\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(b \cos (c+d x))^{5/2}} \, dx\) [134]

Optimal result

Integrand size = 35, antiderivative size = 99 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {A x \sqrt {\cos (c+d x)}}{b^2 \sqrt {b \cos (c+d x)}}+\frac {C x \sqrt {\cos (c+d x)}}{2 b^2 \sqrt {b \cos (c+d x)}}+\frac {C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b^2 d \sqrt {b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {17, 2715, 8} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {A x \sqrt {\cos (c+d x)}}{b^2 \sqrt {b \cos (c+d x)}}+\frac {C x \sqrt {\cos (c+d x)}}{2 b^2 \sqrt {b \cos (c+d x)}}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b^2 d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.56 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos (c+d x)} (2 (2 A+C) (c+d x)+C \sin (2 (c+d x)))}{4 b^2 d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 7.49 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (C \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 A \left (d x +c \right )+C \left (d x +c \right )\right )}{2 b^{2} d \sqrt {\cos \left (d x +c \right ) b}}\)	\(57\)
risch	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) x \left (4 A +2 C \right )}{4 b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) C \sin \left (2 d x +2 c \right )}{4 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(69\)
parts	\(\frac {A \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (d x +c \right )}{d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {C \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}\)	\(78\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\left [\frac {2 \, \sqrt {b \cos \left (d x + c\right )} C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (2 \, A + C\right )} \sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{4 \, b^{3} d}, \frac {\sqrt {b \cos \left (d x + c\right )} C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (2 \, A + C\right )} \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right )}{2 \, b^{3} d}\right ] \]

[In]

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

[Out]

[1/4*(2*sqrt(b*cos(d*x + c))*C*sqrt(cos(d*x + c))*sin(d*x + c) - (2*A + C)*sqrt(-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), 1/2*(sqrt(b*cos(d*x + c))*C*sqrt
(cos(d*x + c))*sin(d*x + c) + (2*A + C)*sqrt(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. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(5/2)*(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 = 0.38 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.53 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C}{b^{\frac {5}{2}}} + \frac {8 \, A \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac {5}{2}}}}{4 \, d} \]

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*C/b^(5/2) + 8*A*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/b^(5/2))/d

Giac [F]

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

[In]

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

Mupad [B] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (C\,\sin \left (c+d\,x\right )+C\,\sin \left (3\,c+3\,d\,x\right )+8\,A\,d\,x\,\cos \left (c+d\,x\right )+4\,C\,d\,x\,\cos \left (c+d\,x\right )\right )}{4\,b^3\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

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

[Out]

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