\(\int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \, dx\) [107]

Optimal result

Integrand size = 35, antiderivative size = 125 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {b^2 (A+C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {b^2 (A+2 C) \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 C \sqrt {b \cos (c+d x)} \sin ^5(c+d x)}{5 d \sqrt {\cos (c+d x)}} \]

[Out]

b^2*(A+C)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)-1/3*b^2*(A+2*C)*sin(d*x+c)^3*(b*cos(d*x+c))^(1/2)
/d/cos(d*x+c)^(1/2)+1/5*b^2*C*sin(d*x+c)^5*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 125, 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, 3092, 380} \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {b^2 (A+2 C) \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 (A+C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {b^2 C \sin ^5(c+d x) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

(b^2*(A + C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) - (b^2*(A + 2*C)*Sqrt[b*Cos[c + d*x]]*S
in[c + d*x]^3)/(3*d*Sqrt[Cos[c + d*x]]) + (b^2*C*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^5)/(5*d*Sqrt[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 380

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

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 {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \int \cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx}{\sqrt {\cos (c+d x)}} \\ & = -\frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \text {Subst}\left (\int \left (1-x^2\right ) \left (A+C-C x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d \sqrt {\cos (c+d x)}} \\ & = -\frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \text {Subst}\left (\int \left (A \left (1+\frac {C}{A}\right )-(A+2 C) x^2+C x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d \sqrt {\cos (c+d x)}} \\ & = \frac {b^2 (A+C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {b^2 (A+2 C) \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 C \sqrt {b \cos (c+d x)} \sin ^5(c+d x)}{5 d \sqrt {\cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.56 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(b \cos (c+d x))^{5/2} (100 A+89 C+4 (5 A+7 C) \cos (2 (c+d x))+3 C \cos (4 (c+d x))) \sin (c+d x)}{120 d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(100*A + 89*C + 4*(5*A + 7*C)*Cos[2*(c + d*x)] + 3*C*Cos[4*(c + d*x)])*Sin[c + d*x])/(
120*d*Cos[c + d*x]^(5/2))

Maple [A] (verified)

Time = 8.43 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {b^{2} \left (3 C \left (\cos ^{4}\left (d x +c \right )\right )+5 A \left (\cos ^{2}\left (d x +c \right )\right )+4 C \left (\cos ^{2}\left (d x +c \right )\right )+10 A +8 C \right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{15 d \sqrt {\cos \left (d x +c \right )}}\)	\(73\)
parts	\(\frac {A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{3 d \sqrt {\cos \left (d x +c \right )}}+\frac {C \,b^{2} \left (3 \left (\cos ^{4}\left (d x +c \right )\right )+4 \left (\cos ^{2}\left (d x +c \right )\right )+8\right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{15 d \sqrt {\cos \left (d x +c \right )}}\)	\(100\)
risch	\(-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{6 i \left (d x +c \right )} C}{80 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{2 i \left (d x +c \right )} \left (6 A +5 C \right )}{8 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (6 A +5 C \right )}{8 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-2 i \left (d x +c \right )} \left (4 A +5 C \right )}{48 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (10 A +11 C \right ) \cos \left (4 d x +4 c \right )}{120 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (5 A +7 C \right ) \sin \left (4 d x +4 c \right )}{60 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\)	\(322\)

[In]

int((cos(d*x+c)*b)^(5/2)*(A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15*b^2/d*(3*C*cos(d*x+c)^4+5*A*cos(d*x+c)^2+4*C*cos(d*x+c)^2+10*A+8*C)*sin(d*x+c)*(cos(d*x+c)*b)^(1/2)/cos(d
*x+c)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {{\left (3 \, C b^{2} \cos \left (d x + c\right )^{4} + {\left (5 \, A + 4 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/15*(3*C*b^2*cos(d*x + c)^4 + (5*A + 4*C)*b^2*cos(d*x + c)^2 + 2*(5*A + 4*C)*b^2)*sqrt(b*cos(d*x + c))*sin(d*
x + c)/(d*sqrt(cos(d*x + c)))

Sympy [F(-1)]

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

[In]

integrate((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*cos(d*x+c)**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.57 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {20 \, {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} A \sqrt {b} + {\left (3 \, b^{2} \sin \left (5 \, d x + 5 \, c\right ) + 25 \, b^{2} \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right ) + 150 \, b^{2} \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right )\right )} C \sqrt {b}}{240 \, d} \]

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/240*(20*(b^2*sin(3*d*x + 3*c) + 9*b^2*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*A*sqrt(b) + (3*b
^2*sin(5*d*x + 5*c) + 25*b^2*sin(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 150*b^2*sin(1/5*arctan2(si
n(5*d*x + 5*c), cos(5*d*x + 5*c))))*C*sqrt(b))/d

Giac [F(-1)]

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

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 2.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.80 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {b^2\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (200\,A\,\sin \left (2\,c+2\,d\,x\right )+20\,A\,\sin \left (4\,c+4\,d\,x\right )+175\,C\,\sin \left (2\,c+2\,d\,x\right )+28\,C\,\sin \left (4\,c+4\,d\,x\right )+3\,C\,\sin \left (6\,c+6\,d\,x\right )\right )}{240\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

int(cos(c + d*x)^(1/2)*(A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2),x)

[Out]

(b^2*cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(200*A*sin(2*c + 2*d*x) + 20*A*sin(4*c + 4*d*x) + 175*C*sin(2*c
+ 2*d*x) + 28*C*sin(4*c + 4*d*x) + 3*C*sin(6*c + 6*d*x)))/(240*d*(cos(2*c + 2*d*x) + 1))