Integrand size = 25, antiderivative size = 95 \[ \int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx=-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}+\frac {2 e \cos (c+d x)}{a d \sqrt {e \sin (c+d x)}}+\frac {4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}} \]
-2*e/a/d/(e*sin(d*x+c))^(1/2)+2*e*cos(d*x+c)/a/d/(e*sin(d*x+c))^(1/2)-4*(s in(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos( 1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/a/d/sin(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.75 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.21 \[ \int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx=\frac {2 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {e \sin (c+d x)} \left (\sec \left (\frac {c}{2}\right ) \sec (c) \left (3 \sin \left (\frac {c}{2}\right )+\sin \left (\frac {3 c}{2}\right )\right )-2 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )-2 \sqrt {\csc ^2(c)} \csc (c+d x) \csc (d x-\arctan (\cot (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x-\arctan (\cot (c)))\right ) \sin (c) \sqrt {\sin ^2(d x-\arctan (\cot (c)))}-\frac {\csc (c) \csc (c+d x) \sec (c) (\sin (c+d x-\arctan (\cot (c)))+3 \sin (c-d x+\arctan (\cot (c))))}{\sqrt {\csc ^2(c)}}\right )}{a d (1+\sec (c+d x))} \]
(2*Cos[(c + d*x)/2]^2*Sec[c + d*x]*Sqrt[e*Sin[c + d*x]]*(Sec[c/2]*Sec[c]*( 3*Sin[c/2] + Sin[(3*c)/2]) - 2*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x)/2] - 2* Sqrt[Csc[c]^2]*Csc[c + d*x]*Csc[d*x - ArcTan[Cot[c]]]*HypergeometricPFQ[{- 1/2, -1/4}, {3/4}, Cos[d*x - ArcTan[Cot[c]]]^2]*Sin[c]*Sqrt[Sin[d*x - ArcT an[Cot[c]]]^2] - (Csc[c]*Csc[c + d*x]*Sec[c]*(Sin[c + d*x - ArcTan[Cot[c]] ] + 3*Sin[c - d*x + ArcTan[Cot[c]]]))/Sqrt[Csc[c]^2]))/(a*d*(1 + Sec[c + d *x]))
Time = 0.69 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4360, 25, 25, 3042, 3318, 3042, 3044, 15, 3047, 3042, 3121, 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.
\(\displaystyle \int \frac {\sqrt {e \sin (c+d x)}}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{a-a \csc \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\frac {\cos (c+d x) \sqrt {e \sin (c+d x)}}{a (-\cos (c+d x))-a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {\cos (c+d x) \sqrt {e \sin (c+d x)}}{\cos (c+d x) a+a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\cos (c+d x) \sqrt {e \sin (c+d x)}}{a \cos (c+d x)+a}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {-e \cos \left (c+d x+\frac {\pi }{2}\right )}}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 3318 |
\(\displaystyle \frac {e^2 \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{3/2}}dx}{a}-\frac {e^2 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^2 \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{3/2}}dx}{a}-\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {e \int \frac {1}{(e \sin (c+d x))^{3/2}}d(e \sin (c+d x))}{a d}-\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {e^2 \int \frac {\cos (c+d x)^2}{(e \sin (c+d x))^{3/2}}dx}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\) |
\(\Big \downarrow \) 3047 |
\(\displaystyle -\frac {e^2 \left (-\frac {2 \int \sqrt {e \sin (c+d x)}dx}{e^2}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {e^2 \left (-\frac {2 \int \sqrt {e \sin (c+d x)}dx}{e^2}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {e^2 \left (-\frac {2 \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {e^2 \left (-\frac {2 \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {e^2 \left (-\frac {4 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a}-\frac {2 e}{a d \sqrt {e \sin (c+d x)}}\) |
(-2*e)/(a*d*Sqrt[e*Sin[c + d*x]]) - (e^2*((-2*Cos[c + d*x])/(d*e*Sqrt[e*Si n[c + d*x]]) - (4*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/( d*e^2*Sqrt[Sin[c + d*x]])))/a
3.2.23.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[a*(a*Cos[e + f*x])^(m - 1)*((b*Sin[e + f*x])^(n + 1)/ (b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1))) Int[(a*Cos[e + f*x] )^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ [m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
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]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d) Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 4.59 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.57
method | result | size |
default | \(-\frac {2 e \left (2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )\right )}{a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(149\) |
-2/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e*(2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x +c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2)) -(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF(( -sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-cos(d*x+c)^2+cos(d*x+c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx=-\frac {2 \, {\left ({\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} \sqrt {-i \, e} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} \sqrt {i \, e} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{a d \cos \left (d x + c\right ) + a d} \]
-2*((-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*sqrt(-I*e)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + (I*sqrt(2)*co s(d*x + c) + I*sqrt(2))*sqrt(I*e)*weierstrassZeta(4, 0, weierstrassPInvers e(4, 0, cos(d*x + c) - I*sin(d*x + c))) + sqrt(e*sin(d*x + c))*sin(d*x + c ))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\sqrt {e \sin {\left (c + d x \right )}}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
\[ \int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx=\int { \frac {\sqrt {e \sin \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx=\int { \frac {\sqrt {e \sin \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sqrt {e \sin (c+d x)}}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\sqrt {e\,\sin \left (c+d\,x\right )}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]