Integrand size = 43, antiderivative size = 217 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {(5 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {(115 A-43 B+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \]
-(5*A-2*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d+1/ 32*(115*A-43*B+3*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c) )^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(A-B+C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(5/2) -1/16*(15*A-7*B-C)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+1/16*(35*A-11*B+3 *C)*tan(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)
Time = 4.00 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.87 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (4 (115 A-43 B+3 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x)-64 \sqrt {2} (5 A-2 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x)+(67 A-11 B+3 C+2 (55 A-15 B+7 C) \cos (c+d x)+(35 A-11 B+3 C) \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{32 a d (a (1+\cos (c+d x)))^{3/2}} \]
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + a* Cos[c + d*x])^(5/2),x]
(Sec[(c + d*x)/2]*Sec[c + d*x]*(4*(115*A - 43*B + 3*C)*ArcTanh[Sin[(c + d* x)/2]]*Cos[(c + d*x)/2]^4*Cos[c + d*x] - 64*Sqrt[2]*(5*A - 2*B)*ArcTanh[Sq rt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^4*Cos[c + d*x] + (67*A - 11*B + 3 *C + 2*(55*A - 15*B + 7*C)*Cos[c + d*x] + (35*A - 11*B + 3*C)*Cos[2*(c + d *x)])*Sin[(c + d*x)/2]))/(32*a*d*(a*(1 + Cos[c + d*x]))^(3/2))
Time = 1.50 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.07, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.372, Rules used = {3042, 3520, 27, 3042, 3457, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {(2 a (5 A-B+C)-a (5 A-5 B-3 C) \cos (c+d x)) \sec ^2(c+d x)}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(2 a (5 A-B+C)-a (5 A-5 B-3 C) \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a (5 A-B+C)-a (5 A-5 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (2 a^2 (35 A-11 B+3 C)-3 a^2 (15 A-7 B-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\left (2 a^2 (35 A-11 B+3 C)-3 a^2 (15 A-7 B-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 a^2 (35 A-11 B+3 C)-3 a^2 (15 A-7 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {\left (16 a^3 (5 A-2 B)-a^3 (35 A-11 B+3 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^2 (35 A-11 B+3 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (16 a^3 (5 A-2 B)-a^3 (35 A-11 B+3 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {16 a^3 (5 A-2 B)-a^3 (35 A-11 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3464 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {16 a^2 (5 A-2 B) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-a^3 (115 A-43 B+3 C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {16 a^2 (5 A-2 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^3 (115 A-43 B+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {16 a^2 (5 A-2 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^3 (115 A-43 B+3 C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {16 a^2 (5 A-2 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {\sqrt {2} a^{5/2} (115 A-43 B+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {32 a^3 (5 A-2 B) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {\sqrt {2} a^{5/2} (115 A-43 B+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {32 a^{5/2} (5 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{5/2} (115 A-43 B+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \tan (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
-1/4*((A - B + C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^(5/2)) + (-1/2*(a* (15*A - 7*B - C)*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^(3/2)) + (-(((32*a^ (5/2)*(5*A - 2*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]] )/d - (Sqrt[2]*a^(5/2)*(115*A - 43*B + 3*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x]) /(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d)/a) + (2*a^2*(35*A - 11*B + 3*C)*T an[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(4*a^2))/(8*a^2)
3.5.23.3.1 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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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. \(1089\) vs. \(2(188)=376\).
Time = 16.28 (sec) , antiderivative size = 1090, normalized size of antiderivative = 5.02
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1090\) |
| default | \(\text {Expression too large to display}\) | \(1292\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+cos(d*x+c)*a)^(5/2),x, method=_RETURNVERBOSE)
1/16*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(230*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))*cos(1/2*d*x+1/2*c)^6*a-16 0*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)^6*a-160* 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)^6*a-115*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+70*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1 /2)*cos(1/2*d*x+1/2*c)^4+80*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))*c os(1/2*d*x+1/2*c)^4*a+80*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*c os(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-15*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)-2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(7/2) /cos(1/2*d*x+1/2*c)^3/(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+1/32*B/a^(7/ 2)/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*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)^4*a+16*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...
Leaf count of result is larger than twice the leaf count of optimal. 421 vs. \(2 (188) = 376\).
Time = 0.34 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.94 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 16 \, {\left ({\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left ({\left (35 \, A - 11 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (55 \, A - 15 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^(5 /2),x, algorithm="fricas")
1/64*(sqrt(2)*((115*A - 43*B + 3*C)*cos(d*x + c)^4 + 3*(115*A - 43*B + 3*C )*cos(d*x + c)^3 + 3*(115*A - 43*B + 3*C)*cos(d*x + c)^2 + (115*A - 43*B + 3*C)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(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)) - 16*((5*A - 2*B)*cos(d*x + c)^4 + 3*(5*A - 2*B) *cos(d*x + c)^3 + 3*(5*A - 2*B)*cos(d*x + c)^2 + (5*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 + co s(d*x + c)^2)) + 4*((35*A - 11*B + 3*C)*cos(d*x + c)^2 + (55*A - 15*B + 7* C)*cos(d*x + c) + 16*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos( d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d *x + c))
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^(5 /2),x, algorithm="maxima")
Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (188) = 376\).
Time = 1.79 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.77 \[ \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))^{5/2}} \, dx=\frac {\frac {\sqrt {2} {\left (115 \, A \sqrt {a} - 43 \, B \sqrt {a} + 3 \, C \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (115 \, A \sqrt {a} - 43 \, B \sqrt {a} + 3 \, C \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {32 \, {\left (5 \, 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 {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {32 \, {\left (5 \, 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 {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {64 \, \sqrt {2} A \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (19 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \sqrt {2} C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, \sqrt {2} C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, 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))^(5 /2),x, algorithm="giac")
1/64*(sqrt(2)*(115*A*sqrt(a) - 43*B*sqrt(a) + 3*C*sqrt(a))*log(sin(1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(1/2*d*x + 1/2*c))) - sqrt(2)*(115*A*sqrt(a) - 43*B*sqrt(a) + 3*C*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(1/ 2*d*x + 1/2*c))) - 32*(5*A - 2*B)*log(abs(1/2*sqrt(2) + sin(1/2*d*x + 1/2* c)))/(a^(5/2)*sgn(cos(1/2*d*x + 1/2*c))) + 32*(5*A - 2*B)*log(abs(-1/2*sqr t(2) + sin(1/2*d*x + 1/2*c)))/(a^(5/2)*sgn(cos(1/2*d*x + 1/2*c))) - 64*sqr t(2)*A*sin(1/2*d*x + 1/2*c)/((2*sin(1/2*d*x + 1/2*c)^2 - 1)*a^(5/2)*sgn(co s(1/2*d*x + 1/2*c))) - 2*(19*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 11 *sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 + 3*sqrt(2)*C*sqrt(a)*sin(1/2*d* x + 1/2*c)^3 - 21*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + 13*sqrt(2)*B*sq rt(a)*sin(1/2*d*x + 1/2*c) - 5*sqrt(2)*C*sqrt(a)*sin(1/2*d*x + 1/2*c))/((s in(1/2*d*x + 1/2*c)^2 - 1)^2*a^3*sgn(cos(1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]