Integrand size = 25, antiderivative size = 105 \[ \int (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^{3/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 (3 A+4 B) \tan (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a B \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3 d} \]
2*a^(3/2)*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/3*a^2*(3 *A+4*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/3*a*B*(a+a*sec(d*x+c))^(1/2) *tan(d*x+c)/d
Time = 0.63 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.97 \[ \int (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {a (1+\sec (c+d x))} \left (3 \sqrt {2} A \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+2 (B+(3 A+5 B) \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d} \]
(a*Sec[(c + d*x)/2]*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*A*A rcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(3/2) + 2*(B + (3*A + 5*B)*Co s[c + d*x])*Sin[(c + d*x)/2]))/(3*d)
Time = 0.60 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {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.
| \(\displaystyle \int (a \sec (c+d x)+a)^{3/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {2}{3} \int \frac {1}{2} \sqrt {\sec (c+d x) a+a} (3 a A+a (3 A+4 B) \sec (c+d x))dx+\frac {2 a B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \sqrt {\sec (c+d x) a+a} (3 a A+a (3 A+4 B) \sec (c+d x))dx+\frac {2 a B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a A+a (3 A+4 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 a B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 4403 |
| \(\displaystyle \frac {1}{3} \left (a (3 A+4 B) \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+3 a A \int \sqrt {\sec (c+d x) a+a}dx\right )+\frac {2 a B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (a (3 A+4 B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+3 a A \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\right )+\frac {2 a B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {1}{3} \left (a (3 A+4 B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {6 a^2 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {2 a B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{3} \left (a (3 A+4 B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {6 a^{3/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}\right )+\frac {2 a B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {1}{3} \left (\frac {6 a^{3/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^2 (3 A+4 B) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
(2*a*B*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) + ((6*a^(3/2)*A*ArcTan [(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(3*A + 4*B)* Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/3
3.2.30.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(205\) vs. \(2(91)=182\).
Time = 1.65 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.96
| method | result | size |
| default | \(\frac {2 A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right )\right )}{d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (5 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}\) | \(206\) |
| parts | \(\frac {2 A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right )\right )}{d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (5 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}\) | \(206\) |
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)+2/3*B/d*a*(a*( 1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(5*sin(d*x+c)+tan(d*x+c))
Time = 0.29 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.99 \[ \int (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left (A a \cos \left (d x + c\right )^{2} + A a \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left ({\left (3 \, A + 5 \, B\right )} a \cos \left (d x + c\right ) + B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac {2 \, {\left (3 \, {\left (A a \cos \left (d x + c\right )^{2} + A a \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left ({\left (3 \, A + 5 \, B\right )} a \cos \left (d x + c\right ) + B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]
[1/3*(3*(A*a*cos(d*x + c)^2 + A*a*cos(d*x + c))*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)*s in(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*((3*A + 5*B)*a*c os(d*x + c) + B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/( d*cos(d*x + c)^2 + d*cos(d*x + c)), -2/3*(3*(A*a*cos(d*x + c)^2 + A*a*cos( d*x + c))*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - ((3*A + 5*B)*a*cos(d*x + c) + B*a)*sqrt((a*c os(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]
\[ \int (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )}\right )\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 998 vs. \(2 (91) = 182\).
Time = 0.41 (sec) , antiderivative size = 998, normalized size of antiderivative = 9.50 \[ \int (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]
1/2*((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)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1 /2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin( 2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos (2*d*x + 2*c)))), (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)) *cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(si n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), c os(2*d*x + 2*c)))) + 1) - 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)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos( 2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(s in(2*d*x + 2*c), cos(2*d*x + 2*c)))), (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), co s(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2 (sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) - 1) - 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),...
\[ \int (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]