\(\int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\) [871]

Optimal result

Integrand size = 33, antiderivative size = 107 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {A \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.61 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {A \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+(A+2 B \cos (c+d x)) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.66, 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, 24, 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.

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

\(\Big \downarrow \) 2032

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

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 = 8.43 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {A \cos \left (d x +c \right )^{2} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-A \cos \left (d x +c \right )^{2} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+2 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+A \sin \left (d x +c \right )}{2 d \cos \left (d x +c \right )^{\frac {3}{2}} \sqrt {\cos \left (d x +c \right ) b}}\)	\(103\)
parts	\(\frac {A \left (\cos \left (d x +c \right )^{2} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-\cos \left (d x +c \right )^{2} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\sin \left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{\frac {3}{2}} \sqrt {\cos \left (d x +c \right ) b}}+\frac {B \sin \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b}}\)	\(115\)
risch	\(-\frac {i \left (A \,{\mathrm e}^{2 i \left (d x +c \right )}-A -4 B \cos \left (d x +c \right )\right )}{2 \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {\sqrt {\cos \left (d x +c \right )}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\sqrt {\cos \left (d x +c \right )}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(137\)

Input:

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(cos(d*x+c)*b)^(1/2),x,method=_RETUR 
NVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.16 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\left [\frac {A \sqrt {b} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b d \cos \left (d x + c\right )^{3}}, -\frac {A \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - {\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b d \cos \left (d x + c\right )^{3}}\right ] \] Input:

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2),x, algori 
thm="fricas")
 

Output:

[1/4*(A*sqrt(b)*cos(d*x + c)^3*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*(2*B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))* 
sin(d*x + c))/(b*d*cos(d*x + c)^3), -1/2*(A*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)^3 - (2*B 
*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/( 
b*d*cos(d*x + c)^3)]
 

Sympy [F(-1)]

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

integrate((A+B*cos(d*x+c))/cos(d*x+c)**(5/2)/(b*cos(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 722 vs. \(2 (91) = 182\).

Time = 0.30 (sec) , antiderivative size = 722, normalized size of antiderivative = 6.75 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2),x, algori 
thm="maxima")
 

Output:

1/4*(8*B*sqrt(b)*sin(2*d*x + 2*c)/(b*cos(2*d*x + 2*c)^2 + b*sin(2*d*x + 2* 
c)^2 + 2*b*cos(2*d*x + 2*c) + b) - (4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2* 
c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(sin(4*d*x + 
4*c) + 2*sin(2*d*x + 2*c))*cos(1/2*arctan2(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*arcta 
n2(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*arctan2(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)*si 
n(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*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c))) + 1) - 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(3/ 
2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*c 
os(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) 
*A/((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)*sqrt(b)))/d
 

Giac [F(-2)]

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

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

int((A + B*cos(c + d*x))/(cos(c + d*x)^(5/2)*(b*cos(c + d*x))^(1/2)),x)
 

Output:

int((A + B*cos(c + d*x))/(cos(c + d*x)^(5/2)*(b*cos(c + d*x))^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.13 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {b}\, \left (-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -\sin \left (d x +c \right ) a \right )}{2 b d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2),x)
 

Output:

(sqrt(b)*( - 2*cos(c + d*x)*sin(c + d*x)*b - log(tan((c + d*x)/2) - 1)*sin 
(c + d*x)**2*a + log(tan((c + d*x)/2) - 1)*a + log(tan((c + d*x)/2) + 1)*s 
in(c + d*x)**2*a - log(tan((c + d*x)/2) + 1)*a - sin(c + d*x)*a))/(2*b*d*( 
sin(c + d*x)**2 - 1))