Integrand size = 25, antiderivative size = 104 \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=-\frac {4 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a d \sqrt {\sin (c+d x)}}+\frac {2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a d} \]
2/3*e*(e*sin(d*x+c))^(3/2)/a/d-2/5*e*cos(d*x+c)*(e*sin(d*x+c))^(3/2)/a/d+4 /5*e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*Ellip ticE(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 = 3.19 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.86 \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\frac {2 e^3 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (10 \sqrt {\csc ^2(c)} \sin ^2(c+d x)+6 \csc (c) \csc (d x-\arctan (\cot (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x-\arctan (\cot (c)))\right ) \sqrt {\sin ^2(d x-\arctan (\cot (c)))}+3 \csc (c) \sec (c) (\sin (c+d x-\arctan (\cot (c)))+3 \sin (c-d x+\arctan (\cot (c))))-3 \sqrt {\csc ^2(c)} \sin (c+d x) (\sin (2 (c+d x))+4 \tan (c))\right )}{15 a d \sqrt {\csc ^2(c)} (1+\sec (c+d x)) \sqrt {e \sin (c+d x)}} \]
(2*e^3*Cos[(c + d*x)/2]^2*Sec[c + d*x]*(10*Sqrt[Csc[c]^2]*Sin[c + d*x]^2 + 6*Csc[c]*Csc[d*x - ArcTan[Cot[c]]]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Sin[d*x - ArcTan[Cot[c]]]^2] + 3*Csc[c] *Sec[c]*(Sin[c + d*x - ArcTan[Cot[c]]] + 3*Sin[c - d*x + ArcTan[Cot[c]]]) - 3*Sqrt[Csc[c]^2]*Sin[c + d*x]*(Sin[2*(c + d*x)] + 4*Tan[c])))/(15*a*d*Sq rt[Csc[c]^2]*(1 + Sec[c + d*x])*Sqrt[e*Sin[c + d*x]])
Time = 0.67 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.02, 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, 3049, 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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^{5/2}}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}{a-a \csc \left (c+d x-\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\cos (c+d x) (e \sin (c+d x))^{5/2}}{a (-\cos (c+d x))-a}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\cos (c+d x) (e \sin (c+d x))^{5/2}}{\cos (c+d x) a+a}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\cos (c+d x) (e \sin (c+d x))^{5/2}}{a \cos (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 3318 |
| \(\displaystyle \frac {e^2 \int \cos (c+d x) \sqrt {e \sin (c+d x)}dx}{a}-\frac {e^2 \int \cos ^2(c+d x) \sqrt {e \sin (c+d x)}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \int \cos (c+d x) \sqrt {e \sin (c+d x)}dx}{a}-\frac {e^2 \int \cos (c+d x)^2 \sqrt {e \sin (c+d x)}dx}{a}\) |
| \(\Big \downarrow \) 3044 |
| \(\displaystyle \frac {e \int \sqrt {e \sin (c+d x)}d(e \sin (c+d x))}{a d}-\frac {e^2 \int \cos (c+d x)^2 \sqrt {e \sin (c+d x)}dx}{a}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac {e^2 \int \cos (c+d x)^2 \sqrt {e \sin (c+d x)}dx}{a}\) |
| \(\Big \downarrow \) 3049 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac {e^2 \left (\frac {2}{5} \int \sqrt {e \sin (c+d x)}dx+\frac {2 \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d e}\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac {e^2 \left (\frac {2}{5} \int \sqrt {e \sin (c+d x)}dx+\frac {2 \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d e}\right )}{a}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac {e^2 \left (\frac {2 \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{5 \sqrt {\sin (c+d x)}}+\frac {2 \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d e}\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac {e^2 \left (\frac {2 \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{5 \sqrt {\sin (c+d x)}}+\frac {2 \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d e}\right )}{a}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2}}{3 a d}-\frac {e^2 \left (\frac {2 \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d e}+\frac {4 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}\right )}{a}\) |
(2*e*(e*Sin[c + d*x])^(3/2))/(3*a*d) - (e^2*((4*EllipticE[(c - Pi/2 + d*x) /2, 2]*Sqrt[e*Sin[c + d*x]])/(5*d*Sqrt[Sin[c + d*x]]) + (2*Cos[c + d*x]*(e *Sin[c + d*x])^(3/2))/(5*d*e)))/a
3.2.21.3.1 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*(b*Sin[e + f*x])^(n + 1)*((a*Cos[e + f*x])^(m - 1)/ (b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n)) Int[(b*Sin[e + f*x])^n*(a *Cos[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m, 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]
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 = 5.36 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.66
| method | result | size |
| default | \(\frac {2 e^{3} \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+3 \cos \left (d x +c \right )^{4}-5 \cos \left (d x +c \right )^{3}-3 \cos \left (d x +c \right )^{2}+5 \cos \left (d x +c \right )\right )}{15 a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(173\) |
2/15/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^3*(6*(-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))-3*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellip ticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+3*cos(d*x+c)^4-5*cos(d*x+c)^3-3*co s(d*x+c)^2+5*cos(d*x+c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.07 \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=-\frac {2 \, {\left (3 i \, \sqrt {2} \sqrt {-i \, e} e^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} \sqrt {i \, e} e^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left (3 \, e^{2} \cos \left (d x + c\right ) - 5 \, e^{2}\right )} \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, a d} \]
-2/15*(3*I*sqrt(2)*sqrt(-I*e)*e^2*weierstrassZeta(4, 0, weierstrassPInvers e(4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*sqrt(I*e)*e^2*weiers trassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) + (3*e^2*cos(d*x + c) - 5*e^2)*sqrt(e*sin(d*x + c))*sin(d*x + c))/(a*d)
Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]