Integrand size = 33, antiderivative size = 160 \[ \int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {\sqrt {a} (5 A+6 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a (5 A+6 B) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {a (5 A+6 B) \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}} \]
1/8*(5*A+6*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*a^(1/2)/d+ 1/8*a*(5*A+6*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/12*a*(5*A+6*B)*cos(d *x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/3*a*A*cos(d*x+c)^2*sin(d*x+c)/ d/(a+a*sec(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.44 \[ \int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 \left (B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right )+A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\sec (c+d x)\right )\right ) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{d} \]
(2*(B*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]] + A*Hypergeometric2 F1[1/2, 4, 3/2, 1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x )/2])/d
Time = 0.71 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4503, 3042, 4292, 3042, 4292, 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.
| \(\displaystyle \int \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{6} (5 A+6 B) \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} (5 A+6 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {1}{6} (5 A+6 B) \left (\frac {3}{4} \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} (5 A+6 B) \left (\frac {3}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {1}{6} (5 A+6 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} (5 A+6 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {1}{6} (5 A+6 B) \left (\frac {3}{4} \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{6} (5 A+6 B) \left (\frac {3}{4} \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\) |
(a*A*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + ((5*A + 6*B)*((a*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3*( (Sqrt[a]*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a*S in[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4))/6
3.2.25.3.1 Defintions of rubi rules used
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_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*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}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(365\) vs. \(2(140)=280\).
Time = 31.90 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.29
| method | result | size |
| default | \(\frac {\left (8 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}+15 A \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+10 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+18 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+12 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+15 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+15 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+18 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+18 B \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{24 d \left (\cos \left (d x +c \right )+1\right )}\) | \(366\) |
1/24/d*(8*A*sin(d*x+c)*cos(d*x+c)^3+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)*co s(d*x+c)+10*A*cos(d*x+c)^2*sin(d*x+c)+18*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)+12*B*sin(d*x+c)*cos(d*x+c)^2+15*A*(-cos(d*x+c)/(cos(d*x+c)+1))^ (1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2) )+15*A*cos(d*x+c)*sin(d*x+c)+18*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)+18*B*cos( d*x+c)*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
Time = 0.30 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.16 \[ \int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (5 \, A + 6 \, B\right )} \cos \left (d x + c\right ) + 5 \, A + 6 \, B\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (8 \, A \cos \left (d x + c\right )^{3} + 2 \, {\left (5 \, A + 6 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 6 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (5 \, A + 6 \, B\right )} \cos \left (d x + c\right ) + 5 \, A + 6 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (8 \, A \cos \left (d x + c\right )^{3} + 2 \, {\left (5 \, A + 6 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 6 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
[1/48*(3*((5*A + 6*B)*cos(d*x + c) + 5*A + 6*B)*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*(8*A*cos(d*x + c )^3 + 2*(5*A + 6*B)*cos(d*x + c)^2 + 3*(5*A + 6*B)*cos(d*x + c))*sqrt((a*c os(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/24*( 3*((5*A + 6*B)*cos(d*x + c) + 5*A + 6*B)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (8*A*cos(d*x + c)^3 + 2*(5*A + 6*B)*cos(d*x + c)^2 + 3*(5*A + 6*B)*cos(d*x + c))*sqrt(( a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]
\[ \int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 2981 vs. \(2 (140) = 280\).
Time = 0.67 (sec) , antiderivative size = 2981, normalized size of antiderivative = 18.63 \[ \int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]
1/96*((4*(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3 *arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3* d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(3/4)*(cos(3/2*arctan2(sin(2/3*arctan2 (sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), c os(3*d*x + 3*c))) + 1))*sin(3*d*x + 3*c) - (cos(3*d*x + 3*c) - 1)*sin(3/2* arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arct an2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)))*sqrt(a) + 6*(cos(2/3*arcta n2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3* c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*((sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 5*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*cos(1/2*arctan2( sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin( 3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)) - (cos(2/3*arctan2(sin(3*d*x + 3*c) , cos(3*d*x + 3*c))) + 3*cos(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c ))) - 4)*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c ))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)))*sqrt(a) + 15*sqrt(a)*(arctan2(-(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)) )^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*a rctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*(cos(1/2*arctan2(si n(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin...
\[ \int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]