Integrand size = 23, antiderivative size = 207 \[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \left (3 a^4+30 a^2 b^2-5 b^4\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {8 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {28 a^3 b \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}-\frac {2 b^2 \left (a^2-5 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]
2/5*a^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(3/2)+28/15*a^3*b*sin(d *x+c)/d/sec(d*x+c)^(1/2)-2/5*b^2*(a^2-5*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d +2/5*(3*a^4+30*a^2*b^2-5*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2 *c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2 )/d+8/3*a*b*(a^2+3*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El lipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d
Time = 1.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {\sec (c+d x)} \left (12 \left (3 a^4+30 a^2 b^2-5 b^4\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+80 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 \left (3 a^4+30 b^4+40 a^3 b \cos (c+d x)+3 a^4 \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{30 d} \]
(Sqrt[Sec[c + d*x]]*(12*(3*a^4 + 30*a^2*b^2 - 5*b^4)*Sqrt[Cos[c + d*x]]*El lipticE[(c + d*x)/2, 2] + 80*a*b*(a^2 + 3*b^2)*Sqrt[Cos[c + d*x]]*Elliptic F[(c + d*x)/2, 2] + 2*(3*a^4 + 30*b^4 + 40*a^3*b*Cos[c + d*x] + 3*a^4*Cos[ 2*(c + d*x)])*Sin[c + d*x]))/(30*d)
Time = 1.35 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.739, Rules used = {3042, 4328, 27, 3042, 4562, 27, 3042, 4535, 3042, 4258, 3042, 3120, 4534, 3042, 4258, 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.
| \(\displaystyle \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^4}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4328 |
| \(\displaystyle \frac {2}{5} \int \frac {(a+b \sec (c+d x)) \left (14 b a^2+3 \left (a^2+5 b^2\right ) \sec (c+d x) a-b \left (a^2-5 b^2\right ) \sec ^2(c+d x)\right )}{2 \sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(a+b \sec (c+d x)) \left (14 b a^2+3 \left (a^2+5 b^2\right ) \sec (c+d x) a-b \left (a^2-5 b^2\right ) \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (14 b a^2+3 \left (a^2+5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a-b \left (a^2-5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4562 |
| \(\displaystyle \frac {1}{5} \left (\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2}{3} \int -\frac {3 \left (3 a^2+29 b^2\right ) a^2+20 b \left (a^2+3 b^2\right ) \sec (c+d x) a-3 b^2 \left (a^2-5 b^2\right ) \sec ^2(c+d x)}{2 \sqrt {\sec (c+d x)}}dx\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 \left (3 a^2+29 b^2\right ) a^2+20 b \left (a^2+3 b^2\right ) \sec (c+d x) a-3 b^2 \left (a^2-5 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}}dx+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 \left (3 a^2+29 b^2\right ) a^2+20 b \left (a^2+3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a-3 b^2 \left (a^2-5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a^2 \left (3 a^2+29 b^2\right )-3 b^2 \left (a^2-5 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}}dx+20 a b \left (a^2+3 b^2\right ) \int \sqrt {\sec (c+d x)}dx\right )+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (20 a b \left (a^2+3 b^2\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\int \frac {3 a^2 \left (3 a^2+29 b^2\right )-3 b^2 \left (a^2-5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a^2 \left (3 a^2+29 b^2\right )-3 b^2 \left (a^2-5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+20 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a^2 \left (3 a^2+29 b^2\right )-3 b^2 \left (a^2-5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+20 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a^2 \left (3 a^2+29 b^2\right )-3 b^2 \left (a^2-5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {40 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (3 a^4+30 a^2 b^2-5 b^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-\frac {6 b^2 \left (a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {40 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (3 a^4+30 a^2 b^2-5 b^4\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 b^2 \left (a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {40 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (3 a^4+30 a^2 b^2-5 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-\frac {6 b^2 \left (a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {40 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (3 a^4+30 a^2 b^2-5 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {6 b^2 \left (a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {40 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (\frac {28 a^3 b \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \left (-\frac {6 b^2 \left (a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {40 a b \left (a^2+3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 \left (3 a^4+30 a^2 b^2-5 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\) |
(2*a^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((2 8*a^3*b*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ((6*(3*a^4 + 30*a^2*b^2 - 5*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (40*a*b*(a^2 + 3*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt [Sec[c + d*x]])/d - (6*b^2*(a^2 - 5*b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/ d)/3)/5
3.7.4.3.1 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[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[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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[a^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)* ((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2* (n + 1))*Csc[e + f*x] - b*(b^2*n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && ((Int egerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si mp[1/(d*n) Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
Time = 29.95 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.09
| method | result | size |
| default | \(\frac {\frac {16 a^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5}-\frac {16 a^{4} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {32 a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{3}+\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{4}}{5}+\frac {16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3} b}{3}+4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{4}-\frac {8 a^{3} b \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 {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3}-8 a \,b^{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 {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+\frac {6 a^{4} \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 )}{5}+12 a^{2} b^{2} \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 )-2 \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 ) b^{4}}{\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}\) | \(432\) |
| parts | \(\text {Expression too large to display}\) | \(846\) |
2/15*(24*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-24*a^4*sin(1/2*d*x+1/ 2*c)^4*cos(1/2*d*x+1/2*c)-80*a^3*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)*b +6*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^4+40*cos(1/2*d*x+1/2*c)*sin(1 /2*d*x+1/2*c)^2*a^3*b+30*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^4-20*a^ 3*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2))-60*a*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*a^4*(si n(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))+90*a^2*b^2*(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))-15*(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))*b^4)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {20 \, \sqrt {2} {\left (i \, a^{3} b + 3 i \, a b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 20 \, \sqrt {2} {\left (-i \, a^{3} b - 3 i \, a b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-3 i \, a^{4} - 30 i \, a^{2} b^{2} + 5 i \, b^{4}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (3 i \, a^{4} + 30 i \, a^{2} b^{2} - 5 i \, b^{4}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, a^{4} \cos \left (d x + c\right )^{2} + 20 \, a^{3} b \cos \left (d x + c\right ) + 15 \, b^{4}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d} \]
-1/15*(20*sqrt(2)*(I*a^3*b + 3*I*a*b^3)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 20*sqrt(2)*(-I*a^3*b - 3*I*a*b^3)*weierstrassPIn verse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(-3*I*a^4 - 30*I*a ^2*b^2 + 5*I*b^4)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(3*I*a^4 + 30*I*a^2*b^2 - 5*I*b^4)*w eierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*a^4*cos(d*x + c)^2 + 20*a^3*b*cos(d*x + c) + 15*b^4)*sin(d*x + c)/sqrt(cos(d*x + c)))/d
\[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{4}}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^4}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]