3.160 \(\int \cos ^{\frac{5}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=116 \[ \frac{b^2 \sin ^5(c+d x) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 \sin ^3(c+d x) \sqrt{b \cos (c+d x)}}{3 d \sqrt{\cos (c+d x)}}+\frac{b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}} \]

[Out]

(b^2*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) - (2*b^2*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^3)/(
3*d*Sqrt[Cos[c + d*x]]) + (b^2*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^5)/(5*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.025976, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {17, 2633} \[ \frac{b^2 \sin ^5(c+d x) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 \sin ^3(c+d x) \sqrt{b \cos (c+d x)}}{3 d \sqrt{\cos (c+d x)}}+\frac{b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) - (2*b^2*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^3)/(
3*d*Sqrt[Cos[c + d*x]]) + (b^2*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^5)/(5*d*Sqrt[Cos[c + d*x]])

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m + 1/2)*b^(n - 1/2)*Sqrt[b*v])/Sqrt[a*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \cos ^{\frac{5}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx &=\frac{\left (b^2 \sqrt{b \cos (c+d x)}\right ) \int \cos ^5(c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=-\frac{\left (b^2 \sqrt{b \cos (c+d x)}\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d \sqrt{\cos (c+d x)}}\\ &=\frac{b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-\frac{2 b^2 \sqrt{b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{b^2 \sqrt{b \cos (c+d x)} \sin ^5(c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.125421, size = 57, normalized size = 0.49 \[ \frac{\sin (c+d x) \left (3 \sin ^4(c+d x)-10 \sin ^2(c+d x)+15\right ) (b \cos (c+d x))^{5/2}}{15 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*Sin[c + d*x]*(15 - 10*Sin[c + d*x]^2 + 3*Sin[c + d*x]^4))/(15*d*Cos[c + d*x]^(5/2))

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Maple [A]  time = 0.169, size = 52, normalized size = 0.5 \begin{align*}{\frac{ \left ( 3\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8 \right ) \sin \left ( dx+c \right ) }{15\,d} \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(5/2),x)

[Out]

1/15/d*(3*cos(d*x+c)^4+4*cos(d*x+c)^2+8)*sin(d*x+c)*(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(5/2)

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Maxima [A]  time = 1.86682, size = 104, normalized size = 0.9 \begin{align*} \frac{{\left (3 \, b^{2} \sin \left (5 \, d x + 5 \, c\right ) + 25 \, b^{2} \sin \left (\frac{3}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right ) + 150 \, b^{2} \sin \left (\frac{1}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right )\right )} \sqrt{b}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(3*b^2*sin(5*d*x + 5*c) + 25*b^2*sin(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 150*b^2*sin(1/5*
arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))))*sqrt(b)/d

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Fricas [A]  time = 1.97581, size = 158, normalized size = 1.36 \begin{align*} \frac{{\left (3 \, b^{2} \cos \left (d x + c\right )^{4} + 4 \, b^{2} \cos \left (d x + c\right )^{2} + 8 \, b^{2}\right )} \sqrt{b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out