3.248 \(\int \frac{\csc ^2(c+d x)}{\sqrt{a+b \sin ^4(c+d x)}} \, dx\)

Optimal. Leaf size=493 \[ \frac{\left (\sqrt{a} \sqrt{a+b}+a+b\right ) \cos ^2(c+d x) \left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right ) \sqrt{\frac{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2} \left (1-\frac{\sqrt{a}}{\sqrt{a+b}}\right )\right )}{2 a^{3/4} d (a+b)^{3/4} \sqrt{a+b \sin ^4(c+d x)}}-\frac{\sqrt [4]{a+b} \cos ^2(c+d x) \left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right ) \sqrt{\frac{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2} \left (1-\frac{\sqrt{a}}{\sqrt{a+b}}\right )\right )}{a^{3/4} d \sqrt{a+b \sin ^4(c+d x)}}+\frac{\sqrt{a+b} \sin (c+d x) \cos (c+d x) \left ((a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}{a d \sqrt{a+b \sin ^4(c+d x)} \left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right )}-\frac{\cos ^2(c+d x) \cot (c+d x) \left ((a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}{a d \sqrt{a+b \sin ^4(c+d x)}} \]

[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 + Sqrt[a]*Sqrt[a + b])*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)*(a + b)^(3/4)*d*Sqrt[a
+ b*Sin[c + d*x]^4])

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Rubi [A]  time = 0.419335, antiderivative size = 493, 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 = {3219, 1281, 1197, 1103, 1195} \[ \frac{\left (\sqrt{a} \sqrt{a+b}+a+b\right ) \cos ^2(c+d x) \left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right ) \sqrt{\frac{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2} \left (1-\frac{\sqrt{a}}{\sqrt{a+b}}\right )\right )}{2 a^{3/4} d (a+b)^{3/4} \sqrt{a+b \sin ^4(c+d x)}}-\frac{\sqrt [4]{a+b} \cos ^2(c+d x) \left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right ) \sqrt{\frac{(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2} \left (1-\frac{\sqrt{a}}{\sqrt{a+b}}\right )\right )}{a^{3/4} d \sqrt{a+b \sin ^4(c+d x)}}+\frac{\sqrt{a+b} \sin (c+d x) \cos (c+d x) \left ((a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}{a d \sqrt{a+b \sin ^4(c+d x)} \left (\sqrt{a+b} \tan ^2(c+d x)+\sqrt{a}\right )}-\frac{\cos ^2(c+d x) \cot (c+d x) \left ((a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}{a d \sqrt{a+b \sin ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[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 + Sqrt[a]*Sqrt[a + b])*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)*(a + b)^(3/4)*d*Sqrt[a
+ b*Sin[c + d*x]^4])

Rule 3219

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_), x_Symbol] :> With[{ff = FreeFa
ctors[Tan[e + f*x], x]}, Dist[(ff^(m + 1)*(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[(x^m*ExpandToSum[a*(1 + ff^2*x^2)^2 + b*ff^4*x^4, x]^p)/(1 +
ff^2*x^2)^(m/2 + 2*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integ
erQ[p - 1/2]

Rule 1281

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

Rule 1197

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

Rule 1103

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)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

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)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{\csc ^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 ) \operatorname{Subst}\left (\int \frac{1+x^2}{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 ) \operatorname{Subst}\left (\int \frac{-a+(-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 (\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 ) \operatorname{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{\left (\left (a+b+\sqrt{a} \sqrt{a+b}\right ) \cos ^2(c+d x) \sqrt{a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{\sqrt{a} \sqrt{a+b} 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{\left (a+b+\sqrt{a} \sqrt{a+b}\right ) \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} (a+b)^{3/4} d \sqrt{a+b \sin ^4(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 16.2194, size = 498, normalized size = 1.01 \[ -\frac{\cot (c+d x) \sqrt{8 a-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))+3 b}}{2 \sqrt{2} a d}-\frac{\sqrt{1-\frac{i \sqrt{b}}{\sqrt{a}}} \tan (c+d x) \left (a \left (\tan ^2(c+d x)+1\right )^2+b \tan ^4(c+d x)\right )-\sqrt{a} \sqrt{b} \left (\tan ^2(c+d x)+1\right ) \sqrt{1+\left (1-\frac{i \sqrt{b}}{\sqrt{a}}\right ) \tan ^2(c+d x)} \sqrt{1+\left (1+\frac{i \sqrt{b}}{\sqrt{a}}\right ) \tan ^2(c+d x)} F\left (i \sinh ^{-1}\left (\sqrt{1-\frac{i \sqrt{b}}{\sqrt{a}}} \tan (c+d x)\right )|\frac{\sqrt{a}+i \sqrt{b}}{\sqrt{a}-i \sqrt{b}}\right )+\sqrt{a} \left (\sqrt{b}+i \sqrt{a}\right ) \left (\tan ^2(c+d x)+1\right ) \sqrt{1+\left (1-\frac{i \sqrt{b}}{\sqrt{a}}\right ) \tan ^2(c+d x)} \sqrt{1+\left (1+\frac{i \sqrt{b}}{\sqrt{a}}\right ) \tan ^2(c+d x)} E\left (i \sinh ^{-1}\left (\sqrt{1-\frac{i \sqrt{b}}{\sqrt{a}}} \tan (c+d x)\right )|\frac{\sqrt{a}+i \sqrt{b}}{\sqrt{a}-i \sqrt{b}}\right )}{a d \sqrt{1-\frac{i \sqrt{b}}{\sqrt{a}}} \left (\tan ^2(c+d x)+1\right )^2 \sqrt{\frac{a \left (\tan ^2(c+d x)+1\right )^2+b \tan ^4(c+d x)}{\left (\tan ^2(c+d x)+1\right )^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.707, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \csc \left ( dx+c \right ) \right ) ^{2}{\frac{1}{\sqrt{a+b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (d x + c\right )^{2}}{\sqrt{b \sin \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\csc \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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(csc(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., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{2}{\left (c + d x \right )}}{\sqrt{a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (d x + c\right )^{2}}{\sqrt{b \sin \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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