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

Optimal. Leaf size=115 \[ \frac{2 a \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

[Out]

(16*a*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])
/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.232277, antiderivative size = 115, 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, 3805, 3804} \[ \frac{2 a \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])
/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]])

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 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{5}{2}}(c+d x) \sqrt{a+a \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)}}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{5} \left (4 \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 \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{15} \left (8 \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 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{8 a \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.174736, size = 61, normalized size = 0.53 \[ \frac{\sqrt{\cos (c+d x)} (8 \cos (c+d x)+3 \cos (2 (c+d x))+19) \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(19 + 8*Cos[c + d*x] + 3*Cos[2*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(1
5*d)

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Maple [A]  time = 0.186, size = 70, normalized size = 0.6 \begin{align*} -{\frac{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8\,\cos \left ( dx+c \right ) -16}{15\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{\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))^(1/2),x)

[Out]

-2/15/d*(3*cos(d*x+c)^3+cos(d*x+c)^2+4*cos(d*x+c)-8)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)/sin(
d*x+c)

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Maxima [B]  time = 2.68739, size = 274, normalized size = 2.38 \begin{align*} \frac{\sqrt{2}{\left (30 \, \cos \left (\frac{4}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \cos \left (\frac{2}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) - 30 \, \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) \sin \left (\frac{4}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) - 5 \, \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) \sin \left (\frac{2}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 6 \, \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \sin \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 30 \, \sin \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right )\right )} \sqrt{a}}{60 \, 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))^(1/2),x, algorithm="maxima")

[Out]

1/60*sqrt(2)*(30*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) + 5*cos(2/5
*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 30*cos(5/2*d*x + 5/2*c)*sin(4/5*a
rctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*cos(5/2*d*x + 5/2*c)*sin(2/5*arctan2(sin(5/2*d*x + 5/2
*c), cos(5/2*d*x + 5/2*c))) + 6*sin(5/2*d*x + 5/2*c) + 5*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5
/2*c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*sqrt(a)/d

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Fricas [A]  time = 1.63023, size = 188, normalized size = 1.63 \begin{align*} \frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 8\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))^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*cos(d*x + c)^2 + 4*cos(d*x + c) + 8)*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)**(5/2)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sec \left (d x + c\right ) + a} \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))^(1/2),x, algorithm="giac")

[Out]

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