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]

((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))

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Rubi [A]  time = 0.17485, antiderivative size = 128, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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 31

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

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*}

Mathematica [A]  time = 0.230153, 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]

-((-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]))/(2*(a - b + c)*(a + b + c)*Sqrt[b^2
- 4*a*c])

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Maple [A]  time = 0.135, size = 224, normalized size = 1.8 \begin{align*} -{\frac{b\ln \left ( a+b\sin \left ( x \right ) +c \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }{ \left ( 2\,a-2\,b+2\,c \right ) \left ( a+b+c \right ) }}+2\,{\frac{ca}{ \left ( a-b+c \right ) \left ( a+b+c \right ) \sqrt{4\,ca-{b}^{2}}}\arctan \left ({\frac{b+2\,c\sin \left ( x \right ) }{\sqrt{4\,ca-{b}^{2}}}} \right ) }-{\frac{{b}^{2}}{ \left ( a-b+c \right ) \left ( a+b+c \right ) }\arctan \left ({(b+2\,c\sin \left ( x \right ) ){\frac{1}{\sqrt{4\,ca-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ca-{b}^{2}}}}}+2\,{\frac{{c}^{2}}{ \left ( a-b+c \right ) \left ( a+b+c \right ) \sqrt{4\,ca-{b}^{2}}}\arctan \left ({\frac{b+2\,c\sin \left ( x \right ) }{\sqrt{4\,ca-{b}^{2}}}} \right ) }-{\frac{\ln \left ( -1+\sin \left ( x \right ) \right ) }{2\,a+2\,b+2\,c}}+{\frac{\ln \left ( 1+\sin \left ( x \right ) \right ) }{2\,a-2\,b+2\,c}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 7.88381, size = 1096, normalized size = 8.56 \begin{align*} \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 \left (x\right )^{2} - 2 \, b c \sin \left (x\right ) - b^{2} + 2 \, a c - 2 \, c^{2} + \sqrt{b^{2} - 4 \, a c}{\left (2 \, c \sin \left (x\right ) + b\right )}}{c \cos \left (x\right )^{2} - b \sin \left (x\right ) - a - c}\right ) +{\left (b^{3} - 4 \, a b c\right )} \log \left (-c \cos \left (x\right )^{2} + b \sin \left (x\right ) + 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 \left (x\right ) + 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 \left (x\right ) + 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 \left (x\right ) + b\right )}}{b^{2} - 4 \, a c}\right ) -{\left (b^{3} - 4 \, a b c\right )} \log \left (-c \cos \left (x\right )^{2} + b \sin \left (x\right ) + 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 \left (x\right ) + 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 \left (x\right ) + 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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec{\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)/(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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Giac [A]  time = 1.14211, size = 177, normalized size = 1.38 \begin{align*} -\frac{b \log \left (c \sin \left (x\right )^{2} + b \sin \left (x\right ) + 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 \left (x\right ) + 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 \left (x\right ) + 1\right )}{2 \,{\left (a - b + c\right )}} - \frac{\log \left (-\sin \left (x\right ) + 1\right )}{2 \,{\left (a + b + c\right )}} \end{align*}

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)