Integrand size = 25, antiderivative size = 62 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 a (3 A+B) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 B \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d} \] Output:
2/3*a*(3*A+B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/3*B*(a+a*cos(d*x+c))^( 1/2)*sin(d*x+c)/d
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.74 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 \sqrt {a (1+\cos (c+d x))} (3 A+2 B+B \cos (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )}{3 d} \] Input:
Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]
Output:
(2*Sqrt[a*(1 + Cos[c + d*x])]*(3*A + 2*B + B*Cos[c + d*x])*Tan[(c + d*x)/2 ])/(3*d)
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3230, 3042, 3125}
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.
\(\displaystyle \int \sqrt {a \cos (c+d x)+a} (A+B \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {1}{3} (3 A+B) \int \sqrt {\cos (c+d x) a+a}dx+\frac {2 B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} (3 A+B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {2 a (3 A+B) \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\) |
Input:
Int[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]
Output:
(2*a*(3*A + B)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*B*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d)
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 2.72 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 A +B \right ) \sqrt {2}}{3 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(62\) |
parts | \(\frac {2 A a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}}{\sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sqrt {2}}{3 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(103\) |
Input:
int((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)
Output:
2/3*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(2*B*cos(1/2*d*x+1/2*c)^2+3*A+ B)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.76 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 \, {\left (B \cos \left (d x + c\right ) + 3 \, A + 2 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:
integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="fricas")
Output:
2/3*(B*cos(d*x + c) + 3*A + 2*B)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d* cos(d*x + c) + d)
\[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + B \cos {\left (c + d x \right )}\right )\, dx \] Input:
integrate((a+a*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c)),x)
Output:
Integral(sqrt(a*(cos(c + d*x) + 1))*(A + B*cos(c + d*x)), x)
Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {6 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + {\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{3 \, d} \] Input:
integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="maxima")
Output:
1/3*(6*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + (sqrt(2)*sin(3/2*d*x + 3/2 *c) + 3*sqrt(2)*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d
Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} {\left (B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, {\left (2 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{3 \, d} \] Input:
integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="giac")
Output:
1/3*sqrt(2)*(B*sgn(cos(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c) + 3*(2*A*sgn (cos(1/2*d*x + 1/2*c)) + B*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c) )*sqrt(a)/d
Timed out. \[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int \left (A+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \] Input:
int((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(1/2),x)
Output:
int((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(1/2), x)
\[ \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\cos \left (d x +c \right )+1}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )d x \right ) b \right ) \] Input:
int((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x)
Output:
sqrt(a)*(int(sqrt(cos(c + d*x) + 1),x)*a + int(sqrt(cos(c + d*x) + 1)*cos( c + d*x),x)*b)