3.535 \(\int \frac{\cot (e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=83 \[ \frac{1}{a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{a^{5/2} f}+\frac{1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

-(ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]]/(a^(5/2)*f)) + 1/(3*a*f*(a + b*Sin[e + f*x]^2)^(3/2)) + 1/(a^2*f
*Sqrt[a + b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.0876517, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3194, 51, 63, 208} \[ \frac{1}{a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{a^{5/2} f}+\frac{1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

-(ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]]/(a^(5/2)*f)) + 1/(3*a*f*(a + b*Sin[e + f*x]^2)^(3/2)) + 1/(a^2*f
*Sqrt[a + b*Sin[e + f*x]^2])

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, 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{\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 a f}\\ &=\frac{1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{1}{a^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 a^2 f}\\ &=\frac{1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{1}{a^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{a^2 b f}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{a^{5/2} f}+\frac{1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{1}{a^2 f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.0643929, size = 49, normalized size = 0.59 \[ \frac{\, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \sin ^2(e+f x)}{a}+1\right )}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 2.888, size = 271, normalized size = 3.3 \begin{align*} -{\frac{7}{12\,{a}^{2}f}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+{\frac{ab+{b}^{2}}{b}}}{\frac{1}{\sqrt{-ab}}} \left ( \sin \left ( fx+e \right ) +{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}}+{\frac{7}{12\,{a}^{2}f}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+{\frac{ab+{b}^{2}}{b}}}{\frac{1}{\sqrt{-ab}}} \left ( \sin \left ( fx+e \right ) -{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}}-{\frac{1}{12\,{a}^{2}fb}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+{\frac{ab+{b}^{2}}{b}}} \left ( \sin \left ( fx+e \right ) +{\frac{1}{b}\sqrt{-ab}} \right ) ^{-2}}-{\frac{1}{12\,{a}^{2}fb}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+{\frac{ab+{b}^{2}}{b}}} \left ( \sin \left ( fx+e \right ) -{\frac{1}{b}\sqrt{-ab}} \right ) ^{-2}}-{\frac{1}{f}\ln \left ({\frac{1}{\sin \left ( fx+e \right ) } \left ( 2\,a+2\,\sqrt{a}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \right ) } \right ){a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x)

[Out]

-7/12/f/a^2/(-a*b)^(1/2)/(sin(f*x+e)+(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)+7/12/f/a^2/(-a*b)^(1/
2)/(sin(f*x+e)-(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/12/f/a^2/b/(sin(f*x+e)+(-a*b)^(1/2)/b)^2*
(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/12/f/a^2/b/(sin(f*x+e)-(-a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(
1/2)-1/f/a^(5/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.57794, size = 911, normalized size = 10.98 \begin{align*} \left [\frac{3 \,{\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a} \log \left (\frac{2 \,{\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \,{\left (3 \, a b \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 3 \, a b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{6 \,{\left (a^{3} b^{2} f \cos \left (f x + e\right )^{4} - 2 \,{\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f\right )}}, \frac{3 \,{\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a}}{a}\right ) -{\left (3 \, a b \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 3 \, a b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{3 \,{\left (a^{3} b^{2} f \cos \left (f x + e\right )^{4} - 2 \,{\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(a)*log(2*(b*cos(f*x + e)^
2 + 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) - 2*a - b)/(cos(f*x + e)^2 - 1)) - 2*(3*a*b*cos(f*x + e)^2 - 4*a
^2 - 3*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^3*b^2*f*cos(f*x + e)^4 - 2*(a^4*b + a^3*b^2)*f*cos(f*x + e)^2
+ (a^5 + 2*a^4*b + a^3*b^2)*f), 1/3*(3*(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)
*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - (3*a*b*cos(f*x + e)^2 - 4*a^2 - 3*a*b)*sqrt(-b*
cos(f*x + e)^2 + a + b))/(a^3*b^2*f*cos(f*x + e)^4 - 2*(a^4*b + a^3*b^2)*f*cos(f*x + e)^2 + (a^5 + 2*a^4*b + a
^3*b^2)*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.10856, size = 100, normalized size = 1.2 \begin{align*} \frac{\arctan \left (\frac{\sqrt{b \sin \left (f x + e\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2} f} + \frac{3 \, b \sin \left (f x + e\right )^{2} + 4 \, a}{3 \,{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

arctan(sqrt(b*sin(f*x + e)^2 + a)/sqrt(-a))/(sqrt(-a)*a^2*f) + 1/3*(3*b*sin(f*x + e)^2 + 4*a)/((b*sin(f*x + e)
^2 + a)^(3/2)*a^2*f)