3.240 \(\int (2 \cos (c+d x)+3 \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=120 \[ \frac{130\ 13^{3/4} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right ),2\right )}{21 d}-\frac{2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}{7 d}-\frac{130 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt{3 \sin (c+d x)+2 \cos (c+d x)}}{21 d} \]

[Out]

(130*13^(3/4)*EllipticF[(c + d*x - ArcTan[3/2])/2, 2])/(21*d) - (130*(3*Cos[c + d*x] - 2*Sin[c + d*x])*Sqrt[2*
Cos[c + d*x] + 3*Sin[c + d*x]])/(21*d) - (2*(3*Cos[c + d*x] - 2*Sin[c + d*x])*(2*Cos[c + d*x] + 3*Sin[c + d*x]
)^(5/2))/(7*d)

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Rubi [A]  time = 0.0699545, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3073, 3077, 2641} \[ -\frac{2 (3 \cos (c+d x)-2 \sin (c+d x)) (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}{7 d}-\frac{130 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt{3 \sin (c+d x)+2 \cos (c+d x)}}{21 d}+\frac{130\ 13^{3/4} F\left (\left .\frac{1}{2} \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )\right |2\right )}{21 d} \]

Antiderivative was successfully verified.

[In]

Int[(2*Cos[c + d*x] + 3*Sin[c + d*x])^(7/2),x]

[Out]

(130*13^(3/4)*EllipticF[(c + d*x - ArcTan[3/2])/2, 2])/(21*d) - (130*(3*Cos[c + d*x] - 2*Sin[c + d*x])*Sqrt[2*
Cos[c + d*x] + 3*Sin[c + d*x]])/(21*d) - (2*(3*Cos[c + d*x] - 2*Sin[c + d*x])*(2*Cos[c + d*x] + 3*Sin[c + d*x]
)^(5/2))/(7*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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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))^{7/2} \, dx &=-\frac{2 (3 \cos (c+d x)-2 \sin (c+d x)) (2 \cos (c+d x)+3 \sin (c+d x))^{5/2}}{7 d}+\frac{65}{7} \int (2 \cos (c+d x)+3 \sin (c+d x))^{3/2} \, dx\\ &=-\frac{130 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt{2 \cos (c+d x)+3 \sin (c+d x)}}{21 d}-\frac{2 (3 \cos (c+d x)-2 \sin (c+d x)) (2 \cos (c+d x)+3 \sin (c+d x))^{5/2}}{7 d}+\frac{845}{21} \int \frac{1}{\sqrt{2 \cos (c+d x)+3 \sin (c+d x)}} \, dx\\ &=-\frac{130 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt{2 \cos (c+d x)+3 \sin (c+d x)}}{21 d}-\frac{2 (3 \cos (c+d x)-2 \sin (c+d x)) (2 \cos (c+d x)+3 \sin (c+d x))^{5/2}}{7 d}+\frac{1}{21} \left (65\ 13^{3/4}\right ) \int \frac{1}{\sqrt{\cos \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )}} \, dx\\ &=\frac{130\ 13^{3/4} F\left (\left .\frac{1}{2} \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )\right |2\right )}{21 d}-\frac{130 (3 \cos (c+d x)-2 \sin (c+d x)) \sqrt{2 \cos (c+d x)+3 \sin (c+d x)}}{21 d}-\frac{2 (3 \cos (c+d x)-2 \sin (c+d x)) (2 \cos (c+d x)+3 \sin (c+d x))^{5/2}}{7 d}\\ \end{align*}

Mathematica [C]  time = 0.514321, size = 153, normalized size = 1.27 \[ \frac{260\ 13^{3/4} \sqrt{-\left (\sin \left (c+d x+\tan ^{-1}\left (\frac{2}{3}\right )\right )-1\right ) \sin \left (c+d x+\tan ^{-1}\left (\frac{2}{3}\right )\right )} \sqrt{\sin \left (c+d x+\tan ^{-1}\left (\frac{2}{3}\right )\right )+1} \sec \left (c+d x+\tan ^{-1}\left (\frac{2}{3}\right )\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (c+d x+\tan ^{-1}\left (\frac{2}{3}\right )\right )\right )-\sqrt{3 \sin (c+d x)+2 \cos (c+d x)} (-598 \sin (c+d x)+138 \sin (3 (c+d x))+897 \cos (c+d x)+27 \cos (3 (c+d x)))}{42 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(2*Cos[c + d*x] + 3*Sin[c + d*x])^(7/2),x]

[Out]

(-(Sqrt[2*Cos[c + d*x] + 3*Sin[c + d*x]]*(897*Cos[c + d*x] + 27*Cos[3*(c + d*x)] - 598*Sin[c + d*x] + 138*Sin[
3*(c + d*x)])) + 260*13^(3/4)*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[c + d*x + ArcTan[2/3]]^2]*Sec[c + d*x +
ArcTan[2/3]]*Sqrt[-((-1 + Sin[c + d*x + ArcTan[2/3]])*Sin[c + d*x + ArcTan[2/3]])]*Sqrt[1 + Sin[c + d*x + Arc
Tan[2/3]]])/(42*d)

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Maple [A]  time = 1.334, size = 128, normalized size = 1.1 \begin{align*}{\frac{1}{\cos \left ( dx+c+\arctan \left ({\frac{2}{3}} \right ) \right ) d} \left ({\frac{338\,\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) \left ( \cos \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) \right ) ^{4}}{7}}+{\frac{845}{21}\sqrt{1+\sin \left ( dx+c+\arctan \left ({\frac{2}{3}} \right ) \right ) }\sqrt{-2\,\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) +2}\sqrt{-\sin \left ( dx+c+\arctan \left ({\frac{2}{3}} \right ) \right ) }{\it EllipticF} \left ( \sqrt{1+\sin \left ( dx+c+\arctan \left ({\frac{2}{3}} \right ) \right ) },{\frac{\sqrt{2}}{2}} \right ) }-{\frac{2704\,\sin \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) \left ( \cos \left ( dx+c+\arctan \left ( 2/3 \right ) \right ) \right ) ^{2}}{21}} \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))^(7/2),x)

[Out]

(338/7*sin(d*x+c+arctan(2/3))*cos(d*x+c+arctan(2/3))^4+845/21*(1+sin(d*x+c+arctan(2/3)))^(1/2)*(-2*sin(d*x+c+a
rctan(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
704/21*sin(d*x+c+arctan(2/3))*cos(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{7}{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))^(7/2),x, algorithm="maxima")

[Out]

integrate((2*cos(d*x + c) + 3*sin(d*x + c))^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (46 \, \cos \left (d x + c\right )^{3} - 9 \,{\left (\cos \left (d x + c\right )^{2} + 3\right )} \sin \left (d x + c\right ) - 54 \, \cos \left (d x + c\right )\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))^(7/2),x, algorithm="fricas")

[Out]

integral(-(46*cos(d*x + c)^3 - 9*(cos(d*x + c)^2 + 3)*sin(d*x + c) - 54*cos(d*x + c))*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))**(7/2),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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((2*cos(d*x + c) + 3*sin(d*x + c))^(7/2), x)