3.1.14 \(\int \frac {\sec ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) [14]

Optimal. Leaf size=206 \[ -\frac {\left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \tanh ^{-1}\left (\frac {b+2 c \sin (x)}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a^2-b^2+2 a c+c^2\right )^2}-\frac {(a+2 b+3 c) \log (1-\sin (x))}{4 (a+b+c)^2}+\frac {(a-2 b+3 c) \log (1+\sin (x))}{4 (a-b+c)^2}+\frac {b \left (b^2-2 c (a+c)\right ) \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 \left (a^2-b^2+2 a c+c^2\right )^2}-\frac {\sec ^2(x) (b-(a+c) \sin (x))}{2 (a-b+c) (a+b+c)} \]

[Out]

-1/4*(a+2*b+3*c)*ln(1-sin(x))/(a+b+c)^2+1/4*(a-2*b+3*c)*ln(1+sin(x))/(a-b+c)^2+1/2*b*(b^2-2*c*(a+c))*ln(a+b*si
n(x)+c*sin(x)^2)/(a^2+2*a*c-b^2+c^2)^2-1/2*sec(x)^2*(b-(a+c)*sin(x))/(a-b+c)/(a+b+c)-(b^4+2*c^2*(a+c)^2-2*b^2*
c*(2*a+c))*arctanh((b+2*c*sin(x))/(-4*a*c+b^2)^(1/2))/(a^2+2*a*c-b^2+c^2)^2/(-4*a*c+b^2)^(1/2)

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Rubi [A]
time = 0.32, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3339, 990, 1088, 648, 632, 212, 642, 647, 31} \begin {gather*} \frac {b \left (b^2-2 c (a+c)\right ) \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 \left (a^2+2 a c-b^2+c^2\right )^2}-\frac {\left (-2 b^2 c (2 a+c)+2 c^2 (a+c)^2+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c \sin (x)}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a^2+2 a c-b^2+c^2\right )^2}-\frac {(a+2 b+3 c) \log (1-\sin (x))}{4 (a+b+c)^2}+\frac {(a-2 b+3 c) \log (\sin (x)+1)}{4 (a-b+c)^2}-\frac {\sec ^2(x) (b-(a+c) \sin (x))}{2 (a-b+c) (a+b+c)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3/(a + b*Sin[x] + c*Sin[x]^2),x]

[Out]

-(((b^4 + 2*c^2*(a + c)^2 - 2*b^2*c*(2*a + c))*ArcTanh[(b + 2*c*Sin[x])/Sqrt[b^2 - 4*a*c]])/(Sqrt[b^2 - 4*a*c]
*(a^2 - b^2 + 2*a*c + c^2)^2)) - ((a + 2*b + 3*c)*Log[1 - Sin[x]])/(4*(a + b + c)^2) + ((a - 2*b + 3*c)*Log[1
+ Sin[x]])/(4*(a - b + c)^2) + (b*(b^2 - 2*c*(a + c))*Log[a + b*Sin[x] + c*Sin[x]^2])/(2*(a^2 - b^2 + 2*a*c +
c^2)^2) - (Sec[x]^2*(b - (a + c)*Sin[x]))/(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 212

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

Rule 632

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 642

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 647

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 648

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 990

Int[((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(2*a*c^2*e + c*(2*
c^2*d - c*(2*a*f))*x)*(a + c*x^2)^(p + 1)*((d + e*x + f*x^2)^(q + 1)/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p +
1))), x] - Dist[1/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 1)), Int[(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Si
mp[2*c*((c*d - a*f)^2 - ((-a)*e)*(c*e))*(p + 1) - (2*c^2*d - c*(2*a*f))*(a*f*(p + 1) - c*d*(p + 2)) - e*(-2*a*
c^2*e)*(p + q + 2) + (2*f*(2*a*c^2*e)*(p + q + 2) - (2*c^2*d - c*(2*a*f))*((-c)*e*(2*p + q + 4)))*x + c*f*(2*c
^2*d - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, q}, x] && NeQ[e^2 - 4*d*f, 0] && Lt
Q[p, -1] && NeQ[a*c*e^2 + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &&  !IGtQ[q, 0]

Rule 1088

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((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[(A*c^2*d - a*c*C*d + A*b^2*f - a*b*B*f -
a*A*c*f + a^2*C*f + c*(B*c*d - b*C*d + A*b*f - a*B*f)*x)/(a + b*x + c*x^2), x], x] + Dist[1/q, Int[(c*C*d^2 +
b*B*d*f - A*c*d*f - a*C*d*f + a*A*f^2 - f*(B*c*d - b*C*d + A*b*f - a*B*f)*x)/(d + f*x^2), x], x] /; NeQ[q, 0]]
/; FreeQ[{a, b, c, d, f, A, B, C}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 3339

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 ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx &=\text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2 \left (a+b x+c x^2\right )} \, dx,x,\sin (x)\right )\\ &=-\frac {\sec ^2(x) (b-(a+c) \sin (x))}{2 (a-b+c) (a+b+c)}+\frac {\text {Subst}\left (\int \frac {2 \left (a^2-2 b^2+3 a c+2 c^2\right )+2 b (a-c) x+2 c (a+c) x^2}{\left (1-x^2\right ) \left (a+b x+c x^2\right )} \, dx,x,\sin (x)\right )}{4 (a-b+c) (a+b+c)}\\ &=-\frac {\sec ^2(x) (b-(a+c) \sin (x))}{2 (a-b+c) (a+b+c)}+\frac {\text {Subst}\left (\int \frac {-2 b^2 (a-c)+2 a c (a+c)+2 c^2 (a+c)+2 a \left (a^2-2 b^2+3 a c+2 c^2\right )+2 c \left (a^2-2 b^2+3 a c+2 c^2\right )+\left (2 a b (a-c)+2 b (a-c) c-2 b c (a+c)-2 b \left (a^2-2 b^2+3 a c+2 c^2\right )\right ) x}{1-x^2} \, dx,x,\sin (x)\right )}{4 (a-b+c)^2 (a+b+c)^2}+\frac {\text {Subst}\left (\int \frac {2 a b^2 (a-c)-2 a^2 c (a+c)-2 a c^2 (a+c)-2 b^2 \left (a^2-2 b^2+3 a c+2 c^2\right )+2 a c \left (a^2-2 b^2+3 a c+2 c^2\right )+2 c^2 \left (a^2-2 b^2+3 a c+2 c^2\right )+c \left (2 a b (a-c)+2 b (a-c) c-2 b c (a+c)-2 b \left (a^2-2 b^2+3 a c+2 c^2\right )\right ) x}{a+b x+c x^2} \, dx,x,\sin (x)\right )}{4 (a-b+c)^2 (a+b+c)^2}\\ &=-\frac {\sec ^2(x) (b-(a+c) \sin (x))}{2 (a-b+c) (a+b+c)}-\frac {(a-2 b+3 c) \text {Subst}\left (\int \frac {1}{-1-x} \, dx,x,\sin (x)\right )}{4 (a-b+c)^2}+\frac {(a+2 b+3 c) \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sin (x)\right )}{4 (a+b+c)^2}+\frac {\left (b \left (b^2-2 c (a+c)\right )\right ) \text {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,\sin (x)\right )}{2 (a-b+c)^2 (a+b+c)^2}+\frac {\left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \text {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,\sin (x)\right )}{2 (a-b+c)^2 (a+b+c)^2}\\ &=-\frac {(a+2 b+3 c) \log (1-\sin (x))}{4 (a+b+c)^2}+\frac {(a-2 b+3 c) \log (1+\sin (x))}{4 (a-b+c)^2}+\frac {b \left (b^2-2 c (a+c)\right ) \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 (a-b+c)^2 (a+b+c)^2}-\frac {\sec ^2(x) (b-(a+c) \sin (x))}{2 (a-b+c) (a+b+c)}-\frac {\left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c \sin (x)\right )}{(a-b+c)^2 (a+b+c)^2}\\ &=-\frac {\left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \tanh ^{-1}\left (\frac {b+2 c \sin (x)}{\sqrt {b^2-4 a c}}\right )}{(a-b+c)^2 (a+b+c)^2 \sqrt {b^2-4 a c}}-\frac {(a+2 b+3 c) \log (1-\sin (x))}{4 (a+b+c)^2}+\frac {(a-2 b+3 c) \log (1+\sin (x))}{4 (a-b+c)^2}+\frac {b \left (b^2-2 c (a+c)\right ) \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 (a-b+c)^2 (a+b+c)^2}-\frac {\sec ^2(x) (b-(a+c) \sin (x))}{2 (a-b+c) (a+b+c)}\\ \end {align*}

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Mathematica [A]
time = 0.47, size = 202, normalized size = 0.98 \begin {gather*} \frac {1}{4} \left (-\frac {4 \left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \tanh ^{-1}\left (\frac {b+2 c \sin (x)}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a^2-b^2+2 a c+c^2\right )^2}-\frac {(a+2 b+3 c) \log (1-\sin (x))}{(a+b+c)^2}+\frac {(a-2 b+3 c) \log (1+\sin (x))}{(a-b+c)^2}+\frac {2 b \left (b^2-2 c (a+c)\right ) \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{\left (a^2-b^2+2 a c+c^2\right )^2}-\frac {1}{(a+b+c) (-1+\sin (x))}-\frac {1}{(a-b+c) (1+\sin (x))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^3/(a + b*Sin[x] + c*Sin[x]^2),x]

[Out]

((-4*(b^4 + 2*c^2*(a + c)^2 - 2*b^2*c*(2*a + c))*ArcTanh[(b + 2*c*Sin[x])/Sqrt[b^2 - 4*a*c]])/(Sqrt[b^2 - 4*a*
c]*(a^2 - b^2 + 2*a*c + c^2)^2) - ((a + 2*b + 3*c)*Log[1 - Sin[x]])/(a + b + c)^2 + ((a - 2*b + 3*c)*Log[1 + S
in[x]])/(a - b + c)^2 + (2*b*(b^2 - 2*c*(a + c))*Log[a + b*Sin[x] + c*Sin[x]^2])/(a^2 - b^2 + 2*a*c + c^2)^2 -
1/((a + b + c)*(-1 + Sin[x])) - 1/((a - b + c)*(1 + Sin[x])))/4

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Maple [A]
time = 2.28, size = 236, normalized size = 1.15

method	result	size
default	\(\frac {\frac {\left (-2 a b \,c^{2}+b^{3} c -2 b \,c^{3}\right ) \ln \left (a +b \sin \left (x \right )+c \left (\sin ^{2}\left (x \right )\right )\right )}{2 c}+\frac {2 \left (a^{2} c^{2}-3 a \,b^{2} c +2 a \,c^{3}+b^{4}-2 b^{2} c^{2}+c^{4}-\frac {\left (-2 a b \,c^{2}+b^{3} c -2 b \,c^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {b +2 c \sin \left (x \right )}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a -b +c \right )^{2} \left (a +b +c \right )^{2}}-\frac {1}{\left (4 a +4 b +4 c \right ) \left (\sin \left (x \right )-1\right )}+\frac {\left (-a -2 b -3 c \right ) \ln \left (\sin \left (x \right )-1\right )}{4 \left (a +b +c \right )^{2}}-\frac {1}{\left (4 a -4 b +4 c \right ) \left (1+\sin \left (x \right )\right )}+\frac {\left (a -2 b +3 c \right ) \ln \left (1+\sin \left (x \right )\right )}{4 \left (a -b +c \right )^{2}}\)	\(236\)
risch	\(\text {Expression too large to display}\)	\(6157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^3/(a+b*sin(x)+c*sin(x)^2),x,method=_RETURNVERBOSE)

[Out]

1/(a-b+c)^2/(a+b+c)^2*(1/2*(-2*a*b*c^2+b^3*c-2*b*c^3)/c*ln(a+b*sin(x)+c*sin(x)^2)+2*(a^2*c^2-3*a*b^2*c+2*a*c^3
+b^4-2*b^2*c^2+c^4-1/2*(-2*a*b*c^2+b^3*c-2*b*c^3)*b/c)/(4*a*c-b^2)^(1/2)*arctan((b+2*c*sin(x))/(4*a*c-b^2)^(1/
2)))-1/(4*a+4*b+4*c)/(sin(x)-1)+1/4/(a+b+c)^2*(-a-2*b-3*c)*ln(sin(x)-1)-1/(4*a-4*b+4*c)/(1+sin(x))+1/4*(a-2*b+
3*c)*ln(1+sin(x))/(a-b+c)^2

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(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 deta

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 603 vs. \(2 (195) = 390\).
time = 5.35, size = 1244, normalized size = 6.04 \begin {gather*} \left [-\frac {2 \, a^{2} b^{3} - 2 \, b^{5} - 8 \, a b c^{3} - 2 \, {\left (b^{4} - 4 \, a b^{2} c + 4 \, a c^{3} + 2 \, c^{4} + 2 \, {\left (a^{2} - b^{2}\right )} c^{2}\right )} \sqrt {b^{2} - 4 \, a c} \cos \left (x\right )^{2} \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 ) - 2 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a b c^{3} + 2 \, {\left (4 \, a^{2} b - b^{3}\right )} c^{2}\right )} \cos \left (x\right )^{2} \log \left (-c \cos \left (x\right )^{2} + b \sin \left (x\right ) + a + c\right ) - {\left (a^{3} b^{2} - 3 \, a b^{4} - 2 \, b^{5} - 12 \, a c^{4} - {\left (28 \, a^{2} + 16 \, a b - 3 \, b^{2}\right )} c^{3} - {\left (20 \, a^{3} + 16 \, a^{2} b - 11 \, a b^{2} - 4 \, b^{3}\right )} c^{2} - {\left (4 \, a^{4} - 17 \, a^{2} b^{2} - 12 \, a b^{3} + b^{4}\right )} c\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) + {\left (a^{3} b^{2} - 3 \, a b^{4} + 2 \, b^{5} - 12 \, a c^{4} - {\left (28 \, a^{2} - 16 \, a b - 3 \, b^{2}\right )} c^{3} - {\left (20 \, a^{3} - 16 \, a^{2} b - 11 \, a b^{2} + 4 \, b^{3}\right )} c^{2} - {\left (4 \, a^{4} - 17 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}\right )} c\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \, {\left (8 \, a^{2} b - b^{3}\right )} c^{2} - 4 \, {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} c - 2 \, {\left (a^{3} b^{2} - a b^{4} - 4 \, a c^{4} - {\left (12 \, a^{2} - b^{2}\right )} c^{3} - {\left (12 \, a^{3} - 7 \, a b^{2}\right )} c^{2} - {\left (4 \, a^{4} - 7 \, a^{2} b^{2} + b^{4}\right )} c\right )} \sin \left (x\right )}{4 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} - 4 \, a c^{5} - {\left (16 \, a^{2} - b^{2}\right )} c^{4} - 12 \, {\left (2 \, a^{3} - a b^{2}\right )} c^{3} - 2 \, {\left (8 \, a^{4} - 11 \, a^{2} b^{2} + b^{4}\right )} c^{2} - 4 \, {\left (a^{5} - 3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} c\right )} \cos \left (x\right )^{2}}, -\frac {2 \, a^{2} b^{3} - 2 \, b^{5} - 8 \, a b c^{3} + 4 \, {\left (b^{4} - 4 \, a b^{2} c + 4 \, a c^{3} + 2 \, c^{4} + 2 \, {\left (a^{2} - b^{2}\right )} 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 ) \cos \left (x\right )^{2} - 2 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a b c^{3} + 2 \, {\left (4 \, a^{2} b - b^{3}\right )} c^{2}\right )} \cos \left (x\right )^{2} \log \left (-c \cos \left (x\right )^{2} + b \sin \left (x\right ) + a + c\right ) - {\left (a^{3} b^{2} - 3 \, a b^{4} - 2 \, b^{5} - 12 \, a c^{4} - {\left (28 \, a^{2} + 16 \, a b - 3 \, b^{2}\right )} c^{3} - {\left (20 \, a^{3} + 16 \, a^{2} b - 11 \, a b^{2} - 4 \, b^{3}\right )} c^{2} - {\left (4 \, a^{4} - 17 \, a^{2} b^{2} - 12 \, a b^{3} + b^{4}\right )} c\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) + {\left (a^{3} b^{2} - 3 \, a b^{4} + 2 \, b^{5} - 12 \, a c^{4} - {\left (28 \, a^{2} - 16 \, a b - 3 \, b^{2}\right )} c^{3} - {\left (20 \, a^{3} - 16 \, a^{2} b - 11 \, a b^{2} + 4 \, b^{3}\right )} c^{2} - {\left (4 \, a^{4} - 17 \, a^{2} b^{2} + 12 \, a b^{3} + b^{4}\right )} c\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \, {\left (8 \, a^{2} b - b^{3}\right )} c^{2} - 4 \, {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} c - 2 \, {\left (a^{3} b^{2} - a b^{4} - 4 \, a c^{4} - {\left (12 \, a^{2} - b^{2}\right )} c^{3} - {\left (12 \, a^{3} - 7 \, a b^{2}\right )} c^{2} - {\left (4 \, a^{4} - 7 \, a^{2} b^{2} + b^{4}\right )} c\right )} \sin \left (x\right )}{4 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} - 4 \, a c^{5} - {\left (16 \, a^{2} - b^{2}\right )} c^{4} - 12 \, {\left (2 \, a^{3} - a b^{2}\right )} c^{3} - 2 \, {\left (8 \, a^{4} - 11 \, a^{2} b^{2} + b^{4}\right )} c^{2} - 4 \, {\left (a^{5} - 3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} c\right )} \cos \left (x\right )^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+b*sin(x)+c*sin(x)^2),x, algorithm="fricas")

[Out]

[-1/4*(2*a^2*b^3 - 2*b^5 - 8*a*b*c^3 - 2*(b^4 - 4*a*b^2*c + 4*a*c^3 + 2*c^4 + 2*(a^2 - b^2)*c^2)*sqrt(b^2 - 4*
a*c)*cos(x)^2*log(-(2*c^2*cos(x)^2 - 2*b*c*sin(x) - b^2 + 2*a*c - 2*c^2 + sqrt(b^2 - 4*a*c)*(2*c*sin(x) + b))/
(c*cos(x)^2 - b*sin(x) - a - c)) - 2*(b^5 - 6*a*b^3*c + 8*a*b*c^3 + 2*(4*a^2*b - b^3)*c^2)*cos(x)^2*log(-c*cos
(x)^2 + b*sin(x) + a + c) - (a^3*b^2 - 3*a*b^4 - 2*b^5 - 12*a*c^4 - (28*a^2 + 16*a*b - 3*b^2)*c^3 - (20*a^3 +
16*a^2*b - 11*a*b^2 - 4*b^3)*c^2 - (4*a^4 - 17*a^2*b^2 - 12*a*b^3 + b^4)*c)*cos(x)^2*log(sin(x) + 1) + (a^3*b^
2 - 3*a*b^4 + 2*b^5 - 12*a*c^4 - (28*a^2 - 16*a*b - 3*b^2)*c^3 - (20*a^3 - 16*a^2*b - 11*a*b^2 + 4*b^3)*c^2 -
(4*a^4 - 17*a^2*b^2 + 12*a*b^3 + b^4)*c)*cos(x)^2*log(-sin(x) + 1) - 2*(8*a^2*b - b^3)*c^2 - 4*(2*a^3*b - 3*a*
b^3)*c - 2*(a^3*b^2 - a*b^4 - 4*a*c^4 - (12*a^2 - b^2)*c^3 - (12*a^3 - 7*a*b^2)*c^2 - (4*a^4 - 7*a^2*b^2 + b^4
)*c)*sin(x))/((a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 -
11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c)*cos(x)^2), -1/4*(2*a^2*b^3 - 2*b^5 - 8*a*b*c^3 + 4*(b
^4 - 4*a*b^2*c + 4*a*c^3 + 2*c^4 + 2*(a^2 - b^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))*cos(x)^2 - 2*(b^5 - 6*a*b^3*c + 8*a*b*c^3 + 2*(4*a^2*b - b^3)*c^2)*cos(x)^2*log(-c*cos(x
)^2 + b*sin(x) + a + c) - (a^3*b^2 - 3*a*b^4 - 2*b^5 - 12*a*c^4 - (28*a^2 + 16*a*b - 3*b^2)*c^3 - (20*a^3 + 16
*a^2*b - 11*a*b^2 - 4*b^3)*c^2 - (4*a^4 - 17*a^2*b^2 - 12*a*b^3 + b^4)*c)*cos(x)^2*log(sin(x) + 1) + (a^3*b^2
- 3*a*b^4 + 2*b^5 - 12*a*c^4 - (28*a^2 - 16*a*b - 3*b^2)*c^3 - (20*a^3 - 16*a^2*b - 11*a*b^2 + 4*b^3)*c^2 - (4
*a^4 - 17*a^2*b^2 + 12*a*b^3 + b^4)*c)*cos(x)^2*log(-sin(x) + 1) - 2*(8*a^2*b - b^3)*c^2 - 4*(2*a^3*b - 3*a*b^
3)*c - 2*(a^3*b^2 - a*b^4 - 4*a*c^4 - (12*a^2 - b^2)*c^3 - (12*a^3 - 7*a*b^2)*c^2 - (4*a^4 - 7*a^2*b^2 + b^4)*
c)*sin(x))/((a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11
*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c)*cos(x)^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{3}{\left (x \right )}}{a + b \sin {\left (x \right )} + c \sin ^{2}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**3/(a+b*sin(x)+c*sin(x)**2),x)

[Out]

Integral(sec(x)**3/(a + b*sin(x) + c*sin(x)**2), x)

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Giac [A]
time = 0.51, size = 377, normalized size = 1.83 \begin {gather*} \frac {{\left (b^{3} - 2 \, a b c - 2 \, b c^{2}\right )} \log \left (c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} + 4 \, a^{3} c - 4 \, a b^{2} c + 6 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + c^{4}\right )}} + \frac {{\left (a - 2 \, b + 3 \, c\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2} + 2 \, a c - 2 \, b c + c^{2}\right )}} - \frac {{\left (a + 2 \, b + 3 \, c\right )} \log \left (-\sin \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2} + 2 \, a c + 2 \, b c + c^{2}\right )}} + \frac {{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + 2 \, c^{4}\right )} \arctan \left (\frac {2 \, c \sin \left (x\right ) + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} + 4 \, a^{3} c - 4 \, a b^{2} c + 6 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + c^{4}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {a^{2} b - b^{3} + 2 \, a b c + b c^{2} - {\left (a^{3} - a b^{2} + 3 \, a^{2} c - b^{2} c + 3 \, a c^{2} + c^{3}\right )} \sin \left (x\right )}{2 \, {\left (a + b + c\right )}^{2} {\left (a - b + c\right )}^{2} {\left (\sin \left (x\right ) + 1\right )} {\left (\sin \left (x\right ) - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+b*sin(x)+c*sin(x)^2),x, algorithm="giac")

[Out]

1/2*(b^3 - 2*a*b*c - 2*b*c^2)*log(c*sin(x)^2 + b*sin(x) + a)/(a^4 - 2*a^2*b^2 + b^4 + 4*a^3*c - 4*a*b^2*c + 6*
a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 + c^4) + 1/4*(a - 2*b + 3*c)*log(sin(x) + 1)/(a^2 - 2*a*b + b^2 + 2*a*c - 2*b*c
+ c^2) - 1/4*(a + 2*b + 3*c)*log(-sin(x) + 1)/(a^2 + 2*a*b + b^2 + 2*a*c + 2*b*c + c^2) + (b^4 - 4*a*b^2*c + 2
*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 + 2*c^4)*arctan((2*c*sin(x) + b)/sqrt(-b^2 + 4*a*c))/((a^4 - 2*a^2*b^2 + b^4 +
4*a^3*c - 4*a*b^2*c + 6*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 + c^4)*sqrt(-b^2 + 4*a*c)) + 1/2*(a^2*b - b^3 + 2*a*b*c
+ b*c^2 - (a^3 - a*b^2 + 3*a^2*c - b^2*c + 3*a*c^2 + c^3)*sin(x))/((a + b + c)^2*(a - b + c)^2*(sin(x) + 1)*(s
in(x) - 1))

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Mupad [B]
time = 35.31, size = 2743, normalized size = 13.32 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^3*(a + c*sin(x)^2 + b*sin(x))),x)

[Out]

log(sin(x) + 1)*(1/(4*(a - b + c)) - (b/4 - c/2)/(a - b + c)^2) - (b/(2*(2*a*c + a^2 - b^2 + c^2)) - (sin(x)*(
a + c))/(2*(2*a*c + a^2 - b^2 + c^2)))/cos(x)^2 - log(sin(x) - 1)*((b/4 + c/2)/(a + b + c)^2 + 1/(4*(a + b + c
))) + (log((c^4*(4*a*c + a^2 - 4*b^2 + 3*c^2))/(4*(2*a*c + a^2 - b^2 + c^2)^2) - (((((c*(a*b^4 + 28*a*c^4 + 4*
a^4*c - 5*b^4*c + 8*c^5 - a^3*b^2 + 36*a^2*c^3 + 20*a^3*c^2 + 5*b^2*c^3 - 3*a*b^2*c^2 - 9*a^2*b^2*c))/(2*(2*a*
c + a^2 - b^2 + c^2)) + (b*c*sin(x)*(36*a*c^3 + 4*a^3*c + 3*b^4 + 16*c^4 - a^2*b^2 + 24*a^2*c^2 - 13*b^2*c^2 -
18*a*b^2*c))/(2*a*c + a^2 - b^2 + c^2) - (2*c*((b^4*(b^2 - 4*a*c)^(1/2))/2 - b^5/2 + c^4*(b^2 - 4*a*c)^(1/2)
+ b^3*c^2 + 2*a*c^3*(b^2 - 4*a*c)^(1/2) - 4*a^2*b*c^2 + a^2*c^2*(b^2 - 4*a*c)^(1/2) - b^2*c^2*(b^2 - 4*a*c)^(1
/2) - 4*a*b*c^3 + 3*a*b^3*c - 2*a*b^2*c*(b^2 - 4*a*c)^(1/2))*(3*b^4*sin(x) + 4*c^4*sin(x) + 4*a*b^3 + 2*b*c^3
+ 2*b^3*c + 4*a*c^3*sin(x) - 4*a^3*c*sin(x) + a^2*b^2*sin(x) - 4*a^2*c^2*sin(x) - 3*b^2*c^2*sin(x) - 12*a*b*c^
2 - 14*a^2*b*c - 10*a*b^2*c*sin(x)))/((4*a*c - b^2)*(2*a*c + a^2 - b^2 + c^2)^2))*((b^4*(b^2 - 4*a*c)^(1/2))/2
- b^5/2 + c^4*(b^2 - 4*a*c)^(1/2) + b^3*c^2 + 2*a*c^3*(b^2 - 4*a*c)^(1/2) - 4*a^2*b*c^2 + a^2*c^2*(b^2 - 4*a*
c)^(1/2) - b^2*c^2*(b^2 - 4*a*c)^(1/2) - 4*a*b*c^3 + 3*a*b^3*c - 2*a*b^2*c*(b^2 - 4*a*c)^(1/2)))/((4*a*c - b^2
)*(2*a*c + a^2 - b^2 + c^2)^2) - (b*c*(2*a*b^4 - 20*a*c^4 + 3*a^4*c - 6*b^4*c + 7*c^5 - a^3*b^2 - 26*a^2*c^3 +
4*a^3*c^2 + 23*a*b^2*c^2 - 6*a^2*b^2*c))/(4*(2*a*c + a^2 - b^2 + c^2)^2) + (c*sin(x)*(64*a*c^5 + 26*c^6 + a^2
*b^4 + 52*a^2*c^4 + 16*a^3*c^3 + 2*a^4*c^2 - 18*b^2*c^4 + 9*b^4*c^2 - 32*a*b^2*c^3 - 4*a^3*b^2*c - 2*a^2*b^2*c
^2 - 2*a*b^4*c))/(4*(2*a*c + a^2 - b^2 + c^2)^2))*((b^4*(b^2 - 4*a*c)^(1/2))/2 - b^5/2 + c^4*(b^2 - 4*a*c)^(1/
2) + b^3*c^2 + 2*a*c^3*(b^2 - 4*a*c)^(1/2) - 4*a^2*b*c^2 + a^2*c^2*(b^2 - 4*a*c)^(1/2) - b^2*c^2*(b^2 - 4*a*c)
^(1/2) - 4*a*b*c^3 + 3*a*b^3*c - 2*a*b^2*c*(b^2 - 4*a*c)^(1/2)))/((4*a*c - b^2)*(2*a*c + a^2 - b^2 + c^2)^2) -
(b*c^5*sin(x))/(2*a*c + a^2 - b^2 + c^2)^2)*(b^3*(3*a*c + c^2) - b^2*(c^2*(b^2 - 4*a*c)^(1/2) + 2*a*c*(b^2 -
4*a*c)^(1/2)) - b*(4*a*c^3 + 4*a^2*c^2) - b^5/2 + (b^4*(b^2 - 4*a*c)^(1/2))/2 + c^4*(b^2 - 4*a*c)^(1/2) + 2*a*
c^3*(b^2 - 4*a*c)^(1/2) + a^2*c^2*(b^2 - 4*a*c)^(1/2)))/(4*a*c^5 + 4*a^5*c - b^6 + 2*a^2*b^4 - a^4*b^2 + 16*a^
2*c^4 + 24*a^3*c^3 + 16*a^4*c^2 - b^2*c^4 + 2*b^4*c^2 - 12*a*b^2*c^3 - 12*a^3*b^2*c - 22*a^2*b^2*c^2 + 8*a*b^4
*c) - (log((c^4*(4*a*c + a^2 - 4*b^2 + 3*c^2))/(4*(2*a*c + a^2 - b^2 + c^2)^2) - (((b*c*(2*a*b^4 - 20*a*c^4 +
3*a^4*c - 6*b^4*c + 7*c^5 - a^3*b^2 - 26*a^2*c^3 + 4*a^3*c^2 + 23*a*b^2*c^2 - 6*a^2*b^2*c))/(4*(2*a*c + a^2 -
b^2 + c^2)^2) + (((c*(a*b^4 + 28*a*c^4 + 4*a^4*c - 5*b^4*c + 8*c^5 - a^3*b^2 + 36*a^2*c^3 + 20*a^3*c^2 + 5*b^2
*c^3 - 3*a*b^2*c^2 - 9*a^2*b^2*c))/(2*(2*a*c + a^2 - b^2 + c^2)) + (b*c*sin(x)*(36*a*c^3 + 4*a^3*c + 3*b^4 + 1
6*c^4 - a^2*b^2 + 24*a^2*c^2 - 13*b^2*c^2 - 18*a*b^2*c))/(2*a*c + a^2 - b^2 + c^2) + (2*c*(b^5/2 + (b^4*(b^2 -
4*a*c)^(1/2))/2 + c^4*(b^2 - 4*a*c)^(1/2) - b^3*c^2 + 2*a*c^3*(b^2 - 4*a*c)^(1/2) + 4*a^2*b*c^2 + a^2*c^2*(b^
2 - 4*a*c)^(1/2) - b^2*c^2*(b^2 - 4*a*c)^(1/2) + 4*a*b*c^3 - 3*a*b^3*c - 2*a*b^2*c*(b^2 - 4*a*c)^(1/2))*(3*b^4
*sin(x) + 4*c^4*sin(x) + 4*a*b^3 + 2*b*c^3 + 2*b^3*c + 4*a*c^3*sin(x) - 4*a^3*c*sin(x) + a^2*b^2*sin(x) - 4*a^
2*c^2*sin(x) - 3*b^2*c^2*sin(x) - 12*a*b*c^2 - 14*a^2*b*c - 10*a*b^2*c*sin(x)))/((4*a*c - b^2)*(2*a*c + a^2 -
b^2 + c^2)^2))*(b^5/2 + (b^4*(b^2 - 4*a*c)^(1/2))/2 + c^4*(b^2 - 4*a*c)^(1/2) - b^3*c^2 + 2*a*c^3*(b^2 - 4*a*c
)^(1/2) + 4*a^2*b*c^2 + a^2*c^2*(b^2 - 4*a*c)^(1/2) - b^2*c^2*(b^2 - 4*a*c)^(1/2) + 4*a*b*c^3 - 3*a*b^3*c - 2*
a*b^2*c*(b^2 - 4*a*c)^(1/2)))/((4*a*c - b^2)*(2*a*c + a^2 - b^2 + c^2)^2) - (c*sin(x)*(64*a*c^5 + 26*c^6 + a^2
*b^4 + 52*a^2*c^4 + 16*a^3*c^3 + 2*a^4*c^2 - 18*b^2*c^4 + 9*b^4*c^2 - 32*a*b^2*c^3 - 4*a^3*b^2*c - 2*a^2*b^2*c
^2 - 2*a*b^4*c))/(4*(2*a*c + a^2 - b^2 + c^2)^2))*(b^5/2 + (b^4*(b^2 - 4*a*c)^(1/2))/2 + c^4*(b^2 - 4*a*c)^(1/
2) - b^3*c^2 + 2*a*c^3*(b^2 - 4*a*c)^(1/2) + 4*a^2*b*c^2 + a^2*c^2*(b^2 - 4*a*c)^(1/2) - b^2*c^2*(b^2 - 4*a*c)
^(1/2) + 4*a*b*c^3 - 3*a*b^3*c - 2*a*b^2*c*(b^2 - 4*a*c)^(1/2)))/((4*a*c - b^2)*(2*a*c + a^2 - b^2 + c^2)^2) -
(b*c^5*sin(x))/(2*a*c + a^2 - b^2 + c^2)^2)*(b*(4*a*c^3 + 4*a^2*c^2) - b^3*(3*a*c + c^2) - b^2*(c^2*(b^2 - 4*
a*c)^(1/2) + 2*a*c*(b^2 - 4*a*c)^(1/2)) + b^5/2 + (b^4*(b^2 - 4*a*c)^(1/2))/2 + c^4*(b^2 - 4*a*c)^(1/2) + 2*a*
c^3*(b^2 - 4*a*c)^(1/2) + a^2*c^2*(b^2 - 4*a*c)^(1/2)))/(4*a*c^5 + 4*a^5*c - b^6 + 2*a^2*b^4 - a^4*b^2 + 16*a^
2*c^4 + 24*a^3*c^3 + 16*a^4*c^2 - b^2*c^4 + 2*b^4*c^2 - 12*a*b^2*c^3 - 12*a^3*b^2*c - 22*a^2*b^2*c^2 + 8*a*b^4
*c)

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