Integrand size = 33, antiderivative size = 236 \[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \left (3 a^3 A+15 a A b^2+15 a^2 b B-5 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^2 A b+3 A b^3+a^3 B+9 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 (9 A b+5 a B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}-\frac {2 b^2 (a A-5 b B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a A (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \] Output:
2/5*(3*A*a^3+15*A*a*b^2+15*B*a^2*b-5*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*(3*A*a^2*b+3*A*b^3+B*a^3+9 *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+2/15*a^2*(9*A*b+5*B*a)*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/5*b^2*(A *a-5*B*b)*sec(d*x+c)^(1/2)*sin(d*x+c)/d+2/5*a*A*(a+b*sec(d*x+c))^2*sin(d*x +c)/d/sec(d*x+c)^(3/2)
Time = 6.63 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {\sec (c+d x)} \left (12 \left (3 a^3 A+15 a A b^2+15 a^2 b B-5 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (3 a^2 A b+3 A b^3+a^3 B+9 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 \left (10 a^2 (3 A b+a B) \cos (c+d x)+3 \left (a^3 A+10 b^3 B+a^3 A \cos (2 (c+d x))\right )\right ) \sin (c+d x)\right )}{30 d} \] Input:
Integrate[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2) ,x]
Output:
(Sqrt[Sec[c + d*x]]*(12*(3*a^3*A + 15*a*A*b^2 + 15*a^2*b*B - 5*b^3*B)*Sqrt [Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 20*(3*a^2*A*b + 3*A*b^3 + a^3*B + 9*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(10*a^2*(3* A*b + a*B)*Cos[c + d*x] + 3*(a^3*A + 10*b^3*B + a^3*A*Cos[2*(c + d*x)]))*S in[c + d*x]))/(30*d)
Time = 1.59 (sec) , antiderivative size = 240, 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, 4513, 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.
| \(\displaystyle \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4513 |
| \(\displaystyle \frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2}{5} \int -\frac {(a+b \sec (c+d x)) \left (-b (a A-5 b B) \sec ^2(c+d x)+\left (3 A a^2+10 b B a+5 A b^2\right ) \sec (c+d x)+a (9 A b+5 a B)\right )}{2 \sec ^{\frac {3}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(a+b \sec (c+d x)) \left (-b (a A-5 b B) \sec ^2(c+d x)+\left (3 A a^2+10 b B a+5 A b^2\right ) \sec (c+d x)+a (9 A b+5 a B)\right )}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b (a A-5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 A a^2+10 b B a+5 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a (9 A b+5 a B)\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4562 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2}{3} \int -\frac {-3 b^2 (a A-5 b B) \sec ^2(c+d x)+5 \left (B a^3+3 A b a^2+9 b^2 B a+3 A b^3\right ) \sec (c+d x)+3 a \left (3 A a^2+15 b B a+14 A b^2\right )}{2 \sqrt {\sec (c+d x)}}dx\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {-3 b^2 (a A-5 b B) \sec ^2(c+d x)+5 \left (B a^3+3 A b a^2+9 b^2 B a+3 A b^3\right ) \sec (c+d x)+3 a \left (3 A a^2+15 b B a+14 A b^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {-3 b^2 (a A-5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 \left (B a^3+3 A b a^2+9 b^2 B a+3 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 a \left (3 A a^2+15 b B a+14 A b^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a \left (3 A a^2+15 b B a+14 A b^2\right )-3 b^2 (a A-5 b B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}}dx+5 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\right ) \int \sqrt {\sec (c+d x)}dx\right )+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a \left (3 A a^2+15 b B a+14 A b^2\right )-3 b^2 (a A-5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a \left (3 A a^2+15 b B a+14 A b^2\right )-3 b^2 (a A-5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a \left (3 A a^2+15 b B a+14 A b^2\right )-3 b^2 (a A-5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a \left (3 A a^2+15 b B a+14 A b^2\right )-3 b^2 (a A-5 b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {10 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (3 a^3 A+15 a^2 b B+15 a A b^2-5 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {10 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 b^2 (a A-5 b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (3 a^3 A+15 a^2 b B+15 a A b^2-5 b^3 B\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {10 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 b^2 (a A-5 b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (3 a^3 A+15 a^2 b B+15 a A b^2-5 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {10 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 b^2 (a A-5 b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (3 \left (3 a^3 A+15 a^2 b B+15 a A b^2-5 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {10 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 b^2 (a A-5 b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 a^2 (5 a B+9 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \left (\frac {10 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 \left (3 a^3 A+15 a^2 b B+15 a A b^2-5 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {6 b^2 (a A-5 b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )\right )+\frac {2 a A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x]
Output:
(2*a*A*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((2 *a^2*(9*A*b + 5*a*B)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ((6*(3*a^3*A + 15*a*A*b^2 + 15*a^2*b*B - 5*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d* x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (10*(3*a^2*A*b + 3*A*b^3 + a^3*B + 9*a*b^ 2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (6*b^2*(a*A - 5*b*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[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] + Sim p[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[ a*(a*B*n - A*b*(m - n - 1)) + (2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] & & 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]
Leaf count of result is larger than twice the leaf count of optimal. \(640\) vs. \(2(215)=430\).
Time = 11.90 (sec) , antiderivative size = 641, normalized size of antiderivative = 2.72
| method | result | size |
| default | \(\frac {\frac {16 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a^{3}}{5}-\frac {16 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{3}}{5}-8 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{2} b -\frac {8 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{3}}{3}+\frac {4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3}}{5}+4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2} b -2 A \,a^{2} 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 )-2 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 \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 ) a^{3}}{5}+6 A \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 ) a \,b^{2}+\frac {4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3}}{3}+4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{3}-\frac {2 B \,a^{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 )}{3}-6 B a \,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 {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2} b -2 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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{3}}{\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}\) | \(641\) |
| parts | \(\text {Expression too large to display}\) | \(875\) |
Input:
int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x,method=_RETURNV ERBOSE)
Output:
2/15*(24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*a^3-24*A*cos(1/2*d*x+1/ 2*c)*sin(1/2*d*x+1/2*c)^4*a^3-60*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4 *a^2*b-20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^3+6*A*cos(1/2*d*x+1/ 2*c)*sin(1/2*d*x+1/2*c)^2*a^3+30*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2 *a^2*b-15*A*a^2*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ell ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3+45*A*(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*b^ 2+10*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^3+30*B*cos(1/2*d*x+1/2*c) *sin(1/2*d*x+1/2*c)^2*b^3-5*B*a^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))-45*B*a*b^2*(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)*EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2))+45*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-15*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)*EllipticE(cos(1/2*d*x +1/2*c),2^(1/2))*b^3)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/ d
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a^{3} + 3 i \, A a^{2} b + 9 i \, B a b^{2} + 3 i \, A b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a^{3} - 3 i \, A a^{2} b - 9 i \, B a b^{2} - 3 i \, A b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-3 i \, A a^{3} - 15 i \, B a^{2} b - 15 i \, A a b^{2} + 5 i \, B b^{3}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (3 i \, A a^{3} + 15 i \, B a^{2} b + 15 i \, A a b^{2} - 5 i \, B b^{3}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, A a^{3} \cos \left (d x + c\right )^{2} + 15 \, B b^{3} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d} \] Input:
integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algorith m="fricas")
Output:
-1/15*(5*sqrt(2)*(I*B*a^3 + 3*I*A*a^2*b + 9*I*B*a*b^2 + 3*I*A*b^3)*weierst rassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B*a^3 - 3*I*A*a^2*b - 9*I*B*a*b^2 - 3*I*A*b^3)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(-3*I*A*a^3 - 15*I*B*a^2*b - 15*I*A*a* b^2 + 5*I*B*b^3)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(3*I*A*a^3 + 15*I*B*a^2*b + 15*I*A*a* b^2 - 5*I*B*b^3)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*A*a^3*cos(d*x + c)^2 + 15*B*b^3 + 5*(B*a^3 + 3*A*a^2*b)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d
\[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{3}}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((a+b*sec(d*x+c))**3*(A+B*sec(d*x+c))/sec(d*x+c)**(5/2),x)
Output:
Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))**3/sec(c + d*x)**(5/2), x)
\[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algorith m="maxima")
Output:
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^3/sec(d*x + c)^(5/2), x)
\[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algorith m="giac")
Output:
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^3/sec(d*x + c)^(5/2), x)
Timed out. \[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^3)/(1/cos(c + d*x))^(5/2),x )
Output:
int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^3)/(1/cos(c + d*x))^(5/2), x)
\[ \int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3}}d x \right ) a^{4}+4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right ) a^{3} b +6 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) a^{2} b^{2}+4 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a \,b^{3}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) b^{4} \] Input:
int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x)
Output:
int(sqrt(sec(c + d*x))/sec(c + d*x)**3,x)*a**4 + 4*int(sqrt(sec(c + d*x))/ sec(c + d*x)**2,x)*a**3*b + 6*int(sqrt(sec(c + d*x))/sec(c + d*x),x)*a**2* b**2 + 4*int(sqrt(sec(c + d*x)),x)*a*b**3 + int(sqrt(sec(c + d*x))*sec(c + d*x),x)*b**4