\(\int (a \csc ^2(x))^{5/2} \, dx\) [48]

Optimal result

Integrand size = 10, antiderivative size = 65 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=-\frac {3}{8} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cot (x)}{\sqrt {a \csc ^2(x)}}\right )-\frac {3}{8} a^2 \cot (x) \sqrt {a \csc ^2(x)}-\frac {1}{4} a \cot (x) \left (a \csc ^2(x)\right )^{3/2} \]

[Out]

-3/8*a^(5/2)*arctanh(cot(x)*a^(1/2)/(a*csc(x)^2)^(1/2))-1/4*a*cot(x)*(a*csc(x)^2)^(3/2)-3/8*a^2*cot(x)*(a*csc(
x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4207, 201, 223, 212} \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=-\frac {3}{8} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cot (x)}{\sqrt {a \csc ^2(x)}}\right )-\frac {3}{8} a^2 \cot (x) \sqrt {a \csc ^2(x)}-\frac {1}{4} a \cot (x) \left (a \csc ^2(x)\right )^{3/2} \]

[In]

Int[(a*Csc[x]^2)^(5/2),x]

[Out]

(-3*a^(5/2)*ArcTanh[(Sqrt[a]*Cot[x])/Sqrt[a*Csc[x]^2]])/8 - (3*a^2*Cot[x]*Sqrt[a*Csc[x]^2])/8 - (a*Cot[x]*(a*C
sc[x]^2)^(3/2))/4

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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]

Rule 4207

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[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
]

Rubi steps \begin{align*} \text {integral}& = -\left (a \text {Subst}\left (\int \left (a+a x^2\right )^{3/2} \, dx,x,\cot (x)\right )\right ) \\ & = -\frac {1}{4} a \cot (x) \left (a \csc ^2(x)\right )^{3/2}-\frac {1}{4} \left (3 a^2\right ) \text {Subst}\left (\int \sqrt {a+a x^2} \, dx,x,\cot (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}-\frac {1}{8} \left (3 a^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+a x^2}} \, dx,x,\cot (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}-\frac {1}{8} \left (3 a^3\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\cot (x)}{\sqrt {a \csc ^2(x)}}\right ) \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\frac {1}{64} \left (a \csc ^2(x)\right )^{5/2} \sin (x) \left (-22 \cos (x)+6 \left (\cos (3 x)+4 \left (-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )\right ) \sin ^4(x)\right )\right ) \]

[In]

Integrate[(a*Csc[x]^2)^(5/2),x]

[Out]

((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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.68

method	result	size
default	\(\frac {a^{2} \sqrt {a \csc \left (x \right )^{2}}\, \left (3 \sin \left (x \right ) \ln \left (\csc \left (x \right )-\cot \left (x \right )\right )+3 \cot \left (x \right )^{3}-5 \csc \left (x \right )^{2} \cot \left (x \right )\right ) \sqrt {4}}{16}\)	\(44\)
risch	\(-\frac {i a^{2} \sqrt {-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}\, \left (3 \,{\mathrm e}^{6 i x}-11 \,{\mathrm e}^{4 i x}-11 \,{\mathrm e}^{2 i x}+3\right )}{4 \left ({\mathrm e}^{2 i x}-1\right )^{3}}-\frac {3 a^{2} \sqrt {-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{i x}+1\right ) \sin \left (x \right )}{4}+\frac {3 a^{2} \sqrt {-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{i x}-1\right ) \sin \left (x \right )}{4}\)	\(127\)

[In]

int((a*csc(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/16*a^2*(a*csc(x)^2)^(1/2)*(3*sin(x)*ln(csc(x)-cot(x))+3*cot(x)^3-5*csc(x)^2*cot(x))*4^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.23 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=-\frac {{\left (6 \, a^{2} \cos \left (x\right )^{3} - 10 \, a^{2} \cos \left (x\right ) + 3 \, {\left (a^{2} \cos \left (x\right )^{4} - 2 \, a^{2} \cos \left (x\right )^{2} + a^{2}\right )} \log \left (-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1}\right )\right )} \sqrt {-\frac {a}{\cos \left (x\right )^{2} - 1}}}{16 \, {\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )} \]

[In]

integrate((a*csc(x)^2)^(5/2),x, algorithm="fricas")

[Out]

-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))

Sympy [F]

\[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\int \left (a \csc ^{2}{\left (x \right )}\right )^{\frac {5}{2}}\, dx \]

[In]

integrate((a*csc(x)**2)**(5/2),x)

[Out]

Integral((a*csc(x)**2)**(5/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1113 vs. \(2 (49) = 98\).

Time = 0.46 (sec) , antiderivative size = 1113, normalized size of antiderivative = 17.12 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\text {Too large to display} \]

[In]

integrate((a*csc(x)^2)^(5/2),x, algorithm="maxima")

[Out]

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)*sin(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
*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)*cos(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) - 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*sin(4*x)^2 - 48*a^2*sin(4*x)*sin(2*x) + 1
6*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))*sin(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) - 2*(3*a^2*sin(7*x) - 11*a^2*sin(5*x) - 11*a^2*sin(3*x) + 3*a^2*sin(x))*cos(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^2*cos(x))*sin(8*x) + 6*(4*a^2*cos(6*x) - 6*a^2*cos(4*x) + 4*a^
2*cos(2*x) - a^2)*sin(7*x) + 8*(11*a^2*cos(5*x) + 11*a^2*cos(3*x) - 3*a^2*cos(x))*sin(6*x) + 22*(6*a^2*cos(4*x
) - 4*a^2*cos(2*x) + a^2)*sin(5*x) - 12*(11*a^2*cos(3*x) - 3*a^2*cos(x))*sin(4*x) - 22*(4*a^2*cos(2*x) - a^2)*
sin(3*x))*sqrt(-a)/(2*(4*cos(6*x) - 6*cos(4*x) + 4*cos(2*x) - 1)*cos(8*x) - cos(8*x)^2 + 8*(6*cos(4*x) - 4*cos
(2*x) + 1)*cos(6*x) - 16*cos(6*x)^2 + 12*(4*cos(2*x) - 1)*cos(4*x) - 36*cos(4*x)^2 - 16*cos(2*x)^2 + 4*(2*sin(
6*x) - 3*sin(4*x) + 2*sin(2*x))*sin(8*x) - sin(8*x)^2 + 16*(3*sin(4*x) - 2*sin(2*x))*sin(6*x) - 16*sin(6*x)^2
- 36*sin(4*x)^2 + 48*sin(4*x)*sin(2*x) - 16*sin(2*x)^2 + 8*cos(2*x) - 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (49) = 98\).

Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.91 \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\frac {1}{64} \, {\left (12 \, a^{2} \log \left (-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1}\right ) \mathrm {sgn}\left (\sin \left (x\right )\right ) - \frac {8 \, a^{2} {\left (\cos \left (x\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (x\right )\right )}{\cos \left (x\right ) + 1} + \frac {a^{2} {\left (\cos \left (x\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (x\right )\right )}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {{\left (a^{2} \mathrm {sgn}\left (\sin \left (x\right )\right ) - \frac {8 \, a^{2} {\left (\cos \left (x\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (x\right )\right )}{\cos \left (x\right ) + 1} + \frac {18 \, a^{2} {\left (\cos \left (x\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (x\right )\right )}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (x\right ) + 1\right )}^{2}}{{\left (\cos \left (x\right ) - 1\right )}^{2}}\right )} \sqrt {a} \]

[In]

integrate((a*csc(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/64*(12*a^2*log(-(cos(x) - 1)/(cos(x) + 1))*sgn(sin(x)) - 8*a^2*(cos(x) - 1)*sgn(sin(x))/(cos(x) + 1) + a^2*(
cos(x) - 1)^2*sgn(sin(x))/(cos(x) + 1)^2 - (a^2*sgn(sin(x)) - 8*a^2*(cos(x) - 1)*sgn(sin(x))/(cos(x) + 1) + 18
*a^2*(cos(x) - 1)^2*sgn(sin(x))/(cos(x) + 1)^2)*(cos(x) + 1)^2/(cos(x) - 1)^2)*sqrt(a)

Mupad [F(-1)]

Timed out. \[ \int \left (a \csc ^2(x)\right )^{5/2} \, dx=\int {\left (\frac {a}{{\sin \left (x\right )}^2}\right )}^{5/2} \,d x \]

[In]

int((a/sin(x)^2)^(5/2),x)

[Out]

int((a/sin(x)^2)^(5/2), x)