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

Optimal. Leaf size=220 \[ \frac {2 a (8 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{63 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a (8 A+9 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a (8 A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {32 a (8 A+9 B) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}} \]

[Out]

4/105*a*(8*A+9*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/63*a*(8*A+9*B)*cos(d*x+c)^(5/2)*sin(d
*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/9*a*A*cos(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+32/315*a*(8*A+9*B)
*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/315*a*(8*A+9*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*se
c(d*x+c))^(1/2)

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Rubi [A]  time = 0.48, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {2955, 4015, 3805, 3804} \[ \frac {2 a (8 A+9 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{63 d \sqrt {a \sec (c+d x)+a}}+\frac {4 a (8 A+9 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a (8 A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {32 a (8 A+9 B) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*a*(8*A + 9*B)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a*(8*A + 9*B)*Sqrt[C
os[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (4*a*(8*A + 9*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x]
)/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(8*A + 9*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d*Sqrt[a + a*Sec[c
+ d*x]]) + (2*a*A*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]])

Rule 2955

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

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 3805

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 4015

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]

Rubi steps

\begin {align*} \int \cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 a A \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{9} \left ((8 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a (8 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 A \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{21} \left (2 (8 A+9 B) \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 {4 a (8 A+9 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (8 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 A \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{105} \left (8 (8 A+9 B) \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 {16 a (8 A+9 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {4 a (8 A+9 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (8 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 A \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {1}{315} \left (16 (8 A+9 B) \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 {32 a (8 A+9 B) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a (8 A+9 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {4 a (8 A+9 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (8 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 A \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 119, normalized size = 0.54 \[ \frac {\sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a (\sec (c+d x)+1)} (94 (8 A+9 B) \cos (c+d x)+4 (83 A+54 B) \cos (2 (c+d x))+80 A \cos (3 (c+d x))+35 A \cos (4 (c+d x))+1321 A+90 B \cos (3 (c+d x))+1368 B)}{1260 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(1321*A + 1368*B + 94*(8*A + 9*B)*Cos[c + d*x] + 4*(83*A + 54*B)*Cos[2*(c + d*x)] + 80*A*C
os[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])/(126
0*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.43, size = 116, normalized size = 0.53 \[ \frac {2 \, {\left (35 \, A \cos \left (d x + c\right )^{4} + 5 \, {\left (8 \, A + 9 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (8 \, A + 9 \, B\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, A + 9 \, B\right )} \cos \left (d x + c\right ) + 128 \, A + 144 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(A+B*sec(d*x+c))*(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 + 6*(8*A + 9*B)*cos(d*x + c)^2 + 8*(8*A + 9*B)*cos(d
*x + c) + 128*A + 144*B)*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 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.96, size = 130, normalized size = 0.59 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (35 A \left (\cos ^{4}\left (d x +c \right )\right )+40 A \left (\cos ^{3}\left (d x +c \right )\right )+45 B \left (\cos ^{3}\left (d x +c \right )\right )+48 A \left (\cos ^{2}\left (d x +c \right )\right )+54 B \left (\cos ^{2}\left (d x +c \right )\right )+64 A \cos \left (d x +c \right )+72 B \cos \left (d x +c \right )+128 A +144 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\sqrt {\cos }\left (d x +c \right )\right )}{315 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315/d*(-1+cos(d*x+c))*(35*A*cos(d*x+c)^4+40*A*cos(d*x+c)^3+45*B*cos(d*x+c)^3+48*A*cos(d*x+c)^2+54*B*cos(d*x
+c)^2+64*A*cos(d*x+c)+72*B*cos(d*x+c)+128*A+144*B)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)/sin(d*
x+c)

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maxima [B]  time = 0.76, size = 547, normalized size = 2.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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