Integrand size = 43, antiderivative size = 172 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 a^2 (5 B+4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (3 A+2 B+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {2 a^2 (15 A-5 B-7 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 (5 A-C) \sqrt {\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d} \] Output:
4/5*a^2*(5*B+4*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/3*a^2*(3*A+2*B +C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d-2/15*a^2*(15*A-5*B-7*C)*cos(d *x+c)^(1/2)*sin(d*x+c)/d+2*A*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(1 /2)-2/5*(5*A-C)*cos(d*x+c)^(1/2)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.09 (sec) , antiderivative size = 1039, normalized size of antiderivative = 6.04 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
Integrate[((a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) /Cos[c + d*x]^(3/2),x]
Output:
Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(-1/20*((-5 *A + 10*B + 8*C + 5*A*Cos[2*c] + 10*B*Cos[2*c] + 8*C*Cos[2*c])*Csc[c]*Sec[ c])/d + ((B + 2*C)*Cos[d*x]*Sin[c])/(6*d) + (C*Cos[2*d*x]*Sin[2*c])/(20*d) + ((B + 2*C)*Cos[c]*Sin[d*x])/(6*d) + (A*Sec[c]*Sec[c + d*x]*Sin[d*x])/(2 *d) + (C*Cos[2*c]*Sin[2*d*x])/(20*d)) - (A*(a + a*Cos[c + d*x])^2*Csc[c]*H ypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]* Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[ d*x - ArcTan[Cot[c]]]])/(d*Sqrt[1 + Cot[c]^2]) - (2*B*(a + a*Cos[c + d*x]) ^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2 ]*Sec[c/2 + (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan [Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sq rt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (C*(a + a*Co s[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan [Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d *x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Co t[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - ( B*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ[{ -1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]] *Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan...
Time = 1.27 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 3522, 27, 3042, 3455, 27, 3042, 3447, 3042, 3502, 3042, 3227, 3042, 3119, 3120}
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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^2 (a (4 A+B)-a (5 A-C) \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^2 (a (4 A+B)-a (5 A-C) \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (4 A+B)-a (5 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {2}{5} \int \frac {(\cos (c+d x) a+a) \left (a^2 (15 A+5 B+C)-a^2 (15 A-5 B-7 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {(\cos (c+d x) a+a) \left (a^2 (15 A+5 B+C)-a^2 (15 A-5 B-7 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (15 A+5 B+C)-a^2 (15 A-5 B-7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {-\left ((15 A-5 B-7 C) \cos ^2(c+d x) a^3\right )+(15 A+5 B+C) a^3+\left (a^3 (15 A+5 B+C)-a^3 (15 A-5 B-7 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {-\left ((15 A-5 B-7 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3\right )+(15 A+5 B+C) a^3+\left (a^3 (15 A+5 B+C)-a^3 (15 A-5 B-7 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {5 (3 A+2 B+C) a^3+3 (5 B+4 C) \cos (c+d x) a^3}{\sqrt {\cos (c+d x)}}dx-\frac {2 a^3 (15 A-5 B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {5 (3 A+2 B+C) a^3+3 (5 B+4 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^3 (15 A-5 B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (3 A+2 B+C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^3 (5 B+4 C) \int \sqrt {\cos (c+d x)}dx\right )-\frac {2 a^3 (15 A-5 B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (3 A+2 B+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^3 (5 B+4 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 a^3 (15 A-5 B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (3 A+2 B+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^3 (5 B+4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^3 (15 A-5 B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (\frac {10 a^3 (3 A+2 B+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^3 (5 B+4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^3 (15 A-5 B-7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt {\cos (c+d x)}}\) |
Input:
Int[((a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(3/2),x]
Output:
(2*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + ((-2*(5 *A - C)*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(5*d) + ((2*((6*a^3*(5*B + 4*C)*EllipticE[(c + d*x)/2, 2])/d + (10*a^3*(3*A + 2*B + C)*EllipticF[(c + d*x)/2, 2])/d))/3 - (2*a^3*(15*A - 5*B - 7*C)*Sqrt[Cos [c + d*x]]*Sin[c + d*x])/(3*d))/5)/a
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*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] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m* (c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* (n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ [m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(594\) vs. \(2(161)=322\).
Time = 10.92 (sec) , antiderivative size = 595, normalized size of antiderivative = 3.46
| method | result | size |
| default | \(-\frac {4 a^{2} \left (-12 C \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (5 B +16 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (15 A +5 B +13 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+10 B \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-15 B \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 C \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 C \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(595\) |
| parts | \(\text {Expression too large to display}\) | \(760\) |
Input:
int((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x, method=_RETURNVERBOSE)
Output:
-4/15*a^2*(-12*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*sin( 1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x +1/2*c)^2)^(1/2)*(5*B+16*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-(-2*si n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(15*A+5*B+13*C)*sin(1/2*d*x +1/2*c)^2*cos(1/2*d*x+1/2*c)+15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2* d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d* x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+10*B*(-2*sin(1/2*d*x+1/2*c)^4+sin(1 /2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^ 2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*(-2*sin(1/2*d*x+1/2* c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d *x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+5*C*(-2*sin(1/2 *d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2* sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*C*( -2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2) ^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/ 2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2* c)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.14 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.34 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, C a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, A a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )} \] Input:
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3 /2),x, algorithm="fricas")
Output:
-2/15*(5*I*sqrt(2)*(3*A + 2*B + C)*a^2*cos(d*x + c)*weierstrassPInverse(-4 , 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(3*A + 2*B + C)*a^2*cos( d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*s qrt(2)*(5*B + 4*C)*a^2*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(5*B + 4*C)*a^2* cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c ) - I*sin(d*x + c))) - (3*C*a^2*cos(d*x + c)^2 + 5*(B + 2*C)*a^2*cos(d*x + c) + 15*A*a^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))
Timed out. \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)* *(3/2),x)
Output:
Timed out
\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3 /2),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^2/c os(d*x + c)^(3/2), x)
\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3 /2),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^2/c os(d*x + c)^(3/2), x)
Time = 1.05 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.38 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2\,C\,a^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,B\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+6\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int(((a + a*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(3/2),x)
Output:
(2*C*a^2*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x) /2, 2))/3))/d + (2*B*a^2*(cos(c + d*x)^(1/2)*sin(c + d*x) + 6*ellipticE(c/ 2 + (d*x)/2, 2) + 4*ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*A*a^2*ellipti cE(c/2 + (d*x)/2, 2))/d + (4*A*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*a ^2*ellipticE(c/2 + (d*x)/2, 2))/d + (2*A*a^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) - (2*C*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, co s(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))
\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=a^{2} \left (2 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a +2 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) b +2 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) c \right ) \] Input:
int((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x)
Output:
a**2*(2*int(sqrt(cos(c + d*x))/cos(c + d*x),x)*a + int(sqrt(cos(c + d*x))/ cos(c + d*x),x)*b + int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x)*a + int(sqrt (cos(c + d*x)),x)*a + 2*int(sqrt(cos(c + d*x)),x)*b + int(sqrt(cos(c + d*x )),x)*c + int(sqrt(cos(c + d*x))*cos(c + d*x),x)*b + 2*int(sqrt(cos(c + d* x))*cos(c + d*x),x)*c + int(sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*c)