\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(b \cos (c+d x))^{3/2}} \, dx\) [125]

Optimal result

Integrand size = 35, antiderivative size = 80 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {C \sqrt {\cos (c+d x)} \sin ^3(c+d x)}{3 b d \sqrt {b \cos (c+d x)}} \]

[Out]

(A+C)*sin(d*x+c)*cos(d*x+c)^(1/2)/b/d/(b*cos(d*x+c))^(1/2)-1/3*C*sin(d*x+c)^3*cos(d*x+c)^(1/2)/b/d/(b*cos(d*x+
c))^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {17, 3092} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}}-\frac {C \sin ^3(c+d x) \sqrt {\cos (c+d x)}}{3 b d \sqrt {b \cos (c+d x)}} \]

[In]

Int[(Cos[c + d*x]^(5/2)*(A + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(3/2),x]

[Out]

((A + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*d*Sqrt[b*Cos[c + d*x]]) - (C*Sqrt[Cos[c + d*x]]*Sin[c + d*x]^3)/(
3*b*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 3092

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[-f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.65 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) (6 A+5 C+C \cos (2 (c+d x))) \sin (c+d x)}{6 d (b \cos (c+d x))^{3/2}} \]

[In]

Integrate[(Cos[c + d*x]^(5/2)*(A + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(3/2),x]

[Out]

(Cos[c + d*x]^(3/2)*(6*A + 5*C + C*Cos[2*(c + d*x)])*Sin[c + d*x])/(6*d*(b*Cos[c + d*x])^(3/2))

Maple [A] (verified)

Time = 7.54 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62

method	result	size
default	\(\frac {\left (C \left (\cos ^{2}\left (d x +c \right )\right )+3 A +2 C \right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 b d \sqrt {\cos \left (d x +c \right ) b}}\)	\(50\)
risch	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (4 A +3 C \right ) \sin \left (d x +c \right )}{4 b \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) C \sin \left (3 d x +3 c \right )}{12 b \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(77\)
parts	\(\frac {A \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{b d \sqrt {\cos \left (d x +c \right ) b}}+\frac {C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{3 d b \sqrt {\cos \left (d x +c \right ) b}}\)	\(77\)

[In]

int(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(cos(d*x+c)*b)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {{\left (C \cos \left (d x + c\right )^{2} + 3 \, A + 2 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, b^{2} d \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/3*(C*cos(d*x + c)^2 + 3*A + 2*C)*sqrt(b*cos(d*x + c))*sin(d*x + c)/(b^2*d*sqrt(cos(d*x + c)))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(5/2)*(A+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\frac {C {\left (\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )}}{b^{\frac {3}{2}}} + \frac {12 \, A \sin \left (d x + c\right )}{b^{\frac {3}{2}}}}{12 \, d} \]

[In]

integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/12*(C*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))/b^(3/2) + 12*A*sin(d*x + c
)/b^(3/2))/d

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^(5/2)/(b*cos(d*x + c))^(3/2), x)

Mupad [B] (verification not implemented)

Time = 1.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (12\,A\,\sin \left (2\,c+2\,d\,x\right )+10\,C\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (4\,c+4\,d\,x\right )\right )}{12\,b^2\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

int((cos(c + d*x)^(5/2)*(A + C*cos(c + d*x)^2))/(b*cos(c + d*x))^(3/2),x)

[Out]

(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(12*A*sin(2*c + 2*d*x) + 10*C*sin(2*c + 2*d*x) + C*sin(4*c + 4*d*x)
))/(12*b^2*d*(cos(2*c + 2*d*x) + 1))