3.13 \(\int \frac{\sec ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\)

Optimal. Leaf size=324 \[ -\frac{\sqrt{2} b c \left (\frac{b^2-2 c (a+c)}{b \sqrt{b^2-4 a c}}+1\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (b-\sqrt{b^2-4 a c}\right )+2 c}{\sqrt{2} \sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{(a-b+c) (a+b+c) \sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}-\frac{\sqrt{2} b c \left (1-\frac{b^2-2 c (a+c)}{b \sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (\sqrt{b^2-4 a c}+b\right )+2 c}{\sqrt{2} \sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{(a-b+c) (a+b+c) \sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}+\frac{\cos (x)}{2 (1-\sin (x)) (a+b+c)}-\frac{\cos (x)}{2 (\sin (x)+1) (a-b+c)} \]

[Out]

-((Sqrt[2]*b*c*(1 + (b^2 - 2*c*(a + c))/(b*Sqrt[b^2 - 4*a*c]))*ArcTan[(2*c + (b - Sqrt[b^2 - 4*a*c])*Tan[x/2])
/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]])])/((a - b + c)*(a + b + c)*Sqrt[b^2 - 2*c*(a + c) - b
*Sqrt[b^2 - 4*a*c]])) - (Sqrt[2]*b*c*(1 - (b^2 - 2*c*(a + c))/(b*Sqrt[b^2 - 4*a*c]))*ArcTan[(2*c + (b + Sqrt[b
^2 - 4*a*c])*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]])])/((a - b + c)*(a + b + c)*Sqrt
[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]) + Cos[x]/(2*(a + b + c)*(1 - Sin[x])) - Cos[x]/(2*(a - b + c)*(1 +
Sin[x]))

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Rubi [A]  time = 2.26767, antiderivative size = 324, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3266, 2648, 3292, 2660, 618, 204} \[ -\frac{\sqrt{2} b c \left (\frac{b^2-2 c (a+c)}{b \sqrt{b^2-4 a c}}+1\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (b-\sqrt{b^2-4 a c}\right )+2 c}{\sqrt{2} \sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{(a-b+c) (a+b+c) \sqrt{-b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}-\frac{\sqrt{2} b c \left (1-\frac{b^2-2 c (a+c)}{b \sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right ) \left (\sqrt{b^2-4 a c}+b\right )+2 c}{\sqrt{2} \sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}\right )}{(a-b+c) (a+b+c) \sqrt{b \sqrt{b^2-4 a c}-2 c (a+c)+b^2}}+\frac{\cos (x)}{2 (1-\sin (x)) (a+b+c)}-\frac{\cos (x)}{2 (\sin (x)+1) (a-b+c)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^2/(a + b*Sin[x] + c*Sin[x]^2),x]

[Out]

-((Sqrt[2]*b*c*(1 + (b^2 - 2*c*(a + c))/(b*Sqrt[b^2 - 4*a*c]))*ArcTan[(2*c + (b - Sqrt[b^2 - 4*a*c])*Tan[x/2])
/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]])])/((a - b + c)*(a + b + c)*Sqrt[b^2 - 2*c*(a + c) - b
*Sqrt[b^2 - 4*a*c]])) - (Sqrt[2]*b*c*(1 - (b^2 - 2*c*(a + c))/(b*Sqrt[b^2 - 4*a*c]))*ArcTan[(2*c + (b + Sqrt[b
^2 - 4*a*c])*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]])])/((a - b + c)*(a + b + c)*Sqrt
[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]) + Cos[x]/(2*(a + b + c)*(1 - Sin[x])) - Cos[x]/(2*(a - b + c)*(1 +
Sin[x]))

Rule 3266

Int[cos[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]
^(n2_.))^(p_.), x_Symbol] :> Int[ExpandTrig[(1 - sin[d + e*x]^2)^(m/2)*(a + b*sin[d + e*x]^n + c*sin[d + e*x]^
(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && IntegerQ[m/2] && NeQ[b^2 - 4*a*c, 0] && Integ
ersQ[n, p]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3292

Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)] + (c_.)*sin[(d_.) + (e_.)*(x
_)]^2), x_Symbol] :> Module[{q = Rt[b^2 - 4*a*c, 2]}, Dist[B + (b*B - 2*A*c)/q, Int[1/(b + q + 2*c*Sin[d + e*x
]), x], x] + Dist[B - (b*B - 2*A*c)/q, Int[1/(b - q + 2*c*Sin[d + e*x]), x], x]] /; FreeQ[{a, b, c, d, e, A, B
}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx &=\int \left (-\frac{1}{2 (a+b+c) (-1+\sin (x))}+\frac{1}{2 (a-b+c) (1+\sin (x))}+\frac{-b^2 \left (1-\frac{c (a+c)}{b^2}\right )-b c \sin (x)}{(a-b+c) (a+b+c) \left (a+b \sin (x)+c \sin ^2(x)\right )}\right ) \, dx\\ &=\frac{\int \frac{1}{1+\sin (x)} \, dx}{2 (a-b+c)}-\frac{\int \frac{1}{-1+\sin (x)} \, dx}{2 (a+b+c)}+\frac{\int \frac{-b^2 \left (1-\frac{c (a+c)}{b^2}\right )-b c \sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx}{(a-b+c) (a+b+c)}\\ &=\frac{\cos (x)}{2 (a+b+c) (1-\sin (x))}-\frac{\cos (x)}{2 (a-b+c) (1+\sin (x))}-\frac{\left (c \left (b+\frac{b^2-2 c (a+c)}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{b-\sqrt{b^2-4 a c}+2 c \sin (x)} \, dx}{(a-b+c) (a+b+c)}-\frac{\left (b c \left (1-\frac{b^2-2 c (a+c)}{b \sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{b+\sqrt{b^2-4 a c}+2 c \sin (x)} \, dx}{(a-b+c) (a+b+c)}\\ &=\frac{\cos (x)}{2 (a+b+c) (1-\sin (x))}-\frac{\cos (x)}{2 (a-b+c) (1+\sin (x))}-\frac{\left (2 c \left (b+\frac{b^2-2 c (a+c)}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b-\sqrt{b^2-4 a c}+4 c x+\left (b-\sqrt{b^2-4 a c}\right ) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{(a-b+c) (a+b+c)}-\frac{\left (2 b c \left (1-\frac{b^2-2 c (a+c)}{b \sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b+\sqrt{b^2-4 a c}+4 c x+\left (b+\sqrt{b^2-4 a c}\right ) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{(a-b+c) (a+b+c)}\\ &=\frac{\cos (x)}{2 (a+b+c) (1-\sin (x))}-\frac{\cos (x)}{2 (a-b+c) (1+\sin (x))}+\frac{\left (4 c \left (b+\frac{b^2-2 c (a+c)}{\sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-8 \left (b^2-2 c (a+c)-b \sqrt{b^2-4 a c}\right )-x^2} \, dx,x,4 c+2 \left (b-\sqrt{b^2-4 a c}\right ) \tan \left (\frac{x}{2}\right )\right )}{(a-b+c) (a+b+c)}+\frac{\left (4 b c \left (1-\frac{b^2-2 c (a+c)}{b \sqrt{b^2-4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (4 c^2-\left (b+\sqrt{b^2-4 a c}\right )^2\right )-x^2} \, dx,x,4 c+2 \left (b+\sqrt{b^2-4 a c}\right ) \tan \left (\frac{x}{2}\right )\right )}{(a-b+c) (a+b+c)}\\ &=-\frac{\sqrt{2} c \left (b+\frac{b^2-2 c (a+c)}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{2 c+\left (b-\sqrt{b^2-4 a c}\right ) \tan \left (\frac{x}{2}\right )}{\sqrt{2} \sqrt{b^2-2 c (a+c)-b \sqrt{b^2-4 a c}}}\right )}{(a-b+c) (a+b+c) \sqrt{b^2-2 c (a+c)-b \sqrt{b^2-4 a c}}}-\frac{\sqrt{2} b c \left (1-\frac{b^2-2 c (a+c)}{b \sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{2 c+\left (b+\sqrt{b^2-4 a c}\right ) \tan \left (\frac{x}{2}\right )}{\sqrt{2} \sqrt{b^2-2 c (a+c)+b \sqrt{b^2-4 a c}}}\right )}{(a-b+c) (a+b+c) \sqrt{b^2-2 c (a+c)+b \sqrt{b^2-4 a c}}}+\frac{\cos (x)}{2 (a+b+c) (1-\sin (x))}-\frac{\cos (x)}{2 (a-b+c) (1+\sin (x))}\\ \end{align*}

Mathematica [C]  time = 0.971924, size = 407, normalized size = 1.26 \[ -\frac{c \left (b \sqrt{4 a c-b^2}+2 i c (a+c)-i b^2\right ) \tan ^{-1}\left (\frac{2 c+\tan \left (\frac{x}{2}\right ) \left (b-i \sqrt{4 a c-b^2}\right )}{\sqrt{2} \sqrt{-i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}\right )}{\sqrt{2 a c-\frac{b^2}{2}} \left (a^2+2 a c-b^2+c^2\right ) \sqrt{-i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}-\frac{c \left (b \sqrt{4 a c-b^2}-2 i c (a+c)+i b^2\right ) \tan ^{-1}\left (\frac{2 c+\tan \left (\frac{x}{2}\right ) \left (b+i \sqrt{4 a c-b^2}\right )}{\sqrt{2} \sqrt{i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}\right )}{\sqrt{2 a c-\frac{b^2}{2}} \left (a^2+2 a c-b^2+c^2\right ) \sqrt{i b \sqrt{4 a c-b^2}-2 c (a+c)+b^2}}+\frac{\sin \left (\frac{x}{2}\right )}{(a+b+c) \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )}+\frac{\sin \left (\frac{x}{2}\right )}{(a-b+c) \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^2/(a + b*Sin[x] + c*Sin[x]^2),x]

[Out]

-((c*((-I)*b^2 + (2*I)*c*(a + c) + b*Sqrt[-b^2 + 4*a*c])*ArcTan[(2*c + (b - I*Sqrt[-b^2 + 4*a*c])*Tan[x/2])/(S
qrt[2]*Sqrt[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]])])/(Sqrt[-b^2/2 + 2*a*c]*(a^2 - b^2 + 2*a*c + c^2)*Sqr
t[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]])) - (c*(I*b^2 - (2*I)*c*(a + c) + b*Sqrt[-b^2 + 4*a*c])*ArcTan[(
2*c + (b + I*Sqrt[-b^2 + 4*a*c])*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqrt[-b^2 + 4*a*c]])])/(Sqrt[
-b^2/2 + 2*a*c]*(a^2 - b^2 + 2*a*c + c^2)*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqrt[-b^2 + 4*a*c]]) + Sin[x/2]/((a + b
+ c)*(Cos[x/2] - Sin[x/2])) + Sin[x/2]/((a - b + c)*(Cos[x/2] + Sin[x/2]))

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Maple [B]  time = 0.168, size = 1934, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^2/(a+b*sin(x)+c*sin(x)^2),x)

[Out]

-6/(a-b+c)/(a+b+c)*a/(4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*
a*c+b^2)^(1/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*(-4*a*c+b^2)^(1/2)*b*c+2/(a-b+c)/(a+b+c)/(
4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*c
*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*(-4*a*c+b^2)^(1/2)*b^3-2/(a-b+c)/(a+b+c)/(4*a*c-b^2)/(4*c*a-2*b^
2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+b
^2)^(1/2)+4*a^2)^(1/2))*(-4*a*c+b^2)^(1/2)*b*c^2-8/(a-b+c)/(a+b+c)*a^2/(4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^
2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)
^(1/2))*c^2+10/(a-b+c)/(a+b+c)*a/(4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan
(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*b^2*c-8/(a-b+c)/(a+b+c)*a/(4*a
*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*c*a-
2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*c^3-2/(a-b+c)/(a+b+c)/(4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/
2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)
)*b^4+2/(a-b+c)/(a+b+c)/(4*a*c-b^2)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(
-4*a*c+b^2)^(1/2)-b)/(4*c*a-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*c^2*b^2-6/(a-b+c)/(a+b+c)*a/(4*a*c-b^2)
/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*
b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*(-4*a*c+b^2)^(1/2)*b*c+2/(a-b+c)/(a+b+c)/(4*a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a
*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*
a^2)^(1/2))*(-4*a*c+b^2)^(1/2)*b^3-2/(a-b+c)/(a+b+c)/(4*a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1
/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*(-4*a*c+b^2
)^(1/2)*b*c^2+8/(a-b+c)/(a+b+c)*a^2/(4*a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*t
an(1/2*x)+b+(-4*a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*c^2-10/(a-b+c)/(a+b+c)*a/(4*
a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2))/(4*c*a-
2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*b^2*c+8/(a-b+c)/(a+b+c)*a/(4*a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)
^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1
/2))*c^3+2/(a-b+c)/(a+b+c)/(4*a*c-b^2)/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)
+b+(-4*a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*b^4-2/(a-b+c)/(a+b+c)/(4*a*c-b^2)/(4*
c*a-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2)^(1/2))/(4*c*a-2*b^2-2*b*(-
4*a*c+b^2)^(1/2)+4*a^2)^(1/2))*c^2*b^2-2/(2*a-2*b+2*c)/(tan(1/2*x)+1)-2/(2*a+2*b+2*c)/(tan(1/2*x)-1)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(a+b*sin(x)+c*sin(x)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(a+b*sin(x)+c*sin(x)^2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (x \right )}}{a + b \sin{\left (x \right )} + c \sin ^{2}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**2/(a+b*sin(x)+c*sin(x)**2),x)

[Out]

Integral(sec(x)**2/(a + b*sin(x) + c*sin(x)**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(a+b*sin(x)+c*sin(x)^2),x, algorithm="giac")

[Out]

Timed out