\(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx\) [196]

Optimal result

Integrand size = 23, antiderivative size = 27 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {x \sqrt {\cos (c+d x)}}{b^2 \sqrt {b \cos (c+d x)}} \]

[Out]

x*cos(d*x+c)^(1/2)/b^2/(b*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {17, 8} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {x \sqrt {\cos (c+d x)}}{b^2 \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(x*Sqrt[Cos[c + d*x]])/(b^2*Sqrt[b*Cos[c + d*x]])

Rule 8

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

Rule 17

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\cos (c+d x)} \int 1 \, dx}{b^2 \sqrt {b \cos (c+d x)}} \\ & = \frac {x \sqrt {\cos (c+d x)}}{b^2 \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {x \sqrt {b \cos (c+d x)}}{b^3 \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

(x*Sqrt[b*Cos[c + d*x]])/(b^3*Sqrt[Cos[c + d*x]])

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {x \left (\sqrt {\cos }\left (d x +c \right )\right )}{b^{2} \sqrt {\cos \left (d x +c \right ) b}}\)	\(24\)
default	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (d x +c \right )}{d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}\)	\(31\)

[In]

int(cos(d*x+c)^(5/2)/(cos(d*x+c)*b)^(5/2),x,method=_RETURNVERBOSE)

[Out]

x*cos(d*x+c)^(1/2)/b^2/(cos(d*x+c)*b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.59 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\left [-\frac {\sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{2 \, b^{3} d}, \frac {\arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right )}{b^{\frac {5}{2}} d}\right ] \]

[In]

integrate(cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/2*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)/(
b^3*d), arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))/(b^(5/2)*d)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac {5}{2}} d} \]

[In]

integrate(cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/(b^(5/2)*d)

Giac [F]

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

[In]

integrate(cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(5/2)/(b*cos(d*x + c))^(5/2), x)

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {2\,x\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {b\,\cos \left (c+d\,x\right )}}{b^3\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

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

[Out]

(2*x*cos(c + d*x)^(3/2)*(b*cos(c + d*x))^(1/2))/(b^3*(cos(2*c + 2*d*x) + 1))