\(\int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx\) [356]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F(-1)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 22, antiderivative size = 142 \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=-\frac {3 \arctan \left (\frac {\sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {-1+\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}}\right )}{400 \sqrt {10} e}+\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac {3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \] Output:

-3/4000*10^(1/2)*arctan(1/2*sin(d+e*x-arctan(3/4))*2^(1/2)/(-1+cos(d+e*x-a 
rctan(3/4)))^(1/2))/e+1/20*(3*cos(e*x+d)-4*sin(e*x+d))/e/(-5+4*cos(e*x+d)+ 
3*sin(e*x+d))^(5/2)-3/400*(3*cos(e*x+d)-4*sin(e*x+d))/e/(-5+4*cos(e*x+d)+3 
*sin(e*x+d))^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.44 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.25 \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\frac {\left (\frac {1}{10000}+\frac {i}{20000}\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right ) \left ((6+6 i) \sqrt {-20-15 i} \text {arctanh}\left (\left (\frac {1}{10}+\frac {3 i}{10}\right ) \sqrt {-\frac {4}{5}-\frac {3 i}{5}} \left (3+\tan \left (\frac {1}{4} (d+e x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right )^4+(10-5 i) \left (165 \cos \left (\frac {1}{2} (d+e x)\right )-27 \cos \left (\frac {3}{2} (d+e x)\right )+55 \sin \left (\frac {1}{2} (d+e x)\right )-39 \sin \left (\frac {3}{2} (d+e x)\right )\right )\right )}{e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \] Input:

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

Output:

((1/10000 + I/20000)*(Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2])*((6 + 6*I)*Sq 
rt[-20 - 15*I]*ArcTanh[(1/10 + (3*I)/10)*Sqrt[-4/5 - (3*I)/5]*(3 + Tan[(d 
+ e*x)/4])]*(Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2])^4 + (10 - 5*I)*(165*Co 
s[(d + e*x)/2] - 27*Cos[(3*(d + e*x))/2] + 55*Sin[(d + e*x)/2] - 39*Sin[(3 
*(d + e*x))/2])))/(e*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {3042, 3595, 3042, 3595, 3042, 3594, 3042, 3128, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3595

\(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \int \frac {1}{(4 \cos (d+e x)+3 \sin (d+e x)-5)^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \int \frac {1}{(4 \cos (d+e x)+3 \sin (d+e x)-5)^{3/2}}dx\)

\(\Big \downarrow \) 3595

\(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}-\frac {1}{20} \int \frac {1}{\sqrt {4 \cos (d+e x)+3 \sin (d+e x)-5}}dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}-\frac {1}{20} \int \frac {1}{\sqrt {4 \cos (d+e x)+3 \sin (d+e x)-5}}dx\right )\)

\(\Big \downarrow \) 3594

\(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}-\frac {1}{20} \int \frac {1}{\sqrt {5 \cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )-5}}dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}-\frac {1}{20} \int \frac {1}{\sqrt {5 \sin \left (d+e x-\arctan \left (\frac {3}{4}\right )+\frac {\pi }{2}\right )-5}}dx\right )\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {\int \frac {1}{-\frac {5 \sin ^2\left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )-1}-10}d\left (-\frac {\sqrt {5} \sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )-1}}\right )}{10 e}+\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3}{40} \left (\frac {\arctan \left (\frac {\sin \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )}{\sqrt {2} \sqrt {\cos \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )-1}}\right )}{10 \sqrt {10} e}+\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}\right )\)

Input:

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

Output:

(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)) - (3*(ArcTan[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))))/40
 

Defintions of rubi rules used

rule 217
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3594
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 3595
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] + Simp[(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]
 
Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.34

method result size
default \(-\frac {\left (-3 \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right ) \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )^{2}+6 \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right ) \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+6 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-3 \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right )-14 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\right ) \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}}{4000 \left (\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-1\right ) \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {-5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e}\) \(190\)

Input:

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

Output:

-1/4000*(-3*10^(1/2)*arctan(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)*arctan(1/10*(-5*sin(e*x+d+arctan( 
4/3))-5)^(1/2)*10^(1/2))*sin(e*x+d+arctan(4/3))+6*(-5*sin(e*x+d+arctan(4/3 
))-5)^(1/2)*sin(e*x+d+arctan(4/3))-3*10^(1/2)*arctan(1/10*(-5*sin(e*x+d+ar 
ctan(4/3))-5)^(1/2)*10^(1/2))-14*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2))*(-5* 
sin(e*x+d+arctan(4/3))-5)^(1/2)/(sin(e*x+d+arctan(4/3))-1)/cos(e*x+d+arcta 
n(4/3))/(-5+5*sin(e*x+d+arctan(4/3)))^(1/2)/e
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (123) = 246\).

Time = 0.08 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.81 \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\frac {3 \, {\left (79 \, \sqrt {10} \cos \left (e x + d\right )^{3} - 123 \, \sqrt {10} \cos \left (e x + d\right )^{2} + 3 \, {\left (\sqrt {10} \cos \left (e x + d\right )^{2} + 38 \, \sqrt {10} \cos \left (e x + d\right ) - 44 \, \sqrt {10}\right )} \sin \left (e x + d\right ) - 78 \, \sqrt {10} \cos \left (e x + d\right ) + 124 \, \sqrt {10}\right )} \arctan \left (-\frac {\sqrt {10} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{3 \, \cos \left (e x + d\right ) - 4 \, \sin \left (e x + d\right )}\right ) + 10 \, {\left (27 \, \cos \left (e x + d\right )^{2} + {\left (39 \, \cos \left (e x + d\right ) - 8\right )} \sin \left (e x + d\right ) - 69 \, \cos \left (e x + d\right ) - 96\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{4000 \, {\left (79 \, e \cos \left (e x + d\right )^{3} - 123 \, e \cos \left (e x + d\right )^{2} - 78 \, e \cos \left (e x + d\right ) + 3 \, {\left (e \cos \left (e x + d\right )^{2} + 38 \, e \cos \left (e x + d\right ) - 44 \, e\right )} \sin \left (e x + d\right ) + 124 \, e\right )}} \] Input:

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

Output:

1/4000*(3*(79*sqrt(10)*cos(e*x + d)^3 - 123*sqrt(10)*cos(e*x + d)^2 + 3*(s 
qrt(10)*cos(e*x + d)^2 + 38*sqrt(10)*cos(e*x + d) - 44*sqrt(10))*sin(e*x + 
 d) - 78*sqrt(10)*cos(e*x + d) + 124*sqrt(10))*arctan(-sqrt(10)*sqrt(4*cos 
(e*x + d) + 3*sin(e*x + d) - 5)/(3*cos(e*x + d) - 4*sin(e*x + d))) + 10*(2 
7*cos(e*x + d)^2 + (39*cos(e*x + d) - 8)*sin(e*x + d) - 69*cos(e*x + d) - 
96)*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) - 5))/(79*e*cos(e*x + d)^3 - 123* 
e*cos(e*x + d)^2 - 78*e*cos(e*x + d) + 3*(e*cos(e*x + d)^2 + 38*e*cos(e*x 
+ d) - 44*e)*sin(e*x + d) + 124*e)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\int \frac {1}{\left (3 \sin {\left (d + e x \right )} + 4 \cos {\left (d + e x \right )} - 5\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\int { \frac {1}{{\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F(-1)]

Timed out. \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\int \frac {1}{{\left (4\,\cos \left (d+e\,x\right )+3\,\sin \left (d+e\,x\right )-5\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\int \frac {\sqrt {4 \cos \left (e x +d \right )+3 \sin \left (e x +d \right )-5}}{64 \cos \left (e x +d \right )^{3}+144 \cos \left (e x +d \right )^{2} \sin \left (e x +d \right )-240 \cos \left (e x +d \right )^{2}+108 \cos \left (e x +d \right ) \sin \left (e x +d \right )^{2}-360 \cos \left (e x +d \right ) \sin \left (e x +d \right )+300 \cos \left (e x +d \right )+27 \sin \left (e x +d \right )^{3}-135 \sin \left (e x +d \right )^{2}+225 \sin \left (e x +d \right )-125}d x \] Input:

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

Output:

int(sqrt(4*cos(d + e*x) + 3*sin(d + e*x) - 5)/(64*cos(d + e*x)**3 + 144*co 
s(d + e*x)**2*sin(d + e*x) - 240*cos(d + e*x)**2 + 108*cos(d + e*x)*sin(d 
+ e*x)**2 - 360*cos(d + e*x)*sin(d + e*x) + 300*cos(d + e*x) + 27*sin(d + 
e*x)**3 - 135*sin(d + e*x)**2 + 225*sin(d + e*x) - 125),x)