Integrand size = 42, antiderivative size = 154 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (19 B+20 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (9 B-4 C) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (B-4 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac {a B \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d} \]
1/4*a^(5/2)*(19*B+20*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/ d+1/2*a*B*cos(d*x+c)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/4*a^3*(9*B-4*C) *sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/2*a^2*(B-4*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 = 1.10 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.03 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {a^2 \left (-\left ((7 B+20 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )\right )+(3 B-8 C+(B-4 C) \cos (c+d x)+3 B \cos (2 (c+d x))) \sqrt {1-\sec (c+d x)}-32 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{4 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]
-1/4*(a^2*(-((7*B + 20*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]]) + (3*B - 8*C + (B - 4*C)*Cos[c + d*x] + 3*B*Cos[2*(c + d*x)])*Sqrt[1 - Sec[c + d*x]] - 32 *B*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]] )*Sqrt[a*(1 + Sec[c + d*x])]*Sin[c + d*x])/(d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])
Time = 1.07 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4560, 3042, 4505, 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.
| \(\displaystyle \int \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2} (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{2} \cos (c+d x) (\sec (c+d x) a+a)^{3/2} (a (7 B+4 C)-a (B-4 C) \sec (c+d x))dx+\frac {a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \cos (c+d x) (\sec (c+d x) a+a)^{3/2} (a (7 B+4 C)-a (B-4 C) \sec (c+d x))dx+\frac {a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (7 B+4 C)-a (B-4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{4} \left (2 \int \frac {1}{2} \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left ((9 B-4 C) a^2+(5 B+12 C) \sec (c+d x) a^2\right )dx-\frac {2 a^2 (B-4 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\int \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left ((9 B-4 C) a^2+(5 B+12 C) \sec (c+d x) a^2\right )dx-\frac {2 a^2 (B-4 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((9 B-4 C) a^2+(5 B+12 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^2 (B-4 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} a^2 (19 B+20 C) \int \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (9 B-4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a^2 (B-4 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} a^2 (19 B+20 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^3 (9 B-4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a^2 (B-4 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {1}{4} \left (-\frac {a^3 (19 B+20 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {a^3 (9 B-4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a^2 (B-4 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{4} \left (\frac {a^{5/2} (19 B+20 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 (9 B-4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a^2 (B-4 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d}\right )+\frac {a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d}\) |
(a*B*Cos[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d) + ((a^(5/ 2)*(19*B + 20*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/ d + (a^3*(9*B - 4*C)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) - (2*a^2*( B - 4*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d)/4
3.4.80.3.1 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[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
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]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(343\) vs. \(2(134)=268\).
Time = 59.65 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.23
| method | result | size |
| default | \(\frac {a^{2} \left (19 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+2 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+20 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+19 B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+11 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+20 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+4 C \cos \left (d x +c \right ) \sin \left (d x +c \right )+8 C \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d \left (\cos \left (d x +c \right )+1\right )}\) | \(344\) |
int(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me thod=_RETURNVERBOSE)
1/4*a^2/d*(19*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)+2*B*sin(d*x+c)*c os(d*x+c)^2+20*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(co s(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+19*B*(-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)/(c os(d*x+c)+1))^(1/2))+11*B*cos(d*x+c)*sin(d*x+c)+20*C*(-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))+4*C*cos(d*x+c)*sin(d*x+c)+8*C*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/ 2)/(cos(d*x+c)+1)
Time = 0.30 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.26 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {{\left ({\left (19 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (19 \, B + 20 \, C\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + {\left (11 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {{\left ({\left (19 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (19 \, B + 20 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + {\left (11 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="fricas")
[1/8*(((19*B + 20*C)*a^2*cos(d*x + c) + (19*B + 20*C)*a^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*(2*B* a^2*cos(d*x + c)^2 + (11*B + 4*C)*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos( d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/4*(((19 *B + 20*C)*a^2*cos(d*x + c) + (19*B + 20*C)*a^2)*sqrt(a)*arctan(sqrt((a*co s(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (2*B* a^2*cos(d*x + c)^2 + (11*B + 4*C)*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos( d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]
Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="maxima")
\[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{3} \,d x } \]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="giac")
Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^3\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]