3.264 \(\int \cot ^4(x) \sqrt {a+a \tan ^2(x)} \, dx\)

Optimal. Leaf size=34 \[ \cot (x) \sqrt {a \sec ^2(x)}-\frac {1}{3} \cot (x) \csc ^2(x) \sqrt {a \sec ^2(x)} \]

[Out]

cot(x)*(a*sec(x)^2)^(1/2)-1/3*cot(x)*csc(x)^2*(a*sec(x)^2)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3657, 4125, 2606} \[ \cot (x) \sqrt {a \sec ^2(x)}-\frac {1}{3} \cot (x) \csc ^2(x) \sqrt {a \sec ^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^4*Sqrt[a + a*Tan[x]^2],x]

[Out]

Cot[x]*Sqrt[a*Sec[x]^2] - (Cot[x]*Csc[x]^2*Sqrt[a*Sec[x]^2])/3

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3657

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

Rule 4125

Int[(u_.)*((b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sec[e + f*x]^n)^FracPart[p])/(Sec[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sec[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
]])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.65 \[ -\frac {1}{3} \cot (x) \left (\csc ^2(x)-3\right ) \sqrt {a \sec ^2(x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^4*Sqrt[a + a*Tan[x]^2],x]

[Out]

-1/3*(Cot[x]*(-3 + Csc[x]^2)*Sqrt[a*Sec[x]^2])

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fricas [A]  time = 0.42, size = 24, normalized size = 0.71 \[ \frac {\sqrt {a \tan \relax (x)^{2} + a} {\left (2 \, \tan \relax (x)^{2} - 1\right )}}{3 \, \tan \relax (x)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^4*(a+a*tan(x)^2)^(1/2),x, algorithm="fricas")

[Out]

1/3*sqrt(a*tan(x)^2 + a)*(2*tan(x)^2 - 1)/tan(x)^3

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giac [B]  time = 0.27, size = 59, normalized size = 1.74 \[ \frac {4 \, {\left (3 \, {\left (\sqrt {a} \tan \relax (x) - \sqrt {a \tan \relax (x)^{2} + a}\right )}^{2} - a\right )} a^{\frac {5}{2}}}{3 \, {\left ({\left (\sqrt {a} \tan \relax (x) - \sqrt {a \tan \relax (x)^{2} + a}\right )}^{2} - a\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^4*(a+a*tan(x)^2)^(1/2),x, algorithm="giac")

[Out]

4/3*(3*(sqrt(a)*tan(x) - sqrt(a*tan(x)^2 + a))^2 - a)*a^(5/2)/((sqrt(a)*tan(x) - sqrt(a*tan(x)^2 + a))^2 - a)^
3

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maple [A]  time = 0.64, size = 25, normalized size = 0.74 \[ -\frac {\left (3 \left (\cos ^{2}\relax (x )\right )-2\right ) \cos \relax (x ) \sqrt {\frac {a}{\cos \relax (x )^{2}}}}{3 \sin \relax (x )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^4*(a+a*tan(x)^2)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.97, size = 29, normalized size = 0.85 \[ \frac {{\left (2 \, \sqrt {a} \tan \relax (x)^{2} - \sqrt {a}\right )} \sqrt {\tan \relax (x)^{2} + 1}}{3 \, \tan \relax (x)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^4*(a+a*tan(x)^2)^(1/2),x, algorithm="maxima")

[Out]

1/3*(2*sqrt(a)*tan(x)^2 - sqrt(a))*sqrt(tan(x)^2 + 1)/tan(x)^3

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mupad [B]  time = 11.85, size = 40, normalized size = 1.18 \[ \frac {\sqrt {2}\,\sqrt {a}\,\left (2\,\sin \left (2\,x\right )-6\,\sin \left (2\,x\right )\,\left (2\,{\cos \relax (x)}^2-1\right )\right )}{24\,\sqrt {2\,{\cos \relax (x)}^2}\,{\left ({\cos \relax (x)}^2-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^4*(a + a*tan(x)^2)^(1/2),x)

[Out]

(2^(1/2)*a^(1/2)*(2*sin(2*x) - 6*sin(2*x)*(2*cos(x)^2 - 1)))/(24*(2*cos(x)^2)^(1/2)*(cos(x)^2 - 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\tan ^{2}{\relax (x )} + 1\right )} \cot ^{4}{\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)**4*(a+a*tan(x)**2)**(1/2),x)

[Out]

Integral(sqrt(a*(tan(x)**2 + 1))*cot(x)**4, x)

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