\(\int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\) [115]

Optimal result

Integrand size = 33, antiderivative size = 261 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(163 A-283 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(475 A-787 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{240 a^3 d} \] Output:

1/32*(163*A-283*B)*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^(5/2)/d+1/4*(A-B)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x 
+c))^(5/2)+1/16*(13*A-21*B)*cos(d*x+c)^3*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^( 
3/2)-1/120*(985*A-1729*B)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)-1/80*(85 
*A-157*B)*cos(d*x+c)^2*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)+1/240*(475* 
A-787*B)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/a^3/d
 

Mathematica [A] (verified)

Time = 1.70 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.53 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {30 (163 A-283 B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+(-1895 A+3491 B-5 (479 A-887 B) \cos (c+d x)+(-400 A+832 B) \cos (2 (c+d x))+40 A \cos (3 (c+d x))-40 B \cos (3 (c+d x))+12 B \cos (4 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{240 a d (a (1+\cos (c+d x)))^{3/2}} \] Input:

Integrate[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(5/2) 
,x]
 

Output:

(30*(163*A - 283*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + (-1895* 
A + 3491*B - 5*(479*A - 887*B)*Cos[c + d*x] + (-400*A + 832*B)*Cos[2*(c + 
d*x)] + 40*A*Cos[3*(c + d*x)] - 40*B*Cos[3*(c + d*x)] + 12*B*Cos[4*(c + d* 
x)])*Tan[(c + d*x)/2])/(240*a*d*(a*(1 + Cos[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.76 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.10, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 3456, 27, 3042, 3456, 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.

\(\displaystyle \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a \cos (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\int \frac {\cos ^3(c+d x) (8 a (A-B)-a (5 A-13 B) \cos (c+d x))}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cos ^3(c+d x) (8 a (A-B)-a (5 A-13 B) \cos (c+d x))}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (8 a (A-B)-a (5 A-13 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\frac {\int \frac {\cos ^2(c+d x) \left (6 a^2 (13 A-21 B)-a^2 (85 A-157 B) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\cos ^2(c+d x) \left (6 a^2 (13 A-21 B)-a^2 (85 A-157 B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (6 a^2 (13 A-21 B)-a^2 (85 A-157 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3462

\(\displaystyle \frac {\frac {\frac {2 \int -\frac {\cos (c+d x) \left (4 a^3 (85 A-157 B)-a^3 (475 A-787 B) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {\int \frac {\cos (c+d x) \left (4 a^3 (85 A-157 B)-a^3 (475 A-787 B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (4 a^3 (85 A-157 B)-a^3 (475 A-787 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {-\frac {\int \frac {4 a^3 (85 A-157 B) \cos (c+d x)-a^3 (475 A-787 B) \cos ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {\int \frac {4 a^3 (85 A-157 B) \sin \left (c+d x+\frac {\pi }{2}\right )-a^3 (475 A-787 B) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {-\frac {\frac {2 \int -\frac {a^4 (475 A-787 B)-2 a^4 (985 A-1729 B) \cos (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (475 A-787 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {a^4 (475 A-787 B)-2 a^4 (985 A-1729 B) \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (475 A-787 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {a^4 (475 A-787 B)-2 a^4 (985 A-1729 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a^2 (475 A-787 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {-\frac {-\frac {15 a^4 (163 A-283 B) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {4 a^4 (985 A-1729 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 a^2 (475 A-787 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {-\frac {15 a^4 (163 A-283 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^4 (985 A-1729 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 a^2 (475 A-787 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {-\frac {-\frac {-\frac {30 a^4 (163 A-283 B) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {4 a^4 (985 A-1729 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 a^2 (475 A-787 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {-\frac {2 a^2 (85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {2 a^2 (475 A-787 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}-\frac {\frac {15 \sqrt {2} a^{7/2} (163 A-283 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {4 a^4 (985 A-1729 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}}{5 a}}{4 a^2}+\frac {a (13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

Input:

Int[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(5/2),x]
 

Output:

((A - B)*Cos[c + d*x]^4*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ( 
(a*(13*A - 21*B)*Cos[c + d*x]^3*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3 
/2)) + ((-2*a^2*(85*A - 157*B)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*Sqrt[a + 
a*Cos[c + d*x]]) - ((-2*a^2*(475*A - 787*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c 
 + d*x])/(3*d) - ((15*Sqrt[2]*a^(7/2)*(163*A - 283*B)*ArcTanh[(Sqrt[a]*Sin 
[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d - (4*a^4*(985*A - 1729*B 
)*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(5*a))/(4*a^2))/(8*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 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 3456

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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
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 - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + 
 b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B}, 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] && (In 
tegerQ[2*n] || EqQ[c, 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(466\) vs. \(2(230)=460\).

Time = 2.87 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.79

method	result	size
default	\(\frac {\sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (768 B \sqrt {a}\, \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+640 A \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-2176 B \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+2445 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -4245 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2560 A \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+5248 B \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-435 A \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+555 B \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+30 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-30 B \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{480 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(467\)
parts	\(\frac {A \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (128 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+489 \sqrt {2}\, \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 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-512 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-87 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \sqrt {2}\, \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+6 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{96 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{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}+\frac {B \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (768 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2176 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4245 \sqrt {2}\, \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 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+5248 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+555 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \sqrt {2}\, \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-30 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{480 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{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}\)	\(520\)

Input:

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

Output:

1/480*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(768*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x 
+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^8+640*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^ 
2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6-2176*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c) 
^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6+2445*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d 
*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a 
-4245*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/ 
2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a-2560*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^ 
2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4+5248*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c) 
^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-435*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c) 
^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^2+555*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c) 
^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^2+30*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x 
+1/2*c)^2)^(1/2)-30*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/cos( 
1/2*d*x+1/2*c)^3/a^(7/2)/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 = 270, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {15 \, \sqrt {2} {\left ({\left (163 \, A - 283 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (163 \, A - 283 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A - 283 \, B\right )} \cos \left (d x + c\right ) + 163 \, A - 283 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (96 \, B \cos \left (d x + c\right )^{4} + 160 \, {\left (A - B\right )} \cos \left (d x + c\right )^{3} - 32 \, {\left (25 \, A - 49 \, B\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (503 \, A - 911 \, B\right )} \cos \left (d x + c\right ) - 1495 \, A + 2671 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{960 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

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

Output:

-1/960*(15*sqrt(2)*((163*A - 283*B)*cos(d*x + c)^3 + 3*(163*A - 283*B)*cos 
(d*x + c)^2 + 3*(163*A - 283*B)*cos(d*x + c) + 163*A - 283*B)*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*(9 
6*B*cos(d*x + c)^4 + 160*(A - B)*cos(d*x + c)^3 - 32*(25*A - 49*B)*cos(d*x 
 + c)^2 - 5*(503*A - 911*B)*cos(d*x + c) - 1495*A + 2671*B)*sqrt(a*cos(d*x 
 + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 
3*a^3*d*cos(d*x + c) + a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{5}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)**5)/(cos(c + d*x)**3 + 
3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*b + int((sqrt(cos(c + d*x) + 1) 
*cos(c + d*x)**4)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 
1),x)*a))/a**3