Integrand size = 23, antiderivative size = 57 \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:
1/3*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^2/d+1/3*cos(d*x+c)^(1/2)*sin( d*x+c)/d/(a+a*cos(d*x+c))^2
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.44 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\frac {\sqrt {\cos (c+d x)} \csc (c+d x) \left (-\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{3 a^2 d} \] Input:
Integrate[Sqrt[Cos[c + d*x]]/(a + a*Cos[c + d*x])^2,x]
Output:
(Sqrt[Cos[c + d*x]]*Csc[c + d*x]*(-(Hypergeometric2F1[1/4, 1/2, 5/4, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2]) + Tan[(c + d*x)/2]^2))/(3*a^2*d)
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3243, 27, 3042, 3120}
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.
\(\displaystyle \int \frac {\sqrt {\cos (c+d x)}}{(a \cos (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 3243 |
\(\displaystyle \frac {\int \frac {1}{2 \sqrt {\cos (c+d x)}}dx}{3 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {\cos (c+d x)}}dx}{6 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{6 a^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
Input:
Int[Sqrt[Cos[c + d*x]]/(a + a*Cos[c + d*x])^2,x]
Output:
EllipticF[(c + d*x)/2, 2]/(3*a^2*d) + (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3 *d*(a + a*Cos[c + d*x])^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m* ((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a*d*n - b*c *(m + 1) - b*d*(m + n + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e , f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && LtQ[0, n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c , 0]))
Leaf count of result is larger than twice the leaf count of optimal. \(187\) vs. \(2(52)=104\).
Time = 2.84 (sec) , antiderivative size = 188, normalized size of antiderivative = 3.30
method | result | size |
default | \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(188\) |
Input:
int(cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*(sin(1/2*d *x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x +1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+2*cos(1/2*d*x+1/2*c)^4-3*cos(1/2*d*x +1/2*c)^2+1)/a^2/cos(1/2*d*x+1/2*c)^3/(-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*cos(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 = 150, normalized size of antiderivative = 2.63 \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\frac {{\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^2,x, algorithm="fricas")
Output:
1/6*((-I*sqrt(2)*cos(d*x + c)^2 - 2*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*we ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (I*sqrt(2)*cos(d *x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*sqrt(cos(d*x + c))*sin(d*x + c))/(a^ 2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {\sqrt {\cos {\left (c + d x \right )}}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(cos(d*x+c)**(1/2)/(a+a*cos(d*x+c))**2,x)
Output:
Integral(sqrt(cos(c + d*x))/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x)/a** 2
\[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^2,x, algorithm="maxima")
Output:
integrate(sqrt(cos(d*x + c))/(a*cos(d*x + c) + a)^2, x)
\[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^2,x, algorithm="giac")
Output:
integrate(sqrt(cos(d*x + c))/(a*cos(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:
int(cos(c + d*x)^(1/2)/(a + a*cos(c + d*x))^2,x)
Output:
int(cos(c + d*x)^(1/2)/(a + a*cos(c + d*x))^2, x)
\[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x}{a^{2}} \] Input:
int(cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^2,x)
Output:
int(sqrt(cos(c + d*x))/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1),x)/a**2