Integrand size = 31, antiderivative size = 199 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {6 a (A+B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (7 A+5 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {6 a (A+B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (7 A+5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 a (A+B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \] Output:
-6/5*a*(A+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+2/21*a*(7*A+5*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+6/5*a*(A+B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d+2 /21*a*(7*A+5*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a*(A+B)*sec(d*x+c)^(5/2) *sin(d*x+c)/d+2/7*a*B*sec(d*x+c)^(7/2)*sin(d*x+c)/d
Time = 2.04 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.01 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x)) (A+B \sec (c+d x)) \left (-63 (A+B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 (7 A+5 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+63 A \sin (c+d x)+63 B \sin (c+d x)+35 A \tan (c+d x)+25 B \tan (c+d x)+21 A \sec (c+d x) \tan (c+d x)+21 B \sec (c+d x) \tan (c+d x)+15 B \sec ^2(c+d x) \tan (c+d x)\right )}{105 d (B+A \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \] Input:
Integrate[Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*Sec[(c + d*x)/2]^2*(1 + Sec[c + d*x])*(A + B*Sec[c + d*x])*(-63*(A + B) *Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 5*(7*A + 5*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 63*A*Sin[c + d*x] + 63*B*Sin[c + d*x] + 35*A*Tan[c + d*x] + 25*B*Tan[c + d*x] + 21*A*Sec[c + d*x]*Tan[c + d*x] + 21*B*Sec[c + d*x]*Tan[c + d*x] + 15*B*Sec[c + d*x]^2*Tan[c + d*x]))/(105*d *(B + A*Cos[c + d*x])*Sec[c + d*x]^(3/2))
Time = 0.90 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.97, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4485, 27, 3042, 4274, 3042, 4255, 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.
\(\displaystyle \int \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a) (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4485 |
\(\displaystyle \frac {2}{7} \int \frac {1}{2} \sec ^{\frac {5}{2}}(c+d x) (a (7 A+5 B)+7 a (A+B) \sec (c+d x))dx+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \int \sec ^{\frac {5}{2}}(c+d x) (a (7 A+5 B)+7 a (A+B) \sec (c+d x))dx+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (7 A+5 B)+7 a (A+B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {1}{7} \left (a (7 A+5 B) \int \sec ^{\frac {5}{2}}(c+d x)dx+7 a (A+B) \int \sec ^{\frac {7}{2}}(c+d x)dx\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (a (7 A+5 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx+7 a (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}dx\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{7} \left (a (7 A+5 B) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+7 a (A+B) \left (\frac {3}{5} \int \sec ^{\frac {3}{2}}(c+d x)dx+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (a (7 A+5 B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+7 a (A+B) \left (\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{7} \left (7 a (A+B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+a (7 A+5 B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (7 a (A+B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )+a (7 A+5 B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{7} \left (a (7 A+5 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+7 a (A+B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (a (7 A+5 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+7 a (A+B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{7} \left (a (7 A+5 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+7 a (A+B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{7} \left (a (7 A+5 B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )+7 a (A+B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}\) |
Input:
Int[Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(2*a*B*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d) + (a*(7*A + 5*B)*((2*Sqrt[Co s[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d)) + 7*a*(A + B)*((2*Sec[c + d*x]^(5/2)*Si n[c + d*x])/(5*d) + (3*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*S qrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/5))/7
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]
Leaf count of result is larger than twice the leaf count of optimal. \(690\) vs. \(2(174)=348\).
Time = 10.78 (sec) , antiderivative size = 691, normalized size of antiderivative = 3.47
method | result | size |
default | \(\text {Expression too large to display}\) | \(691\) |
parts | \(\text {Expression too large to display}\) | \(824\) |
Input:
int(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVER BOSE)
Output:
-a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(-1/6*co s(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)*Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*B*(-1/56*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)^4- 5/42*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+5/21*(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)))+4/5*(1/2*A+1/2*B)/sin(1/2*d*x+ 1/2*c)^2/(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)*(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))/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 = 0.11 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.17 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {-5 i \, \sqrt {2} {\left (7 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (7 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 63 i \, \sqrt {2} {\left (A + B\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 63 i \, \sqrt {2} {\left (A + B\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (63 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 5 \, {\left (7 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{2} + 21 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 15 \, B a\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm= "fricas")
Output:
1/105*(-5*I*sqrt(2)*(7*A + 5*B)*a*cos(d*x + c)^3*weierstrassPInverse(-4, 0 , cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*(7*A + 5*B)*a*cos(d*x + c)^ 3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 63*I*sqrt(2) *(A + B)*a*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0 , cos(d*x + c) + I*sin(d*x + c))) + 63*I*sqrt(2)*(A + B)*a*cos(d*x + c)^3* weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(63*(A + B)*a*cos(d*x + c)^3 + 5*(7*A + 5*B)*a*cos(d*x + c)^2 + 21*(A + B)*a*cos(d*x + c) + 15*B*a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d* cos(d*x + c)^3)
Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(5/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
Timed out
\[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))*(A+B*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)
\[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))*(A+B*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. \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] 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)
\[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:
int(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
a*(int(sqrt(sec(c + d*x))*sec(c + d*x)**4,x)*b + int(sqrt(sec(c + d*x))*se c(c + d*x)**3,x)*a + int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*b + int(sqr t(sec(c + d*x))*sec(c + d*x)**2,x)*a)