3.247 \(\int \frac {1}{(2 \cos (c+d x)+3 \sin (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=120 \[ -\frac {6 (3 \cos (c+d x)-2 \sin (c+d x))}{845 d \sqrt {3 \sin (c+d x)+2 \cos (c+d x)}}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{65 d (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}-\frac {6 E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right |2\right )}{65\ 13^{3/4} d} \]

[Out]

-6/845*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-2/65*(3*cos(d*x+c)-2*sin(d*x+c))/d/(2*cos(d*x+c)+3*sin(d*x+c))^(5/2)
-6/845*(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 = 0.06, antiderivative size = 120, normalized size of antiderivative = 1.00, 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 = {3076, 3077, 2639} \[ -\frac {6 (3 \cos (c+d x)-2 \sin (c+d x))}{845 d \sqrt {3 \sin (c+d x)+2 \cos (c+d x)}}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{65 d (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}-\frac {6 E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right |2\right )}{65\ 13^{3/4} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*EllipticE[(c + d*x - ArcTan[3/2])/2, 2])/(65*13^(3/4)*d) - (2*(3*Cos[c + d*x] - 2*Sin[c + d*x]))/(65*d*(2*
Cos[c + d*x] + 3*Sin[c + d*x])^(5/2)) - (6*(3*Cos[c + d*x] - 2*Sin[c + d*x]))/(845*d*Sqrt[2*Cos[c + d*x] + 3*S
in[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))^{7/2}} \, dx &=-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{65 d (2 \cos (c+d x)+3 \sin (c+d x))^{5/2}}+\frac {3}{65} \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))}{65 d (2 \cos (c+d x)+3 \sin (c+d x))^{5/2}}-\frac {6 (3 \cos (c+d x)-2 \sin (c+d x))}{845 d \sqrt {2 \cos (c+d x)+3 \sin (c+d x)}}-\frac {3}{845} \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))}{65 d (2 \cos (c+d x)+3 \sin (c+d x))^{5/2}}-\frac {6 (3 \cos (c+d x)-2 \sin (c+d x))}{845 d \sqrt {2 \cos (c+d x)+3 \sin (c+d x)}}-\frac {3 \int \sqrt {\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )} \, dx}{65\ 13^{3/4}}\\ &=-\frac {6 E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right |2\right )}{65\ 13^{3/4} d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{65 d (2 \cos (c+d x)+3 \sin (c+d x))^{5/2}}-\frac {6 (3 \cos (c+d x)-2 \sin (c+d x))}{845 d \sqrt {2 \cos (c+d x)+3 \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 1.95, size = 224, normalized size = 1.87 \[ \frac {\frac {3 \sin \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right )}{13^{3/4} \sqrt {-\left (\left (\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )-1\right ) \cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )\right )} \sqrt {\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )+1}}+\frac {-4 (\sin (c+d x)+3 \sin (3 (c+d x)))-33 \cos (c+d x)+5 \cos (3 (c+d x))}{2 (3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}+\frac {4 \sqrt {\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )}}{13^{3/4}}-\frac {3 \sin \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )}{13^{3/4} \sqrt {\cos \left (c+d x-\tan ^{-1}\left (\frac {3}{2}\right )\right )}}}{65 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((4*Sqrt[Cos[c + d*x - ArcTan[3/2]]])/13^(3/4) + (-33*Cos[c + d*x] + 5*Cos[3*(c + d*x)] - 4*(Sin[c + d*x] + 3*
Sin[3*(c + d*x)]))/(2*(2*Cos[c + d*x] + 3*Sin[c + d*x])^(5/2)) - (3*Sin[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]]]))/(65*d)

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fricas [F]  time = 1.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )}}{119 \, \cos \left (d x + c\right )^{4} - 54 \, \cos \left (d x + c\right )^{2} + 24 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 9 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 81}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*cos(d*x+c)+3*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(-sqrt(2*cos(d*x + c) + 3*sin(d*x + c))/(119*cos(d*x + c)^4 - 54*cos(d*x + c)^2 + 24*(5*cos(d*x + c)^3
- 9*cos(d*x + c))*sin(d*x + c) - 81), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*cos(d*x+c)+3*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.35, size = 205, normalized size = 1.71 \[ \frac {\sqrt {13}\, \left (6 \sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \left (\sin ^{2}\left (d x +c +\arctan \left (\frac {2}{3}\right )\right )\right ) \EllipticE \left (\sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \left (\sin ^{2}\left (d x +c +\arctan \left (\frac {2}{3}\right )\right )\right ) \EllipticF \left (\sqrt {1+\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}, \frac {\sqrt {2}}{2}\right )+6 \left (\sin ^{4}\left (d x +c +\arctan \left (\frac {2}{3}\right )\right )\right )-4 \left (\sin ^{2}\left (d x +c +\arctan \left (\frac {2}{3}\right )\right )\right )-2\right )}{845 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )^{2} \cos \left (d x +c +\arctan \left (\frac {2}{3}\right )\right ) \sqrt {\sqrt {13}\, \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*cos(d*x+c)+3*sin(d*x+c))^(7/2),x)

[Out]

1/845*13^(1/2)/sin(d*x+c+arctan(2/3))^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)*sin(d*x+c+arctan(2/3))^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)*si
n(d*x+c+arctan(2/3))^2*EllipticF((1+sin(d*x+c+arctan(2/3)))^(1/2),1/2*2^(1/2))+6*sin(d*x+c+arctan(2/3))^4-4*si
n(d*x+c+arctan(2/3))^2-2)/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 \[ \int \frac {1}{{\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*cos(d*x+c)+3*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (2\,\cos \left (c+d\,x\right )+3\,\sin \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*cos(c + d*x) + 3*sin(c + d*x))^(7/2),x)

[Out]

int(1/(2*cos(c + d*x) + 3*sin(c + d*x))^(7/2), 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))**(7/2),x)

[Out]

Timed out

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