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

Optimal. Leaf size=142 \[ -\frac{3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}}-\frac{3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}+\frac{3 \tanh ^{-1}\left (\frac{\sin \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}{\sqrt{2} \sqrt{\cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )+1}}\right )}{400 \sqrt{10} e} \]

[Out]

(3*ArcTanh[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[1 + Cos[d + e*x - ArcTan[3/4]]])])/(400*Sqrt[10]*e) - (3*C
os[d + e*x] - 4*Sin[d + e*x])/(20*e*(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(5/2)) - (3*(3*Cos[d + e*x] - 4*Sin[
d + e*x]))/(400*e*(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))

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Rubi [A]  time = 0.0767014, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3116, 3115, 2649, 206} \[ -\frac{3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}}-\frac{3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}+\frac{3 \tanh ^{-1}\left (\frac{\sin \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}{\sqrt{2} \sqrt{\cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )+1}}\right )}{400 \sqrt{10} e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[1 + Cos[d + e*x - ArcTan[3/4]]])])/(400*Sqrt[10]*e) - (3*C
os[d + e*x] - 4*Sin[d + e*x])/(20*e*(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(5/2)) - (3*(3*Cos[d + e*x] - 4*Sin[
d + e*x]))/(400*e*(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))

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]

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 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 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])

Rubi steps

\begin{align*} \int \frac{1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx &=-\frac{3 \cos (d+e x)-4 \sin (d+e x)}{20 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}+\frac{3}{40} \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)}{20 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac{3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}+\frac{3}{800} \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)}{20 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac{3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}+\frac{3}{800} \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)}{20 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac{3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}-\frac{3 \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 )}{400 e}\\ &=\frac{3 \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 )}{400 \sqrt{10} e}-\frac{3 \cos (d+e x)-4 \sin (d+e x)}{20 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac{3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.403138, size = 180, normalized size = 1.27 \[ -\frac{\left (\frac{1}{20000}-\frac{i}{10000}\right ) \left (\sin \left (\frac{1}{2} (d+e x)\right )+3 \cos \left (\frac{1}{2} (d+e x)\right )\right ) \left ((5+10 i) \left (-165 \sin \left (\frac{1}{2} (d+e x)\right )-27 \sin \left (\frac{3}{2} (d+e x)\right )+55 \cos \left (\frac{1}{2} (d+e x)\right )+39 \cos \left (\frac{3}{2} (d+e x)\right )\right )-(6-6 i) \sqrt{20+15 i} \tan ^{-1}\left (\left (\frac{1}{10}+\frac{3 i}{10}\right ) \sqrt{\frac{4}{5}+\frac{3 i}{5}} \left (3 \tan \left (\frac{1}{4} (d+e x)\right )-1\right )\right ) \left (\sin \left (\frac{1}{2} (d+e x)\right )+3 \cos \left (\frac{1}{2} (d+e x)\right )\right )^4\right )}{e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/20000 + I/10000)*(3*Cos[(d + e*x)/2] + Sin[(d + e*x)/2])*((-6 + 6*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])^4 + (5 + 10*I)*(55
*Cos[(d + e*x)/2] + 39*Cos[(3*(d + e*x))/2] - 165*Sin[(d + e*x)/2] - 27*Sin[(3*(d + e*x))/2])))/(e*(5 + 4*Cos[
d + e*x] + 3*Sin[d + e*x])^(5/2))

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Maple [A]  time = 1.46, size = 190, normalized size = 1.3 \begin{align*} -{\frac{1}{ \left ( 4000+4000\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) \right ) \cos \left ( ex+d+\arctan \left ({\frac{4}{3}} \right ) \right ) e} \left ( 3\,\sqrt{10}{\it Artanh} \left ( 1/10\,\sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) +5}\sqrt{10} \right ) \left ( \sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) \right ) ^{2}+6\,\sqrt{10}{\it Artanh} \left ( 1/10\,\sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) +5}\sqrt{10} \right ) \sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) +3\,\sqrt{10}{\it Artanh} \left ( 1/10\,\sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) +5}\sqrt{10} \right ) +6\,\sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) +5}\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) +14\,\sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) +5} \right ) \sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) +5}{\frac{1}{\sqrt{5+5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4000*(3*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))^2+6*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))+3*10^(1/2)*arctanh(1/10
*(-5*sin(e*x+d+arctan(4/3))+5)^(1/2)*10^(1/2))+6*(-5*sin(e*x+d+arctan(4/3))+5)^(1/2)*sin(e*x+d+arctan(4/3))+14
*(-5*sin(e*x+d+arctan(4/3))+5)^(1/2))*(-5*sin(e*x+d+arctan(4/3))+5)^(1/2)/(1+sin(e*x+d+arctan(4/3)))/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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.97268, size = 998, normalized size = 7.03 \begin{align*} \frac{3 \,{\left (3 \, \sqrt{10} \cos \left (e x + d\right )^{3} - 111 \, \sqrt{10} \cos \left (e x + d\right )^{2} -{\left (79 \, \sqrt{10} \cos \left (e x + d\right )^{2} + 202 \, \sqrt{10} \cos \left (e x + d\right ) + 124 \, \sqrt{10}\right )} \sin \left (e x + d\right ) - 246 \, \sqrt{10} \cos \left (e x + d\right ) - 132 \, \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 \,{\left (39 \, \cos \left (e x + d\right )^{2} - 3 \,{\left (9 \, \cos \left (e x + d\right ) + 32\right )} \sin \left (e x + d\right ) + 47 \, \cos \left (e x + d\right ) + 8\right )} \sqrt{4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5}}{8000 \,{\left (3 \, e \cos \left (e x + d\right )^{3} - 111 \, e \cos \left (e x + d\right )^{2} - 246 \, e \cos \left (e x + d\right ) -{\left (79 \, e \cos \left (e x + d\right )^{2} + 202 \, e \cos \left (e x + d\right ) + 124 \, e\right )} \sin \left (e x + d\right ) - 132 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8000*(3*(3*sqrt(10)*cos(e*x + d)^3 - 111*sqrt(10)*cos(e*x + d)^2 - (79*sqrt(10)*cos(e*x + d)^2 + 202*sqrt(10
)*cos(e*x + d) + 124*sqrt(10))*sin(e*x + d) - 246*sqrt(10)*cos(e*x + d) - 132*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*(39*cos(e*x + d)^2 - 3*(9*cos(e*x + d) + 32)*sin(e*x + d) + 47*cos(e*x + d)
+ 8)*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) + 5))/(3*e*cos(e*x + d)^3 - 111*e*cos(e*x + d)^2 - 246*e*cos(e*x +
d) - (79*e*cos(e*x + d)^2 + 202*e*cos(e*x + d) + 124*e)*sin(e*x + d) - 132*e)

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

[Out]

Timed out

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Giac [B]  time = 1.97177, size = 563, normalized size = 3.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4000*(3*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*sqrt(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*(797*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^7 - 7137*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1)
- tan(1/2*x*e + 1/2*d))^6 + 27543*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^5 - 30015*(sqrt(ta
n(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^4 - 27105*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e +
1/2*d))^3 - 7491*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^2 - 859*sqrt(tan(1/2*x*e + 1/2*d)^2
+ 1) + 859*tan(1/2*x*e + 1/2*d) - 69)/(((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)^4*sgn(tan(1/2*x*e + 1/2*d) + 3)))*e^(-1)