Integrand size = 35, antiderivative size = 253 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {4 a^3 (5 A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (35 A+13 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {8 a^3 (70 A+53 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {12 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d}+\frac {2 (5 A+7 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \] Output:
-4/5*a^3*(5*A+7*C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))* sec(d*x+c)^(1/2)/d+4/21*a^3*(35*A+13*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1 /2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+8/105*a^3*(70*A+53*C)*sec(d*x+c)^ (1/2)*sin(d*x+c)/d+2/7*C*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^3*sin(d*x+c)/d+ 12/35*C*sec(d*x+c)^(1/2)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d+2/15*(5*A+7 *C)*sec(d*x+c)^(1/2)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.35 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.11 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^3 e^{-i d x} \sec ^{\frac {7}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-630 i A-882 i C-840 i A \cos (2 (c+d x))-1176 i C \cos (2 (c+d x))-210 i A \cos (4 (c+d x))-294 i C \cos (4 (c+d x))+80 (35 A+13 C) \cos ^{\frac {7}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+14 i (5 A+7 C) e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+70 A \sin (c+d x)+380 C \sin (c+d x)+630 A \sin (2 (c+d x))+840 C \sin (2 (c+d x))+70 A \sin (3 (c+d x))+260 C \sin (3 (c+d x))+315 A \sin (4 (c+d x))+294 C \sin (4 (c+d x))\right )}{420 d} \] Input:
Integrate[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Sqrt[Sec[c + d*x ]],x]
Output:
(a^3*Sec[c + d*x]^(7/2)*(Cos[d*x] + I*Sin[d*x])*((-630*I)*A - (882*I)*C - (840*I)*A*Cos[2*(c + d*x)] - (1176*I)*C*Cos[2*(c + d*x)] - (210*I)*A*Cos[4 *(c + d*x)] - (294*I)*C*Cos[4*(c + d*x)] + 80*(35*A + 13*C)*Cos[c + d*x]^( 7/2)*EllipticF[(c + d*x)/2, 2] + ((14*I)*(5*A + 7*C)*(1 + E^((2*I)*(c + d* x)))^(7/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^((2*I )*(c + d*x)) + 70*A*Sin[c + d*x] + 380*C*Sin[c + d*x] + 630*A*Sin[2*(c + d *x)] + 840*C*Sin[2*(c + d*x)] + 70*A*Sin[3*(c + d*x)] + 260*C*Sin[3*(c + d *x)] + 315*A*Sin[4*(c + d*x)] + 294*C*Sin[4*(c + d*x)]))/(420*d*E^(I*d*x))
Time = 1.63 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.04, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 4577, 27, 3042, 4506, 27, 3042, 4506, 3042, 4485, 27, 3042, 4274, 3042, 4258, 3042, 3119, 3120}
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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4577 |
| \(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^3 (a (7 A-C)+6 a C \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^3 (a (7 A-C)+6 a C \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a (7 A-C)+6 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {2}{5} \int \frac {(\sec (c+d x) a+a)^2 \left ((35 A-11 C) a^2+7 (5 A+7 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {(\sec (c+d x) a+a)^2 \left ((35 A-11 C) a^2+7 (5 A+7 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left ((35 A-11 C) a^2+7 (5 A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {(\sec (c+d x) a+a) \left ((35 A-41 C) a^3+2 (70 A+53 C) \sec (c+d x) a^3\right )}{\sqrt {\sec (c+d x)}}dx+\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((35 A-41 C) a^3+2 (70 A+53 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (2 \int -\frac {21 a^4 (5 A+7 C)-5 a^4 (35 A+13 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)}}dx+\frac {4 a^4 (70 A+53 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (\frac {4 a^4 (70 A+53 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {21 a^4 (5 A+7 C)-5 a^4 (35 A+13 C) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx\right )+\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (\frac {4 a^4 (70 A+53 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {21 a^4 (5 A+7 C)-5 a^4 (35 A+13 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (-21 a^4 (5 A+7 C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+5 a^4 (35 A+13 C) \int \sqrt {\sec (c+d x)}dx+\frac {4 a^4 (70 A+53 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (-21 a^4 (5 A+7 C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 a^4 (35 A+13 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {4 a^4 (70 A+53 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^4 (35 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx-21 a^4 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {4 a^4 (70 A+53 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^4 (35 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-21 a^4 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {4 a^4 (70 A+53 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \left (5 a^4 (35 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {4 a^4 (70 A+53 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {42 a^4 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {14 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {2}{3} \left (\frac {4 a^4 (70 A+53 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 a^4 (35 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {42 a^4 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{5 d}}{7 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d}\) |
Input:
Int[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Sqrt[Sec[c + d*x]],x]
Output:
(2*C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(7*d) + ((12* C*Sqrt[Sec[c + d*x]]*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((14 *(5*A + 7*C)*Sqrt[Sec[c + d*x]]*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(3* d) + (2*((-42*a^4*(5*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] *Sqrt[Sec[c + d*x]])/d + (10*a^4*(35*A + 13*C)*Sqrt[Cos[c + d*x]]*Elliptic F[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (4*a^4*(70*A + 53*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/3)/5)/(7*a)
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_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C) *Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n *Simp[A*b*(m + n + 1) + b*C*n + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m, n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(986\) vs. \(2(228)=456\).
Time = 5.09 (sec) , antiderivative size = 987, normalized size of antiderivative = 3.90
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(987\) |
| parts | \(\text {Expression too large to display}\) | \(1298\) |
Input:
int((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x,method=_RETUR NVERBOSE)
Output:
-a^3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*A*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1 /2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+ 2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin (1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c ),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*C*(-1/56*cos(1/2*d*x+1 /2*c)*(-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)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)*(-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)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c)^2)^ (1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d *x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+6/5*C/sin(1/2*d* x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1 /2*c)^2-1)*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(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))*(sin(1/2*d*x+1/2*c) ^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+ 12*(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))* (sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*c os(1/2*d*x+1/2*c)-3*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+si n(1/2*d*x+1/2*c)^2)^(1/2)+16*(1/8*A+3/8*C)*(-1/6*cos(1/2*d*x+1/2*c)*(-2...
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.02 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (35 \, A + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (35 \, A + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (21 \, {\left (15 \, A + 14 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 5 \, {\left (7 \, A + 26 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 63 \, C a^{3} \cos \left (d x + c\right ) + 15 \, C a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algori thm="fricas")
Output:
-2/105*(5*I*sqrt(2)*(35*A + 13*C)*a^3*cos(d*x + c)^3*weierstrassPInverse(- 4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(35*A + 13*C)*a^3*cos(d *x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*I *sqrt(2)*(5*A + 7*C)*a^3*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrass PInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I*sqrt(2)*(5*A + 7*C) *a^3*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos( d*x + c) - I*sin(d*x + c))) - (21*(15*A + 14*C)*a^3*cos(d*x + c)^3 + 5*(7* A + 26*C)*a^3*cos(d*x + c)^2 + 63*C*a^3*cos(d*x + c) + 15*C*a^3)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algori thm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^3/sqrt(sec(d*x + c)) , x)
\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algori thm="giac")
Output:
integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^3/sqrt(sec(d*x + c)) , x)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3)/(1/cos(c + d*x))^(1/2) ,x)
Output:
int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3)/(1/cos(c + d*x))^(1/2) , x)
\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=a^{3} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) a +3 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}d x \right ) c +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) c +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) c \right ) \] Input:
int((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x)
Output:
a**3*(int(sqrt(sec(c + d*x))/sec(c + d*x),x)*a + 3*int(sqrt(sec(c + d*x)), x)*a + int(sqrt(sec(c + d*x))*sec(c + d*x)**4,x)*c + 3*int(sqrt(sec(c + d* x))*sec(c + d*x)**3,x)*c + int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*a + 3 *int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*c + 3*int(sqrt(sec(c + d*x))*se c(c + d*x),x)*a + int(sqrt(sec(c + d*x))*sec(c + d*x),x)*c)