3.525 \(\int \frac{\cot ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=110 \[ -\frac{2 a+3 b}{2 a^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{(2 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{5/2} f}-\frac{\csc ^2(e+f x)}{2 a f \sqrt{a+b \sin ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.122642, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3194, 78, 51, 63, 208} \[ -\frac{2 a+3 b}{2 a^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{(2 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{5/2} f}-\frac{\csc ^2(e+f x)}{2 a f \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a + 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(2*a^(5/2)*f) - (2*a + 3*b)/(2*a^2*f*Sqrt[a + b*Sin[
e + f*x]^2]) - Csc[e + f*x]^2/(2*a*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 78

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

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 ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x}{x^2 (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac{\csc ^2(e+f x)}{2 a f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a+3 b) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 a f}\\ &=-\frac{2 a+3 b}{2 a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\csc ^2(e+f x)}{2 a f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a+3 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 a^2 f}\\ &=-\frac{2 a+3 b}{2 a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\csc ^2(e+f x)}{2 a f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a+3 b) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{2 a^2 b f}\\ &=\frac{(2 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{5/2} f}-\frac{2 a+3 b}{2 a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\csc ^2(e+f x)}{2 a f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.102085, size = 70, normalized size = 0.64 \[ \frac{-(2 a+3 b) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \sin ^2(e+f x)}{a}+1\right )-a \csc ^2(e+f x)}{2 a^2 f \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 1.547, size = 159, normalized size = 1.5 \begin{align*} -{\frac{1}{af}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \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{3}{2}}}}-{\frac{1}{2\,af \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}}-{\frac{3\,b}{2\,{a}^{2}f}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}}+{\frac{3\,b}{2\,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)^3/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

-1/a/f/(a+b*sin(f*x+e)^2)^(1/2)+1/f/a^(3/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))-1/2/f/a/si
n(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2)-3/2/f/a^2*b/(a+b*sin(f*x+e)^2)^(1/2)+3/2/f/a^(5/2)*b*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)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.6528, size = 956, normalized size = 8.69 \begin{align*} \left [\frac{{\left ({\left (2 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (2 \, a^{2} + 7 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 5 \, a b + 3 \, 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 ({\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 3 \, a^{2} - 3 \, a b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{4 \,{\left (a^{3} b f \cos \left (f x + e\right )^{4} -{\left (a^{4} + 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} + a^{3} b\right )} f\right )}}, -\frac{{\left ({\left (2 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (2 \, a^{2} + 7 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 5 \, a b + 3 \, b^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a}}{a}\right ) -{\left ({\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 3 \, a^{2} - 3 \, a b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{2 \,{\left (a^{3} b f \cos \left (f x + e\right )^{4} -{\left (a^{4} + 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} + a^{3} b\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(((2*a*b + 3*b^2)*cos(f*x + e)^4 - (2*a^2 + 7*a*b + 6*b^2)*cos(f*x + e)^2 + 2*a^2 + 5*a*b + 3*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*((2
*a^2 + 3*a*b)*cos(f*x + e)^2 - 3*a^2 - 3*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^3*b*f*cos(f*x + e)^4 - (a^4
+ 2*a^3*b)*f*cos(f*x + e)^2 + (a^4 + a^3*b)*f), -1/2*(((2*a*b + 3*b^2)*cos(f*x + e)^4 - (2*a^2 + 7*a*b + 6*b^2
)*cos(f*x + e)^2 + 2*a^2 + 5*a*b + 3*b^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - ((2*a^
2 + 3*a*b)*cos(f*x + e)^2 - 3*a^2 - 3*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^3*b*f*cos(f*x + e)^4 - (a^4 + 2
*a^3*b)*f*cos(f*x + e)^2 + (a^4 + a^3*b)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**3/(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Integral(cot(e + f*x)**3/(a + b*sin(e + f*x)**2)**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cot(f*x + e)^3/(b*sin(f*x + e)^2 + a)^(3/2), x)