Integrand size = 27, antiderivative size = 133 \[ \int (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^{3/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 (5 A+4 C) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a C \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{5 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
2*a^(3/2)*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/5*C*(a+a *sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/5*a^2*(5*A+4*C)*tan(d*x+c)/d/(a+a*sec(d* x+c))^(1/2)+2/5*a*C*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 3.38 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92 \[ \int (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (10 A \arctan \left (\sqrt {-1+\sec (c+d x)}\right ) \cos ^2(c+d x)+(5 A+8 C+6 C \cos (c+d x)+(5 A+6 C) \cos (2 (c+d x))) \sqrt {-1+\sec (c+d x)}\right ) \sec ^2(c+d x) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{5 d \sqrt {-1+\sec (c+d x)}} \]
(a*(10*A*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Cos[c + d*x]^2 + (5*A + 8*C + 6*C *Cos[c + d*x] + (5*A + 6*C)*Cos[2*(c + d*x)])*Sqrt[-1 + Sec[c + d*x]])*Sec [c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(5*d*Sqrt[-1 + Se c[c + d*x]])
Time = 0.79 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4543, 27, 3042, 4405, 27, 3042, 4403, 3042, 4261, 216, 4279}
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.
| \(\displaystyle \int (a \sec (c+d x)+a)^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4543 |
| \(\displaystyle \frac {2 \int \frac {1}{2} (\sec (c+d x) a+a)^{3/2} (5 a A+3 a C \sec (c+d x))dx}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\sec (c+d x) a+a)^{3/2} (5 a A+3 a C \sec (c+d x))dx}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (5 a A+3 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {3}{2} \sqrt {\sec (c+d x) a+a} \left (5 A a^2+(5 A+4 C) \sec (c+d x) a^2\right )dx+\frac {2 a^2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {\sec (c+d x) a+a} \left (5 A a^2+(5 A+4 C) \sec (c+d x) a^2\right )dx+\frac {2 a^2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (5 A a^2+(5 A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 a^2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4403 |
| \(\displaystyle \frac {a^2 (5 A+4 C) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+5 a^2 A \int \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 (5 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+5 a^2 A \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {a^2 (5 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {10 a^3 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {a^2 (5 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {10 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {10 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (5 A+4 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{5 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\) |
(2*C*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + ((10*a^(5/2)*A*ArcTa n[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(5*A + 4*C) *Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/d)/(5*a)
3.2.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_ .) + (c_)), x_Symbol] :> Simp[c Int[Sqrt[a + b*Csc[e + f*x]], x], x] + Si mp[d Int[Sqrt[a + b*Csc[e + f*x]]*Csc[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Simp[1/m Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m + (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2 *m]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. ) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^ m/(f*(m + 1))), x] + Simp[1/(b*(m + 1)) Int[(a + b*Csc[e + f*x])^m*Simp[A *b*(m + 1) + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 0.31 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.66
| method | result | size |
| default | \(\frac {2 A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right )\right )}{d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (6 \sin \left (d x +c \right )+3 \tan \left (d x +c \right )+\sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right )}\) | \(221\) |
| parts | \(\frac {2 A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right )\right )}{d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (6 \sin \left (d x +c \right )+3 \tan \left (d x +c \right )+\sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right )}\) | \(221\) |
2*A/d*a*(a*(1+sec(d*x+c)))^(1/2)*((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arcta nh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c )+(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-c os(d*x+c)/(cos(d*x+c)+1))^(1/2))+sin(d*x+c))/(cos(d*x+c)+1)+2/5*C/d*a*(a*( 1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(6*sin(d*x+c)+3*tan(d*x+c)+sec(d*x+c)* tan(d*x+c))
Time = 0.28 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.60 \[ \int (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {5 \, {\left (A a \cos \left (d x + c\right )^{3} + A a \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left ({\left (5 \, A + 6 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, C a \cos \left (d x + c\right ) + C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, -\frac {2 \, {\left (5 \, {\left (A a \cos \left (d x + c\right )^{3} + A a \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left ({\left (5 \, A + 6 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, C a \cos \left (d x + c\right ) + C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{5 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
[1/5*(5*(A*a*cos(d*x + c)^3 + A*a*cos(d*x + c)^2)*sqrt(-a)*log((2*a*cos(d* x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) *sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*((5*A + 6*C)*a *cos(d*x + c)^2 + 3*C*a*cos(d*x + c) + C*a)*sqrt((a*cos(d*x + c) + a)/cos( d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2), -2/5*(5*(A* a*cos(d*x + c)^3 + A*a*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - ((5*A + 6*C)*a* cos(d*x + c)^2 + 3*C*a*cos(d*x + c) + C*a)*sqrt((a*cos(d*x + c) + a)/cos(d *x + c))*sin(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)]
\[ \int (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]
\[ \int (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \]
-1/10*(5*((A*a*cos(2*d*x + 2*c)^2 + A*a*sin(2*d*x + 2*c)^2 + 2*A*a*cos(2*d *x + 2*c) + A*a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos( 2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) ^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - (A* a*cos(2*d*x + 2*c)^2 + A*a*sin(2*d*x + 2*c)^2 + 2*A*a*cos(2*d*x + 2*c) + A *a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos( 2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/ 2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1) - 2*(A*a*d*cos(2*d *x + 2*c)^2 + A*a*d*sin(2*d*x + 2*c)^2 + 2*A*a*d*cos(2*d*x + 2*c) + A*a*d) *integrate((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2* d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*si n(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(7/2*arctan2(sin( 2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2* cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2* cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2 *d*x + 2*c))))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*...
\[ \int (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]