\(\int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [211]

Optimal result

Integrand size = 25, antiderivative size = 121 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {6 a^2 \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {12 a^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \] Output:

2/5*a^2*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2)+6/5*a^2*sin(d 
*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+12/5*a^2*sin(d*x+c)/d/cos( 
d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.51 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a \sqrt {a (1+\cos (c+d x))} (4+3 \cos (c+d x)+3 \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)} \] Input:

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

Output:

(2*a*Sqrt[a*(1 + Cos[c + d*x])]*(4 + 3*Cos[c + d*x] + 3*Cos[2*(c + d*x)])* 
Tan[(c + d*x)/2])/(5*d*Cos[c + d*x]^(5/2))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3241, 27, 3042, 3251, 3042, 3250}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

\(\displaystyle \frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {2}{5} a \int -\frac {9 \sqrt {\cos (c+d x) a+a}}{2 \cos ^{\frac {5}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {9}{5} a \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{5} a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {9}{5} a \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3250

\(\displaystyle \frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {9}{5} a \left (\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )\)

Input:

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

Output:

(2*a^2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + ( 
9*a*((2*a*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) 
+ (4*a*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])))/5
 

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 3042

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

rule 3241

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b 
*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* 
d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d))   Int[(a + b*Sin[e + f*x])^(m - 
 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* 
c*(m - 1) - a*d*(m + 2*n + 1))*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] 
 && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || 
 (IntegerQ[m] && EqQ[c, 0]))
 

rule 3250

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq 
rt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[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 3251

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e 
+ f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim 
p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2)))   Int[Sqrt[a + b*Sin[e 
+ f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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[n, 
-1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
 

Maple [A] (verified)

Time = 5.67 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.57

method	result	size
default	\(\frac {2 \sin \left (d x +c \right ) \left (6 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \sqrt {2}}{5 d \cos \left (d x +c \right )^{\frac {5}{2}} \left (\cos \left (d x +c \right )+1\right )}\)	\(69\)

Input:

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

Output:

2/5/d*sin(d*x+c)*(6*cos(d*x+c)^2+3*cos(d*x+c)+1)*(a*cos(1/2*d*x+1/2*c)^2)^ 
(1/2)/cos(d*x+c)^(5/2)/(cos(d*x+c)+1)*a*2^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.60 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (6 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{5 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \] Input:

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

Output:

2/5*(6*a*cos(d*x + c)^2 + 3*a*cos(d*x + c) + a)*sqrt(a*cos(d*x + c) + a)*s 
qrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (103) = 206\).

Time = 0.14 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.79 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 \, {\left (\frac {5 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {7 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{5 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}} \] Input:

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

Output:

4/5*(5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sqrt(2)*a^(3/2 
)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 7*sqrt(2)*a^(3/2)*sin(d*x + c)^5/( 
cos(d*x + c) + 1)^5 - 2*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^ 
7)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/(d*(sin(d*x + c)/(cos(d*x + 
 c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(2*sin(d* 
x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1))
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 42.34 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4\,a\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\left (8\,\sin \left (c+d\,x\right )+6\,\sin \left (2\,c+2\,d\,x\right )+11\,\sin \left (3\,c+3\,d\,x\right )+3\,\sin \left (4\,c+4\,d\,x\right )+3\,\sin \left (5\,c+5\,d\,x\right )\right )}{5\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (10\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+5\,\cos \left (3\,c+3\,d\,x\right )+2\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )+6\right )} \] Input:

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

Output:

(4*a*(a*(cos(c + d*x) + 1))^(1/2)*(8*sin(c + d*x) + 6*sin(2*c + 2*d*x) + 1 
1*sin(3*c + 3*d*x) + 3*sin(4*c + 4*d*x) + 3*sin(5*c + 5*d*x)))/(5*d*cos(c 
+ d*x)^(1/2)*(10*cos(c + d*x) + 8*cos(2*c + 2*d*x) + 5*cos(3*c + 3*d*x) + 
2*cos(4*c + 4*d*x) + cos(5*c + 5*d*x) + 6))
 

Reduce [F]

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

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

Output:

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