\(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [175]

Optimal result

Integrand size = 23, antiderivative size = 32 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, 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, 2717} \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

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

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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.84 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right ) b}}\)	\(29\)
risch	\(\frac {\sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right ) b}}\)	\(29\)

[In]

int(cos(d*x+c)^(3/2)/(cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{b d \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate(cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

sqrt(b*cos(d*x + c))*sin(d*x + c)/(b*d*sqrt(cos(d*x + c)))

Sympy [A] (verification not implemented)

Time = 18.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\begin {cases} \frac {\sin {\left (c + d x \right )} \sqrt {\cos {\left (c + d x \right )}}}{d \sqrt {b \cos {\left (c + d x \right )}}} & \text {for}\: d \neq 0 \\\frac {x \cos ^{\frac {3}{2}}{\left (c \right )}}{\sqrt {b \cos {\left (c \right )}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sin(c + d*x)*sqrt(cos(c + d*x))/(d*sqrt(b*cos(c + d*x))), Ne(d, 0)), (x*cos(c)**(3/2)/sqrt(b*cos(c)
), True))

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.41 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sin \left (d x + c\right )}{\sqrt {b} d} \]

[In]

integrate(cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

sin(d*x + c)/(sqrt(b)*d)

Giac [F]

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

[In]

integrate(cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.62 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (2\,c+2\,d\,x\right )\,\sqrt {b\,\cos \left (c+d\,x\right )}}{b\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

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

[Out]

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