3.6.25 \(\int \frac {\cot ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [525]

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

[Out]

1/2*(2*a+3*b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(5/2)/f+1/2*(-2*a-3*b)/a^2/f/(a+b*sin(f*x+e)^2)^(1/2
)-1/2*csc(f*x+e)^2/a/f/(a+b*sin(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3273, 79, 53, 65, 214} \begin {gather*} \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 {gather*}

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 53

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 65

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 79

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] || L
tQ[p, n]))))

Rule 214

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

Rule 3273

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]

Rubi steps

\begin {align*} \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\text {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) \text {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) \text {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) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.07, size = 70, normalized size = 0.64 \begin {gather*} \frac {-a \csc ^2(e+f x)-(2 a+3 b) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};1+\frac {b \sin ^2(e+f x)}{a}\right )}{2 a^2 f \sqrt {a+b \sin ^2(e+f x)}} \end {gather*}

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 = 10.36, size = 148, normalized size = 1.35

method	result	size
default	\(\frac {-\frac {1}{2 a \sin \left (f x +e \right )^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {3 b}{2 a^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{2 a^{\frac {5}{2}}}-\frac {1}{a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{a^{\frac {3}{2}}}}{f}\)	\(148\)

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,method=_RETURNVERBOSE)

[Out]

(-1/2/a/sin(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2)-3/2/a^2*b/(a+b*sin(f*x+e)^2)^(1/2)+3/2/a^(5/2)*b*ln((2*a+2*a^(1/
2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))-1/a/(a+b*sin(f*x+e)^2)^(1/2)+1/a^(3/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x
+e)^2)^(1/2))/sin(f*x+e)))/f

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Maxima [A]
time = 0.32, size = 123, normalized size = 1.12 \begin {gather*} \frac {\frac {2 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} + \frac {3 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} - \frac {2}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a} - \frac {3 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {1}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right )^{2}}}{2 \, f} \end {gather*}

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]

1/2*(2*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(3/2) + 3*b*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(5/2)
- 2/(sqrt(b*sin(f*x + e)^2 + a)*a) - 3*b/(sqrt(b*sin(f*x + e)^2 + a)*a^2) - 1/(sqrt(b*sin(f*x + e)^2 + a)*a*s
in(f*x + e)^2))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (94) = 188\).
time = 0.47, size = 406, normalized size = 3.69 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 525 vs. \(2 (98) = 196\).
time = 0.81, size = 525, normalized size = 4.77 \begin {gather*} -\frac {\frac {{\left (\frac {{\left (a^{5} b + a^{4} b^{2}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{6} b + a^{5} b^{2}} + \frac {2 \, {\left (5 \, a^{5} b + 11 \, a^{4} b^{2} + 6 \, a^{3} b^{3}\right )}}{a^{6} b + a^{5} b^{2}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \frac {9 \, a^{5} b + 17 \, a^{4} b^{2} + 8 \, a^{3} b^{3}}{a^{6} b + a^{5} b^{2}}}{\sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}} + \frac {2 \, {\left (2 \, a^{\frac {3}{2}} + 3 \, \sqrt {a} b\right )} \log \left ({\left | -{\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} a - a^{\frac {3}{2}} - 2 \, \sqrt {a} b \right |}\right )}{a^{3}} - \frac {2 \, {\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} a^{\frac {3}{2}} + 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} \sqrt {a} b + a^{2}\right )}}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} \sqrt {a} - a^{\frac {3}{2}}\right )} a^{2}}}{8 \, f} \end {gather*}

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]

-1/8*((((a^5*b + a^4*b^2)*tan(1/2*f*x + 1/2*e)^2/(a^6*b + a^5*b^2) + 2*(5*a^5*b + 11*a^4*b^2 + 6*a^3*b^3)/(a^6
*b + a^5*b^2))*tan(1/2*f*x + 1/2*e)^2 + (9*a^5*b + 17*a^4*b^2 + 8*a^3*b^3)/(a^6*b + a^5*b^2))/sqrt(a*tan(1/2*f
*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) + 2*(2*a^(3/2) + 3*sqrt(a)*b)*log
(abs(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1
/2*f*x + 1/2*e)^2 + a))*a - a^(3/2) - 2*sqrt(a)*b))/a^3 - 2*((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*
f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^(3/2) + 2*(sqrt(a)*tan(1/2*f*
x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*s
qrt(a)*b + a^2)/(((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2
+ 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(a) - a^(3/2))*a^2))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^3}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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