3.270 \(\int \cot ^2(x) (a+a \tan ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=33 \[ a \cos (x) \sqrt {a \sec ^2(x)} \tanh ^{-1}(\sin (x))-a \cot (x) \sqrt {a \sec ^2(x)} \]

[Out]

a*arctanh(sin(x))*cos(x)*(a*sec(x)^2)^(1/2)-a*cot(x)*(a*sec(x)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3657, 4125, 2621, 321, 207} \[ a \cos (x) \sqrt {a \sec ^2(x)} \tanh ^{-1}(\sin (x))-a \cot (x) \sqrt {a \sec ^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^2*(a + a*Tan[x]^2)^(3/2),x]

[Out]

a*ArcTanh[Sin[x]]*Cos[x]*Sqrt[a*Sec[x]^2] - a*Cot[x]*Sqrt[a*Sec[x]^2]

Rule 207

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2621

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

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

________________________________________________________________________________________

Mathematica [C]  time = 0.03, size = 27, normalized size = 0.82 \[ -a \cot (x) \sqrt {a \sec ^2(x)} \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\sin ^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^2*(a + a*Tan[x]^2)^(3/2),x]

[Out]

-(a*Cot[x]*Hypergeometric2F1[-1/2, 1, 1/2, Sin[x]^2]*Sqrt[a*Sec[x]^2])

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 53, normalized size = 1.61 \[ \frac {a^{\frac {3}{2}} \log \left (2 \, a \tan \relax (x)^{2} + 2 \, \sqrt {a \tan \relax (x)^{2} + a} \sqrt {a} \tan \relax (x) + a\right ) \tan \relax (x) - 2 \, \sqrt {a \tan \relax (x)^{2} + a} a}{2 \, \tan \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^2*(a+a*tan(x)^2)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.75, size = 62, normalized size = 1.88 \[ -\frac {1}{2} \, {\left (\sqrt {a} \log \left ({\left (\sqrt {a} \tan \relax (x) - \sqrt {a \tan \relax (x)^{2} + a}\right )}^{2}\right ) - \frac {4 \, a^{\frac {3}{2}}}{{\left (\sqrt {a} \tan \relax (x) - \sqrt {a \tan \relax (x)^{2} + a}\right )}^{2} - a}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^2*(a+a*tan(x)^2)^(3/2),x, algorithm="giac")

[Out]

-1/2*(sqrt(a)*log((sqrt(a)*tan(x) - sqrt(a*tan(x)^2 + a))^2) - 4*a^(3/2)/((sqrt(a)*tan(x) - sqrt(a*tan(x)^2 +
a))^2 - a))*a

________________________________________________________________________________________

maple [A]  time = 0.55, size = 55, normalized size = 1.67 \[ -\frac {\left (\ln \left (-\frac {-1+\cos \relax (x )+\sin \relax (x )}{\sin \relax (x )}\right ) \sin \relax (x )-\ln \left (\frac {1-\cos \relax (x )+\sin \relax (x )}{\sin \relax (x )}\right ) \sin \relax (x )+1\right ) \left (\cos ^{3}\relax (x )\right ) \left (\frac {a}{\cos \relax (x )^{2}}\right )^{\frac {3}{2}}}{\sin \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^2*(a+a*tan(x)^2)^(3/2),x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 1.01, size = 134, normalized size = 4.06 \[ -\frac {{\left (4 \, a \cos \relax (x) \sin \left (2 \, x\right ) - 4 \, a \cos \left (2 \, x\right ) \sin \relax (x) - {\left (a \cos \left (2 \, x\right )^{2} + a \sin \left (2 \, x\right )^{2} - 2 \, a \cos \left (2 \, x\right ) + a\right )} \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \sin \relax (x) + 1\right ) + {\left (a \cos \left (2 \, x\right )^{2} + a \sin \left (2 \, x\right )^{2} - 2 \, a \cos \left (2 \, x\right ) + a\right )} \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \sin \relax (x) + 1\right ) + 4 \, a \sin \relax (x)\right )} \sqrt {a}}{2 \, {\left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} - 2 \, \cos \left (2 \, x\right ) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^2*(a+a*tan(x)^2)^(3/2),x, algorithm="maxima")

[Out]

-1/2*(4*a*cos(x)*sin(2*x) - 4*a*cos(2*x)*sin(x) - (a*cos(2*x)^2 + a*sin(2*x)^2 - 2*a*cos(2*x) + a)*log(cos(x)^
2 + sin(x)^2 + 2*sin(x) + 1) + (a*cos(2*x)^2 + a*sin(2*x)^2 - 2*a*cos(2*x) + a)*log(cos(x)^2 + sin(x)^2 - 2*si
n(x) + 1) + 4*a*sin(x))*sqrt(a)/(cos(2*x)^2 + sin(2*x)^2 - 2*cos(2*x) + 1)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\mathrm {cot}\relax (x)}^2\,{\left (a\,{\mathrm {tan}\relax (x)}^2+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^2*(a + a*tan(x)^2)^(3/2),x)

[Out]

int(cot(x)^2*(a + a*tan(x)^2)^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\tan ^{2}{\relax (x )} + 1\right )\right )^{\frac {3}{2}} \cot ^{2}{\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)**2*(a+a*tan(x)**2)**(3/2),x)

[Out]

Integral((a*(tan(x)**2 + 1))**(3/2)*cot(x)**2, x)

________________________________________________________________________________________