3.1.45 \(\int \cos (c+d x) (b \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \, dx\) [45]

3.1.45.1 Optimal result
3.1.45.2 Mathematica [A] (verified)
3.1.45.3 Rubi [A] (verified)
3.1.45.4 Maple [B] (verified)
3.1.45.5 Fricas [C] (verification not implemented)
3.1.45.6 Sympy [F(-1)]
3.1.45.7 Maxima [F]
3.1.45.8 Giac [F]
3.1.45.9 Mupad [F(-1)]

3.1.45.1 Optimal result

Integrand size = 31, antiderivative size = 110 \[ \int \cos (c+d x) (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 b (9 A+7 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (9 A+7 C) (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}+\frac {2 C (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b^2 d} \]

output
2/45*(9*A+7*C)*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/9*C*(b*cos(d*x+c))^(7/2 
)*sin(d*x+c)/b^2/d+2/15*b*(9*A+7*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d 
*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(b*cos(d*x+c))^(1/2)/d/cos 
(d*x+c)^(1/2)
 
3.1.45.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83 \[ \int \cos (c+d x) (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(b \cos (c+d x))^{5/2} \left (24 (9 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sqrt {\cos (c+d x)} (18 A+19 C+5 C \cos (2 (c+d x))) \sin (2 (c+d x))\right )}{180 b d \cos ^{\frac {5}{2}}(c+d x)} \]

input
Integrate[Cos[c + d*x]*(b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2),x]
 
output
((b*Cos[c + d*x])^(5/2)*(24*(9*A + 7*C)*EllipticE[(c + d*x)/2, 2] + 2*Sqrt 
[Cos[c + d*x]]*(18*A + 19*C + 5*C*Cos[2*(c + d*x)])*Sin[2*(c + d*x)]))/(18 
0*b*d*Cos[c + d*x]^(5/2))
 
3.1.45.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {2030, 3042, 3493, 3042, 3115, 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.

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3493

\(\displaystyle \frac {\frac {1}{9} (9 A+7 C) \int (b \cos (c+d x))^{5/2}dx+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{9} (9 A+7 C) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b d}}{b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {\frac {1}{9} (9 A+7 C) \left (\frac {3}{5} b^2 \int \sqrt {b \cos (c+d x)}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{9} (9 A+7 C) \left (\frac {3}{5} b^2 \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b d}}{b}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {\frac {1}{9} (9 A+7 C) \left (\frac {3 b^2 \sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{9} (9 A+7 C) \left (\frac {3 b^2 \sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b d}}{b}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {1}{9} (9 A+7 C) \left (\frac {6 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{7/2}}{9 b d}}{b}\)

input
Int[Cos[c + d*x]*(b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2),x]
 
output
((2*C*(b*Cos[c + d*x])^(7/2)*Sin[c + d*x])/(9*b*d) + ((9*A + 7*C)*((6*b^2* 
Sqrt[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)))/9)/b
 

3.1.45.3.1 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 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 3493
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( 
x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f 
*(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2)   Int[(b*Sin[e + f*x] 
)^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 
3.1.45.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(323\) vs. \(2(122)=244\).

Time = 12.58 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.95

method result size
default \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{2} \left (-160 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+320 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-72 A -296 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (72 A +136 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-18 A -24 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-27 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-21 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{45 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(324\)
parts \(-\frac {2 A \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{2} \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \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 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 C \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{2} \left (160 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-480 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+616 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-432 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+160 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{45 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(438\)

input
int(cos(d*x+c)*(cos(d*x+c)*b)^(3/2)*(A+C*cos(d*x+c)^2),x,method=_RETURNVER 
BOSE)
 
output
-2/45*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(-160* 
C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+320*C*cos(1/2*d*x+1/2*c)*sin(1/ 
2*d*x+1/2*c)^8+(-72*A-296*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(72*A 
+136*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-18*A-24*C)*sin(1/2*d*x+1 
/2*c)^2*cos(1/2*d*x+1/2*c)-27*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))-21*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)/s 
in(1/2*d*x+1/2*c)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d
 
3.1.45.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.13 \[ \int \cos (c+d x) (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 i \, \sqrt {2} {\left (9 \, A + 7 \, C\right )} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A + 7 \, C\right )} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (5 \, C b \cos \left (d x + c\right )^{3} + {\left (9 \, A + 7 \, C\right )} b \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{45 \, d} \]

input
integrate(cos(d*x+c)*(b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2),x, algorithm= 
"fricas")
 
output
1/45*(3*I*sqrt(2)*(9*A + 7*C)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPI 
nverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(9*A + 7*C)*b^ 
(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*s 
in(d*x + c))) + 2*(5*C*b*cos(d*x + c)^3 + (9*A + 7*C)*b*cos(d*x + c))*sqrt 
(b*cos(d*x + c))*sin(d*x + c))/d
 
3.1.45.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)*(b*cos(d*x+c))**(3/2)*(A+C*cos(d*x+c)**2),x)
 
output
Timed out
 
3.1.45.7 Maxima [F]

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

input
integrate(cos(d*x+c)*(b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2),x, algorithm= 
"maxima")
 
output
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(3/2)*cos(d*x + c), x)
 
3.1.45.8 Giac [F]

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

input
integrate(cos(d*x+c)*(b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2),x, algorithm= 
"giac")
 
output
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(3/2)*cos(d*x + c), x)
 
3.1.45.9 Mupad [F(-1)]

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

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