3.57 \(\int \frac{\tan (x)}{(a+b \cot ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=118 \[ \frac{b (2 a-b)}{a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{b}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{5/2}} \]

[Out]

ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a]]/a^(5/2) - ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/(a - b)^(5/2) + b/(3
*a*(a - b)*(a + b*Cot[x]^2)^(3/2)) + ((2*a - b)*b)/(a^2*(a - b)^2*Sqrt[a + b*Cot[x]^2])

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Rubi [A]  time = 0.187825, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {3670, 446, 85, 152, 156, 63, 208} \[ \frac{b (2 a-b)}{a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{b}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(a + b*Cot[x]^2)^(5/2),x]

[Out]

ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a]]/a^(5/2) - ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/(a - b)^(5/2) + b/(3
*a*(a - b)*(a + b*Cot[x]^2)^(3/2)) + ((2*a - b)*b)/(a^2*(a - b)^2*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 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

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 )^{5/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x \left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\cot (x)\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x (1+x) (a+b x)^{5/2}} \, dx,x,\cot ^2(x)\right )\right )\\ &=\frac{b}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{a-b-b x}{x (1+x) (a+b x)^{3/2}} \, dx,x,\cot ^2(x)\right )}{2 a (a-b)}\\ &=\frac{b}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{(2 a-b) b}{a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} (a-b)^2+\frac{1}{2} (2 a-b) b x}{x (1+x) \sqrt{a+b x}} \, dx,x,\cot ^2(x)\right )}{a^2 (a-b)^2}\\ &=\frac{b}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{(2 a-b) b}{a^2 (a-b)^2 \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^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\cot ^2(x)\right )}{2 (a-b)^2}\\ &=\frac{b}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{(2 a-b) b}{a^2 (a-b)^2 \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^2 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)^2 b}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a}}\right )}{a^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{5/2}}+\frac{b}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{(2 a-b) b}{a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}\\ \end{align*}

Mathematica [C]  time = 0.0578537, size = 78, normalized size = 0.66 \[ \frac{a \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \cot ^2(x)}{a-b}\right )+(b-a) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{b \cot ^2(x)}{a}+1\right )}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(a + b*Cot[x]^2)^(5/2),x]

[Out]

(a*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Cot[x]^2)/(a - b)] + (-a + b)*Hypergeometric2F1[-3/2, 1, -1/2, 1 +
(b*Cot[x]^2)/a])/(3*a*(a - b)*(a + b*Cot[x]^2)^(3/2))

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Maple [F]  time = 0.186, size = 0, normalized size = 0. \begin{align*} \int{\tan \left ( x \right ) \left ( a+b \left ( \cot \left ( x \right ) \right ) ^{2} \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a+b*cot(x)^2)^(5/2),x)

[Out]

int(tan(x)/(a+b*cot(x)^2)^(5/2),x)

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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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+b*cot(x)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(tan(x)/(b*cot(x)^2 + a)^(5/2), x)

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Fricas [B]  time = 4.23297, size = 3457, normalized size = 29.3 \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)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5 + (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*tan(x)^4 + 2*(a^4*b - 3*a
^3*b^2 + 3*a^2*b^3 - a*b^4)*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) + 3*(a^5*tan(x)^4 + 2*a^4*b*tan(x)^2 + a^3*b^2)*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)) + 2*((7*a^4*b - 11*a^3*b^2 + 4*a^2*b^3)*tan(x)^
4 + 3*(2*a^3*b^2 - 3*a^2*b^3 + a*b^4)*tan(x)^2)*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(a^6*b^2 - 3*a^5*b^3 + 3*a^4*
b^4 - a^3*b^5 + (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*tan(x)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*t
an(x)^2), -1/6*(6*(a^5*tan(x)^4 + 2*a^4*b*tan(x)^2 + a^3*b^2)*sqrt(-a + b)*arctan(-sqrt(-a + b)*sqrt((a*tan(x)
^2 + b)/tan(x)^2)/(a - b)) - 3*(a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5 + (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*ta
n(x)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*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) - 2*((7*a^4*b - 11*a^3*b^2 + 4*a^2*b^3)*tan(x)^4 + 3*(2*a^3*b^2 - 3*a^2*b^3
+ a*b^4)*tan(x)^2)*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5 + (a^8 - 3*a^7*
b + 3*a^6*b^2 - a^5*b^3)*tan(x)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*tan(x)^2), -1/6*(6*(a^3*b^2 -
3*a^2*b^3 + 3*a*b^4 - b^5 + (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*tan(x)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3
- a*b^4)*tan(x)^2)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*tan(x)^2 + b)/tan(x)^2)/a) - 3*(a^5*tan(x)^4 + 2*a^4*b*tan
(x)^2 + a^3*b^2)*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)) - 2*((7*a^4*b - 11*a^3*b^2 + 4*a^2*b^3)*tan(x)^4 + 3*(2*a^3*b^2 - 3*a^2*b^3 + a*b^4)*tan(
x)^2)*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5 + (a^8 - 3*a^7*b + 3*a^6*b^2
- a^5*b^3)*tan(x)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*tan(x)^2), -1/3*(3*(a^3*b^2 - 3*a^2*b^3 + 3
*a*b^4 - b^5 + (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*tan(x)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*tan(
x)^2)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*tan(x)^2 + b)/tan(x)^2)/a) + 3*(a^5*tan(x)^4 + 2*a^4*b*tan(x)^2 + a^3*b
^2)*sqrt(-a + b)*arctan(-sqrt(-a + b)*sqrt((a*tan(x)^2 + b)/tan(x)^2)/(a - b)) - ((7*a^4*b - 11*a^3*b^2 + 4*a^
2*b^3)*tan(x)^4 + 3*(2*a^3*b^2 - 3*a^2*b^3 + a*b^4)*tan(x)^2)*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(a^6*b^2 - 3*a^
5*b^3 + 3*a^4*b^4 - a^3*b^5 + (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*tan(x)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^
3 - a^4*b^4)*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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+b*cot(x)**2)**(5/2),x)

[Out]

Integral(tan(x)/(a + b*cot(x)**2)**(5/2), x)

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Giac [B]  time = 1.64964, size = 686, normalized size = 5.81 \begin{align*} -\frac{{\left (2 \, a^{3} \arctan \left (-\frac{a - b}{\sqrt{-a^{2} + a b}}\right ) - 6 \, a^{2} b \arctan \left (-\frac{a - b}{\sqrt{-a^{2} + a b}}\right ) + 6 \, a b^{2} \arctan \left (-\frac{a - b}{\sqrt{-a^{2} + a b}}\right ) - 2 \, b^{3} \arctan \left (-\frac{a - b}{\sqrt{-a^{2} + a b}}\right ) + \sqrt{-a^{2} + a b} a^{2} \log \left (b\right )\right )} \mathrm{sgn}\left (\sin \left (x\right )\right )}{2 \,{\left (\sqrt{-a^{2} + a b} \sqrt{a - b} a^{4} - 2 \, \sqrt{-a^{2} + a b} \sqrt{a - b} a^{3} b + \sqrt{-a^{2} + a b} \sqrt{a - b} a^{2} b^{2}\right )}} + \frac{{\left (\frac{{\left (7 \, a^{5} b^{2} - 17 \, a^{4} b^{3} + 13 \, a^{3} b^{4} - 3 \, a^{2} b^{5}\right )} \sin \left (x\right )^{2}}{a^{7} b \mathrm{sgn}\left (\sin \left (x\right )\right ) - 3 \, a^{6} b^{2} \mathrm{sgn}\left (\sin \left (x\right )\right ) + 3 \, a^{5} b^{3} \mathrm{sgn}\left (\sin \left (x\right )\right ) - a^{4} b^{4} \mathrm{sgn}\left (\sin \left (x\right )\right )} + \frac{3 \,{\left (2 \, a^{4} b^{3} - 3 \, a^{3} b^{4} + a^{2} b^{5}\right )}}{a^{7} b \mathrm{sgn}\left (\sin \left (x\right )\right ) - 3 \, a^{6} b^{2} \mathrm{sgn}\left (\sin \left (x\right )\right ) + 3 \, a^{5} b^{3} \mathrm{sgn}\left (\sin \left (x\right )\right ) - a^{4} b^{4} \mathrm{sgn}\left (\sin \left (x\right )\right )}\right )} \sin \left (x\right )}{3 \,{\left (a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b\right )}^{\frac{3}{2}}} + \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^{2} - 2 \, a b + b^{2}\right )} \sqrt{a - b} \mathrm{sgn}\left (\sin \left (x\right )\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^{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)^(5/2),x, algorithm="giac")

[Out]

-1/2*(2*a^3*arctan(-(a - b)/sqrt(-a^2 + a*b)) - 6*a^2*b*arctan(-(a - b)/sqrt(-a^2 + a*b)) + 6*a*b^2*arctan(-(a
- b)/sqrt(-a^2 + a*b)) - 2*b^3*arctan(-(a - b)/sqrt(-a^2 + a*b)) + sqrt(-a^2 + a*b)*a^2*log(b))*sgn(sin(x))/(
sqrt(-a^2 + a*b)*sqrt(a - b)*a^4 - 2*sqrt(-a^2 + a*b)*sqrt(a - b)*a^3*b + sqrt(-a^2 + a*b)*sqrt(a - b)*a^2*b^2
) + 1/3*((7*a^5*b^2 - 17*a^4*b^3 + 13*a^3*b^4 - 3*a^2*b^5)*sin(x)^2/(a^7*b*sgn(sin(x)) - 3*a^6*b^2*sgn(sin(x))
+ 3*a^5*b^3*sgn(sin(x)) - a^4*b^4*sgn(sin(x))) + 3*(2*a^4*b^3 - 3*a^3*b^4 + a^2*b^5)/(a^7*b*sgn(sin(x)) - 3*a
^6*b^2*sgn(sin(x)) + 3*a^5*b^3*sgn(sin(x)) - a^4*b^4*sgn(sin(x))))*sin(x)/(a*sin(x)^2 - b*sin(x)^2 + b)^(3/2)
+ 1/2*log((sqrt(a - b)*sin(x) - sqrt(a*sin(x)^2 - b*sin(x)^2 + b))^2)/((a^2 - 2*a*b + b^2)*sqrt(a - b)*sgn(sin
(x))) + sqrt(a - b)*arctan(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^2*sgn(sin(x)))