3.421 \(\int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac {\tanh ^{-1}\left (\frac {\sin \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {\cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )+1}}\right )}{10 \sqrt {10} e}-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}} \]

[Out]

1/10*(-3*cos(e*x+d)+4*sin(e*x+d))/e/(5+4*cos(e*x+d)+3*sin(e*x+d))^(3/2)+1/100*arctanh(1/2*sin(d+e*x-arctan(3/4
))*2^(1/2)/(1+cos(d+e*x-arctan(3/4)))^(1/2))*10^(1/2)/e

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Rubi [A]  time = 0.05, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3116, 3115, 2649, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sin \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {\cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )+1}}\right )}{10 \sqrt {10} e}-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(-3/2),x]

[Out]

ArcTanh[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[1 + Cos[d + e*x - ArcTan[3/4]]])]/(10*Sqrt[10]*e) - (3*Cos[d
+ e*x] - 4*Sin[d + e*x])/(10*e*(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3115

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Int[1/Sqrt[a +
Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]

Rule 3116

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> Simp[((c*Cos[d +
e*x] - b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^n)/(a*e*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n +
1)), Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 -
c^2, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx &=-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}+\frac {1}{20} \int \frac {1}{\sqrt {5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx\\ &=-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}+\frac {1}{20} \int \frac {1}{\sqrt {5+5 \cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}} \, dx\\ &=-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{10-x^2} \, dx,x,-\frac {5 \sin \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}{\sqrt {5+5 \cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}}\right )}{10 e}\\ &=\frac {\tanh ^{-1}\left (\frac {\sin \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {1+\cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}}\right )}{10 \sqrt {10} e}-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.29, size = 154, normalized size = 1.60 \[ -\frac {\left (\frac {1}{250}-\frac {i}{125}\right ) \left (\sin \left (\frac {1}{2} (d+e x)\right )+3 \cos \left (\frac {1}{2} (d+e x)\right )\right ) \left ((5+10 i) \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right )-(1-i) \sqrt {20+15 i} \tan ^{-1}\left (\left (\frac {1}{10}+\frac {3 i}{10}\right ) \sqrt {\frac {4}{5}+\frac {3 i}{5}} \left (3 \tan \left (\frac {1}{4} (d+e x)\right )-1\right )\right ) \left (\sin \left (\frac {1}{2} (d+e x)\right )+3 \cos \left (\frac {1}{2} (d+e x)\right )\right )^2\right )}{e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(-3/2),x]

[Out]

((-1/250 + I/125)*(3*Cos[(d + e*x)/2] + Sin[(d + e*x)/2])*((5 + 10*I)*(Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2])
- (1 - I)*Sqrt[20 + 15*I]*ArcTan[(1/10 + (3*I)/10)*Sqrt[4/5 + (3*I)/5]*(-1 + 3*Tan[(d + e*x)/4])]*(3*Cos[(d +
e*x)/2] + Sin[(d + e*x)/2])^2))/(e*(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))

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fricas [B]  time = 0.93, size = 268, normalized size = 2.79 \[ \frac {{\left (9 \, \sqrt {10} \cos \left (e x + d\right )^{2} + {\left (13 \, \sqrt {10} \cos \left (e x + d\right ) + 14 \, \sqrt {10}\right )} \sin \left (e x + d\right ) + 27 \, \sqrt {10} \cos \left (e x + d\right ) + 18 \, \sqrt {10}\right )} \log \left (-\frac {9 \, \cos \left (e x + d\right )^{2} + {\left (13 \, \cos \left (e x + d\right ) - 6\right )} \sin \left (e x + d\right ) + 2 \, {\left (\sqrt {10} \cos \left (e x + d\right ) - 3 \, \sqrt {10} \sin \left (e x + d\right ) + \sqrt {10}\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5} - 33 \, \cos \left (e x + d\right ) - 42}{9 \, \cos \left (e x + d\right )^{2} + {\left (13 \, \cos \left (e x + d\right ) + 14\right )} \sin \left (e x + d\right ) + 27 \, \cos \left (e x + d\right ) + 18}\right ) - 20 \, \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5} {\left (\cos \left (e x + d\right ) - 3 \, \sin \left (e x + d\right ) + 1\right )}}{200 \, {\left (9 \, e \cos \left (e x + d\right )^{2} + 27 \, e \cos \left (e x + d\right ) + {\left (13 \, e \cos \left (e x + d\right ) + 14 \, e\right )} \sin \left (e x + d\right ) + 18 \, e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+4*cos(e*x+d)+3*sin(e*x+d))^(3/2),x, algorithm="fricas")

[Out]

1/200*((9*sqrt(10)*cos(e*x + d)^2 + (13*sqrt(10)*cos(e*x + d) + 14*sqrt(10))*sin(e*x + d) + 27*sqrt(10)*cos(e*
x + d) + 18*sqrt(10))*log(-(9*cos(e*x + d)^2 + (13*cos(e*x + d) - 6)*sin(e*x + d) + 2*(sqrt(10)*cos(e*x + d) -
3*sqrt(10)*sin(e*x + d) + sqrt(10))*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) + 5) - 33*cos(e*x + d) - 42)/(9*cos(
e*x + d)^2 + (13*cos(e*x + d) + 14)*sin(e*x + d) + 27*cos(e*x + d) + 18)) - 20*sqrt(4*cos(e*x + d) + 3*sin(e*x
+ d) + 5)*(cos(e*x + d) - 3*sin(e*x + d) + 1))/(9*e*cos(e*x + d)^2 + 27*e*cos(e*x + d) + (13*e*cos(e*x + d) +
14*e)*sin(e*x + d) + 18*e)

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giac [B]  time = 0.74, size = 284, normalized size = 2.96 \[ \frac {1}{100} \, {\left (\frac {\sqrt {10} \log \left (\frac {{\left | -2 \, \sqrt {10} + 2 \, \sqrt {\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 1} - 2 \, \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 6 \right |}}{{\left | 2 \, \sqrt {10} + 2 \, \sqrt {\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 1} - 2 \, \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 6 \right |}}\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 3\right )} - \frac {20 \, {\left (19 \, {\left (\sqrt {\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 1} - \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )\right )}^{3} - 51 \, {\left (\sqrt {\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 1} - \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )\right )}^{2} - 17 \, \sqrt {\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 1} + 17 \, \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 3\right )}}{{\left ({\left (\sqrt {\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 1} - \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )\right )}^{2} - 6 \, \sqrt {\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 1} + 6 \, \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 1\right )}^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 3\right )}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+4*cos(e*x+d)+3*sin(e*x+d))^(3/2),x, algorithm="giac")

[Out]

1/100*(sqrt(10)*log(abs(-2*sqrt(10) + 2*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - 2*tan(1/2*x*e + 1/2*d) - 6)/abs(2*s
qrt(10) + 2*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - 2*tan(1/2*x*e + 1/2*d) - 6))/sgn(tan(1/2*x*e + 1/2*d) + 3) - 20
*(19*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^3 - 51*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(
1/2*x*e + 1/2*d))^2 - 17*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) + 17*tan(1/2*x*e + 1/2*d) - 3)/(((sqrt(tan(1/2*x*e +
1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^2 - 6*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) + 6*tan(1/2*x*e + 1/2*d) - 1)^2*
sgn(tan(1/2*x*e + 1/2*d) + 3)))*e^(-1)

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maple [A]  time = 0.31, size = 117, normalized size = 1.22 \[ -\frac {\left (\sqrt {10}\, \arctanh \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\, \sqrt {10}}{10}\right ) \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+\sqrt {10}\, \arctanh \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\, \sqrt {10}}{10}\right )+2 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\right ) \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}}{100 \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5+4*cos(e*x+d)+3*sin(e*x+d))^(3/2),x)

[Out]

-1/100*(10^(1/2)*arctanh(1/10*(-5*sin(e*x+d+arctan(4/3))+5)^(1/2)*10^(1/2))*sin(e*x+d+arctan(4/3))+10^(1/2)*ar
ctanh(1/10*(-5*sin(e*x+d+arctan(4/3))+5)^(1/2)*10^(1/2))+2*(-5*sin(e*x+d+arctan(4/3))+5)^(1/2))*(-5*sin(e*x+d+
arctan(4/3))+5)^(1/2)/cos(e*x+d+arctan(4/3))/(5+5*sin(e*x+d+arctan(4/3)))^(1/2)/e

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+4*cos(e*x+d)+3*sin(e*x+d))^(3/2),x, algorithm="maxima")

[Out]

integrate((4*cos(e*x + d) + 3*sin(e*x + d) + 5)^(-3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (4\,\cos \left (d+e\,x\right )+3\,\sin \left (d+e\,x\right )+5\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*cos(d + e*x) + 3*sin(d + e*x) + 5)^(3/2),x)

[Out]

int(1/(4*cos(d + e*x) + 3*sin(d + e*x) + 5)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (3 \sin {\left (d + e x \right )} + 4 \cos {\left (d + e x \right )} + 5\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+4*cos(e*x+d)+3*sin(e*x+d))**(3/2),x)

[Out]

Integral((3*sin(d + e*x) + 4*cos(d + e*x) + 5)**(-3/2), x)

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