Integrand size = 23, antiderivative size = 157 \[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{5/2} \, dx=-\frac {a e^{5/2} \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {a e^{5/2} \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {6 a e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {2 a e (e \sin (c+d x))^{3/2}}{3 d}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d} \]
-a*e^(5/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d+a*e^(5/2)*arctanh((e*sin (d*x+c))^(1/2)/e^(1/2))/d-2/3*a*e*(e*sin(d*x+c))^(3/2)/d-2/5*a*e*cos(d*x+c )*(e*sin(d*x+c))^(3/2)/d-6/5*a*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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin (d*x+c))^(1/2)/d/sin(d*x+c)^(1/2)
Time = 0.52 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.68 \[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{5/2} \, dx=-\frac {a (e \sin (c+d x))^{5/2} \left (15 \arctan \left (\sqrt {\sin (c+d x)}\right )-15 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )+18 E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+10 \sin ^{\frac {3}{2}}(c+d x)+3 \sqrt {\sin (c+d x)} \sin (2 (c+d x))\right )}{15 d \sin ^{\frac {5}{2}}(c+d x)} \]
-1/15*(a*(e*Sin[c + d*x])^(5/2)*(15*ArcTan[Sqrt[Sin[c + d*x]]] - 15*ArcTan h[Sqrt[Sin[c + d*x]]] + 18*EllipticE[(-2*c + Pi - 2*d*x)/4, 2] + 10*Sin[c + d*x]^(3/2) + 3*Sqrt[Sin[c + d*x]]*Sin[2*(c + d*x)]))/(d*Sin[c + d*x]^(5/ 2))
Time = 0.75 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.826, Rules used = {3042, 4360, 25, 25, 3042, 3317, 3042, 3044, 27, 262, 266, 827, 216, 219, 3115, 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.
| \(\displaystyle \int (a \sec (c+d x)+a) (e \sin (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right ) \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\left (\sec (c+d x) (a (-\cos (c+d x))-a) (e \sin (c+d x))^{5/2}\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\left ((\cos (c+d x) a+a) \sec (c+d x) (e \sin (c+d x))^{5/2}\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \sec (c+d x) (a \cos (c+d x)+a) (e \sin (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3317 |
| \(\displaystyle a \int (e \sin (c+d x))^{5/2}dx+a \int \sec (c+d x) (e \sin (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int (e \sin (c+d x))^{5/2}dx+a \int \frac {(e \sin (c+d x))^{5/2}}{\cos (c+d x)}dx\) |
| \(\Big \downarrow \) 3044 |
| \(\displaystyle \frac {a \int \frac {e^2 (e \sin (c+d x))^{5/2}}{e^2-e^2 \sin ^2(c+d x)}d(e \sin (c+d x))}{d e}+a \int (e \sin (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a e \int \frac {(e \sin (c+d x))^{5/2}}{e^2-e^2 \sin ^2(c+d x)}d(e \sin (c+d x))}{d}+a \int (e \sin (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {a e \left (e^2 \int \frac {\sqrt {e \sin (c+d x)}}{e^2-e^2 \sin ^2(c+d x)}d(e \sin (c+d x))-\frac {2}{3} (e \sin (c+d x))^{3/2}\right )}{d}+a \int (e \sin (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {a e \left (2 e^2 \int \frac {e^2 \sin ^2(c+d x)}{e^2-e^4 \sin ^4(c+d x)}d\sqrt {e \sin (c+d x)}-\frac {2}{3} (e \sin (c+d x))^{3/2}\right )}{d}+a \int (e \sin (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {a e \left (2 e^2 \left (\frac {1}{2} \int \frac {1}{e-e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}-\frac {1}{2} \int \frac {1}{e^2 \sin ^2(c+d x)+e}d\sqrt {e \sin (c+d x)}\right )-\frac {2}{3} (e \sin (c+d x))^{3/2}\right )}{d}+a \int (e \sin (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {a e \left (2 e^2 \left (\frac {1}{2} \int \frac {1}{e-e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}-\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}\right )-\frac {2}{3} (e \sin (c+d x))^{3/2}\right )}{d}+a \int (e \sin (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle a \int (e \sin (c+d x))^{5/2}dx+\frac {a e \left (2 e^2 \left (\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}-\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}\right )-\frac {2}{3} (e \sin (c+d x))^{3/2}\right )}{d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a \left (\frac {3}{5} e^2 \int \sqrt {e \sin (c+d x)}dx-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\right )+\frac {a e \left (2 e^2 \left (\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}-\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}\right )-\frac {2}{3} (e \sin (c+d x))^{3/2}\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {3}{5} e^2 \int \sqrt {e \sin (c+d x)}dx-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\right )+\frac {a e \left (2 e^2 \left (\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}-\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}\right )-\frac {2}{3} (e \sin (c+d x))^{3/2}\right )}{d}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle a \left (\frac {3 e^2 \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{5 \sqrt {\sin (c+d x)}}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\right )+\frac {a e \left (2 e^2 \left (\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}-\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}\right )-\frac {2}{3} (e \sin (c+d x))^{3/2}\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {3 e^2 \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{5 \sqrt {\sin (c+d x)}}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\right )+\frac {a e \left (2 e^2 \left (\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}-\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}\right )-\frac {2}{3} (e \sin (c+d x))^{3/2}\right )}{d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {a e \left (2 e^2 \left (\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}-\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 \sqrt {e}}\right )-\frac {2}{3} (e \sin (c+d x))^{3/2}\right )}{d}+a \left (\frac {6 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\right )\) |
(a*e*(2*e^2*(-1/2*ArcTan[Sqrt[e]*Sin[c + d*x]]/Sqrt[e] + ArcTanh[Sqrt[e]*S in[c + d*x]]/(2*Sqrt[e])) - (2*(e*Sin[c + d*x])^(3/2))/3))/d + a*((6*e^2*E llipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*d*Sqrt[Sin[c + d* x]]) - (2*e*Cos[c + d*x]*(e*Sin[c + d*x])^(3/2))/(5*d))
3.2.8.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
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[((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]
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[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
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 = 10.94 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.34
| method | result | size |
| default | \(\frac {-\frac {2 a e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-a \,e^{\frac {5}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+a \,e^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-\frac {a \,e^{3} \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{2}\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(210\) |
| parts | \(-\frac {a \,e^{3} \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{2}\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a \left (-\frac {2 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-e^{\frac {5}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+e^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )\right )}{d}\) | \(212\) |
(-2/3*a*e*(e*sin(d*x+c))^(3/2)-a*e^(5/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/ 2))+a*e^(5/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))-1/5*a*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)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^4+ 2*sin(d*x+c)^2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.36 (sec) , antiderivative size = 649, normalized size of antiderivative = 4.13 \[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\left [\frac {72 i \, \sqrt {2} a \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 ) - 72 i \, \sqrt {2} a \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 ) - 30 \, a \sqrt {-e} e^{2} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {-e}}{4 \, {\left (e \cos \left (d x + c\right )^{2} - e \sin \left (d x + c\right ) - e\right )}}\right ) + 15 \, a \sqrt {-e} e^{2} \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} - 8 \, {\left (7 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 8\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {-e} + 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) - 16 \, {\left (3 \, a e^{2} \cos \left (d x + c\right ) + 5 \, a e^{2}\right )} \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right )}{120 \, d}, \frac {72 i \, \sqrt {2} a \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 ) - 72 i \, \sqrt {2} a \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 ) - 30 \, a e^{\frac {5}{2}} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {e}}{4 \, {\left (e \cos \left (d x + c\right )^{2} + e \sin \left (d x + c\right ) - e\right )}}\right ) + 15 \, a e^{\frac {5}{2}} \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} - 8 \, {\left (7 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 8\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {e} - 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) - 16 \, {\left (3 \, a e^{2} \cos \left (d x + c\right ) + 5 \, a e^{2}\right )} \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right )}{120 \, d}\right ] \]
[1/120*(72*I*sqrt(2)*a*sqrt(-I*e)*e^2*weierstrassZeta(4, 0, weierstrassPIn verse(4, 0, cos(d*x + c) + I*sin(d*x + c))) - 72*I*sqrt(2)*a*sqrt(I*e)*e^2 *weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 30*a*sqrt(-e)*e^2*arctan(1/4*(cos(d*x + c)^2 - 6*sin(d*x + c) - 2 )*sqrt(e*sin(d*x + c))*sqrt(-e)/(e*cos(d*x + c)^2 - e*sin(d*x + c) - e)) + 15*a*sqrt(-e)*e^2*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 - 8*(7*cos( d*x + c)^2 - (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt(e*sin(d*x + c))*s qrt(-e) + 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 16*(3*a* e^2*cos(d*x + c) + 5*a*e^2)*sqrt(e*sin(d*x + c))*sin(d*x + c))/d, 1/120*(7 2*I*sqrt(2)*a*sqrt(-I*e)*e^2*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) - 72*I*sqrt(2)*a*sqrt(I*e)*e^2*weierstr assZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 30*a*e^(5/2)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + c) - 2)*sqrt(e*sin(d *x + c))*sqrt(e)/(e*cos(d*x + c)^2 + e*sin(d*x + c) - e)) + 15*a*e^(5/2)*l og((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 - 8*(7*cos(d*x + c)^2 + (cos(d* x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt(e*sin(d*x + c))*sqrt(e) - 28*(e*cos(d *x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 16*(3*a*e^2*cos(d*x + c) + 5* a*e^2)*sqrt(e*sin(d*x + c))*sin(d*x + c))/d]
Timed out. \[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
\[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \]
\[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right ) \,d x \]