Integrand size = 43, antiderivative size = 180 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (6 a b B+b^2 (3 A+C)+a^2 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a (4 A b+3 a B) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:
-2*(B*a^2-B*b^2+2*a*b*(A-C))*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*( 6*B*a*b+b^2*(3*A+C)+a^2*(A+3*C))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+ 2/3*a*(4*A*b+3*B*a)*sin(d*x+c)/d/cos(d*x+c)^(1/2)-2/3*b^2*(A-C)*cos(d*x+c) ^(1/2)*sin(d*x+c)/d+2/3*A*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(3/2)
Time = 3.25 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-6 \left (a^2 B-b^2 B+2 a b (A-C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (6 a b B+b^2 (3 A+C)+a^2 (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+12 a A b \sin (c+d x)+6 a^2 B \sin (c+d x)+b^2 C \sin (2 (c+d x))+2 a^2 A \tan (c+d x)}{3 d \sqrt {\cos (c+d x)}} \] Input:
Integrate[((a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) /Cos[c + d*x]^(5/2),x]
Output:
(-6*(a^2*B - b^2*B + 2*a*b*(A - C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x) /2, 2] + 2*(6*a*b*B + b^2*(3*A + C) + a^2*(A + 3*C))*Sqrt[Cos[c + d*x]]*El lipticF[(c + d*x)/2, 2] + 12*a*A*b*Sin[c + d*x] + 6*a^2*B*Sin[c + d*x] + b ^2*C*Sin[2*(c + d*x)] + 2*a^2*A*Tan[c + d*x])/(3*d*Sqrt[Cos[c + d*x]])
Time = 1.15 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.99, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 3526, 27, 3042, 3510, 27, 3042, 3502, 27, 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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {2}{3} \int \frac {(a+b \cos (c+d x)) \left (-3 b (A-C) \cos ^2(c+d x)+(3 b B+a (A+3 C)) \cos (c+d x)+4 A b+3 a B\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {(a+b \cos (c+d x)) \left (-3 b (A-C) \cos ^2(c+d x)+(3 b B+a (A+3 C)) \cos (c+d x)+4 A b+3 a B\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-3 b (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(3 b B+a (A+3 C)) \sin \left (c+d x+\frac {\pi }{2}\right )+4 A b+3 a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 a (3 a B+4 A b) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-2 \int -\frac {(A+3 C) a^2+6 b B a+4 A b^2-3 b^2 (A-C) \cos ^2(c+d x)-3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {(A+3 C) a^2+6 b B a+4 A b^2-3 b^2 (A-C) \cos ^2(c+d x)-3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 a (3 a B+4 A b) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {(A+3 C) a^2+6 b B a+4 A b^2-3 b^2 (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a (3 a B+4 A b) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{3} \int \frac {3 \left ((A+3 C) a^2+6 b B a+b^2 (3 A+C)-3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 a (3 a B+4 A b) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {(A+3 C) a^2+6 b B a+b^2 (3 A+C)-3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 a (3 a B+4 A b) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {(A+3 C) a^2+6 b B a+b^2 (3 A+C)-3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a (3 a B+4 A b) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{3} \left (\left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \int \sqrt {\cos (c+d x)}dx+\frac {2 a (3 a B+4 A b) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a (3 a B+4 A b) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{3} \left (\left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A-C)-b^2 B\right )}{d}+\frac {2 a (3 a B+4 A b) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right )}{d}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A-C)-b^2 B\right )}{d}+\frac {2 a (3 a B+4 A b) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[((a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]
Output:
(2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((-6* (a^2*B - b^2*B + 2*a*b*(A - C))*EllipticE[(c + d*x)/2, 2])/d + (2*(6*a*b*B + b^2*(3*A + C) + a^2*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/d + (2*a*(4*A *b + 3*a*B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) - (2*b^2*(A - C)*Sqrt[Cos [c + d*x]]*Sin[c + d*x])/d)/3
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_.)*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_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 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/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(767\) vs. \(2(169)=338\).
Time = 4.22 (sec) , antiderivative size = 768, normalized size of antiderivative = 4.27
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(768\) |
| default | \(\text {Expression too large to display}\) | \(1301\) |
Input:
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, method=_RETURNVERBOSE)
Output:
-2*(2*A*a*b+B*a^2)*(-2*(-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)*sin(1/2*d*x+1/2*c)^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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(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+s in(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+2*(B*b^2+2*C*a*b)*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^ 2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*El lipticE(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+2*(A *b^2+2*B*a*b+C*a^2)/d*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))-2/3*A*a^2*(-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(c os(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-2*cos(1/2*d*x+1/2*c)*sin(1 /2*d*x+1/2*c)^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)))*((-1+2*cos(1/2*d*x+1/2*c)^2)*si n(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)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(3/2)/sin(1/2*d*x+1/2*c)/d-2/3*C*b^2*((-1 +2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d*x+1/2*c) *sin(1/2*d*x+1/2*c)^4-2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^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)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/s...
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left (-i \, {\left (A + 3 \, C\right )} a^{2} - 6 i \, B a b - i \, {\left (3 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, {\left (A + 3 \, C\right )} a^{2} + 6 i \, B a b + i \, {\left (3 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (i \, B a^{2} + 2 i \, {\left (A - C\right )} a b - i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (-i \, B a^{2} - 2 i \, {\left (A - C\right )} a b + i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (C b^{2} \cos \left (d x + c\right )^{2} + A a^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5 /2),x, algorithm="fricas")
Output:
1/3*(sqrt(2)*(-I*(A + 3*C)*a^2 - 6*I*B*a*b - I*(3*A + C)*b^2)*cos(d*x + c) ^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I* (A + 3*C)*a^2 + 6*I*B*a*b + I*(3*A + C)*b^2)*cos(d*x + c)^2*weierstrassPIn verse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(I*B*a^2 + 2*I*(A - C)*a*b - I*B*b^2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInve rse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(-I*B*a^2 - 2*I*(A - C)*a*b + I*B*b^2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInve rse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(C*b^2*cos(d*x + c)^2 + A*a ^2 + 3*(B*a^2 + 2*A*a*b)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d *cos(d*x + c)^2)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)* *(5/2),x)
Output:
Timed out
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5 /2),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2/c os(d*x + c)^(5/2), x)
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5 /2),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2/c os(d*x + c)^(5/2), x)
Time = 1.98 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.49 \[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {C\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,B\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,A\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int(((a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(5/2),x)
Output:
(C*b^2*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2 , 2))/3))/d + (2*A*b^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*b^2*ellipticE (c/2 + (d*x)/2, 2))/d + (2*C*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (4*B*a*b *ellipticF(c/2 + (d*x)/2, 2))/d + (4*C*a*b*ellipticE(c/2 + (d*x)/2, 2))/d + (2*A*a^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d* cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*B*a^2*sin(c + d*x)*hyperge om([-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)) + (4*A*a*b*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))
\[ \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{2} c +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a \,b^{2}+\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a^{3}+3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a^{2} b +2 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a b c +\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) b^{3}+\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) b^{2} c \] Input:
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x)
Output:
int(sqrt(cos(c + d*x))/cos(c + d*x),x)*a**2*c + 3*int(sqrt(cos(c + d*x))/c os(c + d*x),x)*a*b**2 + int(sqrt(cos(c + d*x))/cos(c + d*x)**3,x)*a**3 + 3 *int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x)*a**2*b + 2*int(sqrt(cos(c + d*x )),x)*a*b*c + int(sqrt(cos(c + d*x)),x)*b**3 + int(sqrt(cos(c + d*x))*cos( c + d*x),x)*b**2*c