\(\int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [131]

Optimal result

Integrand size = 23, antiderivative size = 183 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {15 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {31 \sin (c+d x)}{5 a d \sqrt {a+a \cos (c+d x)}}+\frac {9 \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {13 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{10 a^2 d} \] Output:

-15/4*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/ 
2)/a^(3/2)/d-1/2*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+31/5*sin 
(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+9/10*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*c 
os(d*x+c))^(1/2)-13/10*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/a^2/d
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.64 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\left (-75 \sqrt {2} \text {arctanh}\left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) (1+\cos (c+d x))+2 \sqrt {1-\cos (c+d x)} (47+39 \cos (c+d x)-2 \cos (2 (c+d x))+\cos (3 (c+d x)))\right ) \sin (c+d x)}{20 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{3/2}} \] Input:

Integrate[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^(3/2),x]
 

Output:

((-75*Sqrt[2]*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2]]*(1 + Cos[c + d*x]) + 2*Sqr 
t[1 - Cos[c + d*x]]*(47 + 39*Cos[c + d*x] - 2*Cos[2*(c + d*x)] + Cos[3*(c 
+ d*x)]))*Sin[c + d*x])/(20*d*Sqrt[1 - Cos[c + d*x]]*(a*(1 + Cos[c + d*x]) 
)^(3/2))
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.09, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {3042, 3244, 27, 3042, 3462, 27, 3042, 3447, 3042, 3502, 27, 3042, 3230, 3042, 3128, 219}

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.

\(\displaystyle \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3244

\(\displaystyle -\frac {\int \frac {3 \cos ^2(c+d x) (2 a-3 a \cos (c+d x))}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 \int \frac {\cos ^2(c+d x) (2 a-3 a \cos (c+d x))}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a-3 a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3462

\(\displaystyle -\frac {3 \left (\frac {2 \int -\frac {\cos (c+d x) \left (12 a^2-13 a^2 \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 \left (-\frac {\int \frac {\cos (c+d x) \left (12 a^2-13 a^2 \cos (c+d x)\right )}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \left (-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (12 a^2-13 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3447

\(\displaystyle -\frac {3 \left (-\frac {\int \frac {12 a^2 \cos (c+d x)-13 a^2 \cos ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \left (-\frac {\int \frac {12 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )-13 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {3 \left (-\frac {\frac {2 \int -\frac {13 a^3-62 a^3 \cos (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {26 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 \left (-\frac {-\frac {\int \frac {13 a^3-62 a^3 \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {26 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \left (-\frac {-\frac {\int \frac {13 a^3-62 a^3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {26 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3230

\(\displaystyle -\frac {3 \left (-\frac {-\frac {75 a^3 \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {124 a^3 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {26 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \left (-\frac {-\frac {75 a^3 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {124 a^3 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {26 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle -\frac {3 \left (-\frac {-\frac {-\frac {150 a^3 \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {124 a^3 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {26 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {3 \left (-\frac {-\frac {\frac {75 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {124 a^3 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {26 a \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {6 a \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

Input:

Int[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^(3/2),x]
 

Output:

-1/2*(Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(3/2)) - (3*((- 
6*a*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]]) - ((-26*a* 
Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) - ((75*Sqrt[2]*a^(5/2)*ArcTan 
h[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d - (124*a^3 
*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(5*a)))/(4*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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

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

rule 3230

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

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 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3462

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Sin[e + f*x])^m*(c + d*S 
in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m 
 + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
 

rule 3502

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

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.45

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (32 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-32 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+80 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-75 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -85 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+75 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{20 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(265\)

Input:

int(cos(d*x+c)^4/(a+a*cos(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/20/cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(32*a^(1/2 
)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^6-32*a^(1/2)*(a*sin(1/ 
2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4+80*a^(1/2)*(a*sin(1/2*d*x+1/2*c 
)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-75*ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^ 
(1/2)+a)/cos(1/2*d*x+1/2*c))*sin(1/2*d*x+1/2*c)^2*a-85*a^(1/2)*(a*sin(1/2* 
d*x+1/2*c)^2)^(1/2)+75*ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos 
(1/2*d*x+1/2*c))*a)/a^(5/2)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1 
/2)/d
                                                                                    
                                                                                    
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (4 \, \cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )^{2} + 36 \, \cos \left (d x + c\right ) + 49\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

1/40*(75*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*log(-(a*cos 
(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2* 
a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(4*cos(d* 
x + c)^3 - 4*cos(d*x + c)^2 + 36*cos(d*x + c) + 49)*sqrt(a*cos(d*x + c) + 
a)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\frac {75 \, \sqrt {2} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {75 \, \sqrt {2} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {10 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {32 \, \sqrt {2} {\left (2 \, a^{\frac {17}{2}} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5 \, a^{\frac {17}{2}} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{10} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{40 \, d} \] Input:

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

-1/40*(75*sqrt(2)*log(sin(1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos(1/2*d*x + 
 1/2*c))) - 75*sqrt(2)*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos(1/2 
*d*x + 1/2*c))) + 10*sqrt(2)*sin(1/2*d*x + 1/2*c)/((sin(1/2*d*x + 1/2*c)^2 
 - 1)*a^(3/2)*sgn(cos(1/2*d*x + 1/2*c))) - 32*sqrt(2)*(2*a^(17/2)*sin(1/2* 
d*x + 1/2*c)^5 + 5*a^(17/2)*sin(1/2*d*x + 1/2*c))/(a^10*sgn(cos(1/2*d*x + 
1/2*c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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