Integrand size = 33, antiderivative size = 98 \[ \int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {B x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {A \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d} \] Output:
1/2*B*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+A*(b*cos(d*x+c))^(1/2)*sin(d *x+c)/d/cos(d*x+c)^(1/2)+1/2*B*cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)*sin(d *x+c)/d
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {b \cos (c+d x)} (4 A \sin (c+d x)+B (2 (c+d x)+\sin (2 (c+d x))))}{4 d \sqrt {\cos (c+d x)}} \] Input:
Integrate[Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]
Output:
(Sqrt[b*Cos[c + d*x]]*(4*A*Sin[c + d*x] + B*(2*(c + d*x) + Sin[2*(c + d*x) ])))/(4*d*Sqrt[Cos[c + d*x]])
Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.62, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2031, 3042, 3213}
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.
\(\displaystyle \int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \cos (c+d x) (A+B \cos (c+d x))dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {A \sin (c+d x)}{d}+\frac {B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {B x}{2}\right )}{\sqrt {\cos (c+d x)}}\) |
Input:
Int[Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]
Output:
(Sqrt[b*Cos[c + d*x]]*((B*x)/2 + (A*Sin[c + d*x])/d + (B*Cos[c + d*x]*Sin[ c + d*x])/(2*d)))/Sqrt[Cos[c + d*x]]
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 8.52 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.56
method | result | size |
default | \(\frac {\left (B \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 A \sin \left (d x +c \right )+B \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{2 d \sqrt {\cos \left (d x +c \right )}}\) | \(55\) |
parts | \(\frac {A \sqrt {\cos \left (d x +c \right ) b}\, \sin \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}+\frac {B \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{2 d \sqrt {\cos \left (d x +c \right )}}\) | \(73\) |
risch | \(\frac {\sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, {\mathrm e}^{i \left (d x +c \right )} B x}{{\mathrm e}^{2 i \left (d x +c \right )}+1}-\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, {\mathrm e}^{3 i \left (d x +c \right )} B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, {\mathrm e}^{2 i \left (d x +c \right )} A}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, A}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, {\mathrm e}^{-i \left (d x +c \right )} B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\) | \(224\) |
Input:
int(cos(d*x+c)^(1/2)*(cos(d*x+c)*b)^(1/2)*(A+B*cos(d*x+c)),x,method=_RETUR NVERBOSE)
Output:
1/2/d*(B*cos(d*x+c)*sin(d*x+c)+2*A*sin(d*x+c)+B*(d*x+c))*(cos(d*x+c)*b)^(1 /2)/cos(d*x+c)^(1/2)
Time = 0.11 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.08 \[ \int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\left [\frac {B \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )}, \frac {B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}\right ] \] Input:
integrate(cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algori thm="fricas")
Output:
[1/4*(B*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*(B*cos(d*x + c) + 2* A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/2*(B*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) + (B*cos(d*x + c) + 2*A)*sqrt(b*cos(d*x + c))*sq rt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]
Time = 2.78 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.68 \[ \int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\begin {cases} x \sqrt {b \cos {\left (c \right )}} \left (A + B \cos {\left (c \right )}\right ) \sqrt {\cos {\left (c \right )}} & \text {for}\: d = 0 \\0 & \text {for}\: c = - d x + \frac {\pi }{2} \vee c = - d x + \frac {3 \pi }{2} \\\frac {A \sqrt {b \cos {\left (c + d x \right )}} \sin {\left (c + d x \right )}}{d \sqrt {\cos {\left (c + d x \right )}}} + \frac {B x \sqrt {b \cos {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )}}{2 \sqrt {\cos {\left (c + d x \right )}}} + \frac {B x \sqrt {b \cos {\left (c + d x \right )}} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}{2} + \frac {B \sqrt {b \cos {\left (c + d x \right )}} \sin {\left (c + d x \right )} \sqrt {\cos {\left (c + d x \right )}}}{2 d} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**(1/2)*(b*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c)),x)
Output:
Piecewise((x*sqrt(b*cos(c))*(A + B*cos(c))*sqrt(cos(c)), Eq(d, 0)), (0, Eq (c, -d*x + pi/2) | Eq(c, -d*x + 3*pi/2)), (A*sqrt(b*cos(c + d*x))*sin(c + d*x)/(d*sqrt(cos(c + d*x))) + B*x*sqrt(b*cos(c + d*x))*sin(c + d*x)**2/(2* sqrt(cos(c + d*x))) + B*x*sqrt(b*cos(c + d*x))*cos(c + d*x)**(3/2)/2 + B*s qrt(b*cos(c + d*x))*sin(c + d*x)*sqrt(cos(c + d*x))/(2*d), True))
Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.41 \[ \int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B \sqrt {b} + 4 \, A \sqrt {b} \sin \left (d x + c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algori thm="maxima")
Output:
1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*B*sqrt(b) + 4*A*sqrt(b)*sin(d*x + c) )/d
Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.37 \[ \int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {1}{4} \, {\left (2 \, B x + \frac {B \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac {4 \, A \sin \left (d x + c\right )}{d}\right )} \sqrt {b} \] Input:
integrate(cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algori thm="giac")
Output:
1/4*(2*B*x + B*sin(2*d*x + 2*c)/d + 4*A*sin(d*x + c)/d)*sqrt(b)
Time = 42.33 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.81 \[ \int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (B\,\sin \left (c+d\,x\right )+4\,A\,\sin \left (2\,c+2\,d\,x\right )+B\,\sin \left (3\,c+3\,d\,x\right )+4\,B\,d\,x\,\cos \left (c+d\,x\right )\right )}{4\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \] Input:
int(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x)),x)
Output:
(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(B*sin(c + d*x) + 4*A*sin(2*c + 2*d*x) + B*sin(3*c + 3*d*x) + 4*B*d*x*cos(c + d*x)))/(4*d*(cos(2*c + 2*d* x) + 1))
Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.36 \[ \int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +2 \sin \left (d x +c \right ) a +b d x \right )}{2 d} \] Input:
int(cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x)
Output:
(sqrt(b)*(cos(c + d*x)*sin(c + d*x)*b + 2*sin(c + d*x)*a + b*d*x))/(2*d)