Integrand size = 36, antiderivative size = 44 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {6 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:
6/5*B*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/5*B*cos(d*x+c)^(3/2)*sin(d *x+c)/d
Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {B \left (6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} \sin (2 (c+d x))\right )}{5 d} \] Input:
Integrate[(Cos[c + d*x]^(5/2)*(a*B + b*B*Cos[c + d*x]))/(a + b*Cos[c + d*x ]),x]
Output:
(B*(6*EllipticE[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*Sin[2*(c + d*x)]))/(5 *d)
Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {2011, 3042, 3115, 3042, 3119}
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.
\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle B \int \cos ^{\frac {5}{2}}(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle B \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle B \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\) |
Input:
Int[(Cos[c + d*x]^(5/2)*(a*B + b*B*Cos[c + d*x]))/(a + b*Cos[c + d*x]),x]
Output:
B*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x ])/(5*d))
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(202\) vs. \(2(40)=80\).
Time = 5.23 (sec) , antiderivative size = 203, normalized size of antiderivative = 4.61
method | result | size |
default | \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, B \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(203\) |
Input:
int(cos(d*x+c)^(5/2)*(B*a+b*B*cos(d*x+c))/(a+cos(d*x+c)*b),x,method=_RETUR NVERBOSE)
Output:
-2/5*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*B*(-8*cos(1/2 *d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+8*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c) -2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)* (2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(- 2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*c os(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.75 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\frac {2 \, B \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + 3 i \, \sqrt {2} B {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} B {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{5 \, d} \] Input:
integrate(cos(d*x+c)^(5/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algori thm="fricas")
Output:
1/5*(2*B*cos(d*x + c)^(3/2)*sin(d*x + c) + 3*I*sqrt(2)*B*weierstrassZeta(- 4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqr t(2)*B*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I* sin(d*x + c))))/d
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(5/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c)),x)
Output:
Timed out
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (B b \cos \left (d x + c\right ) + B a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algori thm="maxima")
Output:
integrate((B*b*cos(d*x + c) + B*a)*cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a) , x)
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (B b \cos \left (d x + c\right ) + B a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algori thm="giac")
Output:
integrate((B*b*cos(d*x + c) + B*a)*cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a) , x)
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (B\,a+B\,b\,\cos \left (c+d\,x\right )\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:
int((cos(c + d*x)^(5/2)*(B*a + B*b*cos(c + d*x)))/(a + b*cos(c + d*x)),x)
Output:
int((cos(c + d*x)^(5/2)*(B*a + B*b*cos(c + d*x)))/(a + b*cos(c + d*x)), x)
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{a+b \cos (c+d x)} \, dx=\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) b \] Input:
int(cos(d*x+c)^(5/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c)),x)
Output:
int(sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*b