3.424 \(\int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \, dx\)

Optimal. Leaf size=139 \[ -\frac{2 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}{5 e}+\frac{16 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt{3 \sin (d+e x)+4 \cos (d+e x)-5}}{3 e}-\frac{320 (3 \cos (d+e x)-4 \sin (d+e x))}{3 e \sqrt{3 \sin (d+e x)+4 \cos (d+e x)-5}} \]

[Out]

(-320*(3*Cos[d + e*x] - 4*Sin[d + e*x]))/(3*e*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]]) + (16*(3*Cos[d + e*x
] - 4*Sin[d + e*x])*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]])/(3*e) - (2*(3*Cos[d + e*x] - 4*Sin[d + e*x])*(
-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))/(5*e)

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Rubi [A]  time = 0.0739644, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3113, 3112} \[ -\frac{2 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}{5 e}+\frac{16 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt{3 \sin (d+e x)+4 \cos (d+e x)-5}}{3 e}-\frac{320 (3 \cos (d+e x)-4 \sin (d+e x))}{3 e \sqrt{3 \sin (d+e x)+4 \cos (d+e x)-5}} \]

Antiderivative was successfully verified.

[In]

Int[(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(5/2),x]

[Out]

(-320*(3*Cos[d + e*x] - 4*Sin[d + e*x]))/(3*e*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]]) + (16*(3*Cos[d + e*x
] - 4*Sin[d + e*x])*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]])/(3*e) - (2*(3*Cos[d + e*x] - 4*Sin[d + e*x])*(
-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))/(5*e)

Rule 3113

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> -Simp[((c*Cos[d
+ e*x] - b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1))/(e*n), x] + Dist[(a*(2*n - 1))/n, Int[
(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]
&& GtQ[n, 0]

Rule 3112

Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Simp[(-2*(c*Cos[d
+ e*x] - b*Sin[d + e*x]))/(e*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]), x] /; FreeQ[{a, b, c, d, e}, x] && E
qQ[a^2 - b^2 - c^2, 0]

Rubi steps

\begin{align*} \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \, dx &=-\frac{2 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}{5 e}-8 \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2} \, dx\\ &=\frac{16 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt{-5+4 \cos (d+e x)+3 \sin (d+e x)}}{3 e}-\frac{2 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}{5 e}+\frac{160}{3} \int \sqrt{-5+4 \cos (d+e x)+3 \sin (d+e x)} \, dx\\ &=-\frac{320 (3 \cos (d+e x)-4 \sin (d+e x))}{3 e \sqrt{-5+4 \cos (d+e x)+3 \sin (d+e x)}}+\frac{16 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt{-5+4 \cos (d+e x)+3 \sin (d+e x)}}{3 e}-\frac{2 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}{5 e}\\ \end{align*}

Mathematica [A]  time = 0.523038, size = 127, normalized size = 0.91 \[ \frac{(3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2} \left (3750 \sin \left (\frac{1}{2} (d+e x)\right )-1625 \sin \left (\frac{3}{2} (d+e x)\right )+237 \sin \left (\frac{5}{2} (d+e x)\right )+11250 \cos \left (\frac{1}{2} (d+e x)\right )-1125 \cos \left (\frac{3}{2} (d+e x)\right )-9 \cos \left (\frac{5}{2} (d+e x)\right )\right )}{30 e \left (\cos \left (\frac{1}{2} (d+e x)\right )-3 \sin \left (\frac{1}{2} (d+e x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(5/2),x]

[Out]

((-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(5/2)*(11250*Cos[(d + e*x)/2] - 1125*Cos[(3*(d + e*x))/2] - 9*Cos[(5*(
d + e*x))/2] + 3750*Sin[(d + e*x)/2] - 1625*Sin[(3*(d + e*x))/2] + 237*Sin[(5*(d + e*x))/2]))/(30*e*(Cos[(d +
e*x)/2] - 3*Sin[(d + e*x)/2])^5)

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Maple [A]  time = 1.248, size = 74, normalized size = 0.5 \begin{align*}{\frac{ \left ( 50\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) -50 \right ) \left ( 1+\sin \left ( ex+d+\arctan \left ({\frac{4}{3}} \right ) \right ) \right ) \left ( 3\, \left ( \sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) \right ) ^{2}-14\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) +43 \right ) }{3\,\cos \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) e}{\frac{1}{\sqrt{-5+5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5+4*cos(e*x+d)+3*sin(e*x+d))^(5/2),x)

[Out]

50/3*(sin(e*x+d+arctan(4/3))-1)*(1+sin(e*x+d+arctan(4/3)))*(3*sin(e*x+d+arctan(4/3))^2-14*sin(e*x+d+arctan(4/3
))+43)/cos(e*x+d+arctan(4/3))/(-5+5*sin(e*x+d+arctan(4/3)))^(1/2)/e

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5+4*cos(e*x+d)+3*sin(e*x+d))^(5/2),x, algorithm="maxima")

[Out]

integrate((4*cos(e*x + d) + 3*sin(e*x + d) - 5)^(5/2), x)

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Fricas [A]  time = 1.66577, size = 293, normalized size = 2.11 \begin{align*} -\frac{2 \,{\left (9 \, \cos \left (e x + d\right )^{3} + 567 \, \cos \left (e x + d\right )^{2} -{\left (237 \, \cos \left (e x + d\right )^{2} - 694 \, \cos \left (e x + d\right ) + 472\right )} \sin \left (e x + d\right ) - 2538 \, \cos \left (e x + d\right ) - 3096\right )} \sqrt{4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{15 \,{\left (e \cos \left (e x + d\right ) - 3 \, e \sin \left (e x + d\right ) + e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5+4*cos(e*x+d)+3*sin(e*x+d))^(5/2),x, algorithm="fricas")

[Out]

-2/15*(9*cos(e*x + d)^3 + 567*cos(e*x + d)^2 - (237*cos(e*x + d)^2 - 694*cos(e*x + d) + 472)*sin(e*x + d) - 25
38*cos(e*x + d) - 3096)*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) - 5)/(e*cos(e*x + d) - 3*e*sin(e*x + d) + e)

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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((-5+4*cos(e*x+d)+3*sin(e*x+d))**(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((-5+4*cos(e*x+d)+3*sin(e*x+d))^(5/2),x, algorithm="giac")

[Out]

Timed out