3.1264 \(\int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=231 \[ \frac {64 a^3 (13 A+15 B+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (13 A+15 B+21 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 a (13 A+15 B+21 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac {2 (5 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d} \]

[Out]

2/105*a*(13*A+15*B+21*C)*cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/63*(5*A+9*B)*cos(d*x+c)^(5/2)*
(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+2/9*A*cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+64/315*a^3*(13*
A+15*B+21*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/315*a^2*(13*A+15*B+21*C)*sin(d*x+c)*cos(d
*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.71, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4265, 4086, 4013, 3809, 3804} \[ \frac {16 a^2 (13 A+15 B+21 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {64 a^3 (13 A+15 B+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a (13 A+15 B+21 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac {2 (5 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(64*a^3*(13*A + 15*B + 21*C)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(13*A
+ 15*B + 21*C)*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(315*d) + (2*a*(13*A + 15*B + 21*C)*
Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(105*d) + (2*(5*A + 9*B)*Cos[c + d*x]^(5/2)*(a + a
*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(63*d) + (2*A*Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(
9*d)

Rule 3804

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 3809

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*Co
t[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*m), x] + Dist[(b*(2*m - 1))/(d*m), Int[(a + b*C
sc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& EqQ[m + n, 0] && GtQ[m, 1/2] && IntegerQ[2*m]

Rule 4013

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 4086

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 4265

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*} \int \cos ^{\frac {9}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{5/2} \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) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (5 A+9 B)+\frac {1}{2} a (2 A+9 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac {2 (5 A+9 B) \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d}+\frac {1}{21} \left ((13 A+15 B+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a (13 A+15 B+21 C) \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {2 (5 A+9 B) \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d}+\frac {1}{105} \left (8 a (13 A+15 B+21 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {16 a^2 (13 A+15 B+21 C) \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (13 A+15 B+21 C) \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {2 (5 A+9 B) \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d}+\frac {1}{315} \left (32 a^2 (13 A+15 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 {64 a^3 (13 A+15 B+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (13 A+15 B+21 C) \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a (13 A+15 B+21 C) \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {2 (5 A+9 B) \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d}+\frac {2 A \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 1.54, size = 124, normalized size = 0.54 \[ \frac {a^2 \sqrt {\cos (c+d x)} \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} ((3116 A+3030 B+2352 C) \cos (c+d x)+4 (254 A+180 B+63 C) \cos (2 (c+d x))+260 A \cos (3 (c+d x))+35 A \cos (4 (c+d x))+5653 A+90 B \cos (3 (c+d x))+6240 B+7476 C)}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^2*Sqrt[Cos[c + d*x]]*(5653*A + 6240*B + 7476*C + (3116*A + 3030*B + 2352*C)*Cos[c + d*x] + 4*(254*A + 180*B
+ 63*C)*Cos[2*(c + d*x)] + 260*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])]*Tan[(c + d*x)/2])/(1260*d)

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fricas [A]  time = 0.48, size = 141, normalized size = 0.61 \[ \frac {2 \, {\left (35 \, A a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (26 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (73 \, A + 60 \, B + 21 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (292 \, A + 345 \, B + 294 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (584 \, A + 690 \, B + 903 \, C\right )} a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/315*(35*A*a^2*cos(d*x + c)^4 + 5*(26*A + 9*B)*a^2*cos(d*x + c)^3 + 3*(73*A + 60*B + 21*C)*a^2*cos(d*x + c)^2
+ (292*A + 345*B + 294*C)*a^2*cos(d*x + c) + (584*A + 690*B + 903*C)*a^2)*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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(9/2), x)

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maple [A]  time = 1.88, size = 156, normalized size = 0.68 \[ -\frac {2 a^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (35 A \left (\cos ^{4}\left (d x +c \right )\right )+130 A \left (\cos ^{3}\left (d x +c \right )\right )+45 B \left (\cos ^{3}\left (d x +c \right )\right )+219 A \left (\cos ^{2}\left (d x +c \right )\right )+180 B \left (\cos ^{2}\left (d x +c \right )\right )+63 C \left (\cos ^{2}\left (d x +c \right )\right )+292 A \cos \left (d x +c \right )+345 B \cos \left (d x +c \right )+294 C \cos \left (d x +c \right )+584 A +690 B +903 C \right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{315 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/315/d*a^2*(-1+cos(d*x+c))*(35*A*cos(d*x+c)^4+130*A*cos(d*x+c)^3+45*B*cos(d*x+c)^3+219*A*cos(d*x+c)^2+180*B*
cos(d*x+c)^2+63*C*cos(d*x+c)^2+292*A*cos(d*x+c)+345*B*cos(d*x+c)+294*C*cos(d*x+c)+584*A+690*B+903*C)*cos(d*x+c
)^(1/2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)

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maxima [B]  time = 0.79, size = 751, normalized size = 3.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/5040*(sqrt(2)*(8190*a^2*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) +
2100*a^2*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) + 756*a^2*cos(4/9*a
rctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 225*a^2*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) - 8190*a^2*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))) - 2100*a^2*cos(9/2*d*x + 9/2*c)*sin(2/3*arctan2(sin(9/2*d*x + 9/2*c), co
s(9/2*d*x + 9/2*c))) - 756*a^2*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)
)) - 225*a^2*cos(9/2*d*x + 9/2*c)*sin(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 70*a^2*sin(9/
2*d*x + 9/2*c) + 225*a^2*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 756*a^2*sin(5/9*arctan
2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 2100*a^2*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x +
9/2*c))) + 8190*a^2*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))*A*sqrt(a) - 30*sqrt(2)*(77*a
^2*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 42*a^2*sin(5/4*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) - 77*a^2*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 630*a^2*sin(1/4*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (77*a^2*cos(2*d*x + 2*c) + 6*a^2)*sin(7/4*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c))))*B*sqrt(a) - 168*(75*sqrt(2)*a^2*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2
*d*x + 2*c) - 25*sqrt(2)*a^2*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 75*sqrt(2)*a^2*sin(1/4*arc
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(25*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*sin(5/4*arctan2(
sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(a))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^{9/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(9/2)*(a + a/cos(c + d*x))^(5/2)*(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))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(9/2)*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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