Integrand size = 33, antiderivative size = 199 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {4 a^2 (5 A+4 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (2 A+B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^2 (5 A+4 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (5 A+7 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d} \] Output:
-4/5*a^2*(5*A+4*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)/d+4/3*a^2*(2*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)/d+4/5*a^2*(5*A+4*B)*sec(d*x+c)^(1/2)*sin (d*x+c)/d+2/15*a^2*(5*A+7*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*B*sec(d*x+c )^(3/2)*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.88 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.61 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {a^2 e^{-i c} \left (-1+e^{2 i c}\right ) \csc (c) \left (5 A+10 B-30 A e^{i (c+d x)}-18 B e^{i (c+d x)}-60 A e^{3 i (c+d x)}-54 B e^{3 i (c+d x)}-5 A e^{4 i (c+d x)}-10 B e^{4 i (c+d x)}-30 A e^{5 i (c+d x)}-24 B e^{5 i (c+d x)}-10 i (2 A+B) \left (1+e^{2 i (c+d x)}\right )^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 (5 A+4 B) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^2 (A+B \sec (c+d x))}{60 d \left (1+e^{2 i (c+d x)}\right )^2 (B+A \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \] Input:
Integrate[Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x ]
Output:
(a^2*(-1 + E^((2*I)*c))*Csc[c]*(5*A + 10*B - 30*A*E^(I*(c + d*x)) - 18*B*E ^(I*(c + d*x)) - 60*A*E^((3*I)*(c + d*x)) - 54*B*E^((3*I)*(c + d*x)) - 5*A *E^((4*I)*(c + d*x)) - 10*B*E^((4*I)*(c + d*x)) - 30*A*E^((5*I)*(c + d*x)) - 24*B*E^((5*I)*(c + d*x)) - (10*I)*(2*A + B)*(1 + E^((2*I)*(c + d*x)))^2 *Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(5*A + 4*B)*E^(I*(c + d* x))*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^(( 2*I)*(c + d*x))])*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + B*Sec[c + d *x]))/(60*d*E^(I*c)*(1 + E^((2*I)*(c + d*x)))^2*(B + A*Cos[c + d*x])*Sec[c + d*x]^(5/2))
Time = 1.03 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4506, 27, 3042, 4485, 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.
| \(\displaystyle \int \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2 (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\sec (c+d x)} (\sec (c+d x) a+a) (a (5 A+B)+a (5 A+7 B) \sec (c+d x))dx+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\sec (c+d x)} (\sec (c+d x) a+a) (a (5 A+B)+a (5 A+7 B) \sec (c+d x))dx+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (5 A+B)+a (5 A+7 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \sqrt {\sec (c+d x)} \left (5 (2 A+B) a^2+3 (5 A+4 B) \sec (c+d x) a^2\right )dx+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 (2 A+B) a^2+3 (5 A+4 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (3 a^2 (5 A+4 B) \int \sec ^{\frac {3}{2}}(c+d x)dx+5 a^2 (2 A+B) \int \sqrt {\sec (c+d x)}dx\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^2 (2 A+B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+3 a^2 (5 A+4 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^2 (2 A+B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+3 a^2 (5 A+4 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^2 (2 A+B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+3 a^2 (5 A+4 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^2 (2 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^2 (5 A+4 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^2 (2 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^2 (5 A+4 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^2 (2 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^2 (5 A+4 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2}{3} \left (\frac {10 a^2 (2 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+3 a^2 (5 A+4 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
Input:
Int[Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
Output:
(2*B*Sec[c + d*x]^(3/2)*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(5*d) + ((2 *a^2*(5*A + 7*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + (2*((10*a^2*(2*A + B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + 3*a^2*(5*A + 4*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)))/3)/5
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_))*(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[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*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] && GtQ[m, 1/2] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(715\) vs. \(2(178)=356\).
Time = 4.92 (sec) , antiderivative size = 716, normalized size of antiderivative = 3.60
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(716\) |
| parts | \(\text {Expression too large to display}\) | \(916\) |
Input:
int(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x,method=_RETURNV ERBOSE)
Output:
-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*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)/(-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))+ 2/5*B/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c) ^2-1)/sin(1/2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12* 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)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1 /2*d*x+1/2*c)^4+12*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)*sin(1/2*d*x+1/2*c)^2+8*cos(1/ 2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*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))*(-2*sin(1/2 *d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+8*(1/4*A+1/2*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(c os(1/2*d*x+1/2*c),2^(1/2)))+8*(1/2*A+1/4*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-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)))/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 = 239, normalized size of antiderivative = 1.20 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (6 \, {\left (5 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, B a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorith m="fricas")
Output:
-2/15*(5*I*sqrt(2)*(2*A + B)*a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(2*A + B)*a^2*cos(d*x + c)^2 *weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)*( 5*A + 4*B)*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(- 4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(5*A + 4*B)*a^2*cos(d* x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (6*(5*A + 4*B)*a^2*cos(d*x + c)^2 + 5*(A + 2*B)*a^2*cos (d*x + c) + 3*B*a^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(1/2)*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)
Output:
Timed out
\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:
integrate(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorith m="maxima")
Output:
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*sqrt(sec(d*x + c)), x)
\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:
integrate(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorith m="giac")
Output:
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*sqrt(sec(d*x + c)), x)
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \] Input:
int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(1/2),x)
Output:
int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(1/2), x)
\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=a^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) b \right ) \] Input:
int(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)
Output:
a**2*(int(sqrt(sec(c + d*x)),x)*a + int(sqrt(sec(c + d*x))*sec(c + d*x)**3 ,x)*b + int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*a + 2*int(sqrt(sec(c + d *x))*sec(c + d*x)**2,x)*b + 2*int(sqrt(sec(c + d*x))*sec(c + d*x),x)*a + i nt(sqrt(sec(c + d*x))*sec(c + d*x),x)*b)