3.6.63 \(\int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [563]

Optimal. Leaf size=477 \[ -\frac {\cos ^2(c+d x) \cot (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\sqrt {a+b} \cos (c+d x) \sin (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)} \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )}-\frac {\sqrt [4]{a+b} \cos ^2(c+d x) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{a^{3/4} d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\sqrt [4]{a+b} \cos ^2(c+d x) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{2 a^{3/4} d \sqrt {a+b \sin ^4(c+d x)}} \]

[Out]

-(a+b)^(1/4)*cos(d*x+c)^2*(cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))^2)^(1/2)/cos(2*arctan((a+b)^(1/4)*tan
(d*x+c)/a^(1/4)))*EllipticE(sin(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4))),1/2*(2-2*a^(1/2)/(a+b)^(1/2))^(1/2))
*((a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)^2)^(1/2)*(a^(1/2)+(a+b)^(1/2)*tan
(d*x+c)^2)/a^(3/4)/d/(a+b*sin(d*x+c)^4)^(1/2)+1/2*(a+b)^(1/4)*cos(d*x+c)^2*(cos(2*arctan((a+b)^(1/4)*tan(d*x+c
)/a^(1/4)))^2)^(1/2)/cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))*EllipticF(sin(2*arctan((a+b)^(1/4)*tan(d*x+
c)/a^(1/4))),1/2*(2-2*a^(1/2)/(a+b)^(1/2))^(1/2))*((a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/(a^(1/2)+(a+b)^(1/2
)*tan(d*x+c)^2)^2)^(1/2)*(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)/a^(3/4)/d/(a+b*sin(d*x+c)^4)^(1/2)-cos(d*x+c)^2*co
t(d*x+c)*(a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/a/d/(a+b*sin(d*x+c)^4)^(1/2)+cos(d*x+c)*sin(d*x+c)*(a+b)^(1/2
)*(a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/a/d/(a+b*sin(d*x+c)^4)^(1/2)/(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)

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Rubi [A]
time = 0.25, antiderivative size = 477, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3311, 1137, 12, 1153, 1117, 1209} \begin {gather*} \frac {\sqrt [4]{a+b} \cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 a^{3/4} d \sqrt {a+b \sin ^4(c+d x)}}-\frac {\sqrt [4]{a+b} \cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{a^{3/4} d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\sqrt {a+b} \sin (c+d x) \cos (c+d x) \left ((a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}{a d \sqrt {a+b \sin ^4(c+d x)} \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )}-\frac {\cos ^2(c+d x) \cot (c+d x) \left ((a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}{a d \sqrt {a+b \sin ^4(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-((Cos[c + d*x]^2*Cot[c + d*x]*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))/(a*d*Sqrt[a + b*Sin[c + d*x]
^4])) + (Sqrt[a + b]*Cos[c + d*x]*Sin[c + d*x]*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))/(a*d*Sqrt[a
+ b*Sin[c + d*x]^4]*(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)) - ((a + b)^(1/4)*Cos[c + d*x]^2*EllipticE[2*ArcTan
[((a + b)^(1/4)*Tan[c + d*x])/a^(1/4)], (1 - Sqrt[a]/Sqrt[a + b])/2]*(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)*Sq
rt[(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)^2])/(a^(3/4)*d*Sqr
t[a + b*Sin[c + d*x]^4]) + ((a + b)^(1/4)*Cos[c + d*x]^2*EllipticF[2*ArcTan[((a + b)^(1/4)*Tan[c + d*x])/a^(1/
4)], (1 - Sqrt[a]/Sqrt[a + b])/2]*(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)*Sqrt[(a + 2*a*Tan[c + d*x]^2 + (a + b
)*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)^2])/(2*a^(3/4)*d*Sqrt[a + b*Sin[c + d*x]^4])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1117

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(
4*c))], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1137

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^2 +
c*x^4)^(p + 1)/(a*d*(m + 1))), x] - Dist[1/(a*d^2*(m + 1)), Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p +
5)*x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && In
tegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1153

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[1/q, Int[1/Sqrt[
a + b*x^2 + c*x^4], x], x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1209

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(
-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 +
q^2*x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c))], x] /; EqQ[e + d*q^2,
0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 3311

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{ff
= FreeFactors[Tan[e + f*x], x]}, Dist[ff*(a + b*Sin[e + f*x]^4)^p*((Sec[e + f*x]^2)^(2*p)/(f*Apart[a*(1 + Tan[
e + f*x]^2)^2 + b*Tan[e + f*x]^4]^p)), Subst[Int[(d*ff*x)^m*(ExpandToSum[a*(1 + ff^2*x^2)^2 + b*ff^4*x^4, x]^p
/(1 + ff^2*x^2)^(2*p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m}, x] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {\cot ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx &=\frac {\left (\cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\\ &=-\frac {\cos ^2(c+d x) \cot (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\left (\cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \text {Subst}\left (\int \frac {(a+b) x^2}{\sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)}}\\ &=-\frac {\cos ^2(c+d x) \cot (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\left ((a+b) \cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)}}\\ &=-\frac {\cos ^2(c+d x) \cot (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\left (\sqrt {a+b} \cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{\sqrt {a} d \sqrt {a+b \sin ^4(c+d x)}}-\frac {\left (\sqrt {a+b} \cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {a+b} x^2}{\sqrt {a}}}{\sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{\sqrt {a} d \sqrt {a+b \sin ^4(c+d x)}}\\ &=-\frac {\cos ^2(c+d x) \cot (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\sqrt {a+b} \cos (c+d x) \sin (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{a d \sqrt {a+b \sin ^4(c+d x)} \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )}-\frac {\sqrt [4]{a+b} \cos ^2(c+d x) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{a^{3/4} d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\sqrt [4]{a+b} \cos ^2(c+d x) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{2 a^{3/4} d \sqrt {a+b \sin ^4(c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 58.31, size = 372, normalized size = 0.78 \begin {gather*} -\frac {\sqrt {2} \sqrt {8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))} \cot (c+d x)+\frac {4 \cos ^4(c+d x) \left (a \sec ^4(c+d x) \tan (c+d x)+b \tan ^5(c+d x)+\frac {\left (i a+\sqrt {a} \sqrt {b}\right ) \left (E\left (i \sinh ^{-1}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right )|\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )-F\left (i \sinh ^{-1}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right )|\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )\right ) \sec ^2(c+d x) \sqrt {1+\left (1-\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \sqrt {1+\left (1+\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)}}{\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\left (a+b+2 a \cot ^2(c+d x)+a \cot ^4(c+d x)\right ) \sin ^4(c+d x)}}}{4 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-1/4*(Sqrt[2]*Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]]*Cot[c + d*x] + (4*Cos[c + d*x]^4*(a*
Sec[c + d*x]^4*Tan[c + d*x] + b*Tan[c + d*x]^5 + ((I*a + Sqrt[a]*Sqrt[b])*(EllipticE[I*ArcSinh[Sqrt[1 - (I*Sqr
t[b])/Sqrt[a]]*Tan[c + d*x]], (Sqrt[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b])] - EllipticF[I*ArcSinh[Sqrt[1 - (I*S
qrt[b])/Sqrt[a]]*Tan[c + d*x]], (Sqrt[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b])])*Sec[c + d*x]^2*Sqrt[1 + (1 - (I*
Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2]*Sqrt[1 + (1 + (I*Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2])/Sqrt[1 - (I*Sqrt[b])/Sqr
t[a]]))/Sqrt[(a + b + 2*a*Cot[c + d*x]^2 + a*Cot[c + d*x]^4)*Sin[c + d*x]^4])/(a*d)

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Maple [F]
time = 1.86, size = 0, normalized size = 0.00 \[\int \frac {\cot ^{2}\left (d x +c \right )}{\sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2/(a+b*sin(d*x+c)^4)^(1/2),x)

[Out]

int(cot(d*x+c)^2/(a+b*sin(d*x+c)^4)^(1/2),x)

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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(d*x+c)^2/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(cot(d*x + c)^2/sqrt(b*sin(d*x + c)^4 + a), x)

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Fricas [F]
time = 0.12, size = 37, normalized size = 0.08 \begin {gather*} {\rm integral}\left (\frac {\cot \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{2}{\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2/(a+b*sin(d*x+c)**4)**(1/2),x)

[Out]

Integral(cot(c + d*x)**2/sqrt(a + b*sin(c + d*x)**4), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^2}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^2/(a + b*sin(c + d*x)^4)^(1/2),x)

[Out]

int(cot(c + d*x)^2/(a + b*sin(c + d*x)^4)^(1/2), x)

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