Integrand size = 43, antiderivative size = 133 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2} \, dx=-\frac {(A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(2 A+B+2 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {(A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:
-(A-C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d+1/3*(2*A+B+2*C)*Inverse JacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^2/d+(A-C)*cos(d*x+c)^(1/2)*sin(d*x+c)/a^ 2/d/(1+cos(d*x+c))-1/3*(A-B+C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+ c))^2
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.62 (sec) , antiderivative size = 1384, normalized size of antiderivative = 10.41 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^2),x]
Output:
(-8*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^ 2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sq rt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - Arc Tan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt [1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2) - (4*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2 ]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/ 2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[ 1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B *Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x] )^2) - (8*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5 /4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqr t[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x] )*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[ Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((8*(A - C)*Csc[c])/ d + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] - C*Sin[(d*x)/2]))/d - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*S...
Time = 0.98 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.302, Rules used = {3042, 4600, 3042, 3520, 27, 3042, 3457, 25, 3042, 3227, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 4600 |
| \(\displaystyle \int \frac {A \cos ^2(c+d x)+B \cos (c+d x)+C}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int -\frac {a (A-B-5 C)-a (5 A+B-C) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {a (A-B-5 C)-a (5 A+B-C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {a (A-B-5 C)-a (5 A+B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle -\frac {\frac {\int -\frac {a^2 (2 A+B+2 C)-3 a^2 (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-\frac {\int \frac {a^2 (2 A+B+2 C)-3 a^2 (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {a^2 (2 A+B+2 C)-3 a^2 (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {-\frac {a^2 (2 A+B+2 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^2 (A-C) \int \sqrt {\cos (c+d x)}dx}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {a^2 (2 A+B+2 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^2 (A-C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {-\frac {a^2 (2 A+B+2 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {-\frac {\frac {2 a^2 (2 A+B+2 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
Input:
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Sec [c + d*x])^2),x]
Output:
-1/3*((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) ^2) - (-(((-6*a^2*(A - C)*EllipticE[(c + d*x)/2, 2])/d + (2*a^2*(2*A + B + 2*C)*EllipticF[(c + d*x)/2, 2])/d)/a^2) - (6*(A - C)*Sqrt[Cos[c + d*x]]*S in[c + d*x])/(d*(1 + Cos[c + d*x])))/(6*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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-1)]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x _)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) *(x_)]^2), x_Symbol] :> Simp[d^(m + 2) Int[(b + a*Cos[e + f*x])^m*(d*Cos[ e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr eeQ[{a, b, d, e, f, A, B, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(508\) vs. \(2(128)=256\).
Time = 3.69 (sec) , antiderivative size = 509, normalized size of antiderivative = 3.83
| 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 (12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+4 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+6 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 )+2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 )-12 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 )-20 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+16 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+9 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A +B -C \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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}\) | \(509\) |
Input:
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^2,x, method=_RETURNVERBOSE)
Output:
-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/2 *d*x+1/2*c)^6+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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+6*A*cos( 1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*B*cos(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/ 2*d*x+1/2*c),2^(1/2))-12*C*cos(1/2*d*x+1/2*c)^6+4*C*cos(1/2*d*x+1/2*c)^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)*EllipticF(co s(1/2*d*x+1/2*c),2^(1/2))-6*C*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/ 2))-20*A*cos(1/2*d*x+1/2*c)^4+2*B*cos(1/2*d*x+1/2*c)^4+16*C*cos(1/2*d*x+1/ 2*c)^4+9*A*cos(1/2*d*x+1/2*c)^2-3*B*cos(1/2*d*x+1/2*c)^2-3*C*cos(1/2*d*x+1 /2*c)^2-A+B-C)/a^2/cos(1/2*d*x+1/2*c)^3/(-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 = 0.09 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.79 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {2 \, {\left (3 \, {\left (A - C\right )} \cos \left (d x + c\right ) + 2 \, A + B - 4 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-2 i \, A - i \, B - 2 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (2 i \, A + i \, B + 2 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-2 i \, A - i \, B - 2 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (2 i \, A + i \, B + 2 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-2 i \, A - i \, B - 2 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (2 i \, A + i \, B + 2 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c) )^2,x, algorithm="fricas")
Output:
1/6*(2*(3*(A - C)*cos(d*x + c) + 2*A + B - 4*C)*sqrt(cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(-2*I*A - I*B - 2*I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(2*I*A + I*B + 2*I*C)*cos(d*x + c) + sqrt(2)*(-2*I*A - I*B - 2*I*C))*weierstrassP Inverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(2*I*A + I*B + 2* I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(-2*I*A - I*B - 2*I*C)*cos(d*x + c) + sqrt (2)*(2*I*A + I*B + 2*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin (d*x + c)) - 3*(sqrt(2)*(I*A - I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(I*A - I*C) *cos(d*x + c) + sqrt(2)*(I*A - I*C))*weierstrassZeta(-4, 0, weierstrassPIn verse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-I*A + I*C)*cos (d*x + c)^2 + 2*sqrt(2)*(-I*A + I*C)*cos(d*x + c) + sqrt(2)*(-I*A + I*C))* weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A}{\sqrt {\cos {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )} + 2 \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )} + \sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )} + 2 \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )} + \sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )} + 2 \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )} + \sqrt {\cos {\left (c + d x \right )}}}\, dx}{a^{2}} \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(1/2)/(a+a*sec(d*x+ c))**2,x)
Output:
(Integral(A/(sqrt(cos(c + d*x))*sec(c + d*x)**2 + 2*sqrt(cos(c + d*x))*sec (c + d*x) + sqrt(cos(c + d*x))), x) + Integral(B*sec(c + d*x)/(sqrt(cos(c + d*x))*sec(c + d*x)**2 + 2*sqrt(cos(c + d*x))*sec(c + d*x) + sqrt(cos(c + d*x))), x) + Integral(C*sec(c + d*x)**2/(sqrt(cos(c + d*x))*sec(c + d*x)* *2 + 2*sqrt(cos(c + d*x))*sec(c + d*x) + sqrt(cos(c + d*x))), x))/a**2
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c) )^2,x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2* sqrt(cos(d*x + c))), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c) )^2,x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2* sqrt(cos(d*x + c))), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + a/cos (c + d*x))^2),x)
Output:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + a/cos (c + d*x))^2), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )+\cos \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )+\cos \left (d x +c \right )}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )+\cos \left (d x +c \right )}d x \right ) b}{a^{2}} \] Input:
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^2,x)
Output:
(int(sqrt(cos(c + d*x))/(cos(c + d*x)*sec(c + d*x)**2 + 2*cos(c + d*x)*sec (c + d*x) + cos(c + d*x)),x)*a + int((sqrt(cos(c + d*x))*sec(c + d*x)**2)/ (cos(c + d*x)*sec(c + d*x)**2 + 2*cos(c + d*x)*sec(c + d*x) + cos(c + d*x) ),x)*c + int((sqrt(cos(c + d*x))*sec(c + d*x))/(cos(c + d*x)*sec(c + d*x)* *2 + 2*cos(c + d*x)*sec(c + d*x) + cos(c + d*x)),x)*b)/a**2