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

Optimal. Leaf size=81 \[ \sin (x) \sqrt {\cot ^2(x)}+\frac {1}{7} \sin (x) \cos ^6(x) \sqrt {\cot ^2(x)}+\frac {1}{5} \sin (x) \cos ^4(x) \sqrt {\cot ^2(x)}+\frac {1}{3} \sin (x) \cos ^2(x) \sqrt {\cot ^2(x)}-\tan (x) \sqrt {\cot ^2(x)} \tanh ^{-1}(\cos (x)) \]

[Out]

sin(x)*(cot(x)^2)^(1/2)+1/3*cos(x)^2*sin(x)*(cot(x)^2)^(1/2)+1/5*cos(x)^4*sin(x)*(cot(x)^2)^(1/2)+1/7*cos(x)^6
*sin(x)*(cot(x)^2)^(1/2)-arctanh(cos(x))*(cot(x)^2)^(1/2)*tan(x)

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Rubi [A]  time = 0.16, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3175, 4121, 3658, 2592, 302, 206} \[ \sin (x) \sqrt {\cot ^2(x)}+\frac {1}{7} \sin (x) \cos ^6(x) \sqrt {\cot ^2(x)}+\frac {1}{5} \sin (x) \cos ^4(x) \sqrt {\cot ^2(x)}+\frac {1}{3} \sin (x) \cos ^2(x) \sqrt {\cot ^2(x)}-\tan (x) \sqrt {\cot ^2(x)} \tanh ^{-1}(\cos (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[Cot[x]^2]*Sin[x] + (Cos[x]^2*Sqrt[Cot[x]^2]*Sin[x])/3 + (Cos[x]^4*Sqrt[Cot[x]^2]*Sin[x])/5 + (Cos[x]^6*Sq
rt[Cot[x]^2]*Sin[x])/7 - ArcTanh[Cos[x]]*Sqrt[Cot[x]^2]*Tan[x]

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 4121

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

Rubi steps

\begin {align*} \int \cos (x) \sqrt {-1+\csc ^2(x)} \left (1-\sin ^2(x)\right )^3 \, dx &=\int \cos ^7(x) \sqrt {-1+\csc ^2(x)} \, dx\\ &=\int \cos ^7(x) \sqrt {\cot ^2(x)} \, dx\\ &=\left (\sqrt {\cot ^2(x)} \tan (x)\right ) \int \cos ^7(x) \cot (x) \, dx\\ &=-\left (\left (\sqrt {\cot ^2(x)} \tan (x)\right ) \operatorname {Subst}\left (\int \frac {x^8}{1-x^2} \, dx,x,\cos (x)\right )\right )\\ &=-\left (\left (\sqrt {\cot ^2(x)} \tan (x)\right ) \operatorname {Subst}\left (\int \left (-1-x^2-x^4-x^6+\frac {1}{1-x^2}\right ) \, dx,x,\cos (x)\right )\right )\\ &=\sqrt {\cot ^2(x)} \sin (x)+\frac {1}{3} \cos ^2(x) \sqrt {\cot ^2(x)} \sin (x)+\frac {1}{5} \cos ^4(x) \sqrt {\cot ^2(x)} \sin (x)+\frac {1}{7} \cos ^6(x) \sqrt {\cot ^2(x)} \sin (x)-\left (\sqrt {\cot ^2(x)} \tan (x)\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (x)\right )\\ &=\sqrt {\cot ^2(x)} \sin (x)+\frac {1}{3} \cos ^2(x) \sqrt {\cot ^2(x)} \sin (x)+\frac {1}{5} \cos ^4(x) \sqrt {\cot ^2(x)} \sin (x)+\frac {1}{7} \cos ^6(x) \sqrt {\cot ^2(x)} \sin (x)-\tanh ^{-1}(\cos (x)) \sqrt {\cot ^2(x)} \tan (x)\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 55, normalized size = 0.68 \[ \frac {\tan (x) \sqrt {\cot ^2(x)} \left (9765 \cos (x)+1295 \cos (3 x)+189 \cos (5 x)+15 \cos (7 x)+6720 \log \left (\sin \left (\frac {x}{2}\right )\right )-6720 \log \left (\cos \left (\frac {x}{2}\right )\right )\right )}{6720} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cot[x]^2]*(9765*Cos[x] + 1295*Cos[3*x] + 189*Cos[5*x] + 15*Cos[7*x] - 6720*Log[Cos[x/2]] + 6720*Log[Sin[
x/2]])*Tan[x])/6720

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fricas [A]  time = 0.93, size = 41, normalized size = 0.51 \[ -\frac {1}{7} \, \cos \relax (x)^{7} - \frac {1}{5} \, \cos \relax (x)^{5} - \frac {1}{3} \, \cos \relax (x)^{3} - \cos \relax (x) + \frac {1}{2} \, \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - \frac {1}{2} \, \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*cos(x)^7 - 1/5*cos(x)^5 - 1/3*cos(x)^3 - cos(x) + 1/2*log(1/2*cos(x) + 1/2) - 1/2*log(-1/2*cos(x) + 1/2)

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giac [A]  time = 0.15, size = 44, normalized size = 0.54 \[ \frac {1}{210} \, {\left (30 \, \cos \relax (x)^{7} + 42 \, \cos \relax (x)^{5} + 70 \, \cos \relax (x)^{3} + 210 \, \cos \relax (x) - 105 \, \log \left (\cos \relax (x) + 1\right ) + 105 \, \log \left (-\cos \relax (x) + 1\right )\right )} \mathrm {sgn}\left (\sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(30*cos(x)^7 + 42*cos(x)^5 + 70*cos(x)^3 + 210*cos(x) - 105*log(cos(x) + 1) + 105*log(-cos(x) + 1))*sgn(
sin(x))

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maple [A]  time = 0.27, size = 65, normalized size = 0.80 \[ \frac {\left (15 \left (\cos ^{7}\relax (x )\right )+21 \left (\cos ^{5}\relax (x )\right )+35 \left (\cos ^{3}\relax (x )\right )+105 \cos \relax (x )+105 \ln \left (-\frac {-1+\cos \relax (x )}{\sin \relax (x )}\right )+176\right ) \sin \relax (x ) \sqrt {-\frac {\cos ^{2}\relax (x )}{-1+\cos ^{2}\relax (x )}}\, \sqrt {4}}{210 \cos \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(15*cos(x)^7+21*cos(x)^5+35*cos(x)^3+105*cos(x)+105*ln(-(-1+cos(x))/sin(x))+176)*sin(x)*(-cos(x)^2/(-1+c
os(x)^2))^(1/2)/cos(x)*4^(1/2)

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maxima [A]  time = 0.93, size = 86, normalized size = 1.06 \[ \frac {1}{7} \, {\left (\frac {1}{\sin \relax (x)^{2}} - 1\right )}^{\frac {7}{2}} \sin \relax (x)^{7} + \frac {1}{5} \, {\left (\frac {1}{\sin \relax (x)^{2}} - 1\right )}^{\frac {5}{2}} \sin \relax (x)^{5} + \frac {1}{3} \, {\left (\frac {1}{\sin \relax (x)^{2}} - 1\right )}^{\frac {3}{2}} \sin \relax (x)^{3} + \sqrt {\frac {1}{\sin \relax (x)^{2}} - 1} \sin \relax (x) - \frac {1}{2} \, \log \left (\sqrt {\frac {1}{\sin \relax (x)^{2}} - 1} \sin \relax (x) + 1\right ) + \frac {1}{2} \, \log \left (\sqrt {\frac {1}{\sin \relax (x)^{2}} - 1} \sin \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*(1/sin(x)^2 - 1)^(7/2)*sin(x)^7 + 1/5*(1/sin(x)^2 - 1)^(5/2)*sin(x)^5 + 1/3*(1/sin(x)^2 - 1)^(3/2)*sin(x)^
3 + sqrt(1/sin(x)^2 - 1)*sin(x) - 1/2*log(sqrt(1/sin(x)^2 - 1)*sin(x) + 1) + 1/2*log(sqrt(1/sin(x)^2 - 1)*sin(
x) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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