3.5.97 \(\int \cot ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [497]

Optimal. Leaf size=174 \[ -\frac {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}-\frac {2 \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {(a+b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

-cot(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/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)/f/(1+b*sin(f*x+e)^2/a)^(1/2)+(a+b)*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)/f/(a+b*sin(f*x+e)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3275, 484, 538, 437, 435, 432, 430} \begin {gather*} \frac {(a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

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

Rule 430

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

Rule 432

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

Rule 435

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

Rule 437

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

Rule 484

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*((c + d*x^n)^q/(e*(m + 1))), x] - Dist[n/(e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^(p -
1)*(c + d*x^n)^(q - 1)*Simp[b*c*p + a*d*q + b*d*(p + q)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*
c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] && LtQ[m, -1] && GtQ[p, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x
]

Rule 538

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 3275

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

________________________________________________________________________________________

Mathematica [A]
time = 0.39, size = 143, normalized size = 0.82 \begin {gather*} \frac {-((2 a+b-b \cos (2 (e+f x))) \cot (e+f x))-2 \sqrt {2} a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+\sqrt {2} (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )}{\sqrt {2} f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

________________________________________________________________________________________

Maple [A]
time = 9.20, size = 156, normalized size = 0.90

method	result	size
default	\(\frac {\sin \left (f x +e \right ) \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \left (\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -2 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \right )+b \left (\cos ^{4}\left (f x +e \right )\right )+\left (-a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{\sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(156\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(sin(f*x+e)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*(cos(f*x+e)^2)^(1/2)*(EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a+Ell
ipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-2*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a)+b*cos(f*x+e)^4+(-a-b)*cos(f*x+e)
^2)/sin(f*x+e)/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.13, size = 27, normalized size = 0.16 \begin {gather*} {\rm integral}\left (\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cot \left (f x + e\right )^{2}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \cot ^{2}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**2*(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^2*(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(cot(e + f*x)^2*(a + b*sin(e + f*x)^2)^(1/2), x)

________________________________________________________________________________________