\(\int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [165]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {36 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 a^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a^3 \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}+\frac {36 a^3 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \] Output:

-36/5*a^3*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4*a^3*InverseJacobiAM(1/ 
2*d*x+1/2*c,2^(1/2))/d+2/5*a^3*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2*a^3*sin(d*x 
+c)/d/cos(d*x+c)^(3/2)+36/5*a^3*sin(d*x+c)/d/cos(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.69 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.18 \[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a^3 \csc (c+d x) \left (\operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},-\frac {1}{4},\cos ^2(c+d x)\right )+5 \cos (c+d x) \left (\operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},\cos ^2(c+d x)\right )+\cos (c+d x) \left (3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},\cos ^2(c+d x)\right )-\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right )\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)} \] Input:

Integrate[(a + a*Cos[c + d*x])^3/Cos[c + d*x]^(7/2),x]
 

Output:

(2*a^3*Csc[c + d*x]*(Hypergeometric2F1[-5/4, 1/2, -1/4, Cos[c + d*x]^2] + 
5*Cos[c + d*x]*(Hypergeometric2F1[-3/4, 1/2, 1/4, Cos[c + d*x]^2] + Cos[c 
+ d*x]*(3*Hypergeometric2F1[-1/4, 1/2, 3/4, Cos[c + d*x]^2] - Cos[c + d*x] 
*Hypergeometric2F1[1/4, 1/2, 5/4, Cos[c + d*x]^2])))*Sqrt[Sin[c + d*x]^2]) 
/(5*d*Cos[c + d*x]^(5/2))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3236, 2009}

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.

\(\displaystyle \int \frac {(a \cos (c+d x)+a)^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\)

\(\Big \downarrow \) 3236

\(\displaystyle \int \left (\frac {3 a^3}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {3 a^3}{\cos ^{\frac {5}{2}}(c+d x)}+\frac {a^3}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {a^3}{\sqrt {\cos (c+d x)}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {36 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^3 \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {36 a^3 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}\)

Input:

Int[(a + a*Cos[c + d*x])^3/Cos[c + d*x]^(7/2),x]
 

Output:

(-36*a^3*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*EllipticF[(c + d*x)/2, 
2])/d + (2*a^3*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*a^3*Sin[c + d*x 
])/(d*Cos[c + d*x]^(3/2)) + (36*a^3*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3236

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + 
f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt 
Q[m, 0] && RationalQ[n]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(385\) vs. \(2(108)=216\).

Time = 6.13 (sec) , antiderivative size = 386, normalized size of antiderivative = 3.30

method	result	size
default	\(-\frac {16 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{3} \left (\frac {7 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{10 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{160 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{3}}-\frac {9 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{10 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {9 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{20 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{16 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(386\)
parts	\(\text {Expression too large to display}\)	\(781\)

Input:

int((a+a*cos(d*x+c))^3/cos(d*x+c)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-16*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(7/10*(s 
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d 
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/ 
2))-1/160*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2 
)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-9/10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x 
+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)-9/20*(si 
n(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d* 
x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/ 
2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-1/16*cos(1/2*d*x+1/2*c)*(-2*sin 
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2 
)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.71 \[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (18 \, a^{3} \cos \left (d x + c\right )^{2} + 5 \, a^{3} \cos \left (d x + c\right ) + a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{5 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((a+a*cos(d*x+c))^3/cos(d*x+c)^(7/2),x, algorithm="fricas")
 

Output:

-2/5*(5*I*sqrt(2)*a^3*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + 
c) + I*sin(d*x + c)) - 5*I*sqrt(2)*a^3*cos(d*x + c)^3*weierstrassPInverse( 
-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*I*sqrt(2)*a^3*cos(d*x + c)^3*wei 
erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + 
c))) - 9*I*sqrt(2)*a^3*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPI 
nverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (18*a^3*cos(d*x + c)^2 + 5 
*a^3*cos(d*x + c) + a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^ 
3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**3/cos(d*x+c)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^3/cos(d*x+c)^(7/2),x, algorithm="maxima")
 

Output:

integrate((a*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2), x)
 

Giac [F]

\[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^3/cos(d*x+c)^(7/2),x, algorithm="giac")
 

Output:

integrate((a*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2), x)
 

Mupad [B] (verification not implemented)

Time = 41.57 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.32 \[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int((a + a*cos(c + d*x))^3/cos(c + d*x)^(7/2),x)
 

Output:

(2*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*a^3*sin(c + d*x)*hypergeom([-1/ 
4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2 
)) + (2*a^3*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*c 
os(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*a^3*sin(c + d*x)*hypergeom( 
[-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x)^ 
2)^(1/2))
 

Reduce [F]

\[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=a^{3} \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x +\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right )+3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right )\right ) \] Input:

int((a+a*cos(d*x+c))^3/cos(d*x+c)^(7/2),x)
 

Output:

a**3*(int(sqrt(cos(c + d*x))/cos(c + d*x),x) + int(sqrt(cos(c + d*x))/cos( 
c + d*x)**4,x) + 3*int(sqrt(cos(c + d*x))/cos(c + d*x)**3,x) + 3*int(sqrt( 
cos(c + d*x))/cos(c + d*x)**2,x))