\(\int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx\) [879]

Optimal result

Integrand size = 33, antiderivative size = 157 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {B \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 b d \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{2 b d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{b d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {A \sin ^3(c+d x)}{3 b d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \] Output:

1/2*B*arctanh(sin(d*x+c))*cos(d*x+c)^(1/2)/b/d/(b*cos(d*x+c))^(1/2)+1/2*B* 
sin(d*x+c)/b/d/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)+A*sin(d*x+c)/b/d/cos( 
d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)+1/3*A*sin(d*x+c)^3/b/d/cos(d*x+c)^(5/2)/ 
(b*cos(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.48 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {3 B \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+3 B \sin (c+d x)+2 A (2+\cos (2 (c+d x))) \tan (c+d x)}{6 d \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \] Input:

Integrate[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^(3/2)) 
,x]
 

Output:

(3*B*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + 3*B*Sin[c + d*x] + 2*A*(2 + Co 
s[2*(c + d*x)])*Tan[c + d*x])/(6*d*Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^(3/ 
2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.57, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2032, 3042, 3227, 3042, 4254, 2009, 4255, 3042, 4257}

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.

\(\displaystyle \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 2032

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int (A+B \cos (c+d x)) \sec ^4(c+d x)dx}{b \sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{b \sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (A \int \sec ^4(c+d x)dx+B \int \sec ^3(c+d x)dx\right )}{b \sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{b \sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {A \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )}{b \sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {A \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{b \sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (B \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {A \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{b \sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (B \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {A \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{b \sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (B \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {A \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{b \sqrt {b \cos (c+d x)}}\)

Input:

Int[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^(3/2)),x]
 

Output:

(Sqrt[Cos[c + d*x]]*(B*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c 
+ d*x])/(2*d)) - (A*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d))/(b*Sqrt[b*Cos[ 
c + d*x]])
 

Defintions of rubi rules used

rule 2009

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

rule 2032

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m - 1/ 
2)*b^(n + 1/2)*(Sqrt[a*v]/Sqrt[b*v])   Int[v^(m + n)*Fx, x], x] /; FreeQ[{a 
, b, m}, x] &&  !IntegerQ[m] && ILtQ[n - 1/2, 0] && IntegerQ[m + n]
 

rule 3042

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

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 9.54 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {-3 B \cos \left (d x +c \right )^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+3 B \cos \left (d x +c \right )^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\left (4 \cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right ) A +3 B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{6 b d \cos \left (d x +c \right )^{\frac {5}{2}} \sqrt {\cos \left (d x +c \right ) b}}\)	\(119\)
parts	\(\frac {A \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )^{2}+1\right )}{3 d \cos \left (d x +c \right )^{\frac {5}{2}} \sqrt {\cos \left (d x +c \right ) b}\, b}+\frac {B \left (\cos \left (d x +c \right )^{2} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-\cos \left (d x +c \right )^{2} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\sin \left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{\frac {3}{2}} b \sqrt {\cos \left (d x +c \right ) b}}\)	\(134\)
risch	\(-\frac {i \left (3 B \,{\mathrm e}^{4 i \left (d x +c \right )}-3 B -16 A \cos \left (d x +c \right )-8 i A \sin \left (d x +c \right )\right )}{6 b \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} d}+\frac {\sqrt {\cos \left (d x +c \right )}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 b \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\sqrt {\cos \left (d x +c \right )}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 b \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(157\)

Input:

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(cos(d*x+c)*b)^(3/2),x,method=_RETUR 
NVERBOSE)
 

Output:

1/6/b/d*(-3*B*cos(d*x+c)^3*ln(-cot(d*x+c)+csc(d*x+c)-1)+3*B*cos(d*x+c)^3*l 
n(-cot(d*x+c)+csc(d*x+c)+1)+(4*cos(d*x+c)^2+2)*sin(d*x+c)*A+3*B*cos(d*x+c) 
*sin(d*x+c))/cos(d*x+c)^(5/2)/(cos(d*x+c)*b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.65 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\left [\frac {3 \, B \sqrt {b} \cos \left (d x + c\right )^{4} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (4 \, A \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, b^{2} d \cos \left (d x + c\right )^{4}}, -\frac {3 \, B \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{4} - {\left (4 \, A \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, b^{2} d \cos \left (d x + c\right )^{4}}\right ] \] Input:

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algori 
thm="fricas")
 

Output:

[1/12*(3*B*sqrt(b)*cos(d*x + c)^4*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d* 
x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c))/cos(d* 
x + c)^3) + 2*(4*A*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 2*A)*sqrt(b*cos(d*x 
 + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^2*d*cos(d*x + c)^4), -1/6*(3*B* 
sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x 
 + c))))*cos(d*x + c)^4 - (4*A*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 2*A)*sq 
rt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^2*d*cos(d*x + c)^4) 
]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c))/cos(d*x+c)**(5/2)/(b*cos(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 983 vs. \(2 (135) = 270\).

Time = 0.37 (sec) , antiderivative size = 983, normalized size of antiderivative = 6.26 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algori 
thm="maxima")
 

Output:

1/12*(16*((3*cos(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c) + 3*(3*cos(2*d*x + 2*c 
) + 1)*sin(4*d*x + 4*c) - 3*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 9*cos(4*d* 
x + 4*c)*sin(2*d*x + 2*c))*A/((b*cos(6*d*x + 6*c)^2 + 9*b*cos(4*d*x + 4*c) 
^2 + 9*b*cos(2*d*x + 2*c)^2 + b*sin(6*d*x + 6*c)^2 + 9*b*sin(4*d*x + 4*c)^ 
2 + 18*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*b*sin(2*d*x + 2*c)^2 + 2*(3 
*b*cos(4*d*x + 4*c) + 3*b*cos(2*d*x + 2*c) + b)*cos(6*d*x + 6*c) + 6*(3*b* 
cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 6*b*cos(2*d*x + 2*c) + 6*(b*sin(4 
*d*x + 4*c) + b*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + b)*sqrt(b)) - 3*(4*(s 
in(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), co 
s(2*d*x + 2*c))) - 4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(1/2*arcta 
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (2*(2*cos(2*d*x + 2*c) + 1)*cos( 
4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c 
)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2 
*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^ 
2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arc 
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (2*(2*cos(2*d*x + 2*c) + 
1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d* 
x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 
4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*s...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algori 
thm="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. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int((A + B*cos(c + d*x))/(cos(c + d*x)^(5/2)*(b*cos(c + d*x))^(3/2)),x)
 

Output:

int((A + B*cos(c + d*x))/(cos(c + d*x)^(5/2)*(b*cos(c + d*x))^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.06 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {b}\, \left (-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +4 \sin \left (d x +c \right )^{3} a -6 \sin \left (d x +c \right ) a \right )}{6 \cos \left (d x +c \right ) b^{2} d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x)
 

Output:

(sqrt(b)*( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 
3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 3*cos(c + d*x)*log(tan((c + d 
*x)/2) + 1)*sin(c + d*x)**2*b - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b 
 - 3*cos(c + d*x)*sin(c + d*x)*b + 4*sin(c + d*x)**3*a - 6*sin(c + d*x)*a) 
)/(6*cos(c + d*x)*b**2*d*(sin(c + d*x)**2 - 1))