\(\int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx\) [20]

Optimal result

Integrand size = 25, antiderivative size = 74 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=-\frac {2 (A-C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}} \] Output:

-2*(A-C)*(b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/d/ 
cos(d*x+c)^(1/2)+2*A*sin(d*x+c)/b/d/(b*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.77 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\frac {-2 (A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 A \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}} \] Input:

Integrate[(A + C*Cos[c + d*x]^2)/(b*Cos[c + d*x])^(3/2),x]
 

Output:

(-2*(A - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 2*A*Sin[c + d*x 
])/(b*d*Sqrt[b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 74, 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, 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.

\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 3491

\(\displaystyle \frac {2 A \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {(A-C) \int \sqrt {b \cos (c+d x)}dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {(A-C) \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {2 A \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {(A-C) \sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b^2 \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {(A-C) \sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 A \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\)

Input:

Int[(A + C*Cos[c + d*x]^2)/(b*Cos[c + d*x])^(3/2),x]
 

Output:

(-2*(A - C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b^2*d*Sqrt[Co 
s[c + d*x]]) + (2*A*Sin[c + d*x])/(b*d*Sqrt[b*Cos[c + d*x]])
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

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]
 

rule 3121

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]
 

rule 3491

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(215\) vs. \(2(70)=140\).

Time = 1.06 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.92

method	result	size
default	\(\frac {2 \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}\, \left (2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{b \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\)	\(216\)
parts	\(-\frac {2 A \left (-2 \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{b \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}+\frac {2 C \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{b \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\)	\(344\)

Input:

int((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/b*(-2*b*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2*b)^(1/2)*(2*A*cos(1/2* 
d*x+1/2*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(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)))/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/si 
n(1/2*d*x+1/2*c)/(b*(-1+2*cos(1/2*d*x+1/2*c)^2))^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.59 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {\frac {1}{2}} {\left (i \, A - i \, C\right )} \sqrt {b} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {\frac {1}{2}} {\left (-i \, A + i \, C\right )} \sqrt {b} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \sqrt {b \cos \left (d x + c\right )} A \sin \left (d x + c\right )\right )}}{b^{2} d \cos \left (d x + c\right )} \] Input:

integrate((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-2*(sqrt(1/2)*(I*A - I*C)*sqrt(b)*cos(d*x + c)*weierstrassZeta(-4, 0, weie 
rstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + sqrt(1/2)*(-I*A + 
 I*C)*sqrt(b)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 
0, cos(d*x + c) - I*sin(d*x + c))) - sqrt(b*cos(d*x + c))*A*sin(d*x + c))/ 
(b^2*d*cos(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c))^(3/2), x)
 

Giac [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int((A + C*cos(c + d*x)^2)/(b*cos(c + d*x))^(3/2),x)
 

Output:

int((A + C*cos(c + d*x)^2)/(b*cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {b}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) c \right )}{b^{2}} \] Input:

int((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(b)*(int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x)*a + int(sqrt(cos(c + d 
*x)),x)*c))/b**2