Integrand size = 33, antiderivative size = 197 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {3 (7 A-5 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}-\frac {5 (A-B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {(7 A-5 B) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (A-B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \] Output:
3/5*(7*A-5*B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d *x+c)^(1/2)/a/d-5/3*(A-B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2 ^(1/2))*sec(d*x+c)^(1/2)/a/d+1/5*(7*A-5*B)*sin(d*x+c)/a/d/sec(d*x+c)^(3/2) -5/3*(A-B)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)-(A-B)*sin(d*x+c)/d/sec(d*x+c)^( 3/2)/(a+a*sec(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.27 (sec) , antiderivative size = 540, normalized size of antiderivative = 2.74 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) (A+B \sec (c+d x)) \left (-84 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+60 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-200 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+200 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+\sqrt {\sec (c+d x)} \left (-3 (51 A-40 B+(33 A-20 B) \cos (2 c)) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )-40 (A-B) \cos (2 d x) \sin (2 c)+12 A \cos (3 d x) \sin (3 c)+120 (A-B) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )+12 (33 A-20 B) \cos (c) \sin (d x)-40 (A-B) \cos (2 c) \sin (2 d x)+12 A \cos (3 c) \sin (3 d x)+120 (A-B) \tan \left (\frac {c}{2}\right )\right )\right )}{60 a d (B+A \cos (c+d x)) (1+\sec (c+d x))} \] Input:
Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x ]
Output:
(Cos[(c + d*x)/2]^2*(A + B*Sec[c + d*x])*((-84*Sqrt[2]*A*Sqrt[E^(I*(c + d* x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sq rt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeomet ric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (60*Sqrt[2]*B*Sq rt[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) - 2 00*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + 200 *B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + Sqrt[ Sec[c + d*x]]*(-3*(51*A - 40*B + (33*A - 20*B)*Cos[2*c])*Cos[d*x]*Csc[c/2] *Sec[c/2] - 40*(A - B)*Cos[2*d*x]*Sin[2*c] + 12*A*Cos[3*d*x]*Sin[3*c] + 12 0*(A - B)*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x)/2] + 12*(33*A - 20*B)*Cos[c] *Sin[d*x] - 40*(A - B)*Cos[2*c]*Sin[2*d*x] + 12*A*Cos[3*c]*Sin[3*d*x] + 12 0*(A - B)*Tan[c/2])))/(60*a*d*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x]))
Time = 0.86 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.95, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4508, 27, 3042, 4274, 3042, 4256, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \frac {a (7 A-5 B)-5 a (A-B) \sec (c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (7 A-5 B)-5 a (A-B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (7 A-5 B)-5 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {a (7 A-5 B) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)}dx-5 a (A-B) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (7 A-5 B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-5 a (A-B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {a (7 A-5 B) \left (\frac {3}{5} \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-5 a (A-B) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (7 A-5 B) \left (\frac {3}{5} \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-5 a (A-B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {a (7 A-5 B) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-5 a (A-B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (7 A-5 B) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-5 a (A-B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {a (7 A-5 B) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )-5 a (A-B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {a (7 A-5 B) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )-5 a (A-B) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )}{2 a^2}-\frac {(A-B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
Input:
Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x]
Output:
-(((A - B)*Sin[c + d*x])/(d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x]))) + (a *(7*A - 5*B)*((6*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))) - 5*a*(A - B)*( (2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])))/(2*a^2)
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[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(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] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Time = 5.30 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.43
| method | result | size |
| default | \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (25 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+63 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-25 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-45 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+48 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-56 A -40 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-30 A +90 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (23 A -35 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{15 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(282\) |
Input:
int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x,method=_RETURNVER BOSE)
Output:
-1/15*((2*cos(1/2*d*x+1/2*c)^2-1)*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)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(25 *A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+63*A*EllipticE(cos(1/2*d*x+1/2*c) ,2^(1/2))-25*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-45*B*EllipticE(cos(1/ 2*d*x+1/2*c),2^(1/2)))+48*A*sin(1/2*d*x+1/2*c)^8+(-56*A-40*B)*sin(1/2*d*x+ 1/2*c)^6+(-30*A+90*B)*sin(1/2*d*x+1/2*c)^4+(23*A-35*B)*sin(1/2*d*x+1/2*c)^ 2)/a/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)/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 = 0.10 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.41 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=-\frac {25 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-7 i \, A + 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A + 5 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (7 i \, A - 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A - 5 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (6 \, A \cos \left (d x + c\right )^{3} - 2 \, {\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} - 25 \, {\left (A - B\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm= "fricas")
Output:
-1/30*(25*(sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*(-I*A + I*B))*weier strassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 25*(sqrt(2)*(I*A - I*B)*cos(d*x + c) + sqrt(2)*(I*A - I*B))*weierstrassPInverse(-4, 0, cos(d* x + c) - I*sin(d*x + c)) + 9*(sqrt(2)*(-7*I*A + 5*I*B)*cos(d*x + c) + sqrt (2)*(-7*I*A + 5*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co s(d*x + c) + I*sin(d*x + c))) + 9*(sqrt(2)*(7*I*A - 5*I*B)*cos(d*x + c) + sqrt(2)*(7*I*A - 5*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(6*A*cos(d*x + c)^3 - 2*(2*A - 5*B)*c os(d*x + c)^2 - 25*(A - B)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/ (a*d*cos(d*x + c) + a*d)
\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\int \frac {A}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx}{a} \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c)),x)
Output:
(Integral(A/(sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x) + Integral(B*s ec(c + d*x)/(sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x))/a
\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm= "maxima")
Output:
integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*sec(d*x + c)^(5/2)), x)
\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm= "giac")
Output:
integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*sec(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))*(1/cos(c + d*x))^(5/2)),x)
Output:
int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))*(1/cos(c + d*x))^(5/2)), x)
\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4}+\sec \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3}+\sec \left (d x +c \right )^{2}}d x \right ) b}{a} \] Input:
int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x)
Output:
(int(sqrt(sec(c + d*x))/(sec(c + d*x)**4 + sec(c + d*x)**3),x)*a + int(sqr t(sec(c + d*x))/(sec(c + d*x)**3 + sec(c + d*x)**2),x)*b)/a