\(\int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx\) [828]

Optimal result

Integrand size = 31, antiderivative size = 147 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\frac {6 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)}}+\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b d \sqrt {b \cos (c+d x)}}+\frac {2 A \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac {2 B (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b^3 d} \] Output:

6/5*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)+2/3*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)+2/3*A*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/b^2/d+2/5 
*B*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/b^3/d
 

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.60 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\frac {2 \cos ^{\frac {3}{2}}(c+d x) \left (9 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 A \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (5 A+3 B \cos (c+d x)) \sin (c+d x)\right )}{15 d (b \cos (c+d x))^{3/2}} \] Input:

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

Output:

(2*Cos[c + d*x]^(3/2)*(9*B*EllipticE[(c + d*x)/2, 2] + 5*A*EllipticF[(c + 
d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*A + 3*B*Cos[c + d*x])*Sin[c + d*x]))/(1 
5*d*(b*Cos[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {2030, 3042, 3227, 3042, 3115, 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.

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle \frac {A \int (b \cos (c+d x))^{3/2}dx+\frac {B \int (b \cos (c+d x))^{5/2}dx}{b}}{b^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx}{b}}{b^3}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {A \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {B \left (\frac {3}{5} b^2 \int \sqrt {b \cos (c+d x)}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{b}}{b^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {B \left (\frac {3}{5} b^2 \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{b}}{b^3}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {A \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {B \left (\frac {3 b^2 \sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{b}}{b^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {B \left (\frac {3 b^2 \sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{b}}{b^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {A \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {B \left (\frac {6 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{b}}{b^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {A \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {B \left (\frac {6 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{b}}{b^3}\)

Input:

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

Output:

(A*((2*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c 
 + d*x]]) + (2*b*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)) + (B*((6*b^2*Sq 
rt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + ( 
2*b*(b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)))/b)/b^3
 

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 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

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. \(272\) vs. \(2(130)=260\).

Time = 12.50 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.86

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (20 A +24 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-10 A -6 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 b \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}\)	\(273\)
parts	\(-\frac {2 A \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 b \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}-\frac {2 B \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 b \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}\)	\(405\)

Input:

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

Output:

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)/b*(-24*B*c 
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(20*A+24*B)*sin(1/2*d*x+1/2*c)^4*co 
s(1/2*d*x+1/2*c)+(-10*A-6*B)*sin(1/2*d*x+1/2*c)^2*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)*EllipticF(cos 
(1/2*d*x+1/2*c),2^(1/2))-9*B*(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)/((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 = 154, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {\frac {1}{2}} A \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {\frac {1}{2}} A \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 i \, \sqrt {\frac {1}{2}} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 i \, \sqrt {\frac {1}{2}} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, B \cos \left (d x + c\right ) + 5 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, b^{2} d} \] Input:

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

Output:

-2/15*(5*I*sqrt(1/2)*A*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I 
*sin(d*x + c)) - 5*I*sqrt(1/2)*A*sqrt(b)*weierstrassPInverse(-4, 0, cos(d* 
x + c) - I*sin(d*x + c)) - 9*I*sqrt(1/2)*B*sqrt(b)*weierstrassZeta(-4, 0, 
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*I*sqrt(1/2) 
*B*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) 
- I*sin(d*x + c))) - (3*B*cos(d*x + c) + 5*A)*sqrt(b*cos(d*x + c))*sin(d*x 
 + c))/(b^2*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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