3.52 \(\int \frac{\tan (x)}{(a+b \cot ^2(x))^{3/2}} \, dx\)
Optimal. Leaf size=84 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a}}\right )}{a^{3/2}}+\frac{b}{a (a-b) \sqrt{a+b \cot ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2}} \]
[Out]
ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a]]/a^(3/2) - ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/(a - b)^(3/2) + b/(a
*(a - b)*Sqrt[a + b*Cot[x]^2])
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Rubi [A] time = 0.130786, antiderivative size = 84, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used =
{3670, 446, 85, 156, 63, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a}}\right )}{a^{3/2}}+\frac{b}{a (a-b) \sqrt{a+b \cot ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Tan[x]/(a + b*Cot[x]^2)^(3/2),x]
[Out]
ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a]]/a^(3/2) - ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/(a - b)^(3/2) + b/(a
*(a - b)*Sqrt[a + b*Cot[x]^2])
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]))
Rule 446
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Rule 85
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[(f*(e + f*x)^(p +
1))/((p + 1)*(b*e - a*f)*(d*e - c*f)), x] + Dist[1/((b*e - a*f)*(d*e - c*f)), Int[((b*d*e - b*c*f - a*d*f - b
*d*f*x)*(e + f*x)^(p + 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[p, -1]
Rule 156
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{\tan (x)}{\left (a+b \cot ^2(x)\right )^{3/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x \left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\cot (x)\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x (1+x) (a+b x)^{3/2}} \, dx,x,\cot ^2(x)\right )\right )\\ &=\frac{b}{a (a-b) \sqrt{a+b \cot ^2(x)}}-\frac{\operatorname{Subst}\left (\int \frac{a-b-b x}{x (1+x) \sqrt{a+b x}} \, dx,x,\cot ^2(x)\right )}{2 a (a-b)}\\ &=\frac{b}{a (a-b) \sqrt{a+b \cot ^2(x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\cot ^2(x)\right )}{2 a}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\cot ^2(x)\right )}{2 (a-b)}\\ &=\frac{b}{a (a-b) \sqrt{a+b \cot ^2(x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \cot ^2(x)}\right )}{a b}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \cot ^2(x)}\right )}{(a-b) b}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2}}+\frac{b}{a (a-b) \sqrt{a+b \cot ^2(x)}}\\ \end{align*}
Mathematica [C] time = 0.0557251, size = 75, normalized size = 0.89 \[ \frac{a \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \cot ^2(x)}{a-b}\right )+(b-a) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{b \cot ^2(x)}{a}+1\right )}{a (a-b) \sqrt{a+b \cot ^2(x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[x]/(a + b*Cot[x]^2)^(3/2),x]
[Out]
(a*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Cot[x]^2)/(a - b)] + (-a + b)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b
*Cot[x]^2)/a])/(a*(a - b)*Sqrt[a + b*Cot[x]^2])
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Maple [C] time = 0.194, size = 962, normalized size = 11.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(x)/(a+b*cot(x)^2)^(3/2),x)
[Out]
-1/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/(a-b)/a*(cos(x)^2*a-b*cos(x)^2-a)*(2^(1/2)*(1/b*(cos(x)*a^(1/2)*(a-
b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1))^(1/2)*(-2/b*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*
(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)+1))^(1/2)*EllipticF((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/
2)/sin(x),((8*a^(3/2)*(a-b)^(1/2)-4*a^(1/2)*(a-b)^(1/2)*b+8*a^2-8*a*b+b^2)/b^2)^(1/2))*b*sin(x)-2*2^(1/2)*(1/b
*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1))^(1/2)*(-2/b*(cos(x)*a^(1/2)*
(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)+1))^(1/2)*EllipticPi((-1+cos(x))*((2*a^(1/2)*(a-b
)^(1/2)-2*a+b)/b)^(1/2)/sin(x),-1/(2*a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/((2
*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*a*sin(x)+2*2^(1/2)*(1/b*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-
a*cos(x)+b*cos(x)+a)/(cos(x)+1))^(1/2)*(-2/b*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)
-a)/(cos(x)+1))^(1/2)*EllipticPi((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),1/(2*a^(1/2)*(a-b)
^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*a*sin(x)-2*2
^(1/2)*(1/b*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1))^(1/2)*(-2/b*(cos(
x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)+1))^(1/2)*EllipticPi((-1+cos(x))*((2*a
^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),1/(2*a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)
^(1/2)/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*b*sin(x)+((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)*b*cos(x)-((2*
a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)*b)/(-1+cos(x))/sin(x)^2/((cos(x)^2*a-b*cos(x)^2-a)/(cos(x)^2-1))^(3/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (x\right )}{{\left (b \cot \left (x\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(x)/(a+b*cot(x)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate(tan(x)/(b*cot(x)^2 + a)^(3/2), x)
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Fricas [B] time = 3.02513, size = 2087, normalized size = 24.85 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(x)/(a+b*cot(x)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/2*(2*(a^2*b - a*b^2)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2 + (a^2*b - 2*a*b^2 + b^3 + (a^3 - 2*a^2*b + a
*b^2)*tan(x)^2)*sqrt(a)*log(2*a*tan(x)^2 + 2*sqrt(a)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2 + b) - (a^3*tan(
x)^2 + a^2*b)*sqrt(a - b)*log(((2*a - b)*tan(x)^2 + 2*sqrt(a - b)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2 + b
)/(tan(x)^2 + 1)))/(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*(2*(a^2*b - a*b^2)*
sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2 - 2*(a^3*tan(x)^2 + a^2*b)*sqrt(-a + b)*arctan(-sqrt(-a + b)*sqrt((a*
tan(x)^2 + b)/tan(x)^2)/(a - b)) + (a^2*b - 2*a*b^2 + b^3 + (a^3 - 2*a^2*b + a*b^2)*tan(x)^2)*sqrt(a)*log(2*a*
tan(x)^2 + 2*sqrt(a)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2 + b))/(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*(2*(a^2*b - a*b^2)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2 - 2*(a^2*b - 2*a*b^2
+ b^3 + (a^3 - 2*a^2*b + a*b^2)*tan(x)^2)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*tan(x)^2 + b)/tan(x)^2)/a) - (a^3*
tan(x)^2 + a^2*b)*sqrt(a - b)*log(((2*a - b)*tan(x)^2 + 2*sqrt(a - b)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2
+ b)/(tan(x)^2 + 1)))/(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), ((a^2*b - a*b^2)*sq
rt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)^2 - (a^2*b - 2*a*b^2 + b^3 + (a^3 - 2*a^2*b + a*b^2)*tan(x)^2)*sqrt(-a)*a
rctan(sqrt(-a)*sqrt((a*tan(x)^2 + b)/tan(x)^2)/a) - (a^3*tan(x)^2 + a^2*b)*sqrt(-a + b)*arctan(-sqrt(-a + b)*s
qrt((a*tan(x)^2 + b)/tan(x)^2)/(a - b)))/(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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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (x \right )}}{\left (a + b \cot ^{2}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(x)/(a+b*cot(x)**2)**(3/2),x)
[Out]
Integral(tan(x)/(a + b*cot(x)**2)**(3/2), x)
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Giac [B] time = 1.48542, size = 404, normalized size = 4.81 \begin{align*} \frac{b \mathrm{sgn}\left (\sin \left (x\right )\right ) \sin \left (x\right )}{\sqrt{a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}{\left (a^{2} - a b\right )}} - \frac{{\left (2 \, a^{2} \arctan \left (-\frac{a - b}{\sqrt{-a^{2} + a b}}\right ) - 4 \, a b \arctan \left (-\frac{a - b}{\sqrt{-a^{2} + a b}}\right ) + 2 \, b^{2} \arctan \left (-\frac{a - b}{\sqrt{-a^{2} + a b}}\right ) + \sqrt{-a^{2} + a b} a \log \left (b\right )\right )} \mathrm{sgn}\left (\sin \left (x\right )\right )}{2 \,{\left (\sqrt{-a^{2} + a b} \sqrt{a - b} a^{2} - \sqrt{-a^{2} + a b} \sqrt{a - b} a b\right )}} + \frac{\sqrt{a - b} \arctan \left (\frac{{\left (\sqrt{a - b} \sin \left (x\right ) - \sqrt{a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}\right )}^{2} - 2 \, a + b}{2 \, \sqrt{-a^{2} + a b}}\right )}{\sqrt{-a^{2} + a b} a \mathrm{sgn}\left (\sin \left (x\right )\right )} + \frac{\log \left ({\left (\sqrt{a - b} \sin \left (x\right ) - \sqrt{a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}\right )}^{2}\right )}{2 \,{\left (a - b\right )}^{\frac{3}{2}} \mathrm{sgn}\left (\sin \left (x\right )\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(x)/(a+b*cot(x)^2)^(3/2),x, algorithm="giac")
[Out]
b*sgn(sin(x))*sin(x)/(sqrt(a*sin(x)^2 - b*sin(x)^2 + b)*(a^2 - a*b)) - 1/2*(2*a^2*arctan(-(a - b)/sqrt(-a^2 +
a*b)) - 4*a*b*arctan(-(a - b)/sqrt(-a^2 + a*b)) + 2*b^2*arctan(-(a - b)/sqrt(-a^2 + a*b)) + sqrt(-a^2 + a*b)*a
*log(b))*sgn(sin(x))/(sqrt(-a^2 + a*b)*sqrt(a - b)*a^2 - sqrt(-a^2 + a*b)*sqrt(a - b)*a*b) + sqrt(a - b)*arcta
n(1/2*((sqrt(a - b)*sin(x) - sqrt(a*sin(x)^2 - b*sin(x)^2 + b))^2 - 2*a + b)/sqrt(-a^2 + a*b))/(sqrt(-a^2 + a*
b)*a*sgn(sin(x))) + 1/2*log((sqrt(a - b)*sin(x) - sqrt(a*sin(x)^2 - b*sin(x)^2 + b))^2)/((a - b)^(3/2)*sgn(sin
(x)))