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

Optimal. Leaf size=119 \[ \frac{64 a^3 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 a \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[Out]

(64*a^3*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*Sqrt[Cos[c + d*x]]*Sqrt[a +
a*Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a*Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.231381, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4264, 3809, 3804} \[ \frac{64 a^3 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 a \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*Sqrt[Cos[c + d*x]]*Sqrt[a +
a*Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a*Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*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 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 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{5}{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{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{1}{5} \left (8 a \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 \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{1}{15} \left (32 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{64 a^3 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.235726, size = 64, normalized size = 0.54 \[ \frac{a^2 \sqrt{\cos (c+d x)} (28 \cos (c+d x)+3 \cos (2 (c+d x))+89) \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[Cos[c + d*x]]*(89 + 28*Cos[c + d*x] + 3*Cos[2*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2
])/(15*d)

________________________________________________________________________________________

Maple [A]  time = 0.149, size = 75, normalized size = 0.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( 3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+11\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+29\,\cos \left ( dx+c \right ) -43 \right ) }{15\,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)^(5/2)*(a+a*sec(d*x+c))^(5/2),x)

[Out]

-2/15/d*a^2*(3*cos(d*x+c)^3+11*cos(d*x+c)^2+29*cos(d*x+c)-43)*cos(d*x+c)^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(
1/2)/sin(d*x+c)

________________________________________________________________________________________

Maxima [A]  time = 2.51076, size = 81, normalized size = 0.68 \begin{align*} \frac{{\left (3 \, \sqrt{2} a^{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 25 \, \sqrt{2} a^{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 150 \, \sqrt{2} a^{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} \sqrt{a}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x +
1/2*c))*sqrt(a)/d

________________________________________________________________________________________

Fricas [A]  time = 1.64596, size = 207, normalized size = 1.74 \begin{align*} \frac{2 \,{\left (3 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) + 43 \, 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 )}{15 \,{\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)^(5/2)*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*a^2*cos(d*x + c)^2 + 14*a^2*cos(d*x + c) + 43*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)

________________________________________________________________________________________

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)**(5/2)*(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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