3.18 \(\int (c \cos (a+b x))^{5/2} \, dx\)

Optimal. Leaf size=70 \[ \frac{6 c^2 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{c \cos (a+b x)}}{5 b \sqrt{\cos (a+b x)}}+\frac{2 c \sin (a+b x) (c \cos (a+b x))^{3/2}}{5 b} \]

[Out]

(6*c^2*Sqrt[c*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(5*b*Sqrt[Cos[a + b*x]]) + (2*c*(c*Cos[a + b*x])^(3/2)*
Sin[a + b*x])/(5*b)

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Rubi [A]  time = 0.0364114, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2635, 2640, 2639} \[ \frac{6 c^2 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{c \cos (a+b x)}}{5 b \sqrt{\cos (a+b x)}}+\frac{2 c \sin (a+b x) (c \cos (a+b x))^{3/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*Cos[a + b*x])^(5/2),x]

[Out]

(6*c^2*Sqrt[c*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(5*b*Sqrt[Cos[a + b*x]]) + (2*c*(c*Cos[a + b*x])^(3/2)*
Sin[a + b*x])/(5*b)

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int (c \cos (a+b x))^{5/2} \, dx &=\frac{2 c (c \cos (a+b x))^{3/2} \sin (a+b x)}{5 b}+\frac{1}{5} \left (3 c^2\right ) \int \sqrt{c \cos (a+b x)} \, dx\\ &=\frac{2 c (c \cos (a+b x))^{3/2} \sin (a+b x)}{5 b}+\frac{\left (3 c^2 \sqrt{c \cos (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \, dx}{5 \sqrt{\cos (a+b x)}}\\ &=\frac{6 c^2 \sqrt{c \cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{5 b \sqrt{\cos (a+b x)}}+\frac{2 c (c \cos (a+b x))^{3/2} \sin (a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0844326, size = 62, normalized size = 0.89 \[ \frac{(c \cos (a+b x))^{5/2} \left (6 E\left (\left .\frac{1}{2} (a+b x)\right |2\right )+\sin (2 (a+b x)) \sqrt{\cos (a+b x)}\right )}{5 b \cos ^{\frac{5}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Cos[a + b*x])^(5/2),x]

[Out]

((c*Cos[a + b*x])^(5/2)*(6*EllipticE[(a + b*x)/2, 2] + Sqrt[Cos[a + b*x]]*Sin[2*(a + b*x)]))/(5*b*Cos[a + b*x]
^(5/2))

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Maple [B]  time = 1.855, size = 213, normalized size = 3. \begin{align*} -{\frac{2\,{c}^{3}}{5\,b}\sqrt{c \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}} \left ( -8\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}\cos \left ( 1/2\,bx+a/2 \right ) +8\,\cos \left ( 1/2\,bx+a/2 \right ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}-3\,\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) -2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}\cos \left ( 1/2\,bx+a/2 \right ) \right ){\frac{1}{\sqrt{-c \left ( 2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2} \right ) }}} \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{c \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(c*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*c^3*(-8*sin(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)
+8*cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^4-3*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^(1/2)*Ell
ipticE(cos(1/2*b*x+1/2*a),2^(1/2))-2*sin(1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a))/(-c*(2*sin(1/2*b*x+1/2*a)^4-sin(
1/2*b*x+1/2*a)^2))^(1/2)/sin(1/2*b*x+1/2*a)/(c*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*cos(b*x + a))^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{c \cos \left (b x + a\right )} c^{2} \cos \left (b x + a\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*cos(b*x + a))*c^2*cos(b*x + a)^2, x)

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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((c*cos(b*x+a))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*cos(b*x + a))^(5/2), x)