Integrand size = 33, antiderivative size = 158 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 a^2 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {4 a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 a^2 (A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \] Output:
4*a^2*A*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+4/3*a^2*(2*A+3*B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2 ^(1/2))*sec(d*x+c)^(1/2)/d-2/3*a^2*(A-3*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d+2 /3*A*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.92 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.16 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^2 \left (\cos \left (\frac {c}{2}\right )-i \sin \left (\frac {c}{2}\right )\right ) \left (-i \cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (12 A-\frac {24 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+\frac {8 (2 A+3 B) e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+2 i A \sin (c+d x)+6 i B \tan (c+d x)\right )}{3 d \sqrt {\sec (c+d x)}} \] Input:
Integrate[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(3/2) ,x]
Output:
(a^2*(Cos[c/2] - I*Sin[c/2])*((-I)*Cos[c/2] + Sin[c/2])*(12*A - (24*A*Hype rgeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + (8*(2*A + 3*B)*E^(I*(c + d*x))*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + (2*I)*A*Sin[c + d* x] + (6*I)*B*Tan[c + d*x]))/(3*d*Sqrt[Sec[c + d*x]])
Time = 0.90 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4505, 27, 3042, 4485, 3042, 4274, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {2}{3} \int \frac {(\sec (c+d x) a+a) (a (5 A+3 B)-a (A-3 B) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {(\sec (c+d x) a+a) (a (5 A+3 B)-a (A-3 B) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (5 A+3 B)-a (A-3 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {1}{3} \left (2 \int \frac {3 A a^2+(2 A+3 B) \sec (c+d x) a^2}{\sqrt {\sec (c+d x)}}dx-\frac {2 a^2 (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (2 \int \frac {3 A a^2+(2 A+3 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^2 (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{3} \left (2 \left (a^2 (2 A+3 B) \int \sqrt {\sec (c+d x)}dx+3 a^2 A \int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )-\frac {2 a^2 (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (2 \left (a^2 (2 A+3 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+3 a^2 A \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 a^2 (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{3} \left (2 \left (a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )-\frac {2 a^2 (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (2 \left (a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 a^2 (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{3} \left (2 \left (a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^2 (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{3} \left (2 \left (\frac {2 a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^2 (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}\) |
Input:
Int[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(3/2),x]
Output:
(2*A*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2* ((6*a^2*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]) /d + (2*a^2*(2*A + 3*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[ Sec[c + d*x]])/d) - (2*a^2*(A - 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d)/3
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_)), 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_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
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[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + 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 ] && GtQ[m, 1/2] && LtQ[n, -1]
Time = 4.19 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.55
| method | result | size |
| default | \(-\frac {4 a^{2} \left (2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}-3 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}-3 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(245\) |
| parts | \(-\frac {2 \left (A \,a^{2}+2 B \,a^{2}\right ) \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}}\, \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 )}{\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}+\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \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}}\, \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 )}{\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}-\frac {2 A \,a^{2} \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 (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{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 )\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}-\frac {2 B \,a^{2} \left (-2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{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}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\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}\) | \(657\) |
Input:
int((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x,method=_RETURNV ERBOSE)
Output:
-4/3*a^2*(2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-A*sin(1/2*d*x+1/2*c) ^2*cos(1/2*d*x+1/2*c)+2*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d *x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)-3*A*EllipticE(cos(1/2* d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1 )^(1/2)-3*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+3*B*EllipticF(cos(1/2* d*x+1/2*c),2^(1/2))*(sin(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 complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.05 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (i \, \sqrt {2} {\left (2 \, A + 3 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} {\left (2 \, A + 3 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} A a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} A a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (A a^{2} \cos \left (d x + c\right ) + 3 \, B a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, d} \] Input:
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x, algorith m="fricas")
Output:
-2/3*(I*sqrt(2)*(2*A + 3*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - I*sqrt(2)*(2*A + 3*B)*a^2*weierstrassPInverse(-4, 0, cos (d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*A*a^2*weierstrassZeta(-4, 0, wei erstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*A*a^ 2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d *x + c))) - (A*a^2*cos(d*x + c) + 3*B*a^2)*sin(d*x + c)/sqrt(cos(d*x + c)) )/d
\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=a^{2} \left (\int \frac {A}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 A}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int A \sqrt {\sec {\left (c + d x \right )}}\, dx + \int \frac {B}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int 2 B \sqrt {\sec {\left (c + d x \right )}}\, dx + \int B \sec ^{\frac {3}{2}}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sec(d*x+c))**2*(A+B*sec(d*x+c))/sec(d*x+c)**(3/2),x)
Output:
a**2*(Integral(A/sec(c + d*x)**(3/2), x) + Integral(2*A/sqrt(sec(c + d*x)) , x) + Integral(A*sqrt(sec(c + d*x)), x) + Integral(B/sqrt(sec(c + d*x)), x) + Integral(2*B*sqrt(sec(c + d*x)), x) + Integral(B*sec(c + d*x)**(3/2), x))
\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x, algorith m="maxima")
Output:
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2/sec(d*x + c)^(3/2), x)
\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x, algorith m="giac")
Output:
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2/sec(d*x + c)^(3/2), x)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^2)/(1/cos(c + d*x))^(3/2),x )
Output:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^2)/(1/cos(c + d*x))^(3/2), x)
\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=a^{2} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) b \right ) \] Input:
int((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x)
Output:
a**2*(int(sqrt(sec(c + d*x))/sec(c + d*x)**2,x)*a + 2*int(sqrt(sec(c + d*x ))/sec(c + d*x),x)*a + int(sqrt(sec(c + d*x))/sec(c + d*x),x)*b + int(sqrt (sec(c + d*x)),x)*a + 2*int(sqrt(sec(c + d*x)),x)*b + int(sqrt(sec(c + d*x ))*sec(c + d*x),x)*b)