\(\int (b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [57]

Optimal result

Integrand size = 33, antiderivative size = 74 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {2 b^2 (A-C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 A b^3 \sin (c+d x)}{d \sqrt {b \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.77 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {2 b^3 \left (-\left ((A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )+A \sin (c+d x)\right )}{d \sqrt {b \cos (c+d x)}} \] Input:

Integrate[(b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
 

Output:

(2*b^3*(-((A - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]) + A*Sin[c 
+ d*x]))/(d*Sqrt[b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 2030, 3491, 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.

\(\displaystyle \int \sec ^4(c+d x) (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle b^4 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 3491

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle b^4 \left (\frac {2 A \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )\)

Input:

Int[(b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
 

Output:

b^4*((-2*(A - C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b^2*d*Sq 
rt[Cos[c + d*x]]) + (2*A*Sin[c + d*x])/(b*d*Sqrt[b*Cos[c + d*x]]))
 

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 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 3491

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(215\) vs. \(2(70)=140\).

Time = 212.38 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.92

method	result	size
default	\(\frac {2 b^{3} \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}\, \left (2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\)	\(216\)
parts	\(-\frac {2 A \,b^{3} \left (-2 \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}+\frac {2 C \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\)	\(344\)

Input:

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method=_RETURNV 
ERBOSE)
 

Output:

2*b^3*(-2*b*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2*b)^(1/2)*(2*A*cos(1/ 
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-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))+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 = 116, normalized size of antiderivative = 1.57 \[ \int (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {2 \, {\left (i \, \sqrt {\frac {1}{2}} {\left (A - C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - i \, \sqrt {\frac {1}{2}} {\left (A - C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \sqrt {b \cos \left (d x + c\right )} A b^{2} \sin \left (d x + c\right )\right )}}{d \cos \left (d x + c\right )} \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorith 
m="fricas")
 

Output:

-2*(I*sqrt(1/2)*(A - C)*b^(5/2)*cos(d*x + c)*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - I*sqrt(1/2)*(A - C) 
*b^(5/2)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co 
s(d*x + c) - I*sin(d*x + c))) - sqrt(b*cos(d*x + c))*A*b^2*sin(d*x + c))/( 
d*cos(d*x + c))
 

Sympy [F(-1)]

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

integrate((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorith 
m="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^4, x)
 

Giac [F]

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

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorith 
m="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^4, x)
 

Mupad [F(-1)]

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

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^4,x)
 

Output:

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^4, x)
 

Reduce [F]

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

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
 

Output:

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