3.423 \(\int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{7/2} \, dx\)

Optimal. Leaf size=185 \[ -\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}{7 e}+\frac {24 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}{7 e}-\frac {320 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}}{7 e}+\frac {6400 (3 \cos (d+e x)-4 \sin (d+e x))}{7 e \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}} \]

[Out]

24/7*(3*cos(e*x+d)-4*sin(e*x+d))*(-5+4*cos(e*x+d)+3*sin(e*x+d))^(3/2)/e-2/7*(3*cos(e*x+d)-4*sin(e*x+d))*(-5+4*
cos(e*x+d)+3*sin(e*x+d))^(5/2)/e+6400/7*(3*cos(e*x+d)-4*sin(e*x+d))/e/(-5+4*cos(e*x+d)+3*sin(e*x+d))^(1/2)-320
/7*(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 = 0.09, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3113, 3112} \[ -\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}{7 e}+\frac {24 (3 \cos (d+e x)-4 \sin (d+e x)) (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}{7 e}-\frac {320 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}}{7 e}+\frac {6400 (3 \cos (d+e x)-4 \sin (d+e x))}{7 e \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}} \]

Antiderivative was successfully verified.

[In]

Int[(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(7/2),x]

[Out]

(6400*(3*Cos[d + e*x] - 4*Sin[d + e*x]))/(7*e*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]]) - (320*(3*Cos[d + e*
x] - 4*Sin[d + e*x])*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]])/(7*e) + (24*(3*Cos[d + e*x] - 4*Sin[d + e*x])
*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))/(7*e) - (2*(3*Cos[d + e*x] - 4*Sin[d + e*x])*(-5 + 4*Cos[d + e*
x] + 3*Sin[d + e*x])^(5/2))/(7*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))^{7/2} \, dx &=-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}{7 e}-\frac {60}{7} \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2} \, dx\\ &=\frac {24 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}{7 e}-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}{7 e}+\frac {480}{7} \int (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2} \, dx\\ &=-\frac {320 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}}{7 e}+\frac {24 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}{7 e}-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}{7 e}-\frac {3200}{7} \int \sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)} \, dx\\ &=\frac {6400 (3 \cos (d+e x)-4 \sin (d+e x))}{7 e \sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}}-\frac {320 (3 \cos (d+e x)-4 \sin (d+e x)) \sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}}{7 e}+\frac {24 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}{7 e}-\frac {2 (3 \cos (d+e x)-4 \sin (d+e x)) (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}{7 e}\\ \end {align*}

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Mathematica [A]  time = 1.82, size = 151, normalized size = 0.82 \[ \frac {(3 \sin (d+e x)+4 \cos (d+e x)-5)^{7/2} \left (30625 \sin \left (\frac {1}{2} (d+e x)\right )-15925 \sin \left (\frac {3}{2} (d+e x)\right )+3871 \sin \left (\frac {5}{2} (d+e x)\right )-307 \sin \left (\frac {7}{2} (d+e x)\right )+91875 \cos \left (\frac {1}{2} (d+e x)\right )-11025 \cos \left (\frac {3}{2} (d+e x)\right )-147 \cos \left (\frac {5}{2} (d+e x)\right )+249 \cos \left (\frac {7}{2} (d+e x)\right )\right )}{28 e \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(7/2),x]

[Out]

((-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(7/2)*(91875*Cos[(d + e*x)/2] - 11025*Cos[(3*(d + e*x))/2] - 147*Cos[(
5*(d + e*x))/2] + 249*Cos[(7*(d + e*x))/2] + 30625*Sin[(d + e*x)/2] - 15925*Sin[(3*(d + e*x))/2] + 3871*Sin[(5
*(d + e*x))/2] - 307*Sin[(7*(d + e*x))/2]))/(28*e*(Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2])^7)

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fricas [A]  time = 0.85, size = 121, normalized size = 0.65 \[ -\frac {2 \, {\left (249 \, \cos \left (e x + d\right )^{4} + 51 \, \cos \left (e x + d\right )^{3} - 3042 \, \cos \left (e x + d\right )^{2} - {\left (307 \, \cos \left (e x + d\right )^{3} - 1782 \, \cos \left (e x + d\right )^{2} + 2860 \, \cos \left (e x + d\right ) - 1392\right )} \sin \left (e x + d\right ) + 10068 \, \cos \left (e x + d\right ) + 12912\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{7 \, {\left (e \cos \left (e x + d\right ) - 3 \, e \sin \left (e x + d\right ) + e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5+4*cos(e*x+d)+3*sin(e*x+d))^(7/2),x, algorithm="fricas")

[Out]

-2/7*(249*cos(e*x + d)^4 + 51*cos(e*x + d)^3 - 3042*cos(e*x + d)^2 - (307*cos(e*x + d)^3 - 1782*cos(e*x + d)^2
+ 2860*cos(e*x + d) - 1392)*sin(e*x + d) + 10068*cos(e*x + d) + 12912)*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5+4*cos(e*x+d)+3*sin(e*x+d))^(7/2),x, algorithm="giac")

[Out]

integrate((4*cos(e*x + d) + 3*sin(e*x + d) - 5)^(7/2), x)

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maple [A]  time = 0.28, size = 86, normalized size = 0.46 \[ \frac {250 \left (\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-1\right ) \left (1+\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )\right ) \left (5 \left (\sin ^{3}\left (e x +d +\arctan \left (\frac {4}{3}\right )\right )\right )-27 \left (\sin ^{2}\left (e x +d +\arctan \left (\frac {4}{3}\right )\right )\right )+71 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-177\right )}{7 \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {-5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5+4*cos(e*x+d)+3*sin(e*x+d))^(7/2),x)

[Out]

250/7*(sin(e*x+d+arctan(4/3))-1)*(1+sin(e*x+d+arctan(4/3)))*(5*sin(e*x+d+arctan(4/3))^3-27*sin(e*x+d+arctan(4/
3))^2+71*sin(e*x+d+arctan(4/3))-177)/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 \[ \int {\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5+4*cos(e*x+d)+3*sin(e*x+d))^(7/2),x, algorithm="maxima")

[Out]

integrate((4*cos(e*x + d) + 3*sin(e*x + d) - 5)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (4\,\cos \left (d+e\,x\right )+3\,\sin \left (d+e\,x\right )-5\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*cos(d + e*x) + 3*sin(d + e*x) - 5)^(7/2),x)

[Out]

int((4*cos(d + e*x) + 3*sin(d + e*x) - 5)^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5+4*cos(e*x+d)+3*sin(e*x+d))**(7/2),x)

[Out]

Timed out

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