Integrand size = 35, antiderivative size = 116 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(A+C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {(A+2 C) \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {C \sqrt {b \cos (c+d x)} \sin ^5(c+d x)}{5 d \sqrt {\cos (c+d x)}} \] Output:
(A+C)*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)-1/3*(A+2*C)*(b*co s(d*x+c))^(1/2)*sin(d*x+c)^3/d/cos(d*x+c)^(1/2)+1/5*C*(b*cos(d*x+c))^(1/2) *sin(d*x+c)^5/d/cos(d*x+c)^(1/2)
Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.60 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {b \cos (c+d x)} (100 A+89 C+4 (5 A+7 C) \cos (2 (c+d x))+3 C \cos (4 (c+d x))) \sin (c+d x)}{120 d \sqrt {\cos (c+d x)}} \] Input:
Integrate[Cos[c + d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2),x ]
Output:
(Sqrt[b*Cos[c + d*x]]*(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*Sqrt[Cos[c + d*x]])
Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.59, 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.
\(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \cos ^3(c+d x) \left (C \cos ^2(c+d x)+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3492 |
\(\displaystyle -\frac {\sqrt {b \cos (c+d x)} \int \left (1-\sin ^2(c+d x)\right ) \left (-C \sin ^2(c+d x)+A+C\right )d(-\sin (c+d x))}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle -\frac {\sqrt {b \cos (c+d x)} \int \left (C \sin ^4(c+d x)-(A+2 C) \sin ^2(c+d x)+A \left (\frac {C}{A}+1\right )\right )d(-\sin (c+d x))}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt {b \cos (c+d x)} \left (\frac {1}{3} (A+2 C) \sin ^3(c+d x)-(A+C) \sin (c+d x)-\frac {1}{5} C \sin ^5(c+d x)\right )}{d \sqrt {\cos (c+d x)}}\) |
Input:
Int[Cos[c + d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2),x]
Output:
-((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]]))
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 = 0.80 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.60
method | result | size |
default | \(\frac {\sin \left (d x +c \right ) \left (3 C \cos \left (d x +c \right )^{4}+5 A \cos \left (d x +c \right )^{2}+4 C \cos \left (d x +c \right )^{2}+10 A +8 C \right ) \sqrt {b \cos \left (d x +c \right )}}{15 d \sqrt {\cos \left (d x +c \right )}}\) | \(70\) |
parts | \(\frac {A \sin \left (d x +c \right ) \left (2+\cos \left (d x +c \right )^{2}\right ) \sqrt {b \cos \left (d x +c \right )}}{3 d \sqrt {\cos \left (d x +c \right )}}+\frac {C \sin \left (d x +c \right ) \left (3 \cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{2}+8\right ) \sqrt {b \cos \left (d x +c \right )}}{15 d \sqrt {\cos \left (d x +c \right )}}\) | \(94\) |
risch | \(-\frac {i \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, {\mathrm e}^{6 i \left (d x +c \right )} C}{80 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, {\mathrm e}^{2 i \left (d x +c \right )} \left (6 A +5 C \right )}{8 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, \left (6 A +5 C \right )}{8 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, {\mathrm e}^{-2 i \left (d x +c \right )} \left (4 A +5 C \right )}{48 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, \left (10 A +11 C \right ) \cos \left (4 d x +4 c \right )}{120 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {\sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, \left (5 A +7 C \right ) \sin \left (4 d x +4 c \right )}{60 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\) | \(304\) |
Input:
int(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2),x,method=_RET URNVERBOSE)
Output:
1/15/d*sin(d*x+c)*(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)/cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.54 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {{\left (3 \, C \cos \left (d x + c\right )^{4} + {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 10 \, A + 8 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \sqrt {\cos \left (d x + c\right )}} \] Input:
integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2),x, algo rithm="fricas")
Output:
1/15*(3*C*cos(d*x + c)^4 + (5*A + 4*C)*cos(d*x + c)^2 + 10*A + 8*C)*sqrt(b *cos(d*x + c))*sin(d*x + c)/(d*sqrt(cos(d*x + c)))
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(5/2)*(b*cos(d*x+c))**(1/2)*(A+C*cos(d*x+c)**2),x)
Output:
Timed out
Time = 0.23 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {C \sqrt {b} {\left (3 \, \sin \left (5 \, d x + 5 \, c\right ) + 25 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right ) + 150 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right )\right )} + 20 \, A \sqrt {b} {\left (\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )}}{240 \, d} \] Input:
integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2),x, algo rithm="maxima")
Output:
1/240*(C*sqrt(b)*(3*sin(5*d*x + 5*c) + 25*sin(3/5*arctan2(sin(5*d*x + 5*c) , cos(5*d*x + 5*c))) + 150*sin(1/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5 *c)))) + 20*A*sqrt(b)*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3* c), cos(3*d*x + 3*c)))))/d
Time = 0.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.52 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{240} \, \sqrt {b} {\left (\frac {3 \, C \sin \left (5 \, d x + 5 \, c\right )}{d} + \frac {5 \, {\left (4 \, A + 5 \, C\right )} \sin \left (3 \, d x + 3 \, c\right )}{d} + \frac {30 \, {\left (6 \, A + 5 \, C\right )} \sin \left (d x + c\right )}{d}\right )} \] Input:
integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2),x, algo rithm="giac")
Output:
1/240*sqrt(b)*(3*C*sin(5*d*x + 5*c)/d + 5*(4*A + 5*C)*sin(3*d*x + 3*c)/d + 30*(6*A + 5*C)*sin(d*x + c)/d)
Time = 43.69 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (200\,A\,\sin \left (2\,c+2\,d\,x\right )+20\,A\,\sin \left (4\,c+4\,d\,x\right )+175\,C\,\sin \left (2\,c+2\,d\,x\right )+28\,C\,\sin \left (4\,c+4\,d\,x\right )+3\,C\,\sin \left (6\,c+6\,d\,x\right )\right )}{240\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \] Input:
int(cos(c + d*x)^(5/2)*(A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(1/2),x)
Output:
(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*si n(6*c + 6*d*x)))/(240*d*(cos(2*c + 2*d*x) + 1))
Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.46 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {b}\, \sin \left (d x +c \right ) \left (3 \sin \left (d x +c \right )^{4} c -5 \sin \left (d x +c \right )^{2} a -10 \sin \left (d x +c \right )^{2} c +15 a +15 c \right )}{15 d} \] Input:
int(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+C*cos(d*x+c)^2),x)
Output:
(sqrt(b)*sin(c + d*x)*(3*sin(c + d*x)**4*c - 5*sin(c + d*x)**2*a - 10*sin( c + d*x)**2*c + 15*a + 15*c))/(15*d)