\(\int \frac {\cos ^{\frac {9}{2}}(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(b \cos (c+d x))^{5/2}} \, dx\) [331]

Optimal result

Integrand size = 43, antiderivative size = 199 \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {(4 A+3 C) x \sqrt {\cos (c+d x)}}{8 b^2 \sqrt {b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \sin (c+d x)}{b^2 d \sqrt {b \cos (c+d x)}}+\frac {(4 A+3 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 b^2 d \sqrt {b \cos (c+d x)}}+\frac {C \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 b^2 d \sqrt {b \cos (c+d x)}}-\frac {B \sqrt {\cos (c+d x)} \sin ^3(c+d x)}{3 b^2 d \sqrt {b \cos (c+d x)}} \]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {17, 3102, 2827, 2715, 8, 2713} \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {x (4 A+3 C) \sqrt {\cos (c+d x)}}{8 b^2 \sqrt {b \cos (c+d x)}}+\frac {(4 A+3 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{8 b^2 d \sqrt {b \cos (c+d x)}}-\frac {B \sin ^3(c+d x) \sqrt {\cos (c+d x)}}{3 b^2 d \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{b^2 d \sqrt {b \cos (c+d x)}}+\frac {C \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{4 b^2 d \sqrt {b \cos (c+d x)}} \]

[In]

Int[(Cos[c + d*x]^(9/2)*(A + B*Cos[c + d*x] + 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]]) + (B*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b^2*d*S
qrt[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]]) - (B*Sqrt[Cos[c + d*x]]*Sin[c + d*x]^3)/(3*b^2*d*Sq
rt[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 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]

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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, 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+B \cos (c+d x)+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 {\sqrt {\cos (c+d x)} \int \cos ^2(c+d x) (4 A+3 C+4 B \cos (c+d x)) \, dx}{4 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 (B \sqrt {\cos (c+d x)}\right ) \int \cos ^3(c+d x) \, dx}{b^2 \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 {\left (B \sqrt {\cos (c+d x)}\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{b^2 d \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 {B \sqrt {\cos (c+d x)} \sin (c+d x)}{b^2 d \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 {B \sqrt {\cos (c+d x)} \sin ^3(c+d x)}{3 b^2 d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.48 \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos (c+d x)} (48 A c+36 c C+48 A d x+36 C d x+72 B \sin (c+d x)+24 (A+C) \sin (2 (c+d x))+8 B \sin (3 (c+d x))+3 C \sin (4 (c+d x)))}{96 b^2 d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(48*A*c + 36*c*C + 48*A*d*x + 36*C*d*x + 72*B*Sin[c + d*x] + 24*(A + C)*Sin[2*(c + d*x)] +
8*B*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(96*b^2*d*Sqrt[b*Cos[c + d*x]])

Maple [A] (verified)

Time = 9.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.59

method	result	size
default	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (6 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+8 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+12 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+9 C \cos \left (d x +c \right ) \sin \left (d x +c \right )+12 A \left (d x +c \right )+16 B \sin \left (d x +c \right )+9 C \left (d x +c \right )\right )}{24 b^{2} d \sqrt {\cos \left (d x +c \right ) b}}\)	\(117\)
parts	\(\frac {A \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{3 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {C \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}\)	\(155\)
risch	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) x \left (8 A +6 C \right )}{16 b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {3 B \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{4 b^{2} d \sqrt {\cos \left (d x +c \right ) b}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) C \sin \left (4 d x +4 c \right )}{32 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \sin \left (3 d x +3 c \right )}{12 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (A +C \right ) \sin \left (2 d x +2 c \right )}{4 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(176\)

[In]

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

[Out]

1/24/b^2/d*cos(d*x+c)^(1/2)*(6*C*cos(d*x+c)^3*sin(d*x+c)+8*B*sin(d*x+c)*cos(d*x+c)^2+12*A*sin(d*x+c)*cos(d*x+c
)+9*C*cos(d*x+c)*sin(d*x+c)+12*A*(d*x+c)+16*B*sin(d*x+c)+9*C*(d*x+c))/(cos(d*x+c)*b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.42 \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\left [-\frac {3 \, {\left (4 \, A + 3 \, C\right )} \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, {\left (6 \, C \cos \left (d x + c\right )^{3} + 8 \, B \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 16 \, B\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, b^{3} d \cos \left (d x + c\right )}, \frac {3 \, {\left (4 \, A + 3 \, C\right )} \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (6 \, C \cos \left (d x + c\right )^{3} + 8 \, B \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 16 \, B\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, b^{3} d \cos \left (d x + c\right )}\right ] \]

[In]

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

[Out]

[-1/48*(3*(4*A + 3*C)*sqrt(-b)*cos(d*x + c)*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) - 2*(6*C*cos(d*x + c)^3 + 8*B*cos(d*x + c)^2 + 3*(4*A + 3*C)*cos(d*x + c) + 16*B)*
sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)), 1/24*(3*(4*A + 3*C)*sqrt(b)*arctan
(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + (6*C*cos(d*x + c)^3 + 8*B*cos(
d*x + c)^2 + 3*(4*A + 3*C)*cos(d*x + c) + 16*B)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d*c
os(d*x + c))]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.58 \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A}{b^{\frac {5}{2}}} + \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} C}{b^{\frac {5}{2}}} + \frac {8 \, B {\left (\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )}}{b^{\frac {5}{2}}}}{96 \, d} \]

[In]

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

[Out]

1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A/b^(5/2) + 3*(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) + 8*B*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c),
cos(3*d*x + 3*c))))/b^(5/2))/d

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.59 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^{\frac {9}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (24\,A\,\sin \left (c+d\,x\right )+24\,C\,\sin \left (c+d\,x\right )+24\,A\,\sin \left (3\,c+3\,d\,x\right )+80\,B\,\sin \left (2\,c+2\,d\,x\right )+8\,B\,\sin \left (4\,c+4\,d\,x\right )+27\,C\,\sin \left (3\,c+3\,d\,x\right )+3\,C\,\sin \left (5\,c+5\,d\,x\right )+96\,A\,d\,x\,\cos \left (c+d\,x\right )+72\,C\,d\,x\,\cos \left (c+d\,x\right )\right )}{96\,b^3\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

int((cos(c + d*x)^(9/2)*(A + B*cos(c + d*x) + 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)*(24*A*sin(c + d*x) + 24*C*sin(c + d*x) + 24*A*sin(3*c + 3*d*x) + 80
*B*sin(2*c + 2*d*x) + 8*B*sin(4*c + 4*d*x) + 27*C*sin(3*c + 3*d*x) + 3*C*sin(5*c + 5*d*x) + 96*A*d*x*cos(c + d
*x) + 72*C*d*x*cos(c + d*x)))/(96*b^3*d*(cos(2*c + 2*d*x) + 1))