\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\) [877]

Optimal result

Integrand size = 42, antiderivative size = 182 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {2 \left (5 a b B-5 a^2 C-3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac {2 \left (3 a^2+b^2\right ) (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 d}-\frac {2 a^3 (b B-a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^4 (a+b) d}+\frac {2 (b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b d} \] Output:

-2/5*(5*B*a*b-5*C*a^2-3*C*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^3/d 
+2/3*(3*a^2+b^2)*(B*b-C*a)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^4/d-2* 
a^3*(B*b-C*a)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/b^4/(a+b)/d 
+2/3*(B*b-C*a)*cos(d*x+c)^(1/2)*sin(d*x+c)/b^2/d+2/5*C*cos(d*x+c)^(3/2)*si 
n(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 2.61 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\frac {2 b^2 \left (-5 a b B+5 a^2 C+9 b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+2 b^2 (5 b B+4 a C) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )+4 b^2 \sqrt {\cos (c+d x)} (5 b B-5 a C+3 b C \cos (c+d x)) \sin (c+d x)+\frac {6 \left (-5 a b B+5 a^2 C+3 b^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a \sqrt {\sin ^2(c+d x)}}}{30 b^4 d} \] Input:

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

Output:

((2*b^2*(-5*a*b*B + 5*a^2*C + 9*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x) 
/2, 2])/(a + b) + 2*b^2*(5*b*B + 4*a*C)*(2*EllipticF[(c + d*x)/2, 2] - (2* 
a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*b^2*Sqrt[Cos[c + 
 d*x]]*(5*b*B - 5*a*C + 3*b*C*Cos[c + d*x])*Sin[c + d*x] + (6*(-5*a*b*B + 
5*a^2*C + 3*b^2*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a 
*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*Ellipt 
icPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*Sqrt[Sin[c 
+ d*x]^2]))/(30*b^4*d)
 

Rubi [A] (verified)

Time = 1.64 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 3508, 3042, 3469, 27, 3042, 3528, 27, 3042, 3538, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) (B+C \cos (c+d x))}{a+b \cos (c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3469

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

\(\displaystyle \frac {\frac {2 \int \frac {-3 \left (-5 C a^2+5 b B a-3 b^2 C\right ) \cos ^2(c+d x)+b (5 b B+4 a C) \cos (c+d x)+5 a (b B-a C)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {-3 \left (-5 C a^2+5 b B a-3 b^2 C\right ) \cos ^2(c+d x)+b (5 b B+4 a C) \cos (c+d x)+5 a (b B-a C)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {-3 \left (-5 C a^2+5 b B a-3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (5 b B+4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a (b B-a C)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {\frac {-\frac {3 \left (-5 a^2 C+5 a b B-3 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {5 \left (a b (b B-a C)+\left (3 a^2+b^2\right ) \cos (c+d x) (b B-a C)\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {5 \int \frac {a b (b B-a C)+\left (3 a^2+b^2\right ) \cos (c+d x) (b B-a C)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {3 \left (-5 a^2 C+5 a b B-3 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 \int \frac {a b (b B-a C)+\left (3 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) (b B-a C)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {3 \left (-5 a^2 C+5 a b B-3 b^2 C\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {5 \int \frac {a b (b B-a C)+\left (3 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) (b B-a C)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {6 \left (-5 a^2 C+5 a b B-3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\frac {5 \left (\frac {\left (3 a^2+b^2\right ) (b B-a C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {3 a^3 (b B-a C) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}\right )}{b}-\frac {6 \left (-5 a^2 C+5 a b B-3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {5 \left (\frac {\left (3 a^2+b^2\right ) (b B-a C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {3 a^3 (b B-a C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}-\frac {6 \left (-5 a^2 C+5 a b B-3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {5 \left (\frac {2 \left (3 a^2+b^2\right ) (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {3 a^3 (b B-a C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}-\frac {6 \left (-5 a^2 C+5 a b B-3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\frac {\frac {5 \left (\frac {2 \left (3 a^2+b^2\right ) (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {6 a^3 (b B-a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}\right )}{b}-\frac {6 \left (-5 a^2 C+5 a b B-3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {10 (b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}\)

Input:

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

Output:

(2*C*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*b*d) + (((-6*(5*a*b*B - 5*a^2*C - 
 3*b^2*C)*EllipticE[(c + d*x)/2, 2])/(b*d) + (5*((2*(3*a^2 + b^2)*(b*B - a 
*C)*EllipticF[(c + d*x)/2, 2])/(b*d) - (6*a^3*(b*B - a*C)*EllipticPi[(2*b) 
/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d)))/b)/(3*b) + (10*(b*B - a*C)*Sqrt 
[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d))/(5*b)
 

Defintions of rubi rules used

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 3508

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

rule 3528

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ 
.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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. \(1073\) vs. \(2(175)=350\).

Time = 5.16 (sec) , antiderivative size = 1074, normalized size of antiderivative = 5.90

method	result	size
default	\(\text {Expression too large to display}\)	\(1074\)

Input:

int(cos(d*x+c)^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x,meth 
od=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((-24*C*a*b^ 
3+24*C*b^4)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(20*B*a*b^3-20*B*b^4-2 
0*C*a^2*b^2+44*C*a*b^3-24*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+( 
-10*B*a*b^3+10*B*b^4+10*C*a^2*b^2-16*C*a*b^3+6*C*b^4)*sin(1/2*d*x+1/2*c)^2 
*cos(1/2*d*x+1/2*c)+15*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))*a^3*b-15*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))*a^2*b^2+5*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))*a*b^3-5*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^4+15*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^2-15*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^3-15*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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^3*b-15*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(co 
s(1/2*d*x+1/2*c),2^(1/2))*a^4+15*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))*a^3*b-5*C*(si 
n(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))*a^2*b^2+5*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(...
 

Fricas [F]

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

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

Output:

integral((C*cos(d*x + c)^3 + B*cos(d*x + c)^2)*sqrt(cos(d*x + c))/(b*cos(d 
*x + c) + a), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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