\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx\) [327]

Optimal result

Integrand size = 43, antiderivative size = 102 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\frac {C x \sqrt {\cos (c+d x)}}{b \sqrt {b \cos (c+d x)}}+\frac {B \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{b d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.59 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cos (c+d x)} \left (C d x \cos (c+d x)+B \coth ^{-1}(\sin (c+d x)) \cos (c+d x)+A \sin (c+d x)\right )}{d (b \cos (c+d x))^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.52, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {2032, 3042, 3500, 3042, 3214, 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.

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

\(\Big \downarrow \) 2032

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

Defintions of rubi rules used

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 3214

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

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

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 = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {-2 B \,\operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \cos \left (d x +c \right )+C \left (d x +c \right ) \cos \left (d x +c \right )+A \sin \left (d x +c \right )}{b d \sqrt {\cos \left (d x +c \right )}\, \sqrt {b \cos \left (d x +c \right )}}\)	\(73\)
parts	\(\frac {A \sin \left (d x +c \right )}{b d \sqrt {\cos \left (d x +c \right )}\, \sqrt {b \cos \left (d x +c \right )}}-\frac {2 B \,\operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}}{d b \sqrt {b \cos \left (d x +c \right )}}+\frac {C \left (d x +c \right ) \sqrt {\cos \left (d x +c \right )}}{d b \sqrt {b \cos \left (d x +c \right )}}\)	\(108\)
risch	\(\frac {C x \sqrt {\cos \left (d x +c \right )}}{b \sqrt {b \cos \left (d x +c \right )}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{b \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, d}+\frac {\sqrt {\cos \left (d x +c \right )}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b \sqrt {b \cos \left (d x +c \right )}\, d}-\frac {\sqrt {\cos \left (d x +c \right )}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b \sqrt {b \cos \left (d x +c \right )}\, d}\)	\(142\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2), 
x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 317, normalized size of antiderivative = 3.11 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\left [-\frac {2 \, B \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{2} + C \sqrt {-b} \cos \left (d x + c\right )^{2} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{2} d \cos \left (d x + c\right )^{2}}, \frac {2 \, C \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right )^{2} + B \sqrt {b} \cos \left (d x + c\right )^{2} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{2} d \cos \left (d x + c\right )^{2}}\right ] \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^ 
(3/2),x, algorithm="fricas")
 

Output:

[-1/2*(2*B*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*s 
qrt(cos(d*x + c))))*cos(d*x + c)^2 + C*sqrt(-b)*cos(d*x + c)^2*log(2*b*cos 
(d*x + c)^2 + 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*sin(d*x + 
 c) - b) - 2*sqrt(b*cos(d*x + c))*A*sqrt(cos(d*x + c))*sin(d*x + c))/(b^2* 
d*cos(d*x + c)^2), 1/2*(2*C*sqrt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + 
c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c)^2 + B*sqrt(b)*cos(d*x + c)^2 
*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*sqrt(b*cos(d*x + c) 
)*A*sqrt(cos(d*x + c))*sin(d*x + c))/(b^2*d*cos(d*x + c)^2)]
                                                                                    
                                                                                    
 

Sympy [F]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2)/(b*cos(d*x+c) 
)**(3/2),x)
 

Output:

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)/((b*cos(c + d*x))**(3/2) 
*sqrt(cos(c + d*x))), x)
 

Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.54 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\frac {\frac {4 \, A \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}} + \frac {B {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )}}{b^{\frac {3}{2}}} + \frac {4 \, C \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac {3}{2}}}}{2 \, d} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^ 
(3/2),x, algorithm="maxima")
 

Output:

1/2*(4*A*sqrt(b)*sin(2*d*x + 2*c)/(b^2*cos(2*d*x + 2*c)^2 + b^2*sin(2*d*x 
+ 2*c)^2 + 2*b^2*cos(2*d*x + 2*c) + b^2) + B*(log(cos(d*x + c)^2 + sin(d*x 
 + c)^2 + 2*sin(d*x + c) + 1) - log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*si 
n(d*x + c) + 1))/b^(3/2) + 4*C*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/b^( 
3/2))/d
 

Giac [F(-2)]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^ 
(3/2),x, algorithm="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. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(b*cos(c + 
 d*x))^(3/2)),x)
 

Output:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(b*cos(c + 
 d*x))^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {b}\, \left (-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +\cos \left (d x +c \right ) c d x +\sin \left (d x +c \right ) a \right )}{\cos \left (d x +c \right ) b^{2} d} \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2), 
x)
 

Output:

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