3.866 \(\int \cot (x) \sqrt {-1+\csc ^2(x)} (1-\sin ^2(x))^3 \, dx\)

Optimal. Leaf size=76 \[ -\frac {35}{16} \sqrt {\cot ^2(x)}+\frac {1}{6} \cos ^6(x) \sqrt {\cot ^2(x)}+\frac {7}{24} \cos ^4(x) \sqrt {\cot ^2(x)}+\frac {35}{48} \cos ^2(x) \sqrt {\cot ^2(x)}-\frac {35}{16} x \tan (x) \sqrt {\cot ^2(x)} \]

[Out]

-35/16*(cot(x)^2)^(1/2)+35/48*cos(x)^2*(cot(x)^2)^(1/2)+7/24*cos(x)^4*(cot(x)^2)^(1/2)+1/6*cos(x)^6*(cot(x)^2)
^(1/2)-35/16*x*(cot(x)^2)^(1/2)*tan(x)

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Rubi [A]  time = 0.16, antiderivative size = 84, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3175, 4360, 25, 266, 47, 50, 63, 203} \[ -\frac {35}{16} \sqrt {\csc ^2(x)-1}+\frac {1}{6} \sin ^6(x) \left (\csc ^2(x)-1\right )^{7/2}+\frac {7}{24} \sin ^4(x) \left (\csc ^2(x)-1\right )^{5/2}+\frac {35}{48} \sin ^2(x) \left (\csc ^2(x)-1\right )^{3/2}+\frac {35}{16} \tan ^{-1}\left (\sqrt {\csc ^2(x)-1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]*Sqrt[-1 + Csc[x]^2]*(1 - Sin[x]^2)^3,x]

[Out]

(35*ArcTan[Sqrt[-1 + Csc[x]^2]])/16 - (35*Sqrt[-1 + Csc[x]^2])/16 + (35*(-1 + Csc[x]^2)^(3/2)*Sin[x]^2)/48 + (
7*(-1 + Csc[x]^2)^(5/2)*Sin[x]^4)/24 + ((-1 + Csc[x]^2)^(7/2)*Sin[x]^6)/6

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3175

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

Rule 4360

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist[1/(b
*c), Subst[Int[SubstFor[1/x, Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a
+ b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cot] || EqQ[F, cot])

Rubi steps

\begin {align*} \int \cot (x) \sqrt {-1+\csc ^2(x)} \left (1-\sin ^2(x)\right )^3 \, dx &=\int \cos ^6(x) \cot (x) \sqrt {-1+\csc ^2(x)} \, dx\\ &=\operatorname {Subst}\left (\int \frac {\sqrt {-1+\frac {1}{x^2}} \left (1-x^2\right )^3}{x} \, dx,x,\sin (x)\right )\\ &=\operatorname {Subst}\left (\int \left (-1+\frac {1}{x^2}\right )^{7/2} x^5 \, dx,x,\sin (x)\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {(-1+x)^{7/2}}{x^4} \, dx,x,\csc ^2(x)\right )\right )\\ &=\frac {1}{6} \cot ^2(x)^{7/2} \sin ^6(x)-\frac {7}{12} \operatorname {Subst}\left (\int \frac {(-1+x)^{5/2}}{x^3} \, dx,x,\csc ^2(x)\right )\\ &=\frac {7}{24} \cot ^2(x)^{5/2} \sin ^4(x)+\frac {1}{6} \cot ^2(x)^{7/2} \sin ^6(x)-\frac {35}{48} \operatorname {Subst}\left (\int \frac {(-1+x)^{3/2}}{x^2} \, dx,x,\csc ^2(x)\right )\\ &=\frac {35}{48} \cot ^2(x)^{3/2} \sin ^2(x)+\frac {7}{24} \cot ^2(x)^{5/2} \sin ^4(x)+\frac {1}{6} \cot ^2(x)^{7/2} \sin ^6(x)-\frac {35}{32} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x}}{x} \, dx,x,\csc ^2(x)\right )\\ &=-\frac {35}{16} \sqrt {\cot ^2(x)}+\frac {35}{48} \cot ^2(x)^{3/2} \sin ^2(x)+\frac {7}{24} \cot ^2(x)^{5/2} \sin ^4(x)+\frac {1}{6} \cot ^2(x)^{7/2} \sin ^6(x)+\frac {35}{32} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} x} \, dx,x,\csc ^2(x)\right )\\ &=-\frac {35}{16} \sqrt {\cot ^2(x)}+\frac {35}{48} \cot ^2(x)^{3/2} \sin ^2(x)+\frac {7}{24} \cot ^2(x)^{5/2} \sin ^4(x)+\frac {1}{6} \cot ^2(x)^{7/2} \sin ^6(x)+\frac {35}{16} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\cot ^2(x)}\right )\\ &=\frac {35}{16} \tan ^{-1}\left (\sqrt {\cot ^2(x)}\right )-\frac {35}{16} \sqrt {\cot ^2(x)}+\frac {35}{48} \cot ^2(x)^{3/2} \sin ^2(x)+\frac {7}{24} \cot ^2(x)^{5/2} \sin ^4(x)+\frac {1}{6} \cot ^2(x)^{7/2} \sin ^6(x)\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 40, normalized size = 0.53 \[ \frac {1}{384} \sqrt {\cot ^2(x)} \sec (x) (-840 x \sin (x)-525 \cos (x)+126 \cos (3 x)+14 \cos (5 x)+\cos (7 x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]*Sqrt[-1 + Csc[x]^2]*(1 - Sin[x]^2)^3,x]

[Out]

(Sqrt[Cot[x]^2]*Sec[x]*(-525*Cos[x] + 126*Cos[3*x] + 14*Cos[5*x] + Cos[7*x] - 840*x*Sin[x]))/384

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fricas [A]  time = 1.92, size = 34, normalized size = 0.45 \[ -\frac {8 \, \cos \relax (x)^{7} + 14 \, \cos \relax (x)^{5} + 35 \, \cos \relax (x)^{3} - 105 \, x \sin \relax (x) - 105 \, \cos \relax (x)}{48 \, \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(1-sin(x)^2)^3*(-1+csc(x)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/48*(8*cos(x)^7 + 14*cos(x)^5 + 35*cos(x)^3 - 105*x*sin(x) - 105*cos(x))/sin(x)

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giac [A]  time = 0.15, size = 97, normalized size = 1.28 \[ -\frac {1}{48} \, {\left ({\left (2 \, {\left (4 \, \sin \relax (x)^{2} - 19\right )} \sin \relax (x)^{2} + 87\right )} \sqrt {-\sin \relax (x)^{2} + 1} \sin \relax (x) - 105 \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor - x\right )} \left (-1\right )^{\left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor } + \frac {24 \, {\left (\sqrt {-\sin \relax (x)^{2} + 1} - 1\right )}}{\sin \relax (x)} - \frac {24 \, \sin \relax (x)}{\sqrt {-\sin \relax (x)^{2} + 1} - 1}\right )} \mathrm {sgn}\left (\sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(1-sin(x)^2)^3*(-1+csc(x)^2)^(1/2),x, algorithm="giac")

[Out]

-1/48*((2*(4*sin(x)^2 - 19)*sin(x)^2 + 87)*sqrt(-sin(x)^2 + 1)*sin(x) - 105*(pi*floor(x/pi + 1/2) - x)*(-1)^fl
oor(x/pi + 1/2) + 24*(sqrt(-sin(x)^2 + 1) - 1)/sin(x) - 24*sin(x)/(sqrt(-sin(x)^2 + 1) - 1))*sgn(sin(x))

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maple [A]  time = 0.38, size = 54, normalized size = 0.71 \[ -\frac {\left (-8 \left (\cos ^{7}\relax (x )\right )-14 \left (\cos ^{5}\relax (x )\right )-35 \left (\cos ^{3}\relax (x )\right )+105 x \sin \relax (x )+105 \cos \relax (x )\right ) \sqrt {-\frac {\cos ^{2}\relax (x )}{-1+\cos ^{2}\relax (x )}}\, \sqrt {4}}{96 \cos \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)*(1-sin(x)^2)^3*(-1+csc(x)^2)^(1/2),x)

[Out]

-1/96*(-8*cos(x)^7-14*cos(x)^5-35*cos(x)^3+105*x*sin(x)+105*cos(x))*(-cos(x)^2/(-1+cos(x)^2))^(1/2)/cos(x)*4^(
1/2)

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maxima [B]  time = 0.71, size = 136, normalized size = 1.79 \[ -\frac {3}{2} \, \sqrt {\frac {1}{\sin \relax (x)^{2}} - 1} \sin \relax (x)^{2} - \sqrt {\frac {1}{\sin \relax (x)^{2}} - 1} + \frac {3 \, {\left (\frac {1}{\sin \relax (x)^{2}} - 1\right )}^{\frac {5}{2}} + 8 \, {\left (\frac {1}{\sin \relax (x)^{2}} - 1\right )}^{\frac {3}{2}} - 3 \, \sqrt {\frac {1}{\sin \relax (x)^{2}} - 1}}{48 \, {\left ({\left (\frac {1}{\sin \relax (x)^{2}} - 1\right )}^{3} + 3 \, {\left (\frac {1}{\sin \relax (x)^{2}} - 1\right )}^{2} + \frac {3}{\sin \relax (x)^{2}} - 2\right )}} - \frac {3 \, {\left ({\left (\frac {1}{\sin \relax (x)^{2}} - 1\right )}^{\frac {3}{2}} - \sqrt {\frac {1}{\sin \relax (x)^{2}} - 1}\right )}}{8 \, {\left ({\left (\frac {1}{\sin \relax (x)^{2}} - 1\right )}^{2} + \frac {2}{\sin \relax (x)^{2}} - 1\right )}} + \frac {35}{16} \, \arctan \left (\sqrt {\frac {1}{\sin \relax (x)^{2}} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(1-sin(x)^2)^3*(-1+csc(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-3/2*sqrt(1/sin(x)^2 - 1)*sin(x)^2 - sqrt(1/sin(x)^2 - 1) + 1/48*(3*(1/sin(x)^2 - 1)^(5/2) + 8*(1/sin(x)^2 - 1
)^(3/2) - 3*sqrt(1/sin(x)^2 - 1))/((1/sin(x)^2 - 1)^3 + 3*(1/sin(x)^2 - 1)^2 + 3/sin(x)^2 - 2) - 3/8*((1/sin(x
)^2 - 1)^(3/2) - sqrt(1/sin(x)^2 - 1))/((1/sin(x)^2 - 1)^2 + 2/sin(x)^2 - 1) + 35/16*arctan(sqrt(1/sin(x)^2 -
1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\mathrm {cot}\relax (x)\,\sqrt {\frac {1}{{\sin \relax (x)}^2}-1}\,{\left ({\sin \relax (x)}^2-1\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-cot(x)*(1/sin(x)^2 - 1)^(1/2)*(sin(x)^2 - 1)^3,x)

[Out]

int(-cot(x)*(1/sin(x)^2 - 1)^(1/2)*(sin(x)^2 - 1)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(1-sin(x)**2)**3*(-1+csc(x)**2)**(1/2),x)

[Out]

Timed out

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