Integrand size = 43, antiderivative size = 173 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(3 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{3/2} d}+\frac {(9 A-5 B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A-B+C) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \]
-(3*A-2*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d+1/ 4*(9*A-5*B+C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2 ))/a^(3/2)/d*2^(1/2)-1/2*(A-B+C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/2*( 3*A-B+C)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)
Time = 1.71 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.13 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \cos ^2(c+d x) \left (C+B \sec (c+d x)+A \sec ^2(c+d x)\right ) \left (2 (9 A-5 B+C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 \sqrt {2} (3 A-2 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )-2 (3 A-B+C+2 A \sec (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )}{-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{d (a (1+\cos (c+d x)))^{3/2} (2 A+C+2 B \cos (c+d x)+C \cos (2 (c+d x)))} \]
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + a* Cos[c + d*x])^(3/2),x]
(Cos[(c + d*x)/2]^3*Cos[c + d*x]^2*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2) *(2*(9*A - 5*B + C)*ArcTanh[Sin[(c + d*x)/2]] + (4*Sqrt[2]*(3*A - 2*B)*Arc Tanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^2 - 2*(3*A - B + C + 2*A*S ec[c + d*x])*Sin[(c + d*x)/2])/(-1 + Sin[(c + d*x)/2]^2)))/(d*(a*(1 + Cos[ c + d*x]))^(3/2)*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*(c + d*x)]))
Time = 1.16 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.302, Rules used = {3042, 3520, 27, 3042, 3463, 25, 3042, 3464, 3042, 3128, 219, 3252, 219}
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.
\(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3520 |
\(\displaystyle \frac {\int \frac {(2 a (3 A-B+C)-a (3 A-3 B-C) \cos (c+d x)) \sec ^2(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(2 a (3 A-B+C)-a (3 A-3 B-C) \cos (c+d x)) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 a (3 A-B+C)-a (3 A-3 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3463 |
\(\displaystyle \frac {\frac {\int -\frac {\left (2 a^2 (3 A-2 B)-a^2 (3 A-B+C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a (3 A-B+C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {2 a (3 A-B+C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (2 a^2 (3 A-2 B)-a^2 (3 A-B+C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 a (3 A-B+C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {2 a^2 (3 A-2 B)-a^2 (3 A-B+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3464 |
\(\displaystyle \frac {\frac {2 a (3 A-B+C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {2 a (3 A-2 B) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-a^2 (9 A-5 B+C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 a (3 A-B+C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {2 a (3 A-2 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^2 (9 A-5 B+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {\frac {2 a (3 A-B+C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^2 (9 A-5 B+C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}+2 a (3 A-2 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {2 a (3 A-B+C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {2 a (3 A-2 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {\sqrt {2} a^{3/2} (9 A-5 B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 3252 |
\(\displaystyle \frac {\frac {2 a (3 A-B+C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {4 a^2 (3 A-2 B) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {\sqrt {2} a^{3/2} (9 A-5 B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {2 a (3 A-B+C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {4 a^{3/2} (3 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{3/2} (9 A-5 B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
-1/2*((A - B + C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^(3/2)) + (-(((4*a^ (3/2)*(3*A - 2*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]] )/d - (Sqrt[2]*a^(3/2)*(9*A - 5*B + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqr t[2]*Sqrt[a + a*Cos[c + d*x]])])/d)/a) + (2*a*(3*A - B + C)*Tan[c + d*x])/ (d*Sqrt[a + a*Cos[c + d*x]]))/(4*a^2)
3.5.15.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A *b - a*B)/(b*c - a*d) Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c - A*d)/(b*c - a*d) Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(991\) vs. \(2(148)=296\).
Time = 13.47 (sec) , antiderivative size = 992, normalized size of antiderivative = 5.73
method | result | size |
parts | \(\text {Expression too large to display}\) | \(992\) |
default | \(\text {Expression too large to display}\) | \(1193\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+cos(d*x+c)*a)^(3/2),x, method=_RETURNVERBOSE)
1/2*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(18*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/ 2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*a*cos(1/2*d*x+1/2*c)^4-12*l n(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*( a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a-12*ln(- 4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a* sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^4*a-9*2^(1/2) *ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*a *cos(1/2*d*x+1/2*c)^2+6*cos(1/2*d*x+1/2*c)^2*(a*sin(1/2*d*x+1/2*c)^2)^(1/2 )*2^(1/2)*a^(1/2)+6*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2 *d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*cos(1/2*d *x+1/2*c)^2*a+6*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d* x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*cos(1/2*d*x+ 1/2*c)^2*a-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(5/2)/cos(1/2 *d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))/(2*cos(1/2*d*x+1/2*c)+2^(1/2))/ sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+1/4*B*(a*sin(1/2*d*x+1 /2*c)^2)^(1/2)*(2*2^(1/2)*ln(-2/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a* cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*co s(1/2*d*x+1/2*c)^2*a+2*2^(1/2)*ln(2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2 )*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a) )*cos(1/2*d*x+1/2*c)^2*a-5*ln(2*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)...
Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (148) = 296\).
Time = 0.33 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.98 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} {\left ({\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right )^{2} + {\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left ({\left (3 \, A - B + C\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^(3 /2),x, algorithm="fricas")
1/8*(sqrt(2)*((9*A - 5*B + C)*cos(d*x + c)^3 + 2*(9*A - 5*B + C)*cos(d*x + c)^2 + (9*A - 5*B + C)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*s qrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 2*((3*A - 2*B)*cos(d*x + c)^ 3 + 2*(3*A - 2*B)*cos(d*x + c)^2 + (3*A - 2*B)*cos(d*x + c))*sqrt(a)*log(( a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a) *(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*((3*A - B + C)*cos(d*x + c) + 2*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**2/(a*(cos( c + d*x) + 1))**(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 63653 vs. \(2 (148) = 296\).
Time = 2.36 (sec) , antiderivative size = 63653, normalized size of antiderivative = 367.94 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^(3 /2),x, algorithm="maxima")
1/4*((128*cos(3/2*d*x + 3/2*c)*sin(3*d*x + 3*c)^3 + 1152*cos(3/2*d*x + 3/2 *c)*sin(2*d*x + 2*c)^3 - 128*cos(3*d*x + 3*c)^3*sin(3/2*d*x + 3/2*c) - 115 2*cos(2*d*x + 2*c)^3*sin(3/2*d*x + 3/2*c) + 32*(4*cos(3/2*d*x + 3/2*c)*sin (2*d*x + 2*c) - 9*(3*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c) - 28*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + 3*cos(3/2*d*x + 3/2*c)*sin(d*x + c))*cos(3* d*x + 3*c)^2 - 96*(11*(3*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c) - 9*cos(3/ 2*d*x + 3/2*c)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + 32*(28*cos(3/2*d*x + 3/2 *c)*sin(2*d*x + 2*c) - (3*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c) - 4*cos(3 *d*x + 3*c)*sin(3/2*d*x + 3/2*c) - 4*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + 27*cos(3/2*d*x + 3/2*c)*sin(d*x + c))*sin(3*d*x + 3*c)^2 - 288*((3*cos( d*x + c) + 1)*sin(3/2*d*x + 3/2*c) + 4*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2* c) - 11*cos(3/2*d*x + 3/2*c)*sin(d*x + c))*sin(2*d*x + 2*c)^2 - 32*(cos(3* d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + 6*(3*cos(d*x + c) + 1)*cos(2*d*x + 2*c )*sin(3/2*d*x + 3/2*c) + 9*cos(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) + sin(3 *d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + 9*sin(2*d*x + 2*c)^2*sin(3/2*d*x + 3/ 2*c) + 18*sin(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c)*sin(d*x + c) + 2*((3*cos(d *x + c) + 1)*sin(3/2*d*x + 3/2*c) + 3*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c ))*cos(3*d*x + 3*c) + 6*(sin(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + sin(3/2*d *x + 3/2*c)*sin(d*x + c))*sin(3*d*x + 3*c) + (9*cos(d*x + c)^2 + 9*sin(d*x + c)^2 + 6*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c))*cos(5*d*x + 5*c) - ...
Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (148) = 296\).
Time = 0.68 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.02 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\frac {\sqrt {2} {\left (9 \, A \sqrt {a} - 5 \, B \sqrt {a} + C \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (9 \, A \sqrt {a} - 5 \, B \sqrt {a} + C \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, {\left (3 \, A - 2 \, B\right )} \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {4 \, {\left (3 \, A - 2 \, B\right )} \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (6 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, \sqrt {2} C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {2} C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{8 \, d} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^(3 /2),x, algorithm="giac")
1/8*(sqrt(2)*(9*A*sqrt(a) - 5*B*sqrt(a) + C*sqrt(a))*log(sin(1/2*d*x + 1/2 *c) + 1)/(a^2*sgn(cos(1/2*d*x + 1/2*c))) - sqrt(2)*(9*A*sqrt(a) - 5*B*sqrt (a) + C*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(1/2*d*x + 1/2 *c))) - 4*(3*A - 2*B)*log(abs(1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a^(3/2 )*sgn(cos(1/2*d*x + 1/2*c))) + 4*(3*A - 2*B)*log(abs(-1/2*sqrt(2) + sin(1/ 2*d*x + 1/2*c)))/(a^(3/2)*sgn(cos(1/2*d*x + 1/2*c))) - 2*(6*sqrt(2)*A*sqrt (a)*sin(1/2*d*x + 1/2*c)^3 - 2*sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 + 2*sqrt(2)*C*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 5*sqrt(2)*A*sqrt(a)*sin(1/2*d *x + 1/2*c) + sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c) - sqrt(2)*C*sqrt(a)*s in(1/2*d*x + 1/2*c))/((2*sin(1/2*d*x + 1/2*c)^4 - 3*sin(1/2*d*x + 1/2*c)^2 + 1)*a^2*sgn(cos(1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]