3.48 \(\int (a \csc ^2(x))^{5/2} \, dx\)

Optimal. Leaf size=65 \[ -\frac {3}{8} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cot (x)}{\sqrt {a \csc ^2(x)}}\right )-\frac {3}{8} a^2 \cot (x) \sqrt {a \csc ^2(x)}-\frac {1}{4} a \cot (x) \left (a \csc ^2(x)\right )^{3/2} \]

[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)

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Rubi [A]  time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4122, 195, 217, 206} \[ -\frac {3}{8} a^2 \cot (x) \sqrt {a \csc ^2(x)}-\frac {3}{8} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cot (x)}{\sqrt {a \csc ^2(x)}}\right )-\frac {1}{4} a \cot (x) \left (a \csc ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[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 195

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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 4122

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*} \int \left (a \csc ^2(x)\right )^{5/2} \, dx &=-\left (a \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \tanh ^{-1}\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*}

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Mathematica [A]  time = 0.21, size = 51, normalized size = 0.78 \[ \frac {1}{64} \sin (x) \left (a \csc ^2(x)\right )^{5/2} \left (6 \left (\cos (3 x)+4 \sin ^4(x) \left (\log \left (\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )\right )\right )\right )-22 \cos (x)\right ) \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.66, size = 80, normalized size = 1.23 \[ -\frac {{\left (6 \, a^{2} \cos \relax (x)^{3} - 10 \, a^{2} \cos \relax (x) + 3 \, {\left (a^{2} \cos \relax (x)^{4} - 2 \, a^{2} \cos \relax (x)^{2} + a^{2}\right )} \log \left (-\frac {\cos \relax (x) - 1}{\cos \relax (x) + 1}\right )\right )} \sqrt {-\frac {a}{\cos \relax (x)^{2} - 1}}}{16 \, {\left (\cos \relax (x)^{2} - 1\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [B]  time = 0.29, size = 124, normalized size = 1.91 \[ \frac {1}{64} \, {\left (12 \, a^{2} \log \left (-\frac {\cos \relax (x) - 1}{\cos \relax (x) + 1}\right ) \mathrm {sgn}\left (\sin \relax (x)\right ) - \frac {8 \, a^{2} {\left (\cos \relax (x) - 1\right )} \mathrm {sgn}\left (\sin \relax (x)\right )}{\cos \relax (x) + 1} + \frac {a^{2} {\left (\cos \relax (x) - 1\right )}^{2} \mathrm {sgn}\left (\sin \relax (x)\right )}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {{\left (a^{2} \mathrm {sgn}\left (\sin \relax (x)\right ) - \frac {8 \, a^{2} {\left (\cos \relax (x) - 1\right )} \mathrm {sgn}\left (\sin \relax (x)\right )}{\cos \relax (x) + 1} + \frac {18 \, a^{2} {\left (\cos \relax (x) - 1\right )}^{2} \mathrm {sgn}\left (\sin \relax (x)\right )}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )} {\left (\cos \relax (x) + 1\right )}^{2}}{{\left (\cos \relax (x) - 1\right )}^{2}}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.44, size = 79, normalized size = 1.22 \[ \frac {\left (3 \left (\cos ^{4}\relax (x )\right ) \ln \left (-\frac {-1+\cos \relax (x )}{\sin \relax (x )}\right )+3 \left (\cos ^{3}\relax (x )\right )-6 \left (\cos ^{2}\relax (x )\right ) \ln \left (-\frac {-1+\cos \relax (x )}{\sin \relax (x )}\right )-5 \cos \relax (x )+3 \ln \left (-\frac {-1+\cos \relax (x )}{\sin \relax (x )}\right )\right ) \sin \relax (x ) \left (-\frac {a}{-1+\cos ^{2}\relax (x )}\right )^{\frac {5}{2}} \sqrt {4}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*cos(x)^4*ln(-(-1+cos(x))/sin(x))+3*cos(x)^3-6*cos(x)^2*ln(-(-1+cos(x))/sin(x))-5*cos(x)+3*ln(-(-1+cos(
x))/sin(x)))*sin(x)*(-1/(-1+cos(x)^2)*a)^(5/2)*4^(1/2)

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maxima [B]  time = 0.87, size = 1113, normalized size = 17.12 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (\frac {a}{{\sin \relax (x)}^2}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a/sin(x)^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \csc ^{2}{\relax (x )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*csc(x)**2)**(5/2), x)

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