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

Optimal. Leaf size=201 \[ \frac{2 a^2 \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{9 d}+\frac{38 a^3 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{146 a^3 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{584 a^3 \sin (c+d x) \sqrt{\cos (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{1168 a^3 \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

[Out]

(1168*a^3*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (584*a^3*Sqrt[Cos[c + d*x]]*Sin[
c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (146*a^3*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c
+ d*x]]) + (38*a^3*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*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 = 0.406101, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4264, 3813, 4015, 3805, 3804} \[ \frac{2 a^2 \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{9 d}+\frac{38 a^3 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{146 a^3 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{584 a^3 \sin (c+d x) \sqrt{\cos (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{1168 a^3 \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1168*a^3*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (584*a^3*Sqrt[Cos[c + d*x]]*Sin[
c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (146*a^3*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c
+ d*x]]) + (38*a^3*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*Cos[c + d*x]^(7/
2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d)

Rule 4264

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

Rule 3813

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[a/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d,
e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2
*m]

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))^{5/2} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 \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 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{19 a}{2}+\frac{15}{2} a \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{38 a^3 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{1}{21} \left (73 a^2 \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{146 a^3 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{38 a^3 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{1}{105} \left (292 a^2 \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{584 a^3 \sqrt{\cos (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{146 a^3 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{38 a^3 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{1}{315} \left (584 a^2 \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{1168 a^3 \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{584 a^3 \sqrt{\cos (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{146 a^3 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{38 a^3 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 \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 = 0.239832, size = 90, normalized size = 0.45 \[ \frac{2 a^2 \sin (c+d x) \sqrt{\cos (c+d x)} \left (35 \cos ^4(c+d x)+130 \cos ^3(c+d x)+219 \cos ^2(c+d x)+292 \cos (c+d x)+584\right ) \sqrt{a (\sec (c+d x)+1)}}{315 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Sqrt[Cos[c + d*x]]*(584 + 292*Cos[c + d*x] + 219*Cos[c + d*x]^2 + 130*Cos[c + d*x]^3 + 35*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 = 0.183, size = 95, normalized size = 0.5 \begin{align*} -{\frac{2\,{a}^{2} \left ( 35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}+95\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+89\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+73\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+292\,\cos \left ( dx+c \right ) -584 \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))^(5/2),x)

[Out]

-2/315/d*a^2*(35*cos(d*x+c)^5+95*cos(d*x+c)^4+89*cos(d*x+c)^3+73*cos(d*x+c)^2+292*cos(d*x+c)-584)*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 = 2.8915, size = 570, normalized size = 2.84 \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))^(5/2),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) + 2
100*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*ar
ctan2(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), cos
(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*arctan2
(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))))*sqrt(a)/d

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Fricas [A]  time = 1.66525, size = 279, normalized size = 1.39 \begin{align*} \frac{2 \,{\left (35 \, a^{2} \cos \left (d x + c\right )^{4} + 130 \, a^{2} \cos \left (d x + c\right )^{3} + 219 \, a^{2} \cos \left (d x + c\right )^{2} + 292 \, a^{2} \cos \left (d x + c\right ) + 584 \, 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 )}} \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))^(5/2),x, algorithm="fricas")

[Out]

2/315*(35*a^2*cos(d*x + c)^4 + 130*a^2*cos(d*x + c)^3 + 219*a^2*cos(d*x + c)^2 + 292*a^2*cos(d*x + c) + 584*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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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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

Timed out