Integrand size = 33, antiderivative size = 244 \[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b \left (15 a A b+14 a^2 B+3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b^2 (5 A b+9 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d} \] Output:
2/5*(5*A*a^3-15*A*a*b^2-15*B*a^2*b-3*B*b^3)*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+2/3*(9*A*a^2*b+A*b^3+3*B*a^3+3 *B*a*b^2)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+ c)^(1/2)/d+2/5*b*(15*A*a*b+14*B*a^2+3*B*b^2)*sec(d*x+c)^(1/2)*sin(d*x+c)/d +2/15*b^2*(5*A*b+9*B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*b*B*sec(d*x+c)^( 1/2)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d
Time = 10.70 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sec ^{\frac {5}{2}}(c+d x) \left (12 \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 b \left (15 \left (3 a A b+3 a^2 B+b^2 B\right )+10 b (A b+3 a B) \cos (c+d x)+9 \left (5 a A b+5 a^2 B+b^2 B\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{30 d} \] Input:
Integrate[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d*x]] ,x]
Output:
(Sec[c + d*x]^(5/2)*(12*(5*a^3*A - 15*a*A*b^2 - 15*a^2*b*B - 3*b^3*B)*Cos[ c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 20*(9*a^2*A*b + A*b^3 + 3*a^3*B + 3*a*b^2*B)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 2*b*(15*(3*a* A*b + 3*a^2*B + b^2*B) + 10*b*(A*b + 3*a*B)*Cos[c + d*x] + 9*(5*a*A*b + 5* a^2*B + b^2*B)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(30*d)
Time = 1.61 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 4514, 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.
| \(\displaystyle \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4514 |
| \(\displaystyle \frac {2}{5} \int \frac {(a+b \sec (c+d x)) \left (b (5 A b+9 a B) \sec ^2(c+d x)+\left (3 B b^2+5 a (2 A b+a B)\right ) \sec (c+d x)+a (5 a A-b B)\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(a+b \sec (c+d x)) \left (b (5 A b+9 a B) \sec ^2(c+d x)+\left (3 B b^2+5 a (2 A b+a B)\right ) \sec (c+d x)+a (5 a A-b B)\right )}{\sqrt {\sec (c+d x)}}dx+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (5 A b+9 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 B b^2+5 a (2 A b+a B)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a (5 a A-b B)\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4564 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {3 (5 a A-b B) a^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right ) \sec ^2(c+d x)+5 \left (3 B a^3+9 A b a^2+3 b^2 B a+A b^3\right ) \sec (c+d x)}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 (5 a A-b B) a^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right ) \sec ^2(c+d x)+5 \left (3 B a^3+9 A b a^2+3 b^2 B a+A b^3\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 (5 a A-b B) a^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 \left (3 B a^3+9 A b a^2+3 b^2 B a+A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 (5 a A-b B) a^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}}dx+5 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \int \sqrt {\sec (c+d x)}dx\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 (5 a A-b B) a^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 (5 a A-b B) a^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 (5 a A-b B) a^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 (5 a A-b B) a^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {6 b \left (14 a^2 B+15 a A b+3 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 b \left (14 a^2 B+15 a A b+3 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {6 b \left (14 a^2 B+15 a A b+3 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {6 b \left (14 a^2 B+15 a A b+3 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {6 b \left (14 a^2 B+15 a A b+3 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\) |
Input:
Int[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d*x]],x]
Output:
(2*b*B*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((2 *b^2*(5*A*b + 9*a*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + ((6*(5*a^3*A - 15*a*A*b^2 - 15*a^2*b*B - 3*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d* x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (10*(9*a^2*A*b + A*b^3 + 3*a^3*B + 3*a*b^ 2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (6*b*(15*a*A*b + 14*a^2*B + 3*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d)/3 )/5
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_)*(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/(m + n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n* Simp[a^2*A*(m + n) + a*b*B*n + (a*(2*A*b + a*B)*(m + n) + b^2*B*(m + n - 1) )*Csc[e + f*x] + b*(A*b*(m + n) + a*B*(2*m + n - 1))*Csc[e + f*x]^2, x], x] , x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && !(IGtQ[n, 1] && !IntegerQ[m])
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. \(969\) vs. \(2(223)=446\).
Time = 10.80 (sec) , antiderivative size = 970, normalized size of antiderivative = 3.98
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(970\) |
| parts | \(\text {Expression too large to display}\) | \(1065\) |
Input:
int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x,method=_RETURNV ERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*a^3*(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))+ 2/5*B*b^3/(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)/sin(1/2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c) -12*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)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*s in(1/2*d*x+1/2*c)^4+12*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)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*co s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*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)^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)+2*b^2*(A*b+3*B*a)*(-1/6*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+1/3*(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)*Ellip ticF(cos(1/2*d*x+1/2*c),2^(1/2)))-2*a^3*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^3*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))-Elliptic...
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.34 \[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, B a^{3} + 9 i \, A a^{2} b + 3 i \, B a b^{2} + i \, A b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, B a^{3} - 9 i \, A a^{2} b - 3 i \, B a b^{2} - i \, A b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, A a^{3} + 15 i \, B a^{2} b + 15 i \, A a b^{2} + 3 i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, A a^{3} - 15 i \, B a^{2} b - 15 i \, A a b^{2} - 3 i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, B b^{3} + 9 \, {\left (5 \, B a^{2} b + 5 \, A a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorith m="fricas")
Output:
-1/15*(5*sqrt(2)*(3*I*B*a^3 + 9*I*A*a^2*b + 3*I*B*a*b^2 + I*A*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt (2)*(-3*I*B*a^3 - 9*I*A*a^2*b - 3*I*B*a*b^2 - I*A*b^3)*cos(d*x + c)^2*weie rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(-5*I*A* a^3 + 15*I*B*a^2*b + 15*I*A*a*b^2 + 3*I*B*b^3)*cos(d*x + c)^2*weierstrassZ eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3* sqrt(2)*(5*I*A*a^3 - 15*I*B*a^2*b - 15*I*A*a*b^2 - 3*I*B*b^3)*cos(d*x + c) ^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin( d*x + c))) - 2*(3*B*b^3 + 9*(5*B*a^2*b + 5*A*a*b^2 + B*b^3)*cos(d*x + c)^2 + 5*(3*B*a*b^2 + A*b^3)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d *cos(d*x + c)^2)
Timed out. \[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((a+b*sec(d*x+c))**3*(A+B*sec(d*x+c))/sec(d*x+c)**(1/2),x)
Output:
Timed out
\[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorith m="maxima")
Output:
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^3/sqrt(sec(d*x + c)), x)
\[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorith m="giac")
Output:
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^3/sqrt(sec(d*x + c)), x)
Timed out. \[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^3)/(1/cos(c + d*x))^(1/2),x )
Output:
int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^3)/(1/cos(c + d*x))^(1/2), x)
\[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) a^{4}+4 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a^{3} b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b^{4}+4 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a \,b^{3}+6 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a^{2} b^{2} \] Input:
int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x)
Output:
int(sqrt(sec(c + d*x))/sec(c + d*x),x)*a**4 + 4*int(sqrt(sec(c + d*x)),x)* a**3*b + int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*b**4 + 4*int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*a*b**3 + 6*int(sqrt(sec(c + d*x))*sec(c + d*x) ,x)*a**2*b**2