Integrand size = 23, antiderivative size = 208 \[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {8 a b \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 \left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {4 a b \left (a^2-6 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]
-2/3*b^2*(a^2-b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/3*a^2*(a+b*sec(d*x+c))^ 2*sin(d*x+c)/d/sec(d*x+c)^(1/2)-4/3*a*b*(a^2-6*b^2)*sin(d*x+c)*sec(d*x+c)^ (1/2)/d+8*a*b*(a^2-b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El lipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/ 3*(a^4+18*a^2*b^2+b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell ipticF(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 = 1.64 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.62 \[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (24 a b \left (a^2-b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (a^4+18 a^2 b^2+b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {\left (a^4+2 b^4+24 a b^3 \cos (c+d x)+a^4 \cos (2 (c+d x))\right ) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}\right )}{3 d} \]
(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(24*a*b*(a^2 - b^2)*EllipticE[(c + d*x)/2, 2] + 2*(a^4 + 18*a^2*b^2 + b^4)*EllipticF[(c + d*x)/2, 2] + ((a^4 + 2*b^4 + 24*a*b^3*Cos[c + d*x] + a^4*Cos[2*(c + d*x)])*Sin[c + d*x])/Cos[ c + d*x]^(3/2)))/(3*d)
Time = 1.32 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, 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, 4564, 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.
| \(\displaystyle \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^4}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4328 |
| \(\displaystyle \frac {2}{3} \int \frac {(a+b \sec (c+d x)) \left (10 b a^2+\left (a^2+9 b^2\right ) \sec (c+d x) a-3 b \left (a^2-b^2\right ) \sec ^2(c+d x)\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {(a+b \sec (c+d x)) \left (10 b a^2+\left (a^2+9 b^2\right ) \sec (c+d x) a-3 b \left (a^2-b^2\right ) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (10 b a^2+\left (a^2+9 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a-3 b \left (a^2-b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4564 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{3} \int \frac {3 \left (10 b a^3-2 b \left (a^2-6 b^2\right ) \sec ^2(c+d x) a+\left (a^4+18 b^2 a^2+b^4\right ) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {10 b a^3-2 b \left (a^2-6 b^2\right ) \sec ^2(c+d x) a+\left (a^4+18 b^2 a^2+b^4\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {10 b a^3-2 b \left (a^2-6 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2 a+\left (a^4+18 b^2 a^2+b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{3} \left (\left (a^4+18 a^2 b^2+b^4\right ) \int \sqrt {\sec (c+d x)}dx+\int \frac {10 a^3 b-2 a b \left (a^2-6 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\left (a^4+18 a^2 b^2+b^4\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\int \frac {10 a^3 b-2 a b \left (a^2-6 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{3} \left (\left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\int \frac {10 a^3 b-2 a b \left (a^2-6 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {10 a^3 b-2 a b \left (a^2-6 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {10 a^3 b-2 a b \left (a^2-6 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}+\frac {2 \left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{3} \left (12 a b \left (a^2-b^2\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}-\frac {4 a b \left (a^2-6 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {2 \left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (12 a b \left (a^2-b^2\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}-\frac {4 a b \left (a^2-6 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {2 \left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{3} \left (12 a b \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}-\frac {4 a b \left (a^2-6 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {2 \left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (12 a b \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}-\frac {4 a b \left (a^2-6 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {2 \left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \left (-\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}-\frac {4 a b \left (a^2-6 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {24 a b \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )\) |
(2*a^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ((2 4*a*b*(a^2 - b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*(a^4 + 18*a^2*b^2 + b^4)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (4*a*b*(a^2 - 6*b^2)*Sqrt[Sec[c + d*x] ]*Sin[c + d*x])/d - (2*b^2*(a^2 - b^2)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/d) /3
3.7.3.3.1 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[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f*x])^ n/(f*(n + 2))), x] + Simp[1/(n + 2) Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*( n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(776\) vs. \(2(238)=476\).
Time = 30.84 (sec) , antiderivative size = 777, normalized size of antiderivative = 3.74
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(777\) |
| parts | \(\text {Expression too large to display}\) | \(857\) |
-2/3*(-8*a^4*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2* d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+8*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2 *c)^2)^(1/2)*a*(a^3+6*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*(-2*s in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(a^4+12*a*b^3+b^4)*sin(1/2 *d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/ 2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2 )*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^4+18*EllipticF(cos(1/2*d*x+1/2* c),2^(1/2))*a^2*b^2+EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^4-12*EllipticE (cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b+12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2) )*a*b^3)*sin(1/2*d*x+1/2*c)^2+a^4*(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))*(-2*sin(1/2*d* x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+18*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) )*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(-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)^(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))*b^4-12*(-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)^(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))*a^ 3*b+12*(-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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left (-i \, a^{4} - 18 i \, a^{2} b^{2} - i \, b^{4}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, a^{4} + 18 i \, a^{2} b^{2} + i \, b^{4}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 \, \sqrt {2} {\left (-i \, a^{3} b + i \, a b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 \, \sqrt {2} {\left (i \, a^{3} b - i \, a b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (a^{4} \cos \left (d x + c\right )^{2} + 12 \, a b^{3} \cos \left (d x + c\right ) + b^{4}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, d \cos \left (d x + c\right )} \]
1/3*(sqrt(2)*(-I*a^4 - 18*I*a^2*b^2 - I*b^4)*cos(d*x + c)*weierstrassPInve rse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I*a^4 + 18*I*a^2*b^2 + I*b^4)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 12*sqrt(2)*(-I*a^3*b + I*a*b^3)*cos(d*x + c)*weierstrassZeta(-4, 0 , weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12*sqrt(2)* (I*a^3*b - I*a*b^3)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInvers e(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(a^4*cos(d*x + c)^2 + 12*a*b^ 3*cos(d*x + c) + b^4)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c))
\[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{4}}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^4}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]