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*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)

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Rubi [A]  time = 0.0445689, antiderivative size = 75, normalized size of antiderivative = 1., 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 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]

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 (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*}

Mathematica [C]  time = 0.876895, size = 199, normalized size = 2.65 \[ \frac{-\frac{39 \sqrt [4]{13} \sin \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )\right )}{\sqrt{-\left (\cos \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )-1\right ) \cos \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )} \sqrt{\cos \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )+1}}+\sqrt{3 \sin (c+d x)+2 \cos (c+d x)} (-5 \sin (2 (c+d x))-12 \cos (2 (c+d x))+52)-\frac{13 \sqrt [4]{13} \left (4 \cos \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )-3 \sin \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )\right )}{\sqrt{\cos \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )}}}{5 d} \]

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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Maple [A]  time = 1.238, size = 174, normalized size = 2.3 \begin{align*} -{\frac{13\,\sqrt{13}}{5\,\cos \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) d} \left ( 6\,\sqrt{1+\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) }\sqrt{-2\,\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) +2}\sqrt{-\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) }{\it EllipticE} \left ( \sqrt{1+\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) },1/2\,\sqrt{2} \right ) -3\,\sqrt{1+\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) }\sqrt{-2\,\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) +2}\sqrt{-\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) }{\it EllipticF} \left ( \sqrt{1+\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) },1/2\,\sqrt{2} \right ) -2\, \left ( \sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) \right ) ^{4}+2\, \left ( \sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{\sqrt{13}\sin \left ( dx+c+\arctan \left ({\frac{2}{3}} \right ) \right ) }}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac{5}{2}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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

[Out]

Timed out

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

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)