\(\int \frac {\cos ^{\frac {9}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(b \cos (c+d x))^{5/2}} \, dx\) [132]

Optimal result

Integrand size = 35, antiderivative size = 122 \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {(4 A+3 C) x \sqrt {\cos (c+d x)}}{8 b^2 \sqrt {b \cos (c+d x)}}+\frac {(4 A+3 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 b^2 d \sqrt {b \cos (c+d x)}}+\frac {C \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 b^2 d \sqrt {b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {17, 3093, 2715, 8} \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {x (4 A+3 C) \sqrt {\cos (c+d x)}}{8 b^2 \sqrt {b \cos (c+d x)}}+\frac {(4 A+3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{8 b^2 d \sqrt {b \cos (c+d x)}}+\frac {C \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{4 b^2 d \sqrt {b \cos (c+d x)}} \]

[In]

Int[(Cos[c + d*x]^(9/2)*(A + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(5/2),x]

[Out]

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

Rule 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.57 \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos (c+d x)} (4 (4 A+3 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+C \sin (4 (c+d x)))}{32 b^2 d \sqrt {b \cos (c+d x)}} \]

[In]

Integrate[(Cos[c + d*x]^(9/2)*(A + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(5/2),x]

[Out]

(Sqrt[Cos[c + d*x]]*(4*(4*A + 3*C)*(c + d*x) + 8*(A + C)*Sin[2*(c + d*x)] + C*Sin[4*(c + d*x)]))/(32*b^2*d*Sqr
t[b*Cos[c + d*x]])

Maple [A] (verified)

Time = 7.73 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (2 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 C \cos \left (d x +c \right ) \sin \left (d x +c \right )+4 A \left (d x +c \right )+3 C \left (d x +c \right )\right )}{8 b^{2} d \sqrt {\cos \left (d x +c \right ) b}}\)	\(91\)
risch	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) x \left (8 A +6 C \right )}{16 b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) C \sin \left (4 d x +4 c \right )}{32 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (A +C \right ) \sin \left (2 d x +2 c \right )}{4 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(107\)
parts	\(\frac {A \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {C \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}\)	\(112\)

[In]

int(cos(d*x+c)^(9/2)*(A+C*cos(d*x+c)^2)/(cos(d*x+c)*b)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.70 \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\left [\frac {2 \, {\left (2 \, C \cos \left (d x + c\right )^{2} + 4 \, A + 3 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (4 \, A + 3 \, C\right )} \sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{16 \, b^{3} d}, \frac {{\left (2 \, C \cos \left (d x + c\right )^{2} + 4 \, A + 3 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (4 \, A + 3 \, C\right )} \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right )}{8 \, b^{3} d}\right ] \]

[In]

integrate(cos(d*x+c)^(9/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/16*(2*(2*C*cos(d*x + c)^2 + 4*A + 3*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - (4*A + 3*C)*s
qrt(-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^3*d)
, 1/8*((2*C*cos(d*x + c)^2 + 4*A + 3*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (4*A + 3*C)*sqr
t(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2))))/(b^3*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(9/2)*(A+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\frac {8 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A}{b^{\frac {5}{2}}} + \frac {{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} C}{b^{\frac {5}{2}}}}{32 \, d} \]

[In]

integrate(cos(d*x+c)^(9/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/32*(8*(2*d*x + 2*c + sin(2*d*x + 2*c))*A/b^(5/2) + (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))))*C/b^(5/2))/d

Giac [F]

\[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {9}{2}}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(cos(d*x+c)^(9/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^(9/2)/(b*cos(d*x + c))^(5/2), x)

Mupad [B] (verification not implemented)

Time = 2.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (8\,A\,\sin \left (c+d\,x\right )+8\,C\,\sin \left (c+d\,x\right )+8\,A\,\sin \left (3\,c+3\,d\,x\right )+9\,C\,\sin \left (3\,c+3\,d\,x\right )+C\,\sin \left (5\,c+5\,d\,x\right )+32\,A\,d\,x\,\cos \left (c+d\,x\right )+24\,C\,d\,x\,\cos \left (c+d\,x\right )\right )}{32\,b^3\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

int((cos(c + d*x)^(9/2)*(A + C*cos(c + d*x)^2))/(b*cos(c + d*x))^(5/2),x)

[Out]

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