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

Optimal. Leaf size=92 \[ \frac {(a-2 b) \tan (x) \sqrt {a+b \cot ^2(x)}}{a^2 (a-b)}+\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \cot (x)}{\sqrt {a+b \cot ^2(x)}}\right )}{(a-b)^{3/2}}+\frac {b \tan (x)}{a (a-b) \sqrt {a+b \cot ^2(x)}} \]

[Out]

arctan(cot(x)*(a-b)^(1/2)/(a+b*cot(x)^2)^(1/2))/(a-b)^(3/2)+b*tan(x)/a/(a-b)/(a+b*cot(x)^2)^(1/2)+(a-2*b)*(a+b
*cot(x)^2)^(1/2)*tan(x)/a^2/(a-b)

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Rubi [A]  time = 0.15, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3670, 472, 583, 12, 377, 203} \[ \frac {(a-2 b) \tan (x) \sqrt {a+b \cot ^2(x)}}{a^2 (a-b)}+\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \cot (x)}{\sqrt {a+b \cot ^2(x)}}\right )}{(a-b)^{3/2}}+\frac {b \tan (x)}{a (a-b) \sqrt {a+b \cot ^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[a - b]*Cot[x])/Sqrt[a + b*Cot[x]^2]]/(a - b)^(3/2) + (b*Tan[x])/(a*(a - b)*Sqrt[a + b*Cot[x]^2])
+ ((a - 2*b)*Sqrt[a + b*Cot[x]^2]*Tan[x])/(a^2*(a - b))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

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

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Mathematica [C]  time = 6.90, size = 674, normalized size = 7.33 \[ \frac {\sin ^2(x) \tan (x) \left (\frac {8 b^2 (a-b) \cos ^2(x) \cot ^4(x) \, _3F_2\left (2,2,2;1,\frac {7}{2};\frac {(a-b) \cos ^2(x)}{a}\right )}{15 a^3}+\frac {16 b (a-b) \cos ^2(x) \cot ^2(x) \, _3F_2\left (2,2,2;1,\frac {7}{2};\frac {(a-b) \cos ^2(x)}{a}\right )}{15 a^2}+\frac {8 (a-b) \cos ^2(x) \, _3F_2\left (2,2,2;1,\frac {7}{2};\frac {(a-b) \cos ^2(x)}{a}\right )}{15 a}+\frac {8 b^2 (a-b) \cos ^2(x) \cot ^4(x) \, _2F_1\left (2,2;\frac {7}{2};\frac {(a-b) \cos ^2(x)}{a}\right )}{5 a^3}-\frac {8 b^2 \cot ^4(x) \sin ^{-1}\left (\sqrt {\frac {(a-b) \cos ^2(x)}{a}}\right )}{a^2 \left (\frac {(a-b) \cos ^2(x)}{a}\right )^{3/2} \sqrt {\frac {\sin ^2(x) \left (a+b \cot ^2(x)\right )}{a}}}+\frac {8 b^2 \cot ^4(x) \sin ^{-1}\left (\sqrt {\frac {(a-b) \cos ^2(x)}{a}}\right )}{a^2 \sqrt {\frac {(a-b) \sin ^2(x) \cos ^2(x) \left (a+b \cot ^2(x)\right )}{a^2}}}+\frac {8 b (a-b) \cos ^2(x) \cot ^2(x) \, _2F_1\left (2,2;\frac {7}{2};\frac {(a-b) \cos ^2(x)}{a}\right )}{3 a^2}+\frac {12 b \cot ^2(x) \sin ^{-1}\left (\sqrt {\frac {(a-b) \cos ^2(x)}{a}}\right )}{a \sqrt {\frac {(a-b) \sin ^2(x) \cos ^2(x) \left (a+b \cot ^2(x)\right )}{a^2}}}+\frac {3 \sin ^{-1}\left (\sqrt {\frac {(a-b) \cos ^2(x)}{a}}\right )}{\sqrt {\frac {(a-b) \sin ^2(x) \cos ^2(x) \left (a+b \cot ^2(x)\right )}{a^2}}}+\frac {8 b^2 \cot ^2(x) \csc ^2(x)}{a (a-b)}+\frac {16 (a-b) \cos ^2(x) \, _2F_1\left (2,2;\frac {7}{2};\frac {(a-b) \cos ^2(x)}{a}\right )}{15 a}+\frac {12 b \csc ^2(x)}{a-b}+\frac {3 a \sec ^2(x)}{a-b}-\frac {12 b \cot ^2(x) \sin ^{-1}\left (\sqrt {\frac {(a-b) \cos ^2(x)}{a}}\right )}{a \left (\frac {(a-b) \cos ^2(x)}{a}\right )^{3/2} \sqrt {\frac {\sin ^2(x) \left (a+b \cot ^2(x)\right )}{a}}}-\frac {3 \sin ^{-1}\left (\sqrt {\frac {(a-b) \cos ^2(x)}{a}}\right )}{\left (\frac {(a-b) \cos ^2(x)}{a}\right )^{3/2} \sqrt {\frac {\sin ^2(x) \left (a+b \cot ^2(x)\right )}{a}}}\right )}{a \sqrt {a+b \cot ^2(x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sin[x]^2*((12*b*Csc[x]^2)/(a - b) + (8*b^2*Cot[x]^2*Csc[x]^2)/(a*(a - b)) + (16*(a - b)*Cos[x]^2*Hypergeometr
ic2F1[2, 2, 7/2, ((a - b)*Cos[x]^2)/a])/(15*a) + (8*(a - b)*b*Cos[x]^2*Cot[x]^2*Hypergeometric2F1[2, 2, 7/2, (
(a - b)*Cos[x]^2)/a])/(3*a^2) + (8*(a - b)*b^2*Cos[x]^2*Cot[x]^4*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Cos[x]^
2)/a])/(5*a^3) + (8*(a - b)*Cos[x]^2*HypergeometricPFQ[{2, 2, 2}, {1, 7/2}, ((a - b)*Cos[x]^2)/a])/(15*a) + (1
6*(a - b)*b*Cos[x]^2*Cot[x]^2*HypergeometricPFQ[{2, 2, 2}, {1, 7/2}, ((a - b)*Cos[x]^2)/a])/(15*a^2) + (8*(a -
b)*b^2*Cos[x]^2*Cot[x]^4*HypergeometricPFQ[{2, 2, 2}, {1, 7/2}, ((a - b)*Cos[x]^2)/a])/(15*a^3) + (3*a*Sec[x]
^2)/(a - b) - (3*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]])/((((a - b)*Cos[x]^2)/a)^(3/2)*Sqrt[((a + b*Cot[x]^2)*Sin[
x]^2)/a]) - (12*b*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^2)/(a*(((a - b)*Cos[x]^2)/a)^(3/2)*Sqrt[((a + b*Co
t[x]^2)*Sin[x]^2)/a]) - (8*b^2*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^4)/(a^2*(((a - b)*Cos[x]^2)/a)^(3/2)*
Sqrt[((a + b*Cot[x]^2)*Sin[x]^2)/a]) + (3*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]])/Sqrt[((a - b)*Cos[x]^2*(a + b*Co
t[x]^2)*Sin[x]^2)/a^2] + (12*b*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^2)/(a*Sqrt[((a - b)*Cos[x]^2*(a + b*C
ot[x]^2)*Sin[x]^2)/a^2]) + (8*b^2*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^4)/(a^2*Sqrt[((a - b)*Cos[x]^2*(a
+ b*Cot[x]^2)*Sin[x]^2)/a^2]))*Tan[x])/(a*Sqrt[a + b*Cot[x]^2])

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fricas [B]  time = 0.90, size = 393, normalized size = 4.27 \[ \left [\frac {{\left (a^{3} \tan \relax (x)^{2} + a^{2} b\right )} \sqrt {-a + b} \log \left (-\frac {a^{2} \tan \relax (x)^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \relax (x)^{2} + a^{2} - 8 \, a b + 8 \, b^{2} - 4 \, {\left (a \tan \relax (x)^{3} - {\left (a - 2 \, b\right )} \tan \relax (x)\right )} \sqrt {-a + b} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}}}{\tan \relax (x)^{4} + 2 \, \tan \relax (x)^{2} + 1}\right ) + 4 \, {\left ({\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \tan \relax (x)^{3} + {\left (a^{2} b - 3 \, a b^{2} + 2 \, b^{3}\right )} \tan \relax (x)\right )} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}}}{4 \, {\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3} + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} \tan \relax (x)^{2}\right )}}, \frac {{\left (a^{3} \tan \relax (x)^{2} + a^{2} b\right )} \sqrt {a - b} \arctan \left (\frac {2 \, \sqrt {a - b} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}} \tan \relax (x)}{a \tan \relax (x)^{2} - a + 2 \, b}\right ) + 2 \, {\left ({\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \tan \relax (x)^{3} + {\left (a^{2} b - 3 \, a b^{2} + 2 \, b^{3}\right )} \tan \relax (x)\right )} \sqrt {\frac {a \tan \relax (x)^{2} + b}{\tan \relax (x)^{2}}}}{2 \, {\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3} + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} \tan \relax (x)^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((a^3*tan(x)^2 + a^2*b)*sqrt(-a + b)*log(-(a^2*tan(x)^4 - 2*(3*a^2 - 4*a*b)*tan(x)^2 + a^2 - 8*a*b + 8*b^
2 - 4*(a*tan(x)^3 - (a - 2*b)*tan(x))*sqrt(-a + b)*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(tan(x)^4 + 2*tan(x)^2 + 1
)) + 4*((a^3 - 2*a^2*b + a*b^2)*tan(x)^3 + (a^2*b - 3*a*b^2 + 2*b^3)*tan(x))*sqrt((a*tan(x)^2 + b)/tan(x)^2))/
(a^4*b - 2*a^3*b^2 + a^2*b^3 + (a^5 - 2*a^4*b + a^3*b^2)*tan(x)^2), 1/2*((a^3*tan(x)^2 + a^2*b)*sqrt(a - b)*ar
ctan(2*sqrt(a - b)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)/(a*tan(x)^2 - a + 2*b)) + 2*((a^3 - 2*a^2*b + a*b^2)
*tan(x)^3 + (a^2*b - 3*a*b^2 + 2*b^3)*tan(x))*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(a^4*b - 2*a^3*b^2 + a^2*b^3 +
(a^5 - 2*a^4*b + a^3*b^2)*tan(x)^2)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(sin(t_nostep))]Discontinuities at zeroes of sin(t_nostep) were not checkedWarning, integration of abs or sig
n assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep)]Evaluation time: 0
.48Error: Bad Argument Type

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maple [B]  time = 0.96, size = 421, normalized size = 4.58 \[ \frac {\left (-1+\cos \relax (x )\right )^{2} \left (\cos \relax (x )+1\right )^{2} \left (a \left (\cos ^{2}\relax (x )\right )-b \left (\cos ^{2}\relax (x )\right )-a \right ) \left (-\left (\cos ^{2}\relax (x )\right ) \sqrt {-\frac {a \left (\cos ^{2}\relax (x )\right )-b \left (\cos ^{2}\relax (x )\right )-a}{\left (\cos \relax (x )+1\right )^{2}}}\, \ln \left (4 \cos \relax (x ) \sqrt {-a +b}\, \sqrt {-\frac {a \left (\cos ^{2}\relax (x )\right )-b \left (\cos ^{2}\relax (x )\right )-a}{\left (\cos \relax (x )+1\right )^{2}}}-4 a \cos \relax (x )+4 b \cos \relax (x )+4 \sqrt {-a +b}\, \sqrt {-\frac {a \left (\cos ^{2}\relax (x )\right )-b \left (\cos ^{2}\relax (x )\right )-a}{\left (\cos \relax (x )+1\right )^{2}}}\right ) a^{2}+\left (\cos ^{2}\relax (x )\right ) \sqrt {-a +b}\, a^{2}-2 \left (\cos ^{2}\relax (x )\right ) \sqrt {-a +b}\, a b +2 \left (\cos ^{2}\relax (x )\right ) \sqrt {-a +b}\, b^{2}-\cos \relax (x ) \sqrt {-\frac {a \left (\cos ^{2}\relax (x )\right )-b \left (\cos ^{2}\relax (x )\right )-a}{\left (\cos \relax (x )+1\right )^{2}}}\, \ln \left (4 \cos \relax (x ) \sqrt {-a +b}\, \sqrt {-\frac {a \left (\cos ^{2}\relax (x )\right )-b \left (\cos ^{2}\relax (x )\right )-a}{\left (\cos \relax (x )+1\right )^{2}}}-4 a \cos \relax (x )+4 b \cos \relax (x )+4 \sqrt {-a +b}\, \sqrt {-\frac {a \left (\cos ^{2}\relax (x )\right )-b \left (\cos ^{2}\relax (x )\right )-a}{\left (\cos \relax (x )+1\right )^{2}}}\right ) a^{2}-\sqrt {-a +b}\, a^{2}+\sqrt {-a +b}\, a b \right ) b}{\cos \relax (x ) \left (\frac {a \left (\cos ^{2}\relax (x )\right )-b \left (\cos ^{2}\relax (x )\right )-a}{-1+\cos ^{2}\relax (x )}\right )^{\frac {3}{2}} \sin \relax (x )^{7} \sqrt {-a +b}\, \left (\sqrt {a \left (a -b \right )}+a -b \right ) \left (\sqrt {a \left (a -b \right )}-a +b \right ) a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1+cos(x))^2*(cos(x)+1)^2*(a*cos(x)^2-b*cos(x)^2-a)*(-cos(x)^2*(-(a*cos(x)^2-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2
)*ln(4*cos(x)*(-a+b)^(1/2)*(-(a*cos(x)^2-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x)+4*b*cos(x)+4*(-a+b)^(1/2
)*(-(a*cos(x)^2-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2))*a^2+cos(x)^2*(-a+b)^(1/2)*a^2-2*cos(x)^2*(-a+b)^(1/2)*a*b+2
*cos(x)^2*(-a+b)^(1/2)*b^2-cos(x)*(-(a*cos(x)^2-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*ln(4*cos(x)*(-a+b)^(1/2)*(-(
a*cos(x)^2-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x)+4*b*cos(x)+4*(-a+b)^(1/2)*(-(a*cos(x)^2-b*cos(x)^2-a)/
(cos(x)+1)^2)^(1/2))*a^2-(-a+b)^(1/2)*a^2+(-a+b)^(1/2)*a*b)*b/cos(x)/((a*cos(x)^2-b*cos(x)^2-a)/(-1+cos(x)^2))
^(3/2)/sin(x)^7/(-a+b)^(1/2)/((a*(a-b))^(1/2)+a-b)/((a*(a-b))^(1/2)-a+b)/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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