Integrand size = 35, antiderivative size = 172 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a^{5/2} B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 (32 A+35 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (8 A+5 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \] Output:
2*a^(5/2)*B*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/15*a^3* (32*A+35*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/15*a^2* (8*A+5*B)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*a*A*(a+ a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(3/2)
Time = 1.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.74 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^3 \left ((89 A+80 B+2 (14 A+5 B) \cos (c+d x)+3 A \cos (2 (c+d x))) \sqrt {1-\sec (c+d x)}+30 B \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{15 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^( 5/2),x]
Output:
(a^3*((89*A + 80*B + 2*(14*A + 5*B)*Cos[c + d*x] + 3*A*Cos[2*(c + d*x)])*S qrt[1 - Sec[c + d*x]] + 30*B*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Sqrt[Sec[c + d *x]])*Tan[c + d*x])/(15*d*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*Sqrt[a *(1 + Sec[c + d*x])])
Time = 1.03 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4505, 27, 3042, 4505, 27, 3042, 4503, 3042, 4288, 222}
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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {2}{5} \int \frac {(\sec (c+d x) a+a)^{3/2} (a (8 A+5 B)+5 a B \sec (c+d x))}{2 \sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(\sec (c+d x) a+a)^{3/2} (a (8 A+5 B)+5 a B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (8 A+5 B)+5 a B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((32 A+35 B) a^2+15 B \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((32 A+35 B) a^2+15 B \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((32 A+35 B) a^2+15 B \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a^2 B \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^3 (32 A+35 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a^2 B \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^3 (32 A+35 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 a^3 (32 A+35 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {30 a^2 B \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 a^2 (8 A+5 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \left (\frac {30 a^{5/2} B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (32 A+35 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x ]
Output:
(2*a*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((2*a^2*(8*A + 5*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Sec[ c + d*x]]) + ((30*a^(5/2)*B*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[ c + d*x]]])/d + (2*a^3*(32*A + 35*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*S qrt[a + a*Sec[c + d*x]]))/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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a *B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
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 = 4.30 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.22
| method | result | size |
| default | \(\frac {a^{2} \left (\left (-15 \cos \left (d x +c \right )-15\right ) \sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (-15 \cos \left (d x +c \right )-15\right ) \sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (12 \cos \left (d x +c \right )^{2}+56 \cos \left (d x +c \right )+172\right ) \sin \left (d x +c \right ) A +\left (20 \cos \left (d x +c \right )+160\right ) \sin \left (d x +c \right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{30 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}}\) | \(210\) |
| parts | \(\frac {A \left (6 \sin \left (d x +c \right )+28 \tan \left (d x +c \right )+86 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}-\frac {B \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-4 \sin \left (d x +c \right )-32 \tan \left (d x +c \right )+\sqrt {2}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (3+3 \sec \left (d x +c \right )\right )+\sqrt {2}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (3+3 \sec \left (d x +c \right )\right )\right )}{6 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(251\) |
Input:
int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x,method=_RET URNVERBOSE)
Output:
1/30/d*a^2*((-15*cos(d*x+c)-15)*2^(1/2)*B*(-2/(1+cos(d*x+c)))^(1/2)*arctan (1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(1+cos(d*x+c)))^(1/2))+(-15*cos(d*x+c)- 15)*2^(1/2)*B*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(cot(d*x+c)-csc(d*x+c)+ 1)/(-1/(1+cos(d*x+c)))^(1/2))+(12*cos(d*x+c)^2+56*cos(d*x+c)+172)*sin(d*x+ c)*A+(20*cos(d*x+c)+160)*sin(d*x+c)*B)*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x +c))/sec(d*x+c)^(1/2)
Time = 0.12 (sec) , antiderivative size = 421, normalized size of antiderivative = 2.45 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\left [\frac {15 \, {\left (B a^{2} \cos \left (d x + c\right ) + B a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (3 \, A a^{2} \cos \left (d x + c\right )^{3} + {\left (14 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (43 \, A + 40 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {15 \, {\left (B a^{2} \cos \left (d x + c\right ) + B a^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (3 \, A a^{2} \cos \left (d x + c\right )^{3} + {\left (14 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (43 \, A + 40 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algo rithm="fricas")
Output:
[1/30*(15*(B*a^2*cos(d*x + c) + B*a^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a *cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos( d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d* x + c)^3 + cos(d*x + c)^2)) + 4*(3*A*a^2*cos(d*x + c)^3 + (14*A + 5*B)*a^2 *cos(d*x + c)^2 + (43*A + 40*B)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a )/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d), 1/1 5*(15*(B*a^2*cos(d*x + c) + B*a^2)*sqrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2 *cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))/(a*sqrt(co s(d*x + c))*sin(d*x + c))) + 2*(3*A*a^2*cos(d*x + c)^3 + (14*A + 5*B)*a^2* cos(d*x + c)^2 + (43*A + 40*B)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a) /cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(5/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 655 vs. \(2 (148) = 296\).
Time = 0.27 (sec) , antiderivative size = 655, normalized size of antiderivative = 3.81 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algo rithm="maxima")
Output:
1/60*(5*sqrt(2)*(30*a^2*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 30*a^2*cos(3/2*d*x + 3/2*c)*sin(2/3*arct an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 3*sqrt(2)*a^2*log(2*cos (1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*ar ctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*a rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*ar ctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 3*sqrt(2)*a^2*lo g(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin (1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*co s(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin (1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 3*sqrt(2) *a^2*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqr t(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt (2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 3* sqrt(2)*a^2*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 4*a^2*sin(3/2*d*x + 3/2*c) + 30*a^2*sin(1/3*arctan2(sin(3/2*d*x + ...
Time = 1.87 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.60 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\frac {15 \, B a^{\frac {7}{2}} \log \left (\frac {{\left | 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} + \frac {2 \, {\left (60 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 45 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (80 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 80 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (32 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 35 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}}}}{15 \, d} \] Input:
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algo rithm="giac")
Output:
1/15*(15*B*a^(7/2)*log(abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/ 2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(a)*tan(1/2* d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2)*abs(a) - 6*a))*sgn(cos(d*x + c))/abs(a) + 2*(60*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 4 5*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + (80*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 80*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + (32*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 35*sqrt(2)*B*a^5*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2))/d
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(5/ 2),x)
Output:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(5/ 2), x)
\[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\sqrt {a}\, a^{2} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}d x \right ) b \right ) \] Input:
int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x)
Output:
sqrt(a)*a**2*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x) **3,x)*a + 2*int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)* *2,x)*a + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**2, x)*b + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x),x)*a + 2*int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x),x)*b + int (sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1),x)*b)