\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx\) [830]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 29, antiderivative size = 85 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\frac {2 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}+\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {b \cos (c+d x)}} \] Output:

2*B*(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*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b/ 
d/(b*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.67 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {\cos (c+d x)} \left (B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+A \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )\right )}{b d \sqrt {b \cos (c+d x)}} \] Input:

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

Output:

(2*Sqrt[Cos[c + d*x]]*(B*EllipticE[(c + d*x)/2, 2] + A*EllipticF[(c + d*x) 
/2, 2]))/(b*d*Sqrt[b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2030, 3042, 3227, 3042, 3121, 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.

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

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int \frac {A+B \cos (c+d x)}{\sqrt {b \cos (c+d x)}}dx}{b}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle \frac {A \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {B \int \sqrt {b \cos (c+d x)}dx}{b}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {B \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {\frac {A \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{\sqrt {b \cos (c+d x)}}+\frac {B \sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b \sqrt {\cos (c+d x)}}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {A \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {b \cos (c+d x)}}+\frac {B \sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \sqrt {\cos (c+d x)}}}{b}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {A \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {b \cos (c+d x)}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {b \cos (c+d x)}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b}\)

Input:

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

Output:

((2*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b*d*Sqrt[Cos[c + d* 
x]]) + (2*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(d*Sqrt[b*Cos[c 
+ d*x]]))/b
 

Defintions of rubi rules used

rule 2030
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]
 

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 3120
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]
 

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 3227
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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(162\) vs. \(2(80)=160\).

Time = 6.72 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.92

method result size
default \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{b \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}\) \(163\)
parts \(-\frac {2 A \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{b \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}+\frac {2 B \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{b \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}\) \(290\)
risch \(-\frac {i B \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {2}\, {\mathrm e}^{-i \left (d x +c \right )}}{d b \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b \,{\mathrm e}^{-i \left (d x +c \right )}}}-\frac {i \left (\frac {i A \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {b \,{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )} b}}+B \left (-\frac {2 \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}{b \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i \operatorname {EllipticE}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {b \,{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )} b}}\right )\right ) \sqrt {2}\, \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b \,{\mathrm e}^{i \left (d x +c \right )}}\, {\mathrm e}^{-i \left (d x +c \right )}}{d b \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b \,{\mathrm e}^{-i \left (d x +c \right )}}}\) \(434\)

Input:

int(cos(d*x+c)*(A+B*cos(d*x+c))/(cos(d*x+c)*b)^(3/2),x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.45 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (i \, \sqrt {\frac {1}{2}} A \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {\frac {1}{2}} A \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {\frac {1}{2}} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {\frac {1}{2}} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )\right )}}{b^{2} d} \] Input:

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(3/2),x, algorithm="f 
ricas")
 

Output:

-2*(I*sqrt(1/2)*A*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin( 
d*x + c)) - I*sqrt(1/2)*A*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) 
- I*sin(d*x + c)) - I*sqrt(1/2)*B*sqrt(b)*weierstrassZeta(-4, 0, weierstra 
ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + I*sqrt(1/2)*B*sqrt(b)* 
weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x 
 + c))))/(b^2*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(3/2),x, algorithm="m 
axima")
                                                                                    
                                                                                    
 

Output:

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

Giac [F]

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

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(3/2),x, algorithm="g 
iac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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