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

Optimal. Leaf size=73 \[ -\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{13 d \sqrt {3 \sin (c+d x)+2 \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right |2\right )}{13^{3/4} d} \]

[Out]

-2/13*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-2/13*(3*cos(d*x+c)-2*sin(d*x+c))/d/(2*cos(d*x+c)+3*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 73, 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, 2639} \[ -\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{13 d \sqrt {3 \sin (c+d x)+2 \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right |2\right )}{13^{3/4} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 1.03, size = 190, normalized size = 2.60 \[ \frac {\frac {3 \sin \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right )}{13^{3/4} \sqrt {-\left (\left (\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )-1\right ) \cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right )} \sqrt {\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )+1}}-\frac {2 \cos (c+d x)}{\sqrt {3 \sin (c+d x)+2 \cos (c+d x)}}+\frac {4 \sqrt {\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )}}{13^{3/4}}-\frac {3 \sin \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )}{13^{3/4} \sqrt {\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )}}}{3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((4*Sqrt[Cos[c + d*x - ArcTan[3/2]]])/13^(3/4) - (2*Cos[c + d*x])/Sqrt[2*Cos[c + d*x] + 3*Sin[c + d*x]] - (3*S
in[c + d*x - ArcTan[3/2]])/(13^(3/4)*Sqrt[Cos[c + d*x - ArcTan[3/2]]]) + (3*HypergeometricPFQ[{-1/2, -1/4}, {3
/4}, Cos[c + d*x - ArcTan[3/2]]^2]*Sin[c + d*x - ArcTan[3/2]])/(13^(3/4)*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]]]))/(3*d)

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fricas [F]  time = 1.09, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )}}{5 \, \cos \left (d x + c\right )^{2} - 12 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 9}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.34, size = 162, normalized size = 2.22 \[ \frac {\sqrt {13}\, \left (2 \sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \EllipticE \left (\sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \EllipticF \left (\sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\cos ^{2}\left (d x +c +\arctan \left (\frac {2}{3}\right )\right )\right )\right )}{13 \cos \left (d x +c +\arctan \left (\frac {2}{3}\right )\right ) \sqrt {\sqrt {13}\, \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13*13^(1/2)*(2*(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))-(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*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 \[ \int \frac {1}{{\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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