Integrand size = 19, antiderivative size = 206 \[ \int \frac {\sec ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=-\frac {\left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \text {arctanh}\left (\frac {b+2 c \sin (x)}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a^2-b^2+2 a c+c^2\right )^2}-\frac {(a+2 b+3 c) \log (1-\sin (x))}{4 (a+b+c)^2}+\frac {(a-2 b+3 c) \log (1+\sin (x))}{4 (a-b+c)^2}+\frac {b \left (b^2-2 c (a+c)\right ) \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 \left (a^2-b^2+2 a c+c^2\right )^2}-\frac {\sec ^2(x) (b-(a+c) \sin (x))}{2 (a-b+c) (a+b+c)} \]
-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*sin(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)
Time = 0.76 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.98 \[ \int \frac {\sec ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {1}{4} \left (-\frac {4 \left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \text {arctanh}\left (\frac {b+2 c \sin (x)}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a^2-b^2+2 a c+c^2\right )^2}-\frac {(a+2 b+3 c) \log (1-\sin (x))}{(a+b+c)^2}+\frac {(a-2 b+3 c) \log (1+\sin (x))}{(a-b+c)^2}+\frac {2 b \left (b^2-2 c (a+c)\right ) \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{\left (a^2-b^2+2 a c+c^2\right )^2}-\frac {1}{(a+b+c) (-1+\sin (x))}-\frac {1}{(a-b+c) (1+\sin (x))}\right ) \]
((-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 + Sin [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
Time = 0.58 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3739, 1301, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sec ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (x)^3 \left (a+b \sin (x)+c \sin (x)^2\right )}dx\) |
\(\Big \downarrow \) 3739 |
\(\displaystyle \int \frac {1}{\left (1-\sin ^2(x)\right )^2 \left (a+b \sin (x)+c \sin ^2(x)\right )}d\sin (x)\) |
\(\Big \downarrow \) 1301 |
\(\displaystyle \int \left (\frac {b c \sin (x) \left (b^2-2 c (a+c)\right )-b^2 c (3 a+2 c)+c^2 (a+c)^2+b^4}{(a-b+c)^2 (a+b+c)^2 \left (a+b \sin (x)+c \sin ^2(x)\right )}+\frac {a-2 b+3 c}{4 (\sin (x)+1) (a-b+c)^2}+\frac {a+2 b+3 c}{4 (1-\sin (x)) (a+b+c)^2}+\frac {1}{4 (1-\sin (x))^2 (a+b+c)}+\frac {1}{4 (\sin (x)+1)^2 (a-b+c)}\right )d\sin (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (-2 b^2 c (2 a+c)+2 c^2 (a+c)^2+b^4\right ) \text {arctanh}\left (\frac {b+2 c \sin (x)}{\sqrt {b^2-4 a c}}\right )}{(a-b+c)^2 (a+b+c)^2 \sqrt {b^2-4 a c}}+\frac {b \left (b^2-2 c (a+c)\right ) \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 (a-b+c)^2 (a+b+c)^2}+\frac {1}{4 (1-\sin (x)) (a+b+c)}-\frac {1}{4 (\sin (x)+1) (a-b+c)}-\frac {(a+2 b+3 c) \log (1-\sin (x))}{4 (a+b+c)^2}+\frac {(a-2 b+3 c) \log (\sin (x)+1)}{4 (a-b+c)^2}\) |
-(((b^4 + 2*c^2*(a + c)^2 - 2*b^2*c*(2*a + c))*ArcTanh[(b + 2*c*Sin[x])/Sq rt[b^2 - 4*a*c]])/((a - b + c)^2*(a + b + c)^2*Sqrt[b^2 - 4*a*c])) - ((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*S in[x]^2])/(2*(a - b + c)^2*(a + b + c)^2) + 1/(4*(a + b + c)*(1 - Sin[x])) - 1/(4*(a - b + c)*(1 + Sin[x]))
3.1.14.3.1 Defintions of rubi rules used
Int[((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x _Symbol] :> With[{r = Rt[(-a)*c, 2]}, Simp[1/c^p Int[ExpandIntegrand[(-r + c*x)^p*(r + c*x)^p*(d + e*x + f*x^2)^q, x], x], x] /; EqQ[p, -1] || !Fra ctionalPowerFactorQ[r]] /; FreeQ[{a, c, d, e, f}, x] && ILtQ[p, 0] && Integ erQ[q] && NiceSqrtQ[(-a)*c]
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]}, Simp[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] && Intege rQ[(m - 1)/2]
Time = 2.19 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.15
method | result | size |
default | \(\frac {\frac {\left (-2 a b \,c^{2}+b^{3} c -2 b \,c^{3}\right ) \ln \left (a +b \sin \left (x \right )+c \left (\sin ^{2}\left (x \right )\right )\right )}{2 c}+\frac {2 \left (a^{2} c^{2}-3 a \,b^{2} c +2 a \,c^{3}+b^{4}-2 b^{2} c^{2}+c^{4}-\frac {\left (-2 a b \,c^{2}+b^{3} c -2 b \,c^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {b +2 \sin \left (x \right ) c}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a -b +c \right )^{2} \left (a +b +c \right )^{2}}-\frac {1}{\left (4 a -4 b +4 c \right ) \left (1+\sin \left (x \right )\right )}+\frac {\left (a -2 b +3 c \right ) \ln \left (1+\sin \left (x \right )\right )}{4 \left (a -b +c \right )^{2}}-\frac {1}{\left (4 a +4 b +4 c \right ) \left (\sin \left (x \right )-1\right )}+\frac {\left (-a -2 b -3 c \right ) \ln \left (\sin \left (x \right )-1\right )}{4 \left (a +b +c \right )^{2}}\) | \(236\) |
risch | \(\text {Expression too large to display}\) | \(6157\) |
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*si n(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*sin(x)*c)/(4*a*c-b^2)^(1/2) ))-1/(4*a-4*b+4*c)/(1+sin(x))+1/4*(a-2*b+3*c)*ln(1+sin(x))/(a-b+c)^2-1/(4* a+4*b+4*c)/(sin(x)-1)+1/4/(a+b+c)^2*(-a-2*b-3*c)*ln(sin(x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 603 vs. \(2 (195) = 390\).
Time = 5.87 (sec) , antiderivative size = 1244, normalized size of antiderivative = 6.04 \[ \int \frac {\sec ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \]
[-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)*co s(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(-sqr t(-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^...
\[ \int \frac {\sec ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int \frac {\sec ^{3}{\left (x \right )}}{a + b \sin {\left (x \right )} + c \sin ^{2}{\left (x \right )}}\, dx \]
Exception generated. \[ \int \frac {\sec ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.36 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.83 \[ \int \frac {\sec ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {{\left (b^{3} - 2 \, a b c - 2 \, b c^{2}\right )} \log \left (c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} + 4 \, a^{3} c - 4 \, a b^{2} c + 6 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + c^{4}\right )}} + \frac {{\left (a - 2 \, b + 3 \, c\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2} + 2 \, a c - 2 \, b c + c^{2}\right )}} - \frac {{\left (a + 2 \, b + 3 \, c\right )} \log \left (-\sin \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2} + 2 \, a c + 2 \, b c + c^{2}\right )}} + \frac {{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + 2 \, c^{4}\right )} \arctan \left (\frac {2 \, c \sin \left (x\right ) + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} + 4 \, a^{3} c - 4 \, a b^{2} c + 6 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + c^{4}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {a^{2} b - b^{3} + 2 \, a b c + b c^{2} - {\left (a^{3} - a b^{2} + 3 \, a^{2} c - b^{2} c + 3 \, a c^{2} + c^{3}\right )} \sin \left (x\right )}{2 \, {\left (a + b + c\right )}^{2} {\left (a - b + c\right )}^{2} {\left (\sin \left (x\right ) + 1\right )} {\left (\sin \left (x\right ) - 1\right )}} \]
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)*(sin(x) - 1))
Time = 34.64 (sec) , antiderivative size = 2743, normalized size of antiderivative = 13.32 \[ \int \frac {\sec ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \]
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...