Integrand size = 31, antiderivative size = 143 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (5 A+2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {a^3 (3 A+14 B) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (A+2 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a B (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \]
a^(5/2)*(5*A+2*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/3* a*B*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d-1/3*a^3*(3*A+14*B)*sin(d*x+c)/d/(a +a*sec(d*x+c))^(1/2)+2*a^2*(A+2*B)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
Time = 4.03 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.82 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^3 \left (3 (5 A+2 B) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)+\sqrt {1-\sec (c+d x)} (3 A \sin (c+d x)+2 (3 A+8 B+B \sec (c+d x)) \tan (c+d x))\right )}{3 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
(a^3*(3*(5*A + 2*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + Sqrt[1 - Sec[c + d*x]]*(3*A*Sin[c + d*x] + 2*(3*A + 8*B + B*Sec[c + d*x])*Tan[c + d*x])))/(3*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.91 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4506, 27, 3042, 4506, 27, 3042, 4503, 3042, 4261, 216}
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.
| \(\displaystyle \int \cos (c+d x) (a \sec (c+d x)+a)^{5/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {2}{3} \int \frac {1}{2} \cos (c+d x) (\sec (c+d x) a+a)^{3/2} (a (3 A-2 B)+3 a (A+2 B) \sec (c+d x))dx+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \cos (c+d x) (\sec (c+d x) a+a)^{3/2} (a (3 A-2 B)+3 a (A+2 B) \sec (c+d x))dx+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (3 A-2 B)+3 a (A+2 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{3} \left (2 \int -\frac {1}{2} \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left (a^2 (3 A+14 B)-a^2 (9 A+10 B) \sec (c+d x)\right )dx+\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-\int \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left (a^2 (3 A+14 B)-a^2 (9 A+10 B) \sec (c+d x)\right )dx\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^2 (3 A+14 B)-a^2 (9 A+10 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} a^2 (5 A+2 B) \int \sqrt {\sec (c+d x) a+a}dx-\frac {a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}+\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} a^2 (5 A+2 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}+\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {1}{3} \left (-\frac {3 a^3 (5 A+2 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}+\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 a^{5/2} (5 A+2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {a^3 (3 A+14 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}+\frac {6 a^2 (A+2 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {2 a B \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\) |
(2*a*B*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + ((3*a^(5/2)*(5*A + 2*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (a^3*(3 *A + 14*B)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) + (6*a^2*(A + 2*B)*S qrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d)/3
3.2.39.3.1 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_)]*(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[(-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. \(329\) vs. \(2(127)=254\).
Time = 45.57 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.31
| method | result | size |
| default | \(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (15 A \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+6 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+15 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+3 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+6 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+6 A \sin \left (d x +c \right )+16 B \sin \left (d x +c \right )+2 B \tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}\) | \(330\) |
1/3*a^2/d*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(15*A*arctanh(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)+1))^(1/2)*cos(d*x+c)+6*B*arctanh(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)+1))^(1/2)*cos(d*x+c)+15*A *(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-co s(d*x+c)/(cos(d*x+c)+1))^(1/2))+3*A*cos(d*x+c)*sin(d*x+c)+6*B*arctanh(sin( d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(co s(d*x+c)+1))^(1/2)+6*A*sin(d*x+c)+16*B*sin(d*x+c)+2*B*tan(d*x+c))
Time = 0.31 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.70 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (3 \, A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac {3 \, {\left ({\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (3 \, A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]
[1/6*(3*((5*A + 2*B)*a^2*cos(d*x + c)^2 + (5*A + 2*B)*a^2*cos(d*x + c))*sq rt(-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*(3*A*a^2*cos(d*x + c)^2 + 2*(3*A + 8*B)*a^2*cos(d*x + c) + 2*B*a^2 )*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)), -1/3*(3*((5*A + 2*B)*a^2*cos(d*x + c)^2 + (5*A + 2*B)*a ^2*cos(d*x + c))*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*co s(d*x + c)/(sqrt(a)*sin(d*x + c))) - (3*A*a^2*cos(d*x + c)^2 + 2*(3*A + 8* B)*a^2*cos(d*x + c) + 2*B*a^2)*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))]
Timed out. \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 2780 vs. \(2 (127) = 254\).
Time = 0.53 (sec) , antiderivative size = 2780, normalized size of antiderivative = 19.44 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]
1/12*(3*(18*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*a^(5/2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1) ) + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^( 1/4)*((4*a^2*sin(3*d*x + 3*c) + 5*a^2*sin(2*d*x + 2*c) + 4*a^2*sin(d*x + c ))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (a^2*cos(2*d *x + 2*c)^2*sin(d*x + c) + a^2*sin(2*d*x + 2*c)^2*sin(d*x + c) + 2*a^2*cos (2*d*x + 2*c)*sin(d*x + c) + a^2*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (4*a^2*cos(3*d*x + 3*c) + 5*a^2*cos(2*d*x + 2*c) + 4*a^2*cos(d*x + c) + 5*a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos (2*d*x + 2*c) + 1)) - ((a^2*cos(d*x + c) - a^2)*cos(2*d*x + 2*c)^2 + a^2*c os(d*x + c) + (a^2*cos(d*x + c) - a^2)*sin(2*d*x + 2*c)^2 - a^2 + 2*(a^2*c os(d*x + c) - a^2)*cos(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos (2*d*x + 2*c) + 1)))*sqrt(a) + 5*((a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2(-(cos(2*d*x + 2*c)^2 + si n(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))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arcta n2(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(d*x + c)*cos(1/2*arctan2 (sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(s in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - (a^2*cos(2*d*x + 2*c)^2...
\[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]