3.63 \(\int \frac{\cot (x)}{(a+b \cot ^4(x))^{3/2}} \, dx\)

Optimal. Leaf size=74 \[ \frac{\tanh ^{-1}\left (\frac{a-b \cot ^2(x)}{\sqrt{a+b} \sqrt{a+b \cot ^4(x)}}\right )}{2 (a+b)^{3/2}}-\frac{a+b \cot ^2(x)}{2 a (a+b) \sqrt{a+b \cot ^4(x)}} \]

[Out]

ArcTanh[(a - b*Cot[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Cot[x]^4])]/(2*(a + b)^(3/2)) - (a + b*Cot[x]^2)/(2*a*(a + b)
*Sqrt[a + b*Cot[x]^4])

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Rubi [A]  time = 0.11073, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3670, 1248, 741, 12, 725, 206} \[ \frac{\tanh ^{-1}\left (\frac{a-b \cot ^2(x)}{\sqrt{a+b} \sqrt{a+b \cot ^4(x)}}\right )}{2 (a+b)^{3/2}}-\frac{a+b \cot ^2(x)}{2 a (a+b) \sqrt{a+b \cot ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]/(a + b*Cot[x]^4)^(3/2),x]

[Out]

ArcTanh[(a - b*Cot[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Cot[x]^4])]/(2*(a + b)^(3/2)) - (a + b*Cot[x]^2)/(2*a*(a + b)
*Sqrt[a + b*Cot[x]^4])

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 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 12

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

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.294488, size = 73, normalized size = 0.99 \[ \frac{1}{2} \left (\frac{\tanh ^{-1}\left (\frac{a-b \cot ^2(x)}{\sqrt{a+b} \sqrt{a+b \cot ^4(x)}}\right )}{(a+b)^{3/2}}-\frac{a+b \cot ^2(x)}{a (a+b) \sqrt{a+b \cot ^4(x)}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]/(a + b*Cot[x]^4)^(3/2),x]

[Out]

(ArcTanh[(a - b*Cot[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Cot[x]^4])]/(a + b)^(3/2) - (a + b*Cot[x]^2)/(a*(a + b)*Sqrt
[a + b*Cot[x]^4]))/2

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Maple [B]  time = 0.055, size = 248, normalized size = 3.4 \begin{align*} -{\frac{b}{2}\ln \left ({\frac{1}{1+ \left ( \cot \left ( x \right ) \right ) ^{2}} \left ( 2\,a+2\,b-2\, \left ( 1+ \left ( \cot \left ( x \right ) \right ) ^{2} \right ) b+2\,\sqrt{a+b}\sqrt{ \left ( 1+ \left ( \cot \left ( x \right ) \right ) ^{2} \right ) ^{2}b-2\, \left ( 1+ \left ( \cot \left ( x \right ) \right ) ^{2} \right ) b+a+b} \right ) } \right ) \left ( \sqrt{-ab}+b \right ) ^{-1} \left ( \sqrt{-ab}-b \right ) ^{-1}{\frac{1}{\sqrt{a+b}}}}-{\frac{1}{4\,a}\sqrt{ \left ( \left ( \cot \left ( x \right ) \right ) ^{2}-{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}b+2\,\sqrt{-ab} \left ( \left ( \cot \left ( x \right ) \right ) ^{2}-{\frac{\sqrt{-ab}}{b}} \right ) } \left ( \sqrt{-ab}+b \right ) ^{-1} \left ( \left ( \cot \left ( x \right ) \right ) ^{2}-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}}+{\frac{1}{4\,a}\sqrt{ \left ( \left ( \cot \left ( x \right ) \right ) ^{2}+{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}b-2\,\sqrt{-ab} \left ( \left ( \cot \left ( x \right ) \right ) ^{2}+{\frac{\sqrt{-ab}}{b}} \right ) } \left ( \sqrt{-ab}-b \right ) ^{-1} \left ( \left ( \cot \left ( x \right ) \right ) ^{2}+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)/(a+b*cot(x)^4)^(3/2),x)

[Out]

-1/2*b/((-a*b)^(1/2)+b)/((-a*b)^(1/2)-b)/(a+b)^(1/2)*ln((2*a+2*b-2*(1+cot(x)^2)*b+2*(a+b)^(1/2)*((1+cot(x)^2)^
2*b-2*(1+cot(x)^2)*b+a+b)^(1/2))/(1+cot(x)^2))-1/4/((-a*b)^(1/2)+b)/a/(cot(x)^2-(-a*b)^(1/2)/b)*((cot(x)^2-(-a
*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2)*(cot(x)^2-(-a*b)^(1/2)/b))^(1/2)+1/4/((-a*b)^(1/2)-b)/a/(cot(x)^2+(-a*b)^(1/2)
/b)*((cot(x)^2+(-a*b)^(1/2)/b)^2*b-2*(-a*b)^(1/2)*(cot(x)^2+(-a*b)^(1/2)/b))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (x\right )}{{\left (b \cot \left (x\right )^{4} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*cot(x)^4)^(3/2),x, algorithm="maxima")

[Out]

integrate(cot(x)/(b*cot(x)^4 + a)^(3/2), x)

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Fricas [B]  time = 3.8069, size = 1628, normalized size = 22. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*cot(x)^4)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(((a^2 + a*b)*cos(2*x)^2 + a^2 + a*b - 2*(a^2 - a*b)*cos(2*x))*sqrt(a + b)*log(1/2*(a^2 + 2*a*b + b^2)*co
s(2*x)^2 + 1/2*a^2 + 1/2*b^2 + 1/2*((a + b)*cos(2*x)^2 - 2*a*cos(2*x) + a - b)*sqrt(a + b)*sqrt(((a + b)*cos(2
*x)^2 - 2*(a - b)*cos(2*x) + a + b)/(cos(2*x)^2 - 2*cos(2*x) + 1)) - (a^2 - b^2)*cos(2*x)) - 2*((a^2 - b^2)*co
s(2*x)^2 + a^2 + 2*a*b + b^2 - 2*(a^2 + a*b)*cos(2*x))*sqrt(((a + b)*cos(2*x)^2 - 2*(a - b)*cos(2*x) + a + b)/
(cos(2*x)^2 - 2*cos(2*x) + 1)))/(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(2
*x)^2 - 2*(a^4 + a^3*b - a^2*b^2 - a*b^3)*cos(2*x)), -1/2*(((a^2 + a*b)*cos(2*x)^2 + a^2 + a*b - 2*(a^2 - a*b)
*cos(2*x))*sqrt(-a - b)*arctan(((a + b)*cos(2*x)^2 - 2*a*cos(2*x) + a - b)*sqrt(-a - b)*sqrt(((a + b)*cos(2*x)
^2 - 2*(a - b)*cos(2*x) + a + b)/(cos(2*x)^2 - 2*cos(2*x) + 1))/((a^2 + 2*a*b + b^2)*cos(2*x)^2 + a^2 + 2*a*b
+ b^2 - 2*(a^2 - b^2)*cos(2*x))) + ((a^2 - b^2)*cos(2*x)^2 + a^2 + 2*a*b + b^2 - 2*(a^2 + a*b)*cos(2*x))*sqrt(
((a + b)*cos(2*x)^2 - 2*(a - b)*cos(2*x) + a + b)/(cos(2*x)^2 - 2*cos(2*x) + 1)))/(a^4 + 3*a^3*b + 3*a^2*b^2 +
a*b^3 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(2*x)^2 - 2*(a^4 + a^3*b - a^2*b^2 - a*b^3)*cos(2*x))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot{\left (x \right )}}{\left (a + b \cot ^{4}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*cot(x)**4)**(3/2),x)

[Out]

Integral(cot(x)/(a + b*cot(x)**4)**(3/2), x)

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Giac [B]  time = 5.86939, size = 211, normalized size = 2.85 \begin{align*} \frac{a b^{2} \log \left ({\left | -{\left (\sqrt{a + b} \cos \left (x\right )^{2} - \sqrt{a \cos \left (x\right )^{4} + b \cos \left (x\right )^{4} - 2 \, a \cos \left (x\right )^{2} + a}\right )}{\left (a + b\right )} + \sqrt{a + b} a \right |}\right )}{2 \,{\left (a^{2} b^{2} + a b^{3}\right )} \sqrt{a + b}} - \frac{\frac{a b^{2}}{a^{2} b^{2} + a b^{3}} - \frac{{\left (a b^{2} - b^{3}\right )} \cos \left (x\right )^{2}}{a^{2} b^{2} + a b^{3}}}{2 \, \sqrt{a \cos \left (x\right )^{4} + b \cos \left (x\right )^{4} - 2 \, a \cos \left (x\right )^{2} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*cot(x)^4)^(3/2),x, algorithm="giac")

[Out]

1/2*a*b^2*log(abs(-(sqrt(a + b)*cos(x)^2 - sqrt(a*cos(x)^4 + b*cos(x)^4 - 2*a*cos(x)^2 + a))*(a + b) + sqrt(a
+ b)*a))/((a^2*b^2 + a*b^3)*sqrt(a + b)) - 1/2*(a*b^2/(a^2*b^2 + a*b^3) - (a*b^2 - b^3)*cos(x)^2/(a^2*b^2 + a*
b^3))/sqrt(a*cos(x)^4 + b*cos(x)^4 - 2*a*cos(x)^2 + a)