3.524 \(\int \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=228 \[ \frac{2 a^2 (10 A+9 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (34 A+39 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a^2 (34 A+39 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (34 A+39 B) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a A \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{9 d} \]

[Out]

(16*a^2*(34*A + 39*B)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a^2*(34*A + 39*B)
*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(34*A + 39*B)*Cos[c + d*x]^(3/2)*S
in[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(10*A + 9*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d*Sqr
t[a + a*Sec[c + d*x]]) + (2*a*A*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d)

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Rubi [A]  time = 0.686358, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2955, 4017, 4015, 3805, 3804} \[ \frac{2 a^2 (10 A+9 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (34 A+39 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a^2 (34 A+39 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (34 A+39 B) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a A \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]

[Out]

(16*a^2*(34*A + 39*B)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a^2*(34*A + 39*B)
*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(34*A + 39*B)*Cos[c + d*x]^(3/2)*S
in[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(10*A + 9*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d*Sqr
t[a + a*Sec[c + d*x]]) + (2*a*A*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d)

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 4017

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]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[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]

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]

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 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]

Rubi steps

\begin{align*} \int \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{3/2} (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) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{1}{9} \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 (10 A+9 B)+\frac{3}{2} a (2 A+3 B) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (10 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) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{1}{21} \left (a (34 A+39 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{2 a^2 (34 A+39 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (10 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) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{1}{105} \left (4 a (34 A+39 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{8 a^2 (34 A+39 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (34 A+39 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (10 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) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{1}{315} \left (8 a (34 A+39 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{16 a^2 (34 A+39 B) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{8 a^2 (34 A+39 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (34 A+39 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (10 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) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}\\ \end{align*}

Mathematica [A]  time = 0.422066, size = 118, normalized size = 0.52 \[ \frac{2 a \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a (\sec (c+d x)+1)} \left (5 (17 A+9 B) \cos ^3(c+d x)+3 (34 A+39 B) \cos ^2(c+d x)+4 (34 A+39 B) \cos (c+d x)+8 (34 A+39 B)+35 A \cos ^4(c+d x)\right )}{315 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]

[Out]

(2*a*Sqrt[Cos[c + d*x]]*(8*(34*A + 39*B) + 4*(34*A + 39*B)*Cos[c + d*x] + 3*(34*A + 39*B)*Cos[c + d*x]^2 + 5*(
17*A + 9*B)*Cos[c + d*x]^3 + 35*A*Cos[c + d*x]^4)*Sqrt[a*(1 + Sec[c + d*x])]*Sin[c + d*x])/(315*d*(1 + Cos[c +
d*x]))

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Maple [A]  time = 0.315, size = 131, normalized size = 0.6 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 35\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+85\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+45\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+102\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+117\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+136\,A\cos \left ( dx+c \right ) +156\,B\cos \left ( dx+c \right ) +272\,A+312\,B \right ) }{315\,d\sin \left ( dx+c \right ) }\sqrt{\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x)

[Out]

-2/315/d*a*(-1+cos(d*x+c))*(35*A*cos(d*x+c)^4+85*A*cos(d*x+c)^3+45*B*cos(d*x+c)^3+102*A*cos(d*x+c)^2+117*B*cos
(d*x+c)^2+136*A*cos(d*x+c)+156*B*cos(d*x+c)+272*A+312*B)*cos(d*x+c)^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/
sin(d*x+c)

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Maxima [B]  time = 2.09384, size = 753, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/5040*(sqrt(2)*(3780*a*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) + 10
50*a*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) + 378*a*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) + 135*a*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) - 3780*a*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))) - 1050*a*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))) - 378*a*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))) - 135*a*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*sin(9/2*d*x + 9/2*c) + 1
35*a*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 378*a*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c)
, cos(9/2*d*x + 9/2*c))) + 1050*a*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 3780*a*sin(1/
9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))*A*sqrt(a) - 6*sqrt(2)*(175*a*cos(7/4*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 5*(35*a*cos(2*d*x + 2*c) + 6*a)*sin(7/4*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c))) - 126*a*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 175*a*sin(3/4*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1470*a*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*B*sqrt(a))/d

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Fricas [A]  time = 0.48523, size = 332, normalized size = 1.46 \begin{align*} \frac{2 \,{\left (35 \, A a \cos \left (d x + c\right )^{4} + 5 \,{\left (17 \, A + 9 \, B\right )} a \cos \left (d x + c\right )^{3} + 3 \,{\left (34 \, A + 39 \, B\right )} a \cos \left (d x + c\right )^{2} + 4 \,{\left (34 \, A + 39 \, B\right )} a \cos \left (d x + c\right ) + 8 \,{\left (34 \, A + 39 \, B\right )} a\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/315*(35*A*a*cos(d*x + c)^4 + 5*(17*A + 9*B)*a*cos(d*x + c)^3 + 3*(34*A + 39*B)*a*cos(d*x + c)^2 + 4*(34*A +
39*B)*a*cos(d*x + c) + 8*(34*A + 39*B)*a)*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(9/2)*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

Timed out