\(\int \frac {\cos ^{\frac {11}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [171]

Optimal result

Integrand size = 23, antiderivative size = 107 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {b \cos (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \sin ^5(c+d x)}{5 d \sqrt {b \cos (c+d x)}} \]

[Out]

sin(d*x+c)*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)-2/3*sin(d*x+c)^3*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)+1/
5*sin(d*x+c)^5*cos(d*x+c)^(1/2)/d/(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 = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {17, 2713} \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sin ^5(c+d x) \sqrt {\cos (c+d x)}}{5 d \sqrt {b \cos (c+d x)}}-\frac {2 \sin ^3(c+d x) \sqrt {\cos (c+d x)}}{3 d \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[b*Cos[c + d*x]]) - (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x]^3)/(3*d*Sqrt[b
*Cos[c + d*x]]) + (Sqrt[Cos[c + d*x]]*Sin[c + d*x]^5)/(5*d*Sqrt[b*Cos[c + d*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 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\cos (c+d x)} \int \cos ^5(c+d x) \, dx}{\sqrt {b \cos (c+d x)}} \\ & = -\frac {\sqrt {\cos (c+d x)} \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d \sqrt {b \cos (c+d x)}} \\ & = \frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {b \cos (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \sin ^5(c+d x)}{5 d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.53 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \sin (c+d x) \left (15-10 \sin ^2(c+d x)+3 \sin ^4(c+d x)\right )}{15 d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*Sin[c + d*x]*(15 - 10*Sin[c + d*x]^2 + 3*Sin[c + d*x]^4))/(15*d*Sqrt[b*Cos[c + d*x]])

Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.49

method	result	size
default	\(\frac {\left (3 \left (\cos ^{4}\left (d x +c \right )\right )+4 \left (\cos ^{2}\left (d x +c \right )\right )+8\right ) \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{15 d \sqrt {\cos \left (d x +c \right ) b}}\)	\(52\)
risch	\(\frac {5 \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{8 d \sqrt {\cos \left (d x +c \right ) b}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (5 d x +5 c \right )}{80 \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {5 \left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (3 d x +3 c \right )}{48 \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(95\)

[In]

int(cos(d*x+c)^(11/2)/(cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15/d*(3*cos(d*x+c)^4+4*cos(d*x+c)^2+8)*sin(d*x+c)*cos(d*x+c)^(1/2)/(cos(d*x+c)*b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.50 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {{\left (3 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} + 8\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, b d \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/15*(3*cos(d*x + c)^4 + 4*cos(d*x + c)^2 + 8)*sqrt(b*cos(d*x + c))*sin(d*x + c)/(b*d*sqrt(cos(d*x + c)))

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**(11/2)/(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.64 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {3 \, \sin \left (5 \, d x + 5 \, c\right ) + 25 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right ) + 150 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right )}{240 \, \sqrt {b} d} \]

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/240*(3*sin(5*d*x + 5*c) + 25*sin(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 150*sin(1/5*arctan2(sin(
5*d*x + 5*c), cos(5*d*x + 5*c))))/(sqrt(b)*d)

Giac [F]

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

[In]

integrate(cos(d*x+c)^(11/2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.70 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^{\frac {11}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (175\,\sin \left (2\,c+2\,d\,x\right )+28\,\sin \left (4\,c+4\,d\,x\right )+3\,\sin \left (6\,c+6\,d\,x\right )\right )}{240\,b\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

int(cos(c + d*x)^(11/2)/(b*cos(c + d*x))^(1/2),x)

[Out]

(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(175*sin(2*c + 2*d*x) + 28*sin(4*c + 4*d*x) + 3*sin(6*c + 6*d*x)))/
(240*b*d*(cos(2*c + 2*d*x) + 1))