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

Optimal. Leaf size=209 \[ -\frac {2 \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a^2 f}-\frac {2 \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{a^2 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {\cot (e+f x)}{a f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{a f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

cot(f*x+e)/a/f/(a+b*sin(f*x+e)^2)^(1/2)-2*cot(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/a^2/f-2*EllipticE(sin(f*x+e),(-b
/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(a+b*sin(f*x+e)^2)^(1/2)/a^2/f/(1+b*sin(f*x+e)^2/a)^(1/2)+EllipticF
(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(1/2)/a/f/(a+b*sin(f*x+e)^2)^(1
/2)

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Rubi [A]  time = 0.24, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3196, 469, 583, 524, 426, 424, 421, 419} \[ -\frac {2 \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a^2 f}-\frac {2 \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{a^2 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {\cot (e+f x)}{a f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{a f \sqrt {a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Cot[e + f*x]/(a*f*Sqrt[a + b*Sin[e + f*x]^2]) - (2*Cot[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(a^2*f) - (2*Sqrt[
Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(a^2*f*Sqrt[1
+ (b*Sin[e + f*x]^2)/a]) + (Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[1
+ (b*Sin[e + f*x]^2)/a])/(a*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 469

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[((e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*e*n*(p + 1)), x] + Dist[1/(a*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)
^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m + n*(p + 1) + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{
a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[0, q, 1] && IntBinomialQ[a, b, c
, d, e, m, n, p, q, x]

Rule 524

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

Rule 583

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

Rule 3196

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

Rubi steps

\begin {align*} \int \frac {\cot ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-x^2}}{x^2 \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\cot (e+f x)}{a f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {-2+x^2}{x^2 \sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{a f}\\ &=\frac {\cot (e+f x)}{a f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a^2 f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {-a-2 b x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{a^2 f}\\ &=\frac {\cot (e+f x)}{a f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a^2 f}-\frac {\left (2 \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{a^2 f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{a f}\\ &=\frac {\cot (e+f x)}{a f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a^2 f}-\frac {\left (2 \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{a^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{a f \sqrt {a+b \sin ^2(e+f x)}}\\ &=\frac {\cot (e+f x)}{a f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a^2 f}-\frac {2 \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{a f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.75, size = 142, normalized size = 0.68 \[ \frac {-2 \cot (e+f x) (a-b \cos (2 (e+f x))+b)+\sqrt {2} a \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )-2 \sqrt {2} a \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{\sqrt {2} a^2 f \sqrt {2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b - b*Cos[2*(e + f*x)])*Cot[e + f*x] - 2*Sqrt[2]*a*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e
+ f*x, -(b/a)] + Sqrt[2]*a*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)])/(Sqrt[2]*a^2*f*
Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cot \left (f x + e\right )^{2}}{b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-b*cos(f*x + e)^2 + a + b)*cot(f*x + e)^2/(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a
^2 + 2*a*b + b^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.61, size = 141, normalized size = 0.67 \[ \frac {\sin \left (f x +e \right ) \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, a \left (\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-2 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )\right )+2 b \left (\cos ^{4}\left (f x +e \right )\right )+\left (-a -2 b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{\sin \left (f x +e \right ) a^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(f*x+e)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*(cos(f*x+e)^2)^(1/2)*a*(EllipticF(sin(f*x+e),(-1/a*b)^(1/2))-2*E
llipticE(sin(f*x+e),(-1/a*b)^(1/2)))+2*b*cos(f*x+e)^4+(-a-2*b)*cos(f*x+e)^2)/sin(f*x+e)/a^2/cos(f*x+e)/(a+b*si
n(f*x+e)^2)^(1/2)/f

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{2}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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