3.2.79 \(\int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\) [179]

3.2.79.1 Optimal result
3.2.79.2 Mathematica [A] (verified)
3.2.79.3 Rubi [A] (verified)
3.2.79.4 Maple [A] (verified)
3.2.79.5 Fricas [A] (verification not implemented)
3.2.79.6 Sympy [F(-1)]
3.2.79.7 Maxima [B] (verification not implemented)
3.2.79.8 Giac [F]
3.2.79.9 Mupad [F(-1)]

3.2.79.1 Optimal result

Integrand size = 23, antiderivative size = 72 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {\text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 d \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \]

output
1/2*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)+1/2*arctanh(sin(d*x 
+c))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)
 
3.2.79.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {\text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+\sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \]

input
Integrate[1/(Cos[c + d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]),x]
 
output
(ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + Sin[c + d*x])/(2*d*Cos[c + d*x]^(3 
/2)*Sqrt[b*Cos[c + d*x]])
 
3.2.79.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2032, 3042, 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.

\(\displaystyle \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 2032

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

input
Int[1/(Cos[c + d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]),x]
 
output
(Sqrt[Cos[c + d*x]]*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d 
*x])/(2*d)))/Sqrt[b*Cos[c + d*x]]
 

3.2.79.3.1 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 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]
 
3.2.79.4 Maple [A] (verified)

Time = 3.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.17

method result size
default \(\frac {-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\sin \left (d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {3}{2}}}\) \(84\)
risch \(-\frac {i \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{2 \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 \sqrt {\cos \left (d x +c \right ) b}\, d}\) \(122\)

input
int(1/cos(d*x+c)^(5/2)/(cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)
 
output
1/2/d*(-cos(d*x+c)^2*ln(-cot(d*x+c)+csc(d*x+c)-1)+cos(d*x+c)^2*ln(-cot(d*x 
+c)+csc(d*x+c)+1)+sin(d*x+c))/(cos(d*x+c)*b)^(1/2)/cos(d*x+c)^(3/2)
 
3.2.79.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.88 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\left [\frac {\sqrt {b} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b d \cos \left (d x + c\right )^{3}}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b d \cos \left (d x + c\right )^{3}}\right ] \]

input
integrate(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")
 
output
[1/4*(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*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b*d*cos(d* 
x + c)^3), -1/2*(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 - sqrt(b*cos(d*x + c))*sqrt(cos(d 
*x + c))*sin(d*x + c))/(b*d*cos(d*x + c)^3)]
 
3.2.79.6 Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Timed out} \]

input
integrate(1/cos(d*x+c)**(5/2)/(b*cos(d*x+c))**(1/2),x)
 
output
Timed out
 
3.2.79.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 661 vs. \(2 (60) = 120\).

Time = 0.41 (sec) , antiderivative size = 661, normalized size of antiderivative = 9.18 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=-\frac {4 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + 2 \, \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - 4 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + 2 \, \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - {\left (2 \, {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) + {\left (2 \, {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) - 4 \, {\left (\cos \left (4 \, d x + 4 \, c\right ) + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 4 \, {\left (\cos \left (4 \, d x + 4 \, c\right ) + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )}{4 \, {\left (2 \, {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sqrt {b} d} \]

input
integrate(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")
 
output
-1/4*(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*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 
in(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)*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), co 
s(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), co 
s(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*a 
rctan2(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*c 
os(2*d*x + 2*c) + 1)*sqrt(b)*d)
 
3.2.79.8 Giac [F]

\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

input
integrate(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")
 
output
integrate(1/(sqrt(b*cos(d*x + c))*cos(d*x + c)^(5/2)), x)
 
3.2.79.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \]

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