^2*sin(4*x) + 2*a^2*sin(2*x))*s in(8*x) - 16*(3*a^2*sin(4*x) - 2*a^2*sin(2*x))*sin(6*x))*arctan2(sin(x), c os(x) + 1) - 3*(a^2*cos(8*x)^2 + 16*a^2*cos(6*x)^2 + 36*a^2*cos(4*x)^2 + 1 6*a^2*cos(2*x)^2 + a^2*sin(8*x)^2 + 16*a^2*sin(6*x)^2 + 36*a^2*sin(4*x)^2 - 48*a^2*sin(4*x)*sin(2*x) + 16*a^2*sin(2*x)^2 - 8*a^2*cos(2*x) + a^2 - 2* (4*a^2*cos(6*x) - 6*a^2*cos(4*x) + 4*a^2*cos(2*x) - a^2)*cos(8*x) - 8*(6*a ^2*cos(4*x) - 4*a^2*cos(2*x) + a^2)*cos(6*x) - 12*(4*a^2*cos(2*x) - a^2)*c os(4*x) - 4*(2*a^2*sin(6*x) - 3*a^2*sin(4*x) + 2*a^2*sin(2*x))*sin(8*x) - 16*(3*a^2*sin(4*x) - 2*a^2*sin(2*x))*sin(6*x))*arctan2(sin(x), cos(x) - 1) - 2*(3*a^2*sin(7*x) - 11*a^2*sin(5*x) - 11*a^2*sin(3*x) + 3*a^2*sin(x))*c os(8*x) - 12*(2*a^2*sin(6*x) - 3*a^2*sin(4*x) + 2*a^2*sin(2*x))*cos(7*x) - 8*(11*a^2*sin(5*x) + 11*a^2*sin(3*x) - 3*a^2*sin(x))*cos(6*x) - 44*(3*a^2 *sin(4*x) - 2*a^2*sin(2*x))*cos(5*x) + 12*(11*a^2*sin(3*x) - 3*a^2*sin(x)) *cos(4*x) + 2*(3*a^2*cos(7*x) - 11*a^2*cos(5*x) - 11*a^2*cos(3*x) + 3*a...
Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (49) = 98\).
Time = 0.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.63 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\frac {1}{64} \, {\left (12 \, \log \left (-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1}\right ) \mathrm {sgn}\left (\sin \left (x\right )\right ) + \frac {{\left (\frac {8 \, {\left (\cos \left (x\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (x\right )\right )}{\cos \left (x\right ) + 1} - \frac {18 \, {\left (\cos \left (x\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (x\right )\right )}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \mathrm {sgn}\left (\sin \left (x\right )\right )\right )} {\left (\cos \left (x\right ) + 1\right )}^{2}}{{\left (\cos \left (x\right ) - 1\right )}^{2}} - \frac {8 \, {\left (\cos \left (x\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (x\right )\right )}{\cos \left (x\right ) + 1} + \frac {{\left (\cos \left (x\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (x\right )\right )}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )} a^{\frac {5}{2}} \] Input:
integrate((a*csc(x)^2)^(5/2),x, algorithm="giac")
Output:
1/64*(12*log(-(cos(x) - 1)/(cos(x) + 1))*sgn(sin(x)) + (8*(cos(x) - 1)*sgn (sin(x))/(cos(x) + 1) - 18*(cos(x) - 1)^2*sgn(sin(x))/(cos(x) + 1)^2 - sgn (sin(x)))*(cos(x) + 1)^2/(cos(x) - 1)^2 - 8*(cos(x) - 1)*sgn(sin(x))/(cos( x) + 1) + (cos(x) - 1)^2*sgn(sin(x))/(cos(x) + 1)^2)*a^(5/2)
Timed out. \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\int {\left (\frac {a}{{\sin \left (x\right )}^2}\right )}^{5/2} \,d x \] Input:
int((a/sin(x)^2)^(5/2),x)
Output:
int((a/sin(x)^2)^(5/2), x)
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.54 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\frac {\sqrt {a}\, a^{2} \left (-3 \cos \left (x \right ) \sin \left (x \right )^{2}-2 \cos \left (x \right )+3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{4}\right )}{8 \sin \left (x \right )^{4}} \] Input:
int((a*csc(x)^2)^(5/2),x)
Output:
(sqrt(a)*a**2*( - 3*cos(x)*sin(x)**2 - 2*cos(x) + 3*log(tan(x/2))*sin(x)** 4))/(8*sin(x)**4)