Integrand size = 33, antiderivative size = 78 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 b^2 (5 A+3 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b C (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d} \] Output:
2/5*b^2*(5*A+3*C)*(b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2 ))/d/cos(d*x+c)^(1/2)+2/5*b*C*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/d
Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {b^2 \sqrt {b \cos (c+d x)} \left (2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+C \sqrt {\cos (c+d x)} \sin (2 (c+d x))\right )}{5 d \sqrt {\cos (c+d x)}} \] Input:
Integrate[(b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]
Output:
(b^2*Sqrt[b*Cos[c + d*x]]*(2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2] + C*Sqr t[Cos[c + d*x]]*Sin[2*(c + d*x)]))/(5*d*Sqrt[Cos[c + d*x]])
Time = 0.42 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 2030, 3493, 3042, 3121, 3042, 3119}
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.
\(\displaystyle \int \sec ^2(c+d x) (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^2 \int \sqrt {b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \left (C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A\right )dx\) |
\(\Big \downarrow \) 3493 |
\(\displaystyle b^2 \left (\frac {1}{5} (5 A+3 C) \int \sqrt {b \cos (c+d x)}dx+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {1}{5} (5 A+3 C) \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle b^2 \left (\frac {(5 A+3 C) \sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {(5 A+3 C) \sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle b^2 \left (\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\) |
Input:
Int[(b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]
Output:
b^2*((2*(5*A + 3*C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*S qrt[Cos[c + d*x]]) + (2*C*(b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*b*d))
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
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[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f *(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2) Int[(b*Sin[e + f*x] )^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(262\) vs. \(2(70)=140\).
Time = 13.04 (sec) , antiderivative size = 263, normalized size of antiderivative = 3.37
method | result | size |
default | \(\frac {2 \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b^{3} \left (8 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-8 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\) | \(263\) |
parts | \(\frac {2 A \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}-\frac {2 C \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b^{3} \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\) | \(359\) |
Input:
int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x,method=_RETURNV ERBOSE)
Output:
2/5*(b*(-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*(8*C*si n(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-8*C*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x +1/2*c)+5*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)* EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*C*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x +1/2*c)+3*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)))/(-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)/(b*(-1+2*cos(1/2*d*x+1/2*c)^2))^( 1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.35 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \, {\left (\sqrt {b \cos \left (d x + c\right )} C b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + i \, \sqrt {\frac {1}{2}} {\left (5 \, A + 3 \, C\right )} b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - i \, \sqrt {\frac {1}{2}} {\left (5 \, A + 3 \, C\right )} b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )\right )}}{5 \, d} \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorith m="fricas")
Output:
2/5*(sqrt(b*cos(d*x + c))*C*b^2*cos(d*x + c)*sin(d*x + c) + I*sqrt(1/2)*(5 *A + 3*C)*b^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* x + c) + I*sin(d*x + c))) - I*sqrt(1/2)*(5*A + 3*C)*b^(5/2)*weierstrassZet a(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
Timed out. \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)
Output:
Timed out
\[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorith m="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^2, x)
\[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorith m="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^2, x)
Timed out. \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \] Input:
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^2,x)
Output:
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^2, x)
\[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\sqrt {b}\, b^{2} \left (\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:
int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x)
Output:
sqrt(b)*b**2*(int(sqrt(cos(c + d*x))*cos(c + d*x)**4*sec(c + d*x)**2,x)*c + int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**2,x)*a)