3.9.46 \(\int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [846]

3.9.46.1 Optimal result
3.9.46.2 Mathematica [A] (verified)
3.9.46.3 Rubi [A] (verified)
3.9.46.4 Maple [A] (verified)
3.9.46.5 Fricas [A] (verification not implemented)
3.9.46.6 Sympy [F]
3.9.46.7 Maxima [A] (verification not implemented)
3.9.46.8 Giac [F]
3.9.46.9 Mupad [F(-1)]

3.9.46.1 Optimal result

Integrand size = 33, antiderivative size = 60 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {B x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {A \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]

output
B*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+A*arctanh(sin(d*x+c))*(b*cos(d*x 
+c))^(1/2)/d/cos(d*x+c)^(1/2)
 
3.9.46.2 Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(B d x+A \text {arctanh}(\sin (c+d x))) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]

input
Integrate[(Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2),x 
]
 
output
((B*d*x + A*ArcTanh[Sin[c + d*x]])*Sqrt[b*Cos[c + d*x]])/(d*Sqrt[Cos[c + d 
*x]])
 
3.9.46.3 Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.65, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2031, 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.

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

\(\Big \downarrow \) 2031

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

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

3.9.46.3.1 Defintions of rubi rules used

rule 2031
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 
2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v])   Int[v^(m + n)*Fx, x], x] /; FreeQ[{a 
, b, m}, x] &&  !IntegerQ[m] && IGtQ[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 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
3.9.46.4 Maple [A] (verified)

Time = 4.68 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87

method result size
default \(-\frac {\sqrt {\cos \left (d x +c \right ) b}\, \left (2 A \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-B \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right )}}\) \(52\)
parts \(-\frac {2 A \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {B \sqrt {\cos \left (d x +c \right ) b}\, \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}\) \(70\)
risch \(\frac {B x \sqrt {\cos \left (d x +c \right ) b}}{\sqrt {\cos \left (d x +c \right )}}+\frac {\sqrt {\cos \left (d x +c \right ) b}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}-\frac {\sqrt {\cos \left (d x +c \right ) b}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}\) \(96\)

input
int((cos(d*x+c)*b)^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x,method=_RETUR 
NVERBOSE)
 
output
-1/d*(cos(d*x+c)*b)^(1/2)*(2*A*arctanh(cot(d*x+c)-csc(d*x+c))-B*(d*x+c))/c 
os(d*x+c)^(1/2)
 
3.9.46.5 Fricas [A] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (52) = 104\).

Time = 0.36 (sec) , antiderivative size = 210, normalized size of antiderivative = 3.50 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\left [-\frac {2 \, A \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) - B \sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{2 \, d}, \frac {2 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) + A \sqrt {b} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right )}{2 \, d}\right ] \]

input
integrate((b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algori 
thm="fricas")
 
output
[-1/2*(2*A*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*s 
qrt(cos(d*x + c)))) - B*sqrt(-b)*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))/d, 1/2*(2*B*sqrt(b)* 
arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2))) + A 
*sqrt(b)*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))/d]
 
3.9.46.6 Sympy [F]

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

input
integrate((b*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(3/2),x)
 
output
Integral(sqrt(b*cos(c + d*x))*(A + B*cos(c + d*x))/cos(c + d*x)**(3/2), x)
 
3.9.46.7 Maxima [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.53 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {A \sqrt {b} {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} + 4 \, B \sqrt {b} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{2 \, d} \]

input
integrate((b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algori 
thm="maxima")
 
output
1/2*(A*sqrt(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*sin(d*x + c) + 1)) + 4*B*sqrt(b) 
*arctan(sin(d*x + c)/(cos(d*x + c) + 1)))/d
 
3.9.46.8 Giac [F]

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

input
integrate((b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algori 
thm="giac")
 
output
integrate((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))/cos(d*x + c)^(3/2), x)
 
3.9.46.9 Mupad [F(-1)]

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

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