\(\int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx\) [182]

Optimal result

Integrand size = 23, antiderivative size = 107 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\frac {3 x \sqrt {\cos (c+d x)}}{8 b \sqrt {b \cos (c+d x)}}+\frac {3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 b d \sqrt {b \cos (c+d x)}}+\frac {\cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 b d \sqrt {b \cos (c+d x)}} \]

[Out]

3/8*cos(d*x+c)^(3/2)*sin(d*x+c)/b/d/(b*cos(d*x+c))^(1/2)+1/4*cos(d*x+c)^(7/2)*sin(d*x+c)/b/d/(b*cos(d*x+c))^(1
/2)+3/8*x*cos(d*x+c)^(1/2)/b/(b*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 2715, 8} \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\frac {3 x \sqrt {\cos (c+d x)}}{8 b \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{4 b d \sqrt {b \cos (c+d x)}}+\frac {3 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{8 b d \sqrt {b \cos (c+d x)}} \]

[In]

Int[Cos[c + d*x]^(11/2)/(b*Cos[c + d*x])^(3/2),x]

[Out]

(3*x*Sqrt[Cos[c + d*x]])/(8*b*Sqrt[b*Cos[c + d*x]]) + (3*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(8*b*d*Sqrt[b*Cos[c
+ d*x]]) + (Cos[c + d*x]^(7/2)*Sin[c + d*x])/(4*b*d*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]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\cos (c+d x)} \int \cos ^4(c+d x) \, dx}{b \sqrt {b \cos (c+d x)}} \\ & = \frac {\cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 b d \sqrt {b \cos (c+d x)}}+\frac {\left (3 \sqrt {\cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 b \sqrt {b \cos (c+d x)}} \\ & = \frac {3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 b d \sqrt {b \cos (c+d x)}}+\frac {\cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 b d \sqrt {b \cos (c+d x)}}+\frac {\left (3 \sqrt {\cos (c+d x)}\right ) \int 1 \, dx}{8 b \sqrt {b \cos (c+d x)}} \\ & = \frac {3 x \sqrt {\cos (c+d x)}}{8 b \sqrt {b \cos (c+d x)}}+\frac {3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 b d \sqrt {b \cos (c+d x)}}+\frac {\cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 b d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.51 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{32 d (b \cos (c+d x))^{3/2}} \]

[In]

Integrate[Cos[c + d*x]^(11/2)/(b*Cos[c + d*x])^(3/2),x]

[Out]

(Cos[c + d*x]^(3/2)*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d*(b*Cos[c + d*x])^(3/2))

Maple [A] (verified)

Time = 2.85 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.61

method	result	size
default	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d b \sqrt {\cos \left (d x +c \right ) b}}\)	\(65\)
risch	\(\frac {3 x \left (\sqrt {\cos }\left (d x +c \right )\right )}{8 b \sqrt {\cos \left (d x +c \right ) b}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (4 d x +4 c \right )}{32 b \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (2 d x +2 c \right )}{4 b \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(96\)

[In]

int(cos(d*x+c)^(11/2)/(cos(d*x+c)*b)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/8/d*cos(d*x+c)^(1/2)*(2*sin(d*x+c)*cos(d*x+c)^3+3*cos(d*x+c)*sin(d*x+c)+3*d*x+3*c)/b/(cos(d*x+c)*b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.70 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\left [\frac {2 \, \sqrt {b \cos \left (d x + c\right )} {\left (2 \, \cos \left (d x + c\right )^{2} + 3\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \, \sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{16 \, b^{2} d}, \frac {\sqrt {b \cos \left (d x + c\right )} {\left (2 \, \cos \left (d x + c\right )^{2} + 3\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right )}{8 \, b^{2} d}\right ] \]

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/16*(2*sqrt(b*cos(d*x + c))*(2*cos(d*x + c)^2 + 3)*sqrt(cos(d*x + c))*sin(d*x + c) - 3*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^2*d), 1/8*(sqrt(b*cos(d*
x + c))*(2*cos(d*x + c)^2 + 3)*sqrt(cos(d*x + c))*sin(d*x + c) + 3*sqrt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x
+ c)/(sqrt(b)*cos(d*x + c)^(3/2))))/(b^2*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(11/2)/(b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.46 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\frac {12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )}{32 \, b^{\frac {3}{2}} d} \]

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/32*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))/(b^(3/2)*d)

Giac [F]

\[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {11}{2}}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(11/2)/(b*cos(d*x + c))^(3/2), x)

Mupad [B] (verification not implemented)

Time = 15.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (8\,\sin \left (c+d\,x\right )+9\,\sin \left (3\,c+3\,d\,x\right )+\sin \left (5\,c+5\,d\,x\right )+24\,d\,x\,\cos \left (c+d\,x\right )\right )}{32\,b^2\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

int(cos(c + d*x)^(11/2)/(b*cos(c + d*x))^(3/2),x)

[Out]

(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(8*sin(c + d*x) + 9*sin(3*c + 3*d*x) + sin(5*c + 5*d*x) + 24*d*x*co
s(c + d*x)))/(32*b^2*d*(cos(2*c + 2*d*x) + 1))