3.400 \(\int \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=77 \[ \frac{2 a \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{4 a \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

[Out]

(4*a*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(
3*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.173543, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4264, 3805, 3804} \[ \frac{2 a \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{4 a \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(
3*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{3}{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{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{3} \left (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{4 a \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 a \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.116315, size = 49, normalized size = 0.64 \[ \frac{2 \sqrt{\cos (c+d x)} (\cos (c+d x)+2) \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.178, size = 58, normalized size = 0.8 \begin{align*} -{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2\,\cos \left ( dx+c \right ) -4}{3\,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)^(3/2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/3/d*(cos(d*x+c)^2+cos(d*x+c)-2)*(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 [A]  time = 2.80224, size = 153, normalized size = 1.99 \begin{align*} \frac{\sqrt{2}{\left (3 \, \cos \left (\frac{2}{3} \, \arctan \left (\sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ), \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )\right )\right ) \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) - 3 \, \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) \sin \left (\frac{2}{3} \, \arctan \left (\sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ), \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )\right )\right ) + 2 \, \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 3 \, \sin \left (\frac{1}{3} \, \arctan \left (\sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ), \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )\right )\right )\right )} \sqrt{a}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/6*sqrt(2)*(3*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 3*cos(3/2*d
*x + 3/2*c)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sin(3/2*d*x + 3/2*c) + 3*sin(1/3*
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sqrt(a)/d

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Fricas [A]  time = 1.62224, size = 158, normalized size = 2.05 \begin{align*} \frac{2 \, \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}{\left (\cos \left (d x + c\right ) + 2\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \,{\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)^(3/2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/3*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) + 2)*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)**(3/2)*(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{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*sec(d*x + c) + a)*cos(d*x + c)^(3/2), x)