\(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx\) [254]

Optimal result

Integrand size = 25, antiderivative size = 177 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx=\frac {7 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{64 \sqrt {2} a^{7/2} d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac {3 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}+\frac {17 \sqrt {\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}} \] Output:

7/128*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+ 
c))^(1/2))*2^(1/2)/a^(7/2)/d-1/6*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d* 
x+c))^(7/2)+3/16*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(5/2)+17 
/192*cos(d*x+c)^(1/2)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(3/2)
 

Mathematica [A] (warning: unable to verify)

Time = 2.10 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx=\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (672 \text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos ^6\left (\frac {1}{2} (c+d x)\right )+(140+135 \cos (c+d x)+140 \cos (2 (c+d x))+17 \cos (3 (c+d x))) \sqrt {2-2 \sec (c+d x)}\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{3072 \sqrt {2} a^3 d \sqrt {-1+\cos (c+d x)} \sqrt {a (1+\cos (c+d x))}} \] Input:

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

Output:

(Sec[(c + d*x)/2]^4*(672*ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]] 
*Cos[(c + d*x)/2]^6 + (140 + 135*Cos[c + d*x] + 140*Cos[2*(c + d*x)] + 17* 
Cos[3*(c + d*x)])*Sqrt[2 - 2*Sec[c + d*x]])*Tan[(c + d*x)/2])/(3072*Sqrt[2 
]*a^3*d*Sqrt[-1 + Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3244, 27, 3042, 3457, 27, 3042, 3457, 27, 3042, 3261, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3244

\(\displaystyle -\frac {\int \frac {a-8 a \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{5/2}}dx}{6 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {a-8 a \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{5/2}}dx}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {a-8 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

\(\Big \downarrow \) 3457

\(\displaystyle -\frac {\frac {\int -\frac {18 \cos (c+d x) a^2+a^2}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\int \frac {18 \cos (c+d x) a^2+a^2}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \frac {18 \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

\(\Big \downarrow \) 3457

\(\displaystyle -\frac {-\frac {\frac {\int \frac {21 a^3}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}+\frac {17 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\frac {21}{4} a \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx+\frac {17 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {21}{4} a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {17 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

\(\Big \downarrow \) 3261

\(\displaystyle -\frac {-\frac {\frac {17 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {21 a^2 \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}}{8 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {-\frac {\frac {17 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac {21 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} d}}{8 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\)

Input:

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

Output:

-1/6*(Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(7/2)) - (( 
-9*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ( 
(21*Sqrt[a]*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt 
[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*d) + (17*a^2*Sqrt[Cos[c + d*x]]*Sin[c + 
 d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)))/(8*a^2))/(12*a^2)
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3244

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e 
+ f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* 
(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* 
Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) 
 + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && 
 NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] 
&& GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
 

rule 3261

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 
 EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3457

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

Maple [A] (verified)

Time = 3.60 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {\left (\sin \left (d x +c \right ) \left (17 \cos \left (d x +c \right )^{2}+70 \cos \left (d x +c \right )+21\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (-21 \cos \left (d x +c \right )^{3}-63 \cos \left (d x +c \right )^{2}-63 \cos \left (d x +c \right )-21\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}}{384 d \left (\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{4}}\)	\(188\)

Input:

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

Output:

1/384/d*(sin(d*x+c)*(17*cos(d*x+c)^2+70*cos(d*x+c)+21)*2^(1/2)*(cos(d*x+c) 
/(cos(d*x+c)+1))^(1/2)+(-21*cos(d*x+c)^3-63*cos(d*x+c)^2-63*cos(d*x+c)-21) 
*arcsin(cot(d*x+c)-csc(d*x+c)))*cos(d*x+c)^(1/2)*2^(1/2)*(a*(cos(d*x+c)+1) 
)^(1/2)/(cos(d*x+c)^4+4*cos(d*x+c)^3+6*cos(d*x+c)^2+4*cos(d*x+c)+1)/(cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)/a^4
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.21 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx=\frac {21 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) + 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (17 \, \cos \left (d x + c\right )^{2} + 70 \, \cos \left (d x + c\right ) + 21\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:

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

Output:

1/384*(21*sqrt(2)*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 
4*cos(d*x + c) + 1)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sq 
rt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) 
 + 2*sqrt(a*cos(d*x + c) + a)*(17*cos(d*x + c)^2 + 70*cos(d*x + c) + 21)*s 
qrt(cos(d*x + c))*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + 
c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1}d x \right )}{a^{4}} \] Input:

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

Output:

(sqrt(a)*int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x))/(cos 
(c + d*x)**4 + 4*cos(c + d*x)**3 + 6*cos(c + d*x)**2 + 4*cos(c + d*x) + 1) 
,x))/a**4