3.241 \(\int (2 \cos (c+d x)+3 \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=75 \[ \frac {78 \sqrt [4]{13} E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right |2\right )}{5 d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}{5 d} \]

[Out]

78/5*13^(1/4)*(cos(1/2*c+1/2*d*x-1/2*arctan(3/2))^2)^(1/2)/cos(1/2*c+1/2*d*x-1/2*arctan(3/2))*EllipticE(sin(1/
2*c+1/2*d*x-1/2*arctan(3/2)),2^(1/2))/d-2/5*(3*cos(d*x+c)-2*sin(d*x+c))*(2*cos(d*x+c)+3*sin(d*x+c))^(3/2)/d

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Rubi [A]  time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3073, 3077, 2639} \[ \frac {78 \sqrt [4]{13} E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right |2\right )}{5 d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(78*13^(1/4)*EllipticE[(c + d*x - ArcTan[3/2])/2, 2])/(5*d) - (2*(3*Cos[c + d*x] - 2*Sin[c + d*x])*(2*Cos[c +
d*x] + 3*Sin[c + d*x])^(3/2))/(5*d)

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]

Rule 3073

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

Rule 3077

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

Rubi steps

\begin {align*} \int (2 \cos (c+d x)+3 \sin (c+d x))^{5/2} \, dx &=-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (2 \cos (c+d x)+3 \sin (c+d x))^{3/2}}{5 d}+\frac {39}{5} \int \sqrt {2 \cos (c+d x)+3 \sin (c+d x)} \, dx\\ &=-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (2 \cos (c+d x)+3 \sin (c+d x))^{3/2}}{5 d}+\frac {1}{5} \left (39 \sqrt [4]{13}\right ) \int \sqrt {\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )} \, dx\\ &=\frac {78 \sqrt [4]{13} E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right |2\right )}{5 d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x)) (2 \cos (c+d x)+3 \sin (c+d x))^{3/2}}{5 d}\\ \end {align*}

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Mathematica [C]  time = 0.82, size = 199, normalized size = 2.65 \[ \frac {-\frac {39 \sqrt [4]{13} \sin \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right )}{\sqrt {-\left (\left (\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )-1\right ) \cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right )} \sqrt {\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )+1}}+\sqrt {3 \sin (c+d x)+2 \cos (c+d x)} (-5 \sin (2 (c+d x))-12 \cos (2 (c+d x))+52)-\frac {13 \sqrt [4]{13} \left (4 \cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )-3 \sin \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right )}{\sqrt {\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )}}}{5 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[2*Cos[c + d*x] + 3*Sin[c + d*x]]*(52 - 12*Cos[2*(c + d*x)] - 5*Sin[2*(c + d*x)]) - (13*13^(1/4)*(4*Cos[c
+ d*x - ArcTan[3/2]] - 3*Sin[c + d*x - ArcTan[3/2]]))/Sqrt[Cos[c + d*x - ArcTan[3/2]]] - (39*13^(1/4)*Hyperge
ometricPFQ[{-1/2, -1/4}, {3/4}, Cos[c + d*x - ArcTan[3/2]]^2]*Sin[c + d*x - ArcTan[3/2]])/(Sqrt[-((-1 + Cos[c
+ d*x - ArcTan[3/2]])*Cos[c + d*x - ArcTan[3/2]])]*Sqrt[1 + Cos[c + d*x - ArcTan[3/2]]]))/(5*d)

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (5 \, \cos \left (d x + c\right )^{2} - 12 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 9\right )} \sqrt {2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(5*cos(d*x + c)^2 - 12*cos(d*x + c)*sin(d*x + c) - 9)*sqrt(2*cos(d*x + c) + 3*sin(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((2*cos(d*x + c) + 3*sin(d*x + c))^(5/2), x)

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maple [A]  time = 0.37, size = 174, normalized size = 2.32 \[ -\frac {13 \sqrt {13}\, \left (6 \sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \EllipticE \left (\sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \EllipticF \left (\sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{4}\left (d x +c +\arctan \left (\frac {2}{3}\right )\right )\right )+2 \left (\sin ^{2}\left (d x +c +\arctan \left (\frac {2}{3}\right )\right )\right )\right )}{5 \cos \left (d x +c +\arctan \left (\frac {2}{3}\right )\right ) \sqrt {\sqrt {13}\, \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x)

[Out]

-13/5*13^(1/2)*(6*(1+sin(d*x+c+arctan(2/3)))^(1/2)*(-2*sin(d*x+c+arctan(2/3))+2)^(1/2)*(-sin(d*x+c+arctan(2/3)
))^(1/2)*EllipticE((1+sin(d*x+c+arctan(2/3)))^(1/2),1/2*2^(1/2))-3*(1+sin(d*x+c+arctan(2/3)))^(1/2)*(-2*sin(d*
x+c+arctan(2/3))+2)^(1/2)*(-sin(d*x+c+arctan(2/3)))^(1/2)*EllipticF((1+sin(d*x+c+arctan(2/3)))^(1/2),1/2*2^(1/
2))-2*sin(d*x+c+arctan(2/3))^4+2*sin(d*x+c+arctan(2/3))^2)/cos(d*x+c+arctan(2/3))/(13^(1/2)*sin(d*x+c+arctan(2
/3)))^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((2*cos(d*x + c) + 3*sin(d*x + c))^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*cos(c + d*x) + 3*sin(c + d*x))^(5/2),x)

[Out]

int((2*cos(c + d*x) + 3*sin(c + d*x))^(5/2), 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((2*cos(d*x+c)+3*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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