Integrand size = 35, antiderivative size = 218 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \left (3 b^2 (3 A+C)+a^2 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {2 a b^2 (5 A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{d}-\frac {2 b^3 (35 A-3 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {4 A b (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:
-2/5*b*(15*a^2*(A-C)-b^2*(5*A+3*C))*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/ d+2/3*a*(3*b^2*(3*A+C)+a^2*(A+3*C))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)) /d-2*a*b^2*(5*A-C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d-2/15*b^3*(35*A-3*C)*cos(d *x+c)^(3/2)*sin(d*x+c)/d+4*A*b*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^ (1/2)+2/3*A*(a+b*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(3/2)
Time = 3.93 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \left (-45 a^2 b (A-C)+3 b^3 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {6 \left (15 a^2 A b+b^3 C \cos ^2(c+d x)\right ) \sin (c+d x)+5 a \left (3 b^2 C \sin (2 (c+d x))+2 a^2 A \tan (c+d x)\right )}{\sqrt {\cos (c+d x)}}}{15 d} \] Input:
Integrate[((a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/ 2),x]
Output:
(2*(-45*a^2*b*(A - C) + 3*b^3*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 10* (3*a*b^2*(3*A + C) + a^3*(A + 3*C))*EllipticF[(c + d*x)/2, 2] + (6*(15*a^2 *A*b + b^3*C*Cos[c + d*x]^2)*Sin[c + d*x] + 5*a*(3*b^2*C*Sin[2*(c + d*x)] + 2*a^2*A*Tan[c + d*x]))/Sqrt[Cos[c + d*x]])/(15*d)
Time = 1.52 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 3527, 27, 3042, 3526, 27, 3042, 3512, 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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle \frac {2}{3} \int \frac {(a+b \cos (c+d x))^2 \left (-b (5 A-3 C) \cos ^2(c+d x)+a (A+3 C) \cos (c+d x)+6 A b\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {(a+b \cos (c+d x))^2 \left (-b (5 A-3 C) \cos ^2(c+d x)+a (A+3 C) \cos (c+d x)+6 A b\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (-b (5 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+6 A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{3} \left (2 \int \frac {(a+b \cos (c+d x)) \left ((A+3 C) a^2-2 b (5 A-3 C) \cos (c+d x) a+24 A b^2-b^2 (35 A-3 C) \cos ^2(c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {(a+b \cos (c+d x)) \left ((A+3 C) a^2-2 b (5 A-3 C) \cos (c+d x) a+24 A b^2-b^2 (35 A-3 C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left ((A+3 C) a^2-2 b (5 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+24 A b^2-b^2 (35 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{5} \int \frac {-45 a b^2 (5 A-C) \cos ^2(c+d x)-3 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)+5 a \left ((A+3 C) a^2+24 A b^2\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{5} \int \frac {-45 a b^2 (5 A-C) \cos ^2(c+d x)-3 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)+5 a \left ((A+3 C) a^2+24 A b^2\right )}{\sqrt {\cos (c+d x)}}dx+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{5} \int \frac {-45 a b^2 (5 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a \left ((A+3 C) a^2+24 A b^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {3 \left (5 a \left ((A+3 C) a^2+3 b^2 (3 A+C)\right )-3 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {30 a b^2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {5 a \left ((A+3 C) a^2+3 b^2 (3 A+C)\right )-3 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {30 a b^2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {5 a \left ((A+3 C) a^2+3 b^2 (3 A+C)\right )-3 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a b^2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (-3 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx+5 a \left (a^2 (A+3 C)+3 b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-\frac {30 a b^2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (-3 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 a \left (a^2 (A+3 C)+3 b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a b^2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (5 a \left (a^2 (A+3 C)+3 b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {30 a b^2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\frac {10 a \left (a^2 (A+3 C)+3 b^2 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 b \left (15 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {30 a b^2 (5 A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}\right )+\frac {12 A b \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}-\frac {2 b^3 (35 A-3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[((a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]
Output:
(2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((-2* b^3*(35*A - 3*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (12*A*b*(a + b*C os[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + ((-6*b*(15*a^2*(A - C) - b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/d + (10*a*(3*b^2*(3*A + C ) + a^2*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/d - (30*a*b^2*(5*A - C)*Sqrt [Cos[c + d*x]]*Sin[c + d*x])/d)/5)/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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, 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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b *c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( A*d^2*(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, 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. \(966\) vs. \(2(205)=410\).
Time = 22.83 (sec) , antiderivative size = 967, normalized size of antiderivative = 4.44
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(967\) |
| default | \(\text {Expression too large to display}\) | \(1267\) |
Input:
int((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x,method=_RETUR NVERBOSE)
Output:
2*(A*b^3+3*C*a^2*b)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2 )*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*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) ^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d+2*(3*A*a*b^2+ C*a^3)/d*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))-2/3*A*a^3*(-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))*sin(1/2*d*x+1/2*c)^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)*EllipticF (cos(1/2*d*x+1/2*c),2^(1/2)))*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(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)/(2*cos(1/ 2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d-2/5*C*b^3*((2*cos(1/2*d*x+1/2 *c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/ 2*c)^6+8*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c)^2*co s(1/2*d*x+1/2*c)-3*(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)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/ 2)/d-6*A*a^2*b*(-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*co s(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+si...
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.38 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, {\left (A + 3 \, C\right )} a^{3} + 3 i \, {\left (3 \, A + C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (A + 3 \, C\right )} a^{3} - 3 i \, {\left (3 \, A + C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (15 i \, {\left (A - C\right )} a^{2} b - i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-15 i \, {\left (A - C\right )} a^{2} b + i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, C b^{3} \cos \left (d x + c\right )^{3} + 15 \, C a b^{2} \cos \left (d x + c\right )^{2} + 45 \, A a^{2} b \cos \left (d x + c\right ) + 5 \, A a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algori thm="fricas")
Output:
-1/15*(5*sqrt(2)*(I*(A + 3*C)*a^3 + 3*I*(3*A + C)*a*b^2)*cos(d*x + c)^2*we ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*(A + 3*C)*a^3 - 3*I*(3*A + C)*a*b^2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(15*I*(A - C)*a^2*b - I*(5*A + 3*C)*b^3)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-15*I*(A - C)*a^2*b + I*( 5*A + 3*C)*b^3)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse( -4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*C*b^3*cos(d*x + c)^3 + 15*C* a*b^2*cos(d*x + c)^2 + 45*A*a^2*b*cos(d*x + c) + 5*A*a^3)*sqrt(cos(d*x + c ))*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**3*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algori thm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2) , x)
\[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2) , x)
Time = 1.48 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,\left (A\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^3+3\,A\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^2\right )}{d}+\frac {2\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^2\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {3\,C\,a\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,A\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^3)/cos(c + d*x)^(5/2),x)
Output:
(2*(A*b^3*ellipticE(c/2 + (d*x)/2, 2) + 3*A*a*b^2*ellipticF(c/2 + (d*x)/2, 2)))/d + (2*C*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*C*a^2*b*ellipticE(c /2 + (d*x)/2, 2))/d + (3*C*a*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*a^3*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*C*b^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11 /4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) + (6*A*a^2*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))
\[ \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{3} c +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{2} b^{2}+\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a^{4}+3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a^{3} b +3 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a^{2} b c +\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a \,b^{3}+3 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) a \,b^{2} c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) b^{3} c \] Input:
int((a+b*cos(d*x+c))^3*(A+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**3*c + 3*int(sqrt(cos(c + d*x))/c os(c + d*x),x)*a**2*b**2 + int(sqrt(cos(c + d*x))/cos(c + d*x)**3,x)*a**4 + 3*int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x)*a**3*b + 3*int(sqrt(cos(c + d*x)),x)*a**2*b*c + int(sqrt(cos(c + d*x)),x)*a*b**3 + 3*int(sqrt(cos(c + d*x))*cos(c + d*x),x)*a*b**2*c + int(sqrt(cos(c + d*x))*cos(c + d*x)**2,x) *b**3*c