Integrand size = 25, antiderivative size = 78 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}} \] Output:
2/3*(A+3*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^2/d/ (b*cos(d*x+c))^(1/2)+2/3*A*sin(d*x+c)/b/d/(b*cos(d*x+c))^(3/2)
Time = 0.55 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left ((A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+A \tan (c+d x)\right )}{3 b^2 d \sqrt {b \cos (c+d x)}} \] Input:
Integrate[(A + C*Cos[c + d*x]^2)/(b*Cos[c + d*x])^(5/2),x]
Output:
(2*((A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + A*Tan[c + d*x ]))/(3*b^2*d*Sqrt[b*Cos[c + d*x]])
Time = 0.37 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3491, 3042, 3121, 3042, 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.
| \(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3491 |
| \(\displaystyle \frac {(A+3 C) \int \frac {1}{\sqrt {b \cos (c+d x)}}dx}{3 b^2}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+3 C) \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {(A+3 C) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+3 C) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\) |
Input:
Int[(A + C*Cos[c + d*x]^2)/(b*Cos[c + d*x])^(5/2),x]
Output:
(2*(A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*b^2*d*Sqrt[b *Cos[c + d*x]]) + (2*A*Sin[c + d*x])/(3*b*d*(b*Cos[c + d*x])^(3/2))
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[(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[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)) Int[(b*Sin[e + f* x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(293\) vs. \(2(69)=138\).
Time = 0.95 (sec) , antiderivative size = 294, normalized size of antiderivative = 3.77
| method | result | size |
| default | \(-\frac {2 \left (-2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A +3 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 b^{2} \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\) | \(294\) |
| parts | \(-\frac {2 A \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 b^{2} \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}-\frac {2 C \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{b^{2} \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\) | \(387\) |
Input:
int((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*(-2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*EllipticF(cos(1/2*d*x +1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^( 1/2)*(A+3*C)*sin(1/2*d*x+1/2*c)^2+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))+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)*EllipticF(cos(1/2*d* x+1/2*c),2^(1/2)))/b^2*(b*(-1+2*cos(1/2*d*x+1/2*c)^2)*sin(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)/(-1+2*cos (1/2*d*x+1/2*c)^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 = 0.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.49 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (\sqrt {\frac {1}{2}} {\left (i \, A + 3 i \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {\frac {1}{2}} {\left (-i \, A - 3 i \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - \sqrt {b \cos \left (d x + c\right )} A \sin \left (d x + c\right )\right )}}{3 \, b^{3} d \cos \left (d x + c\right )^{2}} \] Input:
integrate((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-2/3*(sqrt(1/2)*(I*A + 3*I*C)*sqrt(b)*cos(d*x + c)^2*weierstrassPInverse(- 4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(1/2)*(-I*A - 3*I*C)*sqrt(b)*co s(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - s qrt(b*cos(d*x + c))*A*sin(d*x + c))/(b^3*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c))^(5/2), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c))^(5/2), x)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int((A + C*cos(c + d*x)^2)/(b*cos(c + d*x))^(5/2),x)
Output:
int((A + C*cos(c + d*x)^2)/(b*cos(c + d*x))^(5/2), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a \right )}{b^{3}} \] Input:
int((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x)
Output:
(sqrt(b)*(int(sqrt(cos(c + d*x))/cos(c + d*x),x)*c + int(sqrt(cos(c + d*x) )/cos(c + d*x)**3,x)*a))/b**3