\(\int (b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \, dx\) [53]

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 = 25, antiderivative size = 113 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 b^2 (9 A+7 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {2 b (9 A+7 C) (b \cos (c+d x))^{3/2} \sin (c+d x)}{45 d}+\frac {2 C (b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d} \] Output:

2/15*b^2*(9*A+7*C)*(b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/ 
2))/d/cos(d*x+c)^(1/2)+2/45*b*(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/d
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.78 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(b \cos (c+d x))^{5/2} \left (24 (9 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sqrt {\cos (c+d x)} (18 A+19 C+5 C \cos (2 (c+d x))) \sin (2 (c+d x))\right )}{180 d \cos ^{\frac {5}{2}}(c+d x)} \] Input:

Integrate[(b*Cos[c + d*x])^(5/2)*(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*d*Cos[c + d*x]^(5/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

Int[(b*Cos[c + d*x])^(5/2)*(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*S 
qrt[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
 

Defintions of rubi rules used

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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(323\) vs. \(2(101)=202\).

Time = 0.00 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.87

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

Input:

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x,method=_RETURNVERBOSE)
 

Output:

-2/45*(b*(-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*(-160 
*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+320*C*sin(1/2*d*x+1/2*c)^8*cos 
(1/2*d*x+1/2*c)+(-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)/ 
sin(1/2*d*x+1/2*c)/(b*(-1+2*cos(1/2*d*x+1/2*c)^2))^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.13 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (-3 i \, \sqrt {\frac {1}{2}} {\left (9 \, A + 7 \, C\right )} b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {\frac {1}{2}} {\left (9 \, A + 7 \, C\right )} b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (5 \, C b^{2} \cos \left (d x + c\right )^{3} + {\left (9 \, A + 7 \, C\right )} b^{2} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{45 \, d} \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x, algorithm="fricas")
 

Output:

-2/45*(-3*I*sqrt(1/2)*(9*A + 7*C)*b^(5/2)*weierstrassZeta(-4, 0, weierstra 
ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(1/2)*(9*A + 7 
*C)*b^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) 
 - I*sin(d*x + c))) - (5*C*b^2*cos(d*x + c)^3 + (9*A + 7*C)*b^2*cos(d*x + 
c))*sqrt(b*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2), x)
 

Giac [F]

\[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:

int((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2),x)
 

Output:

int((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\sqrt {b}\, b^{2} \left (\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{4}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2),x)
 

Output:

sqrt(b)*b**2*(int(sqrt(cos(c + d*x))*cos(c + d*x)**4,x)*c + int(sqrt(cos(c 
 + d*x))*cos(c + d*x)**2,x)*a)