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

Optimal. Leaf size=128 \[ \frac {\left (-2 a c+b^2-2 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c \sin (x)}{\sqrt {b^2-4 a c}}\right )}{(a-b+c) (a+b+c) \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 (a-b+c) (a+b+c)}-\frac {\log (1-\sin (x))}{2 (a+b+c)}+\frac {\log (\sin (x)+1)}{2 (a-b+c)} \]

[Out]

-1/2*ln(1-sin(x))/(a+b+c)+1/2*ln(1+sin(x))/(a-b+c)-1/2*b*ln(a+b*sin(x)+c*sin(x)^2)/(a-b+c)/(a+b+c)+(-2*a*c+b^2
-2*c^2)*arctanh((b+2*c*sin(x))/(-4*a*c+b^2)^(1/2))/(a-b+c)/(a+b+c)/(-4*a*c+b^2)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3258, 981, 634, 618, 206, 628, 633, 31} \[ \frac {\left (-2 a c+b^2-2 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c \sin (x)}{\sqrt {b^2-4 a c}}\right )}{(a-b+c) (a+b+c) \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 (a-b+c) (a+b+c)}-\frac {\log (1-\sin (x))}{2 (a+b+c)}+\frac {\log (\sin (x)+1)}{2 (a-b+c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 2*a*c - 2*c^2)*ArcTanh[(b + 2*c*Sin[x])/Sqrt[b^2 - 4*a*c]])/((a - b + c)*(a + b + c)*Sqrt[b^2 - 4*a*c]
) - Log[1 - Sin[x]]/(2*(a + b + c)) + Log[1 + Sin[x]]/(2*(a - b + c)) - (b*Log[a + b*Sin[x] + c*Sin[x]^2])/(2*
(a - b + c)*(a + b + c))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 628

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

Rule 633

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

Rule 634

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

Rule 981

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

Rule 3258

Int[cos[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*sin[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*sin[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_.), x_Symbol] :> Module[{g = FreeFactors[Sin[d + e*x], x]}, Dist[g/e, Subst[Int[(1
- g^2*x^2)^((m - 1)/2)*(a + b*(f*g*x)^n + c*(f*g*x)^(2*n))^p, x], x, Sin[d + e*x]/g], x]] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] && EqQ[n2, 2*n] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 119, normalized size = 0.93 \[ -\frac {\sqrt {b^2-4 a c} \left (b \log \left (a+b \sin (x)+c \sin ^2(x)\right )+(a-b+c) \log (1-\sin (x))-(a+b+c) \log (\sin (x)+1)\right )+\left (4 c (a+c)-2 b^2\right ) \tanh ^{-1}\left (\frac {b+2 c \sin (x)}{\sqrt {b^2-4 a c}}\right )}{2 (a-b+c) (a+b+c) \sqrt {b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((-2*b^2 + 4*c*(a + c))*ArcTanh[(b + 2*c*Sin[x])/Sqrt[b^2 - 4*a*c]] + Sqrt[b^2 - 4*a*c]*((a - b + c)*Log[
1 - Sin[x]] - (a + b + c)*Log[1 + Sin[x]] + b*Log[a + b*Sin[x] + c*Sin[x]^2]))/((a - b + c)*(a + b + c)*Sqrt[b
^2 - 4*a*c])

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fricas [A]  time = 1.29, size = 482, normalized size = 3.77 \[ \left [-\frac {{\left (b^{2} - 2 \, a c - 2 \, c^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (-\frac {2 \, c^{2} \cos \relax (x)^{2} - 2 \, b c \sin \relax (x) - b^{2} + 2 \, a c - 2 \, c^{2} + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c \sin \relax (x) + b\right )}}{c \cos \relax (x)^{2} - b \sin \relax (x) - a - c}\right ) + {\left (b^{3} - 4 \, a b c\right )} \log \left (-c \cos \relax (x)^{2} + b \sin \relax (x) + a + c\right ) - {\left (a b^{2} + b^{3} - 4 \, a c^{2} - {\left (4 \, a^{2} + 4 \, a b - b^{2}\right )} c\right )} \log \left (\sin \relax (x) + 1\right ) + {\left (a b^{2} - b^{3} - 4 \, a c^{2} - {\left (4 \, a^{2} - 4 \, a b - b^{2}\right )} c\right )} \log \left (-\sin \relax (x) + 1\right )}{2 \, {\left (a^{2} b^{2} - b^{4} - 4 \, a c^{3} - {\left (8 \, a^{2} - b^{2}\right )} c^{2} - 2 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} c\right )}}, \frac {2 \, {\left (b^{2} - 2 \, a c - 2 \, c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c \sin \relax (x) + b\right )}}{b^{2} - 4 \, a c}\right ) - {\left (b^{3} - 4 \, a b c\right )} \log \left (-c \cos \relax (x)^{2} + b \sin \relax (x) + a + c\right ) + {\left (a b^{2} + b^{3} - 4 \, a c^{2} - {\left (4 \, a^{2} + 4 \, a b - b^{2}\right )} c\right )} \log \left (\sin \relax (x) + 1\right ) - {\left (a b^{2} - b^{3} - 4 \, a c^{2} - {\left (4 \, a^{2} - 4 \, a b - b^{2}\right )} c\right )} \log \left (-\sin \relax (x) + 1\right )}{2 \, {\left (a^{2} b^{2} - b^{4} - 4 \, a c^{3} - {\left (8 \, a^{2} - b^{2}\right )} c^{2} - 2 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} c\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((b^2 - 2*a*c - 2*c^2)*sqrt(b^2 - 4*a*c)*log(-(2*c^2*cos(x)^2 - 2*b*c*sin(x) - b^2 + 2*a*c - 2*c^2 + sqr
t(b^2 - 4*a*c)*(2*c*sin(x) + b))/(c*cos(x)^2 - b*sin(x) - a - c)) + (b^3 - 4*a*b*c)*log(-c*cos(x)^2 + b*sin(x)
+ a + c) - (a*b^2 + b^3 - 4*a*c^2 - (4*a^2 + 4*a*b - b^2)*c)*log(sin(x) + 1) + (a*b^2 - b^3 - 4*a*c^2 - (4*a^
2 - 4*a*b - b^2)*c)*log(-sin(x) + 1))/(a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c), 1
/2*(2*(b^2 - 2*a*c - 2*c^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*sin(x) + b)/(b^2 - 4*a*c)) - (b
^3 - 4*a*b*c)*log(-c*cos(x)^2 + b*sin(x) + a + c) + (a*b^2 + b^3 - 4*a*c^2 - (4*a^2 + 4*a*b - b^2)*c)*log(sin(
x) + 1) - (a*b^2 - b^3 - 4*a*c^2 - (4*a^2 - 4*a*b - b^2)*c)*log(-sin(x) + 1))/(a^2*b^2 - b^4 - 4*a*c^3 - (8*a^
2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)]

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giac [A]  time = 1.23, size = 131, normalized size = 1.02 \[ -\frac {b \log \left (c \sin \relax (x)^{2} + b \sin \relax (x) + a\right )}{2 \, {\left (a^{2} - b^{2} + 2 \, a c + c^{2}\right )}} - \frac {{\left (b^{2} - 2 \, a c - 2 \, c^{2}\right )} \arctan \left (\frac {2 \, c \sin \relax (x) + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{2} - b^{2} + 2 \, a c + c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {\log \left (\sin \relax (x) + 1\right )}{2 \, {\left (a - b + c\right )}} - \frac {\log \left (-\sin \relax (x) + 1\right )}{2 \, {\left (a + b + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*log(c*sin(x)^2 + b*sin(x) + a)/(a^2 - b^2 + 2*a*c + c^2) - (b^2 - 2*a*c - 2*c^2)*arctan((2*c*sin(x) + b
)/sqrt(-b^2 + 4*a*c))/((a^2 - b^2 + 2*a*c + c^2)*sqrt(-b^2 + 4*a*c)) + 1/2*log(sin(x) + 1)/(a - b + c) - 1/2*l
og(-sin(x) + 1)/(a + b + c)

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maple [A]  time = 0.29, size = 224, normalized size = 1.75 \[ -\frac {\ln \left (-1+\sin \relax (x )\right )}{2 a +2 b +2 c}-\frac {b \ln \left (a +b \sin \relax (x )+c \left (\sin ^{2}\relax (x )\right )\right )}{2 \left (a -b +c \right ) \left (a +b +c \right )}+\frac {2 \arctan \left (\frac {b +2 c \sin \relax (x )}{\sqrt {4 c a -b^{2}}}\right ) c a}{\left (a -b +c \right ) \left (a +b +c \right ) \sqrt {4 c a -b^{2}}}-\frac {\arctan \left (\frac {b +2 c \sin \relax (x )}{\sqrt {4 c a -b^{2}}}\right ) b^{2}}{\left (a -b +c \right ) \left (a +b +c \right ) \sqrt {4 c a -b^{2}}}+\frac {2 \arctan \left (\frac {b +2 c \sin \relax (x )}{\sqrt {4 c a -b^{2}}}\right ) c^{2}}{\left (a -b +c \right ) \left (a +b +c \right ) \sqrt {4 c a -b^{2}}}+\frac {\ln \left (1+\sin \relax (x )\right )}{2 a -2 b +2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*a+2*b+2*c)*ln(-1+sin(x))-1/2*b*ln(a+b*sin(x)+c*sin(x)^2)/(a-b+c)/(a+b+c)+2/(a-b+c)/(a+b+c)/(4*a*c-b^2)^(
1/2)*arctan((b+2*c*sin(x))/(4*a*c-b^2)^(1/2))*c*a-1/(a-b+c)/(a+b+c)/(4*a*c-b^2)^(1/2)*arctan((b+2*c*sin(x))/(4
*a*c-b^2)^(1/2))*b^2+2/(a-b+c)/(a+b+c)/(4*a*c-b^2)^(1/2)*arctan((b+2*c*sin(x))/(4*a*c-b^2)^(1/2))*c^2+1/(2*a-2
*b+2*c)*ln(1+sin(x))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 17.35, size = 1001, normalized size = 7.82 \[ \frac {\ln \left (\sin \relax (x)+1\right )}{2\,\left (a-b+c\right )}-\frac {\ln \left (\sin \relax (x)-1\right )}{2\,\left (a+b+c\right )}+\frac {\ln \left (4\,c^3\,\sin \relax (x)+b\,c^2+\frac {\left (a\,\left (4\,b\,c-2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3+b^2\,\sqrt {b^2-4\,a\,c}-2\,c^2\,\sqrt {b^2-4\,a\,c}\right )\,\left (8\,a\,c^3+\sin \relax (x)\,\left (-3\,b^3\,c+12\,b\,c^3+12\,a\,b\,c^2\right )+4\,c^4+4\,a^2\,c^2+3\,b^2\,c^2-\frac {\left (a\,\left (4\,b\,c-2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3+b^2\,\sqrt {b^2-4\,a\,c}-2\,c^2\,\sqrt {b^2-4\,a\,c}\right )\,\left (\sin \relax (x)\,\left (-8\,a^3\,c^2+2\,a^2\,b^2\,c-8\,a^2\,c^3-20\,a\,b^2\,c^2+8\,a\,c^4+6\,b^4\,c-6\,b^2\,c^3+8\,c^5\right )+4\,b\,c^4+4\,b^3\,c^2-28\,a^2\,b\,c^2-24\,a\,b\,c^3+8\,a\,b^3\,c\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )}-a\,b^2\,c\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )}\right )\,\left (a\,\left (4\,b\,c-2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3+b^2\,\sqrt {b^2-4\,a\,c}-2\,c^2\,\sqrt {b^2-4\,a\,c}\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )}+\frac {\ln \left (4\,c^3\,\sin \relax (x)+b\,c^2+\frac {\left (a\,\left (4\,b\,c+2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3-b^2\,\sqrt {b^2-4\,a\,c}+2\,c^2\,\sqrt {b^2-4\,a\,c}\right )\,\left (8\,a\,c^3+\sin \relax (x)\,\left (-3\,b^3\,c+12\,b\,c^3+12\,a\,b\,c^2\right )+4\,c^4+4\,a^2\,c^2+3\,b^2\,c^2-\frac {\left (a\,\left (4\,b\,c+2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3-b^2\,\sqrt {b^2-4\,a\,c}+2\,c^2\,\sqrt {b^2-4\,a\,c}\right )\,\left (\sin \relax (x)\,\left (-8\,a^3\,c^2+2\,a^2\,b^2\,c-8\,a^2\,c^3-20\,a\,b^2\,c^2+8\,a\,c^4+6\,b^4\,c-6\,b^2\,c^3+8\,c^5\right )+4\,b\,c^4+4\,b^3\,c^2-28\,a^2\,b\,c^2-24\,a\,b\,c^3+8\,a\,b^3\,c\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )}-a\,b^2\,c\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )}\right )\,\left (a\,\left (4\,b\,c+2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3-b^2\,\sqrt {b^2-4\,a\,c}+2\,c^2\,\sqrt {b^2-4\,a\,c}\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)*(a + c*sin(x)^2 + b*sin(x))),x)

[Out]

log(sin(x) + 1)/(2*(a - b + c)) - log(sin(x) - 1)/(2*(a + b + c)) + (log(4*c^3*sin(x) + b*c^2 + ((a*(4*b*c - 2
*c*(b^2 - 4*a*c)^(1/2)) - b^3 + b^2*(b^2 - 4*a*c)^(1/2) - 2*c^2*(b^2 - 4*a*c)^(1/2))*(8*a*c^3 + sin(x)*(12*b*c
^3 - 3*b^3*c + 12*a*b*c^2) + 4*c^4 + 4*a^2*c^2 + 3*b^2*c^2 - ((a*(4*b*c - 2*c*(b^2 - 4*a*c)^(1/2)) - b^3 + b^2
*(b^2 - 4*a*c)^(1/2) - 2*c^2*(b^2 - 4*a*c)^(1/2))*(sin(x)*(8*a*c^4 + 6*b^4*c + 8*c^5 - 8*a^2*c^3 - 8*a^3*c^2 -
6*b^2*c^3 - 20*a*b^2*c^2 + 2*a^2*b^2*c) + 4*b*c^4 + 4*b^3*c^2 - 28*a^2*b*c^2 - 24*a*b*c^3 + 8*a*b^3*c))/(b^2*
(12*a*c + 2*a^2 - 2*b^2 + 2*c^2) - 4*a*c*(4*a*c + 2*a^2 + 2*c^2)) - a*b^2*c))/(b^2*(12*a*c + 2*a^2 - 2*b^2 + 2
*c^2) - 4*a*c*(4*a*c + 2*a^2 + 2*c^2)))*(a*(4*b*c - 2*c*(b^2 - 4*a*c)^(1/2)) - b^3 + b^2*(b^2 - 4*a*c)^(1/2) -
2*c^2*(b^2 - 4*a*c)^(1/2)))/(b^2*(12*a*c + 2*a^2 - 2*b^2 + 2*c^2) - 4*a*c*(4*a*c + 2*a^2 + 2*c^2)) + (log(4*c
^3*sin(x) + b*c^2 + ((a*(4*b*c + 2*c*(b^2 - 4*a*c)^(1/2)) - b^3 - b^2*(b^2 - 4*a*c)^(1/2) + 2*c^2*(b^2 - 4*a*c
)^(1/2))*(8*a*c^3 + sin(x)*(12*b*c^3 - 3*b^3*c + 12*a*b*c^2) + 4*c^4 + 4*a^2*c^2 + 3*b^2*c^2 - ((a*(4*b*c + 2*
c*(b^2 - 4*a*c)^(1/2)) - b^3 - b^2*(b^2 - 4*a*c)^(1/2) + 2*c^2*(b^2 - 4*a*c)^(1/2))*(sin(x)*(8*a*c^4 + 6*b^4*c
+ 8*c^5 - 8*a^2*c^3 - 8*a^3*c^2 - 6*b^2*c^3 - 20*a*b^2*c^2 + 2*a^2*b^2*c) + 4*b*c^4 + 4*b^3*c^2 - 28*a^2*b*c^
2 - 24*a*b*c^3 + 8*a*b^3*c))/(b^2*(12*a*c + 2*a^2 - 2*b^2 + 2*c^2) - 4*a*c*(4*a*c + 2*a^2 + 2*c^2)) - a*b^2*c)
)/(b^2*(12*a*c + 2*a^2 - 2*b^2 + 2*c^2) - 4*a*c*(4*a*c + 2*a^2 + 2*c^2)))*(a*(4*b*c + 2*c*(b^2 - 4*a*c)^(1/2))
- b^3 - b^2*(b^2 - 4*a*c)^(1/2) + 2*c^2*(b^2 - 4*a*c)^(1/2)))/(b^2*(12*a*c + 2*a^2 - 2*b^2 + 2*c^2) - 4*a*c*(
4*a*c + 2*a^2 + 2*c^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec {\relax (x )}}{a + b \sin {\relax (x )} + c \sin ^{2}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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