Integrand size = 43, antiderivative size = 270 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\frac {(119 A-49 B+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(33 A-13 B+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(33 A-13 B+3 C) \sin (c+d x)}{6 a^3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(119 A-49 B+9 C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac {(2 A-B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {(119 A-49 B+9 C) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
1/10*(119*A-49*B+9*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/6*(33* A-13*B+3*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d+1/6*(33*A-13*B+3* C)*sin(d*x+c)/a^3/d/cos(d*x+c)^(3/2)-1/10*(119*A-49*B+9*C)*sin(d*x+c)/a^3/ d/cos(d*x+c)^(1/2)-1/5*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+ c))^3-1/3*(2*A-B)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2-1/30* (119*A-49*B+9*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a^3+a^3*cos(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.87 (sec) , antiderivative size = 1529, normalized size of antiderivative = 5.66 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^3),x]
Output:
(-22*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - S in[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTa n[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^ 3*Sqrt[1 + Cot[c]^2]) + (26*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Hypergeometric PFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - Arc Tan[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]]]])/( 3*d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2]) - (2*C*Cos[c/2 + (d*x)/2]^6 *Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2 ]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*S qrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d *x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2]) + (Co s[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*((-2*(60*A - 20*B + 59*A*Cos[c] - 29 *B*Cos[c] + 9*C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/(5*d) - (2*Sec[c/2]*Sec[ c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(16*A*Sin[(d*x)/2] - 11*B*Sin[(d*x)/2] + 6*C*Sin[(d*x)/2]))/(15*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(59*A*Sin[(d *x)/2] - 29*B*Sin[(d*x)/2] + 9*C*Sin[(d*x)/2]))/(5*d) + (16*A*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(3*d) + (16*Sec[c]*Sec[c + d*x]*(A*Sin[c] - 9*A*Sin...
Time = 1.40 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 3520, 27, 3042, 3457, 3042, 3457, 27, 3042, 3227, 3042, 3116, 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.
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3520 |
\(\displaystyle \frac {\int \frac {a (13 A-3 B+3 C)-a (7 A-7 B-3 C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a (13 A-3 B+3 C)-a (7 A-7 B-3 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (13 A-3 B+3 C)-a (7 A-7 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\int \frac {3 a^2 (23 A-8 B+3 C)-25 a^2 (2 A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {3 a^2 (23 A-8 B+3 C)-25 a^2 (2 A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\frac {\int \frac {3 \left (5 a^3 (33 A-13 B+3 C)-a^3 (119 A-49 B+9 C) \cos (c+d x)\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {a^2 (119 A-49 B+9 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (33 A-13 B+3 C)-a^3 (119 A-49 B+9 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {a^2 (119 A-49 B+9 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (33 A-13 B+3 C)-a^3 (119 A-49 B+9 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {a^2 (119 A-49 B+9 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B+3 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx-a^3 (119 A-49 B+9 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )}{2 a^2}-\frac {a^2 (119 A-49 B+9 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B+3 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-a^3 (119 A-49 B+9 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )}{2 a^2}-\frac {a^2 (119 A-49 B+9 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (119 A-49 B+9 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )}{2 a^2}-\frac {a^2 (119 A-49 B+9 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (119 A-49 B+9 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )}{2 a^2}-\frac {a^2 (119 A-49 B+9 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (119 A-49 B+9 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}-\frac {a^2 (119 A-49 B+9 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B+3 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (119 A-49 B+9 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}-\frac {a^2 (119 A-49 B+9 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
Input:
Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Cos [c + d*x])^3),x]
Output:
-1/5*((A - B + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x]) ^3) + ((-10*a*(2*A - B)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2) + (-((a^2*(119*A - 49*B + 9*C)*Sin[c + d*x])/(d*Cos[c + d*x]^( 3/2)*(a + a*Cos[c + d*x]))) + (3*(5*a^3*(33*A - 13*B + 3*C)*((2*EllipticF[ (c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))) - a^3* (119*A - 49*B + 9*C)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/ (d*Sqrt[Cos[c + d*x]]))))/(2*a^2))/(3*a^2))/(10*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
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_)])^(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_)*((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. \(1012\) vs. \(2(249)=498\).
Time = 9.52 (sec) , antiderivative size = 1013, normalized size of antiderivative = 3.75
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x, method=_RETURNVERBOSE)
Output:
-1/4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^3*((-24*A +8*B)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2* c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c )-(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)))+1/3*(4*A-2*B)*(2*(2*sin(1/2*d*x+1/2*c)^2-1)^( 1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) -3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x +1/2*c)-2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2 *EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1 /2)))*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7 *sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*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)+(12*A-4*B)*(cos(1/2*d*x+1/2* c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(Elliptic F(cos(1/2*d*x+1/2*c),2^(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)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x +1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(A-B+C)*(1/5*(-2*sin(1/2*d*x+1/2*c)^ 4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)^5+4/5*(-2*sin(1/2*d*x+1/2 *c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)^3+18/5*(-2*sin(1/2*d* x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)-8/5*(sin(1/2*d*x +1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*...
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 605, normalized size of antiderivative = 2.24 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c) )^3,x, algorithm="fricas")
Output:
-1/60*(2*(3*(119*A - 49*B + 9*C)*cos(d*x + c)^4 + 2*(453*A - 188*B + 33*C) *cos(d*x + c)^3 + 5*(139*A - 59*B + 9*C)*cos(d*x + c)^2 + 60*(2*A - B)*cos (d*x + c) - 20*A)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(33*I*A - 1 3*I*B + 3*I*C)*cos(d*x + c)^5 + 3*sqrt(2)*(33*I*A - 13*I*B + 3*I*C)*cos(d* x + c)^4 + 3*sqrt(2)*(33*I*A - 13*I*B + 3*I*C)*cos(d*x + c)^3 + sqrt(2)*(3 3*I*A - 13*I*B + 3*I*C)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-33*I*A + 13*I*B - 3*I*C)*cos(d*x + c)^5 + 3*sqrt(2)*(-33*I*A + 13*I*B - 3*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-3 3*I*A + 13*I*B - 3*I*C)*cos(d*x + c)^3 + sqrt(2)*(-33*I*A + 13*I*B - 3*I*C )*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) ) + 3*(sqrt(2)*(-119*I*A + 49*I*B - 9*I*C)*cos(d*x + c)^5 + 3*sqrt(2)*(-11 9*I*A + 49*I*B - 9*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-119*I*A + 49*I*B - 9* I*C)*cos(d*x + c)^3 + sqrt(2)*(-119*I*A + 49*I*B - 9*I*C)*cos(d*x + c)^2)* weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(119*I*A - 49*I*B + 9*I*C)*cos(d*x + c)^5 + 3*sqrt(2) *(119*I*A - 49*I*B + 9*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(119*I*A - 49*I*B + 9*I*C)*cos(d*x + c)^3 + sqrt(2)*(119*I*A - 49*I*B + 9*I*C)*cos(d*x + c)^2 )*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d *x + c))))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d* x + c)^3 + a^3*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+a*cos(d*x+ c))**3,x)
Output:
Timed out
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c) )^3,x, algorithm="maxima")
Output:
Timed out
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c) )^3,x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^3* cos(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a*cos (c + d*x))^3),x)
Output:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a*cos (c + d*x))^3), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{6}+3 \cos \left (d x +c \right )^{5}+3 \cos \left (d x +c \right )^{4}+\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5}+3 \cos \left (d x +c \right )^{4}+3 \cos \left (d x +c \right )^{3}+\cos \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}+3 \cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right ) c}{a^{3}} \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x)
Output:
(int(sqrt(cos(c + d*x))/(cos(c + d*x)**6 + 3*cos(c + d*x)**5 + 3*cos(c + d *x)**4 + cos(c + d*x)**3),x)*a + int(sqrt(cos(c + d*x))/(cos(c + d*x)**5 + 3*cos(c + d*x)**4 + 3*cos(c + d*x)**3 + cos(c + d*x)**2),x)*b + int(sqrt( cos(c + d*x))/(cos(c + d*x)**4 + 3*cos(c + d*x)**3 + 3*cos(c + d*x)**2 + c os(c + d*x)),x)*c)/a**3