\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\) [366]

Optimal result

Integrand size = 33, antiderivative size = 137 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d}-\frac {2 \left (3 a A b-3 a^2 B-b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^3 d}+\frac {2 a^2 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 (a+b) d}+\frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b d} \] Output:

2*(A*b-B*a)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/d-2/3*(3*A*a*b-3*B*a 
^2-B*b^2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^3/d+2*a^2*(A*b-B*a)*Ell 
ipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/b^3/(a+b)/d+2/3*B*cos(d*x+c) 
^(1/2)*sin(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 1.87 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.51 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {\frac {(3 A b-a B) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+B \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )+2 B \sqrt {\cos (c+d x)} \sin (c+d x)+\frac {3 (A b-a B) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{3 b d} \] Input:

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

Output:

(((3*A*b - a*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + B*(2* 
EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2] 
)/(a + b)) + 2*B*Sqrt[Cos[c + d*x]]*Sin[c + d*x] + (3*(A*b - a*B)*(-2*a*b* 
EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[S 
qrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Co 
s[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x]^2]))/(3*b*d)
 

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3469, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3469

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

((6*(A*b - a*B)*EllipticE[(c + d*x)/2, 2])/(b*d) + ((-2*(3*a*A*b - 3*a^2*B 
 - b^2*B)*EllipticF[(c + d*x)/2, 2])/(b*d) + (6*a^2*(A*b - a*B)*EllipticPi 
[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d))/b)/(3*b) + (2*B*Sqrt[Cos[c 
 + d*x]]*Sin[c + d*x])/(3*b*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3469

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^( 
n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
 f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*( 
m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c - b*d*(m 
+ n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin 
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !(IGt 
Q[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

rule 3481

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d   Int[(a + b* 
Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, 
 B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(821\) vs. \(2(136)=272\).

Time = 8.01 (sec) , antiderivative size = 822, normalized size of antiderivative = 6.00

method	result	size
default	\(\text {Expression too large to display}\)	\(822\)

Input:

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

Output:

2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*B*cos(1/2* 
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2+4*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1 
/2*c)^4*b^3+3*A*a^2*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*a*b^2*(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*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))*a*b^2-3*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) 
)*b^3-3*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)*El 
lipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^2*b+2*B*cos(1/2*d*x+1/2* 
c)*sin(1/2*d*x+1/2*c)^2*a*b^2-2*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2* 
b^3-3*a^3*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)* 
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*a^2*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))- 
B*a*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli 
pticF(cos(1/2*d*x+1/2*c),2^(1/2))+b^3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*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))*a^2*b+3*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))*a*b^2+3*B*(si...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(cos(c + d*x))*cos(c + d*x),x)