\(\int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx\) [355]

Optimal result

Integrand size = 22, antiderivative size = 96 \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {-1+\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}}\right )}{10 \sqrt {10} e}+\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx=\frac {\left (\frac {1}{250}-\frac {i}{125}\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right ) \left ((-1+i) \sqrt {-20-15 i} \text {arctanh}\left (\left (\frac {1}{10}+\frac {3 i}{10}\right ) \sqrt {-\frac {4}{5}-\frac {3 i}{5}} \left (3+\tan \left (\frac {1}{4} (d+e x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right )^2+(5+10 i) \left (3 \cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )\right )}{e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \] Input:

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

Output:

((1/250 - I/125)*(Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2])*((-1 + I)*Sqrt[-2 
0 - 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])^2 + (5 + 10*I)*(3*Cos[(d + 
e*x)/2] + Sin[(d + e*x)/2])))/(e*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3 
/2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3595

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3594

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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))
 

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 = 0.37 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.23

method	result	size
default	\(-\frac {\left (\sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right ) \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-\sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right )-2 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\right ) \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}}{100 \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {-5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e}\)	\(118\)

Input:

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

Output:

-1/100*(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))-10^(1/2)*arctan(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+arct 
an(4/3))-5)^(1/2)/cos(e*x+d+arctan(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. 187 vs. \(2 (81) = 162\).

Time = 0.08 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.95 \[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx=-\frac {{\left (13 \, \sqrt {10} \cos \left (e x + d\right )^{2} - 9 \, {\left (\sqrt {10} \cos \left (e x + d\right ) - 2 \, \sqrt {10}\right )} \sin \left (e x + d\right ) - \sqrt {10} \cos \left (e x + d\right ) - 14 \, \sqrt {10}\right )} \arctan \left (-\frac {\sqrt {10} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{3 \, \cos \left (e x + d\right ) - 4 \, \sin \left (e x + d\right )}\right ) + 10 \, \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5} {\left (3 \, \cos \left (e x + d\right ) + \sin \left (e x + d\right ) + 3\right )}}{100 \, {\left (13 \, e \cos \left (e x + d\right )^{2} - e \cos \left (e x + d\right ) - 9 \, {\left (e \cos \left (e x + d\right ) - 2 \, e\right )} \sin \left (e x + d\right ) - 14 \, e\right )}} \] Input:

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

Output:

-1/100*((13*sqrt(10)*cos(e*x + d)^2 - 9*(sqrt(10)*cos(e*x + d) - 2*sqrt(10 
))*sin(e*x + d) - sqrt(10)*cos(e*x + d) - 14*sqrt(10))*arctan(-sqrt(10)*sq 
rt(4*cos(e*x + d) + 3*sin(e*x + d) - 5)/(3*cos(e*x + d) - 4*sin(e*x + d))) 
 + 10*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) - 5)*(3*cos(e*x + d) + sin(e*x 
+ d) + 3))/(13*e*cos(e*x + d)^2 - e*cos(e*x + d) - 9*(e*cos(e*x + d) - 2*e 
)*sin(e*x + d) - 14*e)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx=\int \frac {\sqrt {4 \cos \left (e x +d \right )+3 \sin \left (e x +d \right )-5}}{16 \cos \left (e x +d \right )^{2}+24 \cos \left (e x +d \right ) \sin \left (e x +d \right )-40 \cos \left (e x +d \right )+9 \sin \left (e x +d \right )^{2}-30 \sin \left (e x +d \right )+25}d x \] Input:

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

Output:

int(sqrt(4*cos(d + e*x) + 3*sin(d + e*x) - 5)/(16*cos(d + e*x)**2 + 24*cos 
(d + e*x)*sin(d + e*x) - 40*cos(d + e*x) + 9*sin(d + e*x)**2 - 30*sin(d + 
e*x) + 25),x)