3.246 \(\int \frac {1}{(2 \cos (c+d x)+3 \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=75 \[ \frac {2 F\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right |2\right )}{39 \sqrt [4]{13} d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{39 d (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}} \]

[Out]

2/507*13^(3/4)*(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))*EllipticF(sin(1
/2*c+1/2*d*x-1/2*arctan(3/2)),2^(1/2))/d-2/39*(3*cos(d*x+c)-2*sin(d*x+c))/d/(2*cos(d*x+c)+3*sin(d*x+c))^(3/2)

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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 = {3076, 3077, 2641} \[ \frac {2 F\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right |2\right )}{39 \sqrt [4]{13} d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{39 d (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*EllipticF[(c + d*x - ArcTan[3/2])/2, 2])/(39*13^(1/4)*d) - (2*(3*Cos[c + d*x] - 2*Sin[c + d*x]))/(39*d*(2*C
os[c + d*x] + 3*Sin[c + d*x])^(3/2))

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]

Rule 3076

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 + 1)*(a^2 + b^2)), x] + Dist[(n + 2)/((n + 1
)*(a^2 + b^2)), 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] && LtQ[n, -1] && NeQ[n, -2]

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

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Mathematica [C]  time = 0.71, size = 157, normalized size = 2.09 \[ \frac {\sqrt {2} 13^{3/4} \sqrt {\sin \left (c+d x+\tan ^{-1}\left (\frac {2}{3}\right )\right )+1} (3 \sin (c+d x)+2 \cos (c+d x))^{3/2} \sec \left (c+d x+\tan ^{-1}\left (\frac {2}{3}\right )\right ) \sqrt {2 \sin \left (c+d x+\tan ^{-1}\left (\frac {2}{3}\right )\right )+\cos \left (2 \left (c+d x+\tan ^{-1}\left (\frac {2}{3}\right )\right )\right )-1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (c+d x+\tan ^{-1}\left (\frac {2}{3}\right )\right )\right )+52 \sin (c+d x)-78 \cos (c+d x)}{507 d (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-78*Cos[c + d*x] + 52*Sin[c + d*x] + Sqrt[2]*13^(3/4)*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[c + d*x + ArcT
an[2/3]]^2]*Sec[c + d*x + ArcTan[2/3]]*(2*Cos[c + d*x] + 3*Sin[c + d*x])^(3/2)*Sqrt[1 + Sin[c + d*x + ArcTan[2
/3]]]*Sqrt[-1 + Cos[2*(c + d*x + ArcTan[2/3])] + 2*Sin[c + d*x + ArcTan[2/3]]])/(507*d*(2*Cos[c + d*x] + 3*Sin
[c + d*x])^(3/2))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )}}{46 \, \cos \left (d x + c\right )^{3} - 9 \, {\left (\cos \left (d x + c\right )^{2} + 3\right )} \sin \left (d x + c\right ) - 54 \, \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(2*cos(d*x + c) + 3*sin(d*x + c))/(46*cos(d*x + c)^3 - 9*(cos(d*x + c)^2 + 3)*sin(d*x + c) - 54*
cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 = 0.33, size = 118, normalized size = 1.57 \[ \frac {\sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \EllipticF \left (\sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )-2 \left (\cos ^{2}\left (d x +c +\arctan \left (\frac {2}{3}\right )\right )\right )}{39 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right ) \cos \left (d x +c +\arctan \left (\frac {2}{3}\right )\right ) \sqrt {\sqrt {13}\, \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x)

[Out]

1/39/sin(d*x+c+arctan(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))*sin(d*x+c+arctan(2/3))-2*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 \[ \int \frac {1}{{\left (2\,\cos \left (c+d\,x\right )+3\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*cos(c + d*x) + 3*sin(c + d*x))^(5/2),x)

[Out]

int(1/(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(1/(2*cos(d*x+c)+3*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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