Integrand size = 43, antiderivative size = 174 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=-\frac {(3 A-3 B+C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(5 A-3 B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {(5 A-3 B+3 C) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \] Output:
-(3*A-3*B+C)*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+1/3*(5*A-3*B+3*C)*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+1/3*(5*A-3*B+3*C)*sin(d*x+c)/a/d/sec(d *x+c)^(1/2)-(A-B+C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.92 (sec) , antiderivative size = 1287, normalized size of antiderivative = 7.40 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx =\text {Too large to display} \] Input:
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])),x]
Output:
(Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2* I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*(-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))])*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])* (a + a*Sec[c + d*x])) - (Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*C sc[c/2]*(-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))])*Sec[c/2]*(A + B *Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x ] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + ( d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*(-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))])*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)* (A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (10*A*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x) /2, 2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + a*Sec [c + d*x])) - (2*B*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*Ell...
Time = 0.86 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.96, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 4572, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {a (5 A-3 B+3 C)-a (3 A-3 B+C) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (5 A-3 B+3 C)-a (3 A-3 B+C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (5 A-3 B+3 C)-a (3 A-3 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {a (5 A-3 B+3 C) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)}dx-a (3 A-3 B+C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (5 A-3 B+3 C) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx-a (3 A-3 B+C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {a (5 A-3 B+3 C) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a (3 A-3 B+C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (5 A-3 B+3 C) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a (3 A-3 B+C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {a (5 A-3 B+3 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a (3 A-3 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (5 A-3 B+3 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a (3 A-3 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {a (5 A-3 B+3 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-\frac {2 a (3 A-3 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {a (5 A-3 B+3 C) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )-\frac {2 a (3 A-3 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}\) |
Input:
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + a*Sec [c + d*x])),x]
Output:
-(((A - B + C)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x]))) + ((-2*a*(3*A - 3*B + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt [Sec[c + d*x]])/d + a*(5*A - 3*B + 3*C)*((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[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(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*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 5.22 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.72
| method | result | size |
| default | \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (5 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-8 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (18 A -6 B +6 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-7 A +3 B -3 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(300\) |
Input:
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x,me thod=_RETURNVERBOSE)
Output:
-1/3*((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)*(5*A* EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*A*EllipticE(cos(1/2*d*x+1/2*c),2^( 1/2))-3*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*B*EllipticE(cos(1/2*d*x+ 1/2*c),2^(1/2))+3*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(co s(1/2*d*x+1/2*c),2^(1/2)))-8*A*sin(1/2*d*x+1/2*c)^6+(18*A-6*B+6*C)*sin(1/2 *d*x+1/2*c)^4+(-7*A+3*B-3*C)*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*c os(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 = 287, normalized size of antiderivative = 1.65 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {{\left (\sqrt {2} {\left (-5 i \, A + 3 i \, B - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 3 i \, B - 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (5 i \, A - 3 i \, B + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 3 i \, B + 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (3 i \, A - 3 i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - 3 i \, B + i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-3 i \, A + 3 i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + 3 i \, B - i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (2 \, A \cos \left (d x + c\right )^{2} + {\left (5 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c) ),x, algorithm="fricas")
Output:
1/6*((sqrt(2)*(-5*I*A + 3*I*B - 3*I*C)*cos(d*x + c) + sqrt(2)*(-5*I*A + 3* I*B - 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(5*I*A - 3*I*B + 3*I*C)*cos(d*x + c) + sqrt(2)*(5*I*A - 3*I*B + 3 *I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sqrt (2)*(3*I*A - 3*I*B + I*C)*cos(d*x + c) + sqrt(2)*(3*I*A - 3*I*B + I*C))*we ierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-3*I*A + 3*I*B - I*C)*cos(d*x + c) + sqrt(2)*(-3*I*A + 3*I*B - I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(2*A*cos(d*x + c)^2 + (5*A - 3*B + 3*C)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\int \frac {A}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a} \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(3/2)/(a+a*sec(d*x+ c)),x)
Output:
(Integral(A/(sec(c + d*x)**(5/2) + sec(c + d*x)**(3/2)), x) + Integral(B*s ec(c + d*x)/(sec(c + d*x)**(5/2) + sec(c + d*x)**(3/2)), x) + Integral(C*s ec(c + d*x)**2/(sec(c + d*x)**(5/2) + sec(c + d*x)**(3/2)), x))/a
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c) ),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*se c(d*x + c)^(3/2)), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c) ),x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*se c(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))*(1/cos(c + d*x))^(3/2)),x)
Output:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))*(1/cos(c + d*x))^(3/2)), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3}+\sec \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )+1}d x \right ) c}{a} \] Input:
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x)
Output:
(int(sqrt(sec(c + d*x))/(sec(c + d*x)**3 + sec(c + d*x)**2),x)*a + int(sqr t(sec(c + d*x))/(sec(c + d*x)**2 + sec(c + d*x)),x)*b + int(sqrt(sec(c + d *x))/(sec(c + d*x) + 1),x)*c)/a