3.2.51 \(\int \frac {\csc ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [151]

Optimal. Leaf size=177 \[ -\frac {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a f}-\frac {\sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 177, 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 = {3267, 491, 12, 507, 437, 435, 432, 430} \begin {gather*} \frac {\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 {\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 )}{a f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {\cot (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[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])/(a*f)) - (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*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) + (Sqrt[Cos[e + f*x]^2]*Ell
ipticF[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 12

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

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 491

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 + 1)/(a*c*e*(m + 1))), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 507

Int[(x_)^(n_)/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[1/b, Int[Sqrt[a +
b*x^n]/Sqrt[c + d*x^n], x], x] - Dist[a/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b, c,
d}, x] && NeQ[b*c - a*d, 0] && (EqQ[n, 2] || EqQ[n, 4]) &&  !(EqQ[n, 2] && SimplerSqrtQ[-b/a, -d/c])

Rule 3267

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), 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/Sqrt[1 - ff^2*x^2]), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !In
tegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.36, size = 138, normalized size = 0.78 \begin {gather*} \frac {-\sqrt {2} (2 a+b-b \cos (2 (e+f x))) \cot (e+f x)-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 )+2 a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )}{2 a f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^2/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

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

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Maple [A]
time = 6.60, size = 140, normalized size = 0.79

method	result	size
default	\(\frac {\sin \left (f x +e \right ) \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, a \left (\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-\EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )\right )+b \left (\cos ^{4}\left (f x +e \right )\right )+\left (-a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{a \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 0.13, size = 643, normalized size = 3.63 \begin {gather*} -\frac {-4 i \, \sqrt {-b} b \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} \sqrt {\frac {a^{2} + a b}{b^{2}}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) \sin \left (f x + e\right ) + 4 i \, \sqrt {-b} b \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} \sqrt {\frac {a^{2} + a b}{b^{2}}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) \sin \left (f x + e\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right ) + {\left (2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} \sin \left (f x + e\right ) + {\left (2 i \, a + i \, b\right )} \sqrt {-b} \sin \left (f x + e\right )\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (-2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} \sin \left (f x + e\right ) + {\left (-2 i \, a - i \, b\right )} \sqrt {-b} \sin \left (f x + e\right )\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}})}{2 \, a b f \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-4*I*sqrt(-b)*b*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*sqrt((a^2 + a*b)/b^2)*elliptic_f(arcsin(sq
rt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b
+ b^2)*sqrt((a^2 + a*b)/b^2))/b^2)*sin(f*x + e) + 4*I*sqrt(-b)*b*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)
*sqrt((a^2 + a*b)/b^2)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) - I*sin(f
*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2)*sin(f*x + e) + 2*sqrt(-b*cos(f*x
+ e)^2 + a + b)*b*cos(f*x + e) + (2*I*sqrt(-b)*b*sqrt((a^2 + a*b)/b^2)*sin(f*x + e) + (2*I*a + I*b)*sqrt(-b)*
sin(f*x + e))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2)
+ 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/
b^2) + (-2*I*sqrt(-b)*b*sqrt((a^2 + a*b)/b^2)*sin(f*x + e) + (-2*I*a - I*b)*sqrt(-b)*sin(f*x + e))*sqrt((2*b*s
qrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x +
e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2))/(a*b*f*sin(f*x + e)
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

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

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