\(\int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx\) [348]

Optimal result

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

1/100*10^(1/2)*arctanh(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*s 
in(e*x+d))^(3/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.60 \[ \int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx=-\frac {\left (\frac {1}{250}-\frac {i}{125}\right ) \left (3 \cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right ) \left ((5+10 i) \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right )-(1-i) \sqrt {20+15 i} \arctan \left (\left (\frac {1}{10}+\frac {3 i}{10}\right ) \sqrt {\frac {4}{5}+\frac {3 i}{5}} \left (-1+3 \tan \left (\frac {1}{4} (d+e x)\right )\right )\right ) \left (3 \cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )^2\right )}{e (5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \] Input:

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

Output:

((-1/250 + I/125)*(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))
 

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, 219}

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

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3594

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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))
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[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 = 117, normalized size of antiderivative = 1.22

method	result	size
default	\(-\frac {\left (\sqrt {10}\, \operatorname {arctanh}\left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\, \sqrt {10}}{10}\right ) \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+\sqrt {10}\, \operatorname {arctanh}\left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\, \sqrt {10}}{10}\right )+2 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\right ) \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}}{100 \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e}\)	\(117\)

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)*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)*arctanh(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+ar 
ctan(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. 268 vs. \(2 (81) = 162\).

Time = 0.08 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.79 \[ \int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx=\frac {{\left (9 \, \sqrt {10} \cos \left (e x + d\right )^{2} + {\left (13 \, \sqrt {10} \cos \left (e x + d\right ) + 14 \, \sqrt {10}\right )} \sin \left (e x + d\right ) + 27 \, \sqrt {10} \cos \left (e x + d\right ) + 18 \, \sqrt {10}\right )} \log \left (-\frac {9 \, \cos \left (e x + d\right )^{2} + {\left (13 \, \cos \left (e x + d\right ) - 6\right )} \sin \left (e x + d\right ) + 2 \, {\left (\sqrt {10} \cos \left (e x + d\right ) - 3 \, \sqrt {10} \sin \left (e x + d\right ) + \sqrt {10}\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5} - 33 \, \cos \left (e x + d\right ) - 42}{9 \, \cos \left (e x + d\right )^{2} + {\left (13 \, \cos \left (e x + d\right ) + 14\right )} \sin \left (e x + d\right ) + 27 \, \cos \left (e x + d\right ) + 18}\right ) - 20 \, \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5} {\left (\cos \left (e x + d\right ) - 3 \, \sin \left (e x + d\right ) + 1\right )}}{200 \, {\left (9 \, e \cos \left (e x + d\right )^{2} + 27 \, e \cos \left (e x + d\right ) + {\left (13 \, e \cos \left (e x + d\right ) + 14 \, e\right )} \sin \left (e x + d\right ) + 18 \, e\right )}} \] Input:

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

Output:

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)
                                                                                    
                                                                                    
 

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)