\(\int \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1245]

Optimal result

Integrand size = 45, antiderivative size = 226 \[ \int \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {16 a (16 A+18 B+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {8 a (16 A+18 B+21 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (16 A+18 B+21 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (A+9 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d} \]

[Out]

2/105*a*(16*A+18*B+21*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/63*a*(A+9*B)*cos(d*x+c)^(5/2)*
sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/315*a*(16*A+18*B+21*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(
1/2)+8/315*a*(16*A+18*B+21*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/9*A*cos(d*x+c)^(7/2)*sin(
d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4350, 4171, 4100, 3890, 3889} \[ \int \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a (16 A+18 B+21 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a (16 A+18 B+21 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a (16 A+18 B+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a (A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{63 d \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d} \]

[In]

Int[Cos[c + d*x]^(9/2)*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(16*a*(16*A + 18*B + 21*C)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a*(16*A + 18
*B + 21*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(16*A + 18*B + 21*C)*Cos[c
+ d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(A + 9*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])
/(63*d*Sqrt[a + a*Sec[c + d*x]]) + (2*A*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d)

Rule 3889

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[-2*a*(Co
t[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rule 3890

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] + Dist[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]

Rule 4100

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*Cot[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] +
Dist[(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] /; Fr
eeQ[{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]

Rule 4171

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*
Csc[e + f*x])^n/(f*n)), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rule 4350

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx \\ & = \frac {2 A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (A+9 B)+\frac {3}{2} a (2 A+3 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{9 a} \\ & = \frac {2 a (A+9 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac {1}{21} \left ((16 A+18 B+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a (16 A+18 B+21 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (A+9 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac {1}{105} \left (4 (16 A+18 B+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {8 a (16 A+18 B+21 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (16 A+18 B+21 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (A+9 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac {1}{315} \left (8 (16 A+18 B+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {16 a (16 A+18 B+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {8 a (16 A+18 B+21 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (16 A+18 B+21 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (A+9 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.42 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.56 \[ \int \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {\cos (c+d x)} (1321 A+1368 B+1596 C+(752 A+846 B+672 C) \cos (c+d x)+4 (83 A+54 B+63 C) \cos (2 (c+d x))+80 A \cos (3 (c+d x))+90 B \cos (3 (c+d x))+35 A \cos (4 (c+d x))) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{1260 d (1+\cos (c+d x))} \]

[In]

Integrate[Cos[c + d*x]^(9/2)*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(Sqrt[Cos[c + d*x]]*(1321*A + 1368*B + 1596*C + (752*A + 846*B + 672*C)*Cos[c + d*x] + 4*(83*A + 54*B + 63*C)*
Cos[2*(c + d*x)] + 80*A*Cos[3*(c + d*x)] + 90*B*Cos[3*(c + d*x)] + 35*A*Cos[4*(c + d*x)])*Sqrt[a*(1 + Sec[c +
d*x])]*Sin[c + d*x])/(1260*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.61

method	result	size
default	\(-\frac {2 \left (\left (35 \cos \left (d x +c \right )^{4}+40 \cos \left (d x +c \right )^{3}+48 \cos \left (d x +c \right )^{2}+64 \cos \left (d x +c \right )+128\right ) A +\left (45 \cos \left (d x +c \right )^{3}+54 \cos \left (d x +c \right )^{2}+72 \cos \left (d x +c \right )+144\right ) B +\left (63 \cos \left (d x +c \right )^{2}+84 \cos \left (d x +c \right )+168\right ) C \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{315 d}\)	\(138\)

[In]

int(cos(d*x+c)^(9/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/315/d*((35*cos(d*x+c)^4+40*cos(d*x+c)^3+48*cos(d*x+c)^2+64*cos(d*x+c)+128)*A+(45*cos(d*x+c)^3+54*cos(d*x+c)
^2+72*cos(d*x+c)+144)*B+(63*cos(d*x+c)^2+84*cos(d*x+c)+168)*C)*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)*(cot(
d*x+c)-csc(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.55 \[ \int \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (35 \, A \cos \left (d x + c\right )^{4} + 5 \, {\left (8 \, A + 9 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (16 \, A + 18 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (16 \, A + 18 \, B + 21 \, C\right )} \cos \left (d x + c\right ) + 128 \, A + 144 \, B + 168 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^(9/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*A*cos(d*x + c)^4 + 5*(8*A + 9*B)*cos(d*x + c)^3 + 3*(16*A + 18*B + 21*C)*cos(d*x + c)^2 + 4*(16*A +
18*B + 21*C)*cos(d*x + c) + 128*A + 144*B + 168*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*
sin(d*x + c)/(d*cos(d*x + c) + d)

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(9/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 668 vs. \(2 (196) = 392\).

Time = 0.50 (sec) , antiderivative size = 668, normalized size of antiderivative = 2.96 \[ \int \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^(9/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/5040*(sqrt(2)*(1890*cos(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 420*
cos(2/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 252*cos(4/9*arctan2(sin(9/
2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 45*cos(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2
*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) - 1890*cos(9/2*d*x + 9/2*c)*sin(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2
*d*x + 9/2*c))) - 420*cos(9/2*d*x + 9/2*c)*sin(2/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 252*
cos(9/2*d*x + 9/2*c)*sin(4/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 45*cos(9/2*d*x + 9/2*c)*si
n(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 70*sin(9/2*d*x + 9/2*c) + 45*sin(7/9*arctan2(sin(
9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 252*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) +
420*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 1890*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c),
cos(9/2*d*x + 9/2*c))))*A*sqrt(a) - 18*sqrt(2)*(7*(15*sin(3*d*x + 3*c) + 5*sin(2*d*x + 2*c) + sin(d*x + c))*co
s(7/2*arctan2(sin(d*x + c), cos(d*x + c))) - (105*cos(3*d*x + 3*c) + 35*cos(2*d*x + 2*c) + 7*cos(d*x + c) + 10
)*sin(7/2*arctan2(sin(d*x + c), cos(d*x + c))) - 7*sin(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - 35*sin(3/2*a
rctan2(sin(d*x + c), cos(d*x + c))) - 105*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*B*sqrt(a) - 84*sqrt(2)
*(5*(6*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - (30*cos(2*d*x + 2*c) +
5*cos(d*x + c) + 6)*sin(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - 5*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c
))) - 30*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*C*sqrt(a))/d

Giac [F]

\[ \int \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac {9}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(9/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{9/2}\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

[In]

int(cos(c + d*x)^(9/2)*(a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

int(cos(c + d*x)^(9/2)*(a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)