Integrand size = 33, antiderivative size = 237 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {(4 A-7 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {5 (A-2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {(4 A-7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac {5 (A-2 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {(4 A-7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]
-5/3*(A-2*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d+1/3*(4*A-7*B)*sec(d*x+c)^(5 /2)*sin(d*x+c)/a^2/d/(1+sec(d*x+c))+1/3*(A-B)*sec(d*x+c)^(7/2)*sin(d*x+c)/ d/(a+a*sec(d*x+c))^2+(4*A-7*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/a^2/d-(4*A-7*B) *(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)/a^2/d-5/3*(A-2*B)*(cos(1/2* d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2 ))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.61 (sec) , antiderivative size = 480, normalized size of antiderivative = 2.03 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (A+B \sec (c+d x)) \left (8 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-14 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-20 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+40 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+2 \sqrt {\sec (c+d x)} \left (6 (4 A-7 B) \cos (d x) \csc (c)-\frac {1}{2} (5 A-14 B+2 (6 A-13 B) \cos (c+d x)+5 (A-2 B) \cos (2 (c+d x))) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{3 a^2 d (B+A \cos (c+d x)) (1+\sec (c+d x))^2} \]
(Cos[(c + d*x)/2]^4*Sec[c + d*x]*(A + B*Sec[c + d*x])*((8*Sqrt[2]*A*Sqrt[E ^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cs c[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))* Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) - (14*S qrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I) *(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^ (I*d*x) - 20*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d *x]] + 40*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x] ] + 2*Sqrt[Sec[c + d*x]]*(6*(4*A - 7*B)*Cos[d*x]*Csc[c] - ((5*A - 14*B + 2 *(6*A - 13*B)*Cos[c + d*x] + 5*(A - 2*B)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2 ]^2*Sec[c + d*x]*Tan[(c + d*x)/2])/2)))/(3*a^2*d*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x])^2)
Time = 1.24 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.99, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4507, 27, 3042, 4507, 27, 3042, 4274, 3042, 4255, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (5 a (A-B)-3 a (A-3 B) \sec (c+d x))}{2 (\sec (c+d x) a+a)}dx}{3 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (5 a (A-B)-3 a (A-3 B) \sec (c+d x))}{\sec (c+d x) a+a}dx}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (5 a (A-B)-3 a (A-3 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int 3 \sec ^{\frac {3}{2}}(c+d x) \left (a^2 (4 A-7 B)-5 a^2 (A-2 B) \sec (c+d x)\right )dx}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \int \sec ^{\frac {3}{2}}(c+d x) \left (a^2 (4 A-7 B)-5 a^2 (A-2 B) \sec (c+d x)\right )dx}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a^2 (4 A-7 B)-5 a^2 (A-2 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (4 A-7 B) \int \sec ^{\frac {3}{2}}(c+d x)dx-5 a^2 (A-2 B) \int \sec ^{\frac {5}{2}}(c+d x)dx\right )}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (4 A-7 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-5 a^2 (A-2 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (4 A-7 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )-5 a^2 (A-2 B) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (4 A-7 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-5 a^2 (A-2 B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (4 A-7 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )-5 a^2 (A-2 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (4 A-7 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-5 a^2 (A-2 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (4 A-7 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-5 a^2 (A-2 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (4 A-7 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-5 a^2 (A-2 B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )\right )}{a^2}+\frac {2 (4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
((A - B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ( (2*(4*A - 7*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])) + ( 3*(a^2*(4*A - 7*B)*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[ Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d) - 5*a^2*(A - 2*B )*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3* d) + (2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d))))/a^2)/(6*a^2)
3.3.9.3.1 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)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
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_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(722\) vs. \(2(265)=530\).
Time = 27.43 (sec) , antiderivative size = 723, normalized size of antiderivative = 3.05
-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*(1/3*(- A+B)*(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)*(2*E llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2 )))*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)^2-1)^( 1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) -3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*sin(1/2*d* x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/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)/(-1+sin(1/2*d*x+1 /2*c)^2)+(-2*A+4*B)*(cos(1/2*d*x+1/2*c)*(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))-Ellipti cE(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)/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) +4*B*(-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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+(4*A-8*B)/sin(1/2*d*x+1/2 *c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2* c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-(sin(1/2*d*x+1/2*c) ^2)^(1/2)*EllipticE(cos(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*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 = 0.14 (sec) , antiderivative size = 426, normalized size of antiderivative = 1.80 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (-i \, A + 2 i \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (-i \, A + 2 i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A + 2 i \, B\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (i \, A - 2 i \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (i \, A - 2 i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A - 2 i \, B\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (4 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (4 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (4 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )\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 \, {\left (\sqrt {2} {\left (-4 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (-4 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-4 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )\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 \, {\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (19 \, A - 32 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - 4 \, B\right )} \cos \left (d x + c\right ) + 2 \, B\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]
-1/6*(5*(sqrt(2)*(-I*A + 2*I*B)*cos(d*x + c)^3 + 2*sqrt(2)*(-I*A + 2*I*B)* cos(d*x + c)^2 + sqrt(2)*(-I*A + 2*I*B)*cos(d*x + c))*weierstrassPInverse( -4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(I*A - 2*I*B)*cos(d*x + c)^3 + 2*sqrt(2)*(I*A - 2*I*B)*cos(d*x + c)^2 + sqrt(2)*(I*A - 2*I*B)*cos (d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*( sqrt(2)*(4*I*A - 7*I*B)*cos(d*x + c)^3 + 2*sqrt(2)*(4*I*A - 7*I*B)*cos(d*x + c)^2 + sqrt(2)*(4*I*A - 7*I*B)*cos(d*x + c))*weierstrassZeta(-4, 0, wei erstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(-4*I *A + 7*I*B)*cos(d*x + c)^3 + 2*sqrt(2)*(-4*I*A + 7*I*B)*cos(d*x + c)^2 + s qrt(2)*(-4*I*A + 7*I*B)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPI nverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*(4*A - 7*B)*cos(d*x + c)^3 + (19*A - 32*B)*cos(d*x + c)^2 + 2*(3*A - 4*B)*cos(d*x + c) + 2*B)*s in(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]