Integrand size = 43, antiderivative size = 232 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(19 A-12 B+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{3/2} d}-\frac {(13 A-9 B+5 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 A-6 B+2 C) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \]
1/4*(19*A-12*B+8*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^( 3/2)/d-1/4*(13*A-9*B+5*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)*sec(d*x+c)*tan(d*x+c)/d/(a+a* cos(d*x+c))^(3/2)-1/4*(7*A-6*B+2*C)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+ 1/2*(2*A-B+C)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)
Time = 1.64 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.80 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (-2 (13 A-9 B+5 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {-\sqrt {2} (19 A-12 B+8 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+\frac {1}{2} (3 A-6 B+2 C+(6 A-8 B) \cos (c+d x)+(7 A-6 B+2 C) \cos (2 (c+d x))) \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )}{-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{2 d (a (1+\cos (c+d x)))^{3/2}} \]
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a* Cos[c + d*x])^(3/2),x]
(Cos[(c + d*x)/2]^3*(-2*(13*A - 9*B + 5*C)*ArcTanh[Sin[(c + d*x)/2]] + (-( Sqrt[2]*(19*A - 12*B + 8*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x )/2]^2) + ((3*A - 6*B + 2*C + (6*A - 8*B)*Cos[c + d*x] + (7*A - 6*B + 2*C) *Cos[2*(c + d*x)])*Sec[c + d*x]^2*Sin[(c + d*x)/2])/2)/(-1 + Sin[(c + d*x) /2]^2)))/(2*d*(a*(1 + Cos[c + d*x]))^(3/2))
Time = 1.52 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.05, 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, 3463, 27, 3042, 3463, 27, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {(4 a (2 A-B+C)-a (5 A-5 B+C) \cos (c+d x)) \sec ^3(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(4 a (2 A-B+C)-a (5 A-5 B+C) \cos (c+d x)) \sec ^3(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {4 a (2 A-B+C)-a (5 A-5 B+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\int -\frac {2 \left (a^2 (7 A-6 B+2 C)-3 a^2 (2 A-B+C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (a^2 (7 A-6 B+2 C)-3 a^2 (2 A-B+C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^2 (7 A-6 B+2 C)-3 a^2 (2 A-B+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {\int -\frac {\left (a^3 (19 A-12 B+8 C)-a^3 (7 A-6 B+2 C) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {a^2 (7 A-6 B+2 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B+2 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (a^3 (19 A-12 B+8 C)-a^3 (7 A-6 B+2 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B+2 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (19 A-12 B+8 C)-a^3 (7 A-6 B+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3464 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B+2 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (19 A-12 B+8 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-2 a^3 (13 A-9 B+5 C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B+2 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (19 A-12 B+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-2 a^3 (13 A-9 B+5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B+2 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (19 A-12 B+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {4 a^3 (13 A-9 B+5 C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B+2 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (19 A-12 B+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 \sqrt {2} a^{5/2} (13 A-9 B+5 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B+2 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {2 a^3 (19 A-12 B+8 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {2 \sqrt {2} a^{5/2} (13 A-9 B+5 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {2 a (2 A-B+C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B+2 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^{5/2} (19 A-12 B+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 \sqrt {2} a^{5/2} (13 A-9 B+5 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
-1/2*((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^(3/2) ) + ((2*a*(2*A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d *x]]) - (-1/2*((2*a^(5/2)*(19*A - 12*B + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x ])/Sqrt[a + a*Cos[c + d*x]]])/d - (2*Sqrt[2]*a^(5/2)*(13*A - 9*B + 5*C)*Ar cTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d)/a + ( a^2*(7*A - 6*B + 2*C)*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/a)/(4*a^ 2)
3.5.16.3.1 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. \(1662\) vs. \(2(201)=402\).
Time = 16.45 (sec) , antiderivative size = 1663, normalized size of antiderivative = 7.17
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1663\) |
| default | \(\text {Expression too large to display}\) | \(2261\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(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)*(104*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-76 *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-76*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)^6*a-104*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+28*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+76*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))*co s(1/2*d*x+1/2*c)^4*a+76*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+26*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)^2-22*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)-19*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-19*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*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...
Time = 0.41 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.65 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {2 \, \sqrt {2} {\left ({\left (13 \, A - 9 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (13 \, A - 9 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A - 9 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + {\left ({\left (19 \, A - 12 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (19 \, A - 12 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (19 \, A - 12 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, {\left ({\left (7 \, A - 6 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - 4 \, B\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{16 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(3 /2),x, algorithm="fricas")
1/16*(2*sqrt(2)*((13*A - 9*B + 5*C)*cos(d*x + c)^4 + 2*(13*A - 9*B + 5*C)* cos(d*x + c)^3 + (13*A - 9*B + 5*C)*cos(d*x + c)^2)*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*c os(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + ((19*A - 12*B + 8*C)*cos(d*x + c)^4 + 2*(19*A - 12*B + 8*C)*cos(d*x + c)^3 + (19*A - 12* B + 8*C)*cos(d*x + c)^2)*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*((7*A - 6*B + 2*C)*cos(d*x + c) ^2 + (3*A - 4*B)*cos(d*x + c) - 2*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) )/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**3/(a*(cos( c + d*x) + 1))**(3/2), x)
Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(3 /2),x, algorithm="maxima")
Exception raised: RuntimeError >> ECL says: Memory limit reached. Please j ump to an outer pointer, quit program and enlarge thememory limits before executing the program again.
Time = 0.70 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.61 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\frac {\sqrt {2} {\left (13 \, A \sqrt {a} - 9 \, B \sqrt {a} + 5 \, 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 (13 \, A \sqrt {a} - 9 \, B \sqrt {a} + 5 \, 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 {{\left (19 \, A - 12 \, B + 8 \, C\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 {{\left (19 \, A - 12 \, B + 8 \, C\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 \, \sqrt {2} {\left (A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 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 )} a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (10 \, \sqrt {2} A \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, \sqrt {2} B \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sqrt {2} A \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, \sqrt {2} B \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 )^{2} - 1\right )}^{2} a^{\frac {3}{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)^3/(a+a*cos(d*x+c))^(3 /2),x, algorithm="giac")
-1/8*(sqrt(2)*(13*A*sqrt(a) - 9*B*sqrt(a) + 5*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)*(13*A*sqrt(a) - 9*B *sqrt(a) + 5*C*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(1/2*d* x + 1/2*c))) - (19*A - 12*B + 8*C)*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))) + (19*A - 12*B + 8*C)*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*sqrt(2)*(A*sqrt(a)*sin(1/2*d*x + 1/2*c) - B*sqrt(a)*sin(1/2*d*x + 1/2*c) + C*sqrt(a)*sin(1/2*d*x + 1/2*c))/((sin(1/2*d*x + 1/2*c)^2 - 1)*a^2*sgn(c os(1/2*d*x + 1/2*c))) - 2*(10*sqrt(2)*A*sin(1/2*d*x + 1/2*c)^3 - 8*sqrt(2) *B*sin(1/2*d*x + 1/2*c)^3 - 3*sqrt(2)*A*sin(1/2*d*x + 1/2*c) + 4*sqrt(2)*B *sin(1/2*d*x + 1/2*c))/((2*sin(1/2*d*x + 1/2*c)^2 - 1)^2*a^(3/2)*sgn(cos(1 /2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]