Integrand size = 35, antiderivative size = 125 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {b^2 (A+C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {b^2 (A+2 C) \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 C \sqrt {b \cos (c+d x)} \sin ^5(c+d x)}{5 d \sqrt {\cos (c+d x)}} \]
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)
Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.56 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(b \cos (c+d x))^{5/2} (100 A+89 C+4 (5 A+7 C) \cos (2 (c+d x))+3 C \cos (4 (c+d x))) \sin (c+d x)}{120 d \cos ^{\frac {5}{2}}(c+d x)} \]
((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))
Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.58, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2031, 3042, 3492, 290, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \cos ^3(c+d x) \left (C \cos ^2(c+d x)+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3492 |
\(\displaystyle -\frac {b^2 \sqrt {b \cos (c+d x)} \int \left (1-\sin ^2(c+d x)\right ) \left (-C \sin ^2(c+d x)+A+C\right )d(-\sin (c+d x))}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle -\frac {b^2 \sqrt {b \cos (c+d x)} \int \left (C \sin ^4(c+d x)-(A+2 C) \sin ^2(c+d x)+A \left (\frac {C}{A}+1\right )\right )d(-\sin (c+d x))}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{3} (A+2 C) \sin ^3(c+d x)-(A+C) \sin (c+d x)-\frac {1}{5} C \sin ^5(c+d x)\right )}{d \sqrt {\cos (c+d x)}}\) |
-((b^2*Sqrt[b*Cos[c + d*x]]*(-((A + C)*Sin[c + d*x]) + ((A + 2*C)*Sin[c + d*x]^3)/3 - (C*Sin[c + d*x]^5)/5))/(d*Sqrt[Cos[c + d*x]]))
3.2.7.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(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]
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 | \(-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{6 i \left (d x +c \right )} C}{80 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{2 i \left (d x +c \right )} \left (6 A +5 C \right )}{8 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (6 A +5 C \right )}{8 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-2 i \left (d x +c \right )} \left (4 A +5 C \right )}{48 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (10 A +11 C \right ) \cos \left (4 d x +4 c \right )}{120 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (5 A +7 C \right ) \sin \left (4 d x +4 c \right )}{60 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\) | \(322\) |
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)*s in(d*x+c)*(cos(d*x+c)*b)^(1/2)/cos(d*x+c)^(1/2)
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {{\left (3 \, C b^{2} \cos \left (d x + c\right )^{4} + {\left (5 \, A + 4 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \sqrt {\cos \left (d x + c\right )}} \]
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)))
Timed out. \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.57 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {20 \, {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} A \sqrt {b} + {\left (3 \, b^{2} \sin \left (5 \, d x + 5 \, c\right ) + 25 \, b^{2} \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right ) + 150 \, b^{2} \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right )\right )} C \sqrt {b}}{240 \, d} \]
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*a rctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 150*b^2*sin(1/5*arctan2(sin( 5*d*x + 5*c), cos(5*d*x + 5*c))))*C*sqrt(b))/d
Timed out. \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 2.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.80 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {b^2\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (200\,A\,\sin \left (2\,c+2\,d\,x\right )+20\,A\,\sin \left (4\,c+4\,d\,x\right )+175\,C\,\sin \left (2\,c+2\,d\,x\right )+28\,C\,\sin \left (4\,c+4\,d\,x\right )+3\,C\,\sin \left (6\,c+6\,d\,x\right )\right )}{240\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]