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

Optimal. Leaf size=93 \[ \frac {40 (3 \cos (d+e x)-4 \sin (d+e x))}{3 e \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}}-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}}{3 e} \]

[Out]

40/3*(3*cos(e*x+d)-4*sin(e*x+d))/e/(-5+4*cos(e*x+d)+3*sin(e*x+d))^(1/2)-2/3*(3*cos(e*x+d)-4*sin(e*x+d))*(-5+4*
cos(e*x+d)+3*sin(e*x+d))^(1/2)/e

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Rubi [A]  time = 0.04, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3113, 3112} \[ \frac {40 (3 \cos (d+e x)-4 \sin (d+e x))}{3 e \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}}-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(40*(3*Cos[d + e*x] - 4*Sin[d + e*x]))/(3*e*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]]) - (2*(3*Cos[d + e*x] -
4*Sin[d + e*x])*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]])/(3*e)

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]

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]

Rubi steps

\begin {align*} \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2} \, dx &=-\frac {2 (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 {20}{3} \int \sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)} \, dx\\ &=\frac {40 (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 {2 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}}{3 e}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 103, normalized size = 1.11 \[ \frac {(3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2} \left (45 \sin \left (\frac {1}{2} (d+e x)\right )-13 \sin \left (\frac {3}{2} (d+e x)\right )+135 \cos \left (\frac {1}{2} (d+e x)\right )-9 \cos \left (\frac {3}{2} (d+e x)\right )\right )}{3 e \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2)*(135*Cos[(d + e*x)/2] - 9*Cos[(3*(d + e*x))/2] + 45*Sin[(d + e*x
)/2] - 13*Sin[(3*(d + e*x))/2]))/(3*e*(Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2])^3)

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fricas [A]  time = 0.73, size = 80, normalized size = 0.86 \[ \frac {2 \, {\left (9 \, \cos \left (e x + d\right )^{2} + {\left (13 \, \cos \left (e x + d\right ) - 16\right )} \sin \left (e x + d\right ) - 63 \, \cos \left (e x + d\right ) - 72\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{3 \, {\left (e \cos \left (e x + d\right ) - 3 \, e \sin \left (e x + d\right ) + e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(9*cos(e*x + d)^2 + (13*cos(e*x + d) - 16)*sin(e*x + d) - 63*cos(e*x + d) - 72)*sqrt(4*cos(e*x + d) + 3*si
n(e*x + d) - 5)/(e*cos(e*x + d) - 3*e*sin(e*x + d) + e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.31, size = 60, normalized size = 0.65 \[ \frac {50 \left (\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-1\right ) \left (1+\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )\right ) \left (\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5\right )}{3 \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {-5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50/3*(sin(e*x+d+arctan(4/3))-1)*(1+sin(e*x+d+arctan(4/3)))*(sin(e*x+d+arctan(4/3))-5)/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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (4\,\cos \left (d+e\,x\right )+3\,\sin \left (d+e\,x\right )-5\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (3 \sin {\left (d + e x \right )} + 4 \cos {\left (d + e x \right )} - 5\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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