Integrand size = 23, antiderivative size = 106 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {19 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {9 a^3 \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d} \]
19/4*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+9/4*a^3*s in(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/2*a^2*cos(d*x+c)*sin(d*x+c)*(a+a*sec( d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.57 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.42 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {a^2 \cos (c+d x) \sqrt {a (1+\sec (c+d x))} \left (\sqrt {1-\sec (c+d x)} (\sin (c+d x)+3 \sin (2 (c+d x)))-7 \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)-32 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)} \tan (c+d x)\right )}{4 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]
-1/4*(a^2*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(Sqrt[1 - Sec[c + d*x]]* (Sin[c + d*x] + 3*Sin[2*(c + d*x)]) - 7*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Ta n[c + d*x] - 32*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]]*Tan[c + d*x]))/(d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])
Time = 0.53 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4300, 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.
| \(\displaystyle \int \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4300 |
| \(\displaystyle \frac {1}{2} a \int \frac {1}{2} \cos (c+d x) \sqrt {\sec (c+d x) a+a} (5 \sec (c+d x) a+9 a)dx+\frac {a^2 \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} a \int \cos (c+d x) \sqrt {\sec (c+d x) a+a} (5 \sec (c+d x) a+9 a)dx+\frac {a^2 \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} a \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (5 \csc \left (c+d x+\frac {\pi }{2}\right ) a+9 a\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{4} a \left (\frac {19}{2} a \int \sqrt {\sec (c+d x) a+a}dx+\frac {9 a^2 \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} a \left (\frac {19}{2} a \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {9 a^2 \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {1}{4} a \left (\frac {9 a^2 \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {19 a^2 \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {a^2 \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}+\frac {1}{4} a \left (\frac {19 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {9 a^2 \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\) |
(a^2*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + (a*((19*a ^(3/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (9*a^2 *Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4
3.2.13.3.1 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_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[b^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)* ((d*Csc[e + f*x])^n/(f*n)), x] - Simp[a/(d*n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f *x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2*m]
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. \(187\) vs. \(2(90)=180\).
Time = 58.66 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.77
| method | result | size |
| default | \(\frac {a^{2} \left (19 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+2 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+19 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+11 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d \left (\cos \left (d x +c \right )+1\right )}\) | \(188\) |
1/4/d*a^2*(19*(-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))*cos(d*x+c)+2*cos(d*x+c)^2*sin (d*x+c)+19*(-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))+11*cos(d*x+c)*sin(d*x+c))*(a*(1+ sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
Time = 0.29 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.77 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {19 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} + 11 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {19 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} + 11 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
[1/8*(19*(a^2*cos(d*x + c) + a^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqr t(-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*(2*a^2*cos(d*x + c)^2 + 11*a^2 *cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*co s(d*x + c) + d), -1/4*(19*(a^2*cos(d*x + c) + a^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))) - (2* a^2*cos(d*x + c)^2 + 11*a^2*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)]
Timed out. \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]
\[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]