\(\int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \, dx\) [152]

Optimal result

Integrand size = 23, antiderivative size = 65 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \, dx=\frac {b x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 2715, 8} \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \, dx=\frac {b x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{2 d} \]

[In]

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

[Out]

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

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 {\left (b \sqrt {b \cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{\sqrt {\cos (c+d x)}} \\ & = \frac {b \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d}+\frac {\left (b \sqrt {b \cos (c+d x)}\right ) \int 1 \, dx}{2 \sqrt {\cos (c+d x)}} \\ & = \frac {b x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \, dx=\frac {(b \cos (c+d x))^{3/2} (2 (c+d x)+\sin (2 (c+d x)))}{4 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}\)	\(43\)
risch	\(\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{i \left (d x +c \right )} x}{{\mathrm e}^{2 i \left (d x +c \right )}+1}-\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{3 i \left (d x +c \right )}}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-i \left (d x +c \right )}}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\)	\(139\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.35 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \, dx=\left [\frac {2 \, \sqrt {b \cos \left (d x + c\right )} b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {-b} b \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{4 \, d}, \frac {\sqrt {b \cos \left (d x + c\right )} b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right )}{2 \, d}\right ] \]

[In]

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

[Out]

[1/4*(2*sqrt(b*cos(d*x + c))*b*sqrt(cos(d*x + c))*sin(d*x + c) + sqrt(-b)*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))/d, 1/2*(sqrt(b*cos(d*x + c))*b*sqrt(cos(d*x + c))
*sin(d*x + c) + b^(3/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2))))/d]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.43 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \, dx=\frac {{\left (2 \, {\left (d x + c\right )} b + b \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {b}}{4 \, d} \]

[In]

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

[Out]

1/4*(2*(d*x + c)*b + b*sin(2*d*x + 2*c))*sqrt(b)/d

Giac [A] (verification not implemented)

none

Time = 1.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.63 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \, dx=\frac {{\left (\sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \sqrt {b} d x + 2 \, \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} b}{2 \, {\left (d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + d\right )}} \]

[In]

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

[Out]

1/2*(sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^4 + 2*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^2 - 2*sqrt(b)*tan(1/2*d*x + 1/2*c
)^3 + sqrt(b)*d*x + 2*sqrt(b)*tan(1/2*d*x + 1/2*c))*b/(d*tan(1/2*d*x + 1/2*c)^4 + 2*d*tan(1/2*d*x + 1/2*c)^2 +
d)

Mupad [B] (verification not implemented)

Time = 14.50 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \, dx=\frac {b\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )+4\,d\,x\,\cos \left (c+d\,x\right )\right )}{4\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

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

[Out]

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