Integrand size = 22, antiderivative size = 142 \[ \int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\frac {3 \text {arctanh}\left (\frac {\sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {1+\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}}\right )}{400 \sqrt {10} e}-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac {3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \] Output:
3/4000*10^(1/2)*arctanh(1/2*sin(d+e*x-arctan(3/4))*2^(1/2)/(1+cos(d+e*x-ar ctan(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*si n(e*x+d))^(3/2)
Result contains complex when optimal does not.
Time = 0.47 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.27 \[ \int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=-\frac {\left (\frac {1}{20000}-\frac {i}{10000}\right ) \left (3 \cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right ) \left ((-6+6 i) \sqrt {20+15 i} \arctan \left (\left (\frac {1}{10}+\frac {3 i}{10}\right ) \sqrt {\frac {4}{5}+\frac {3 i}{5}} \left (-1+3 \tan \left (\frac {1}{4} (d+e x)\right )\right )\right ) \left (3 \cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )^4+(5+10 i) \left (55 \cos \left (\frac {1}{2} (d+e x)\right )+39 \cos \left (\frac {3}{2} (d+e x)\right )-165 \sin \left (\frac {1}{2} (d+e x)\right )-27 \sin \left (\frac {3}{2} (d+e x)\right )\right )\right )}{e (5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \] Input:
Integrate[(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(-5/2),x]
Output:
((-1/20000 + 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*C os[(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))
Time = 0.52 (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, 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.
| \(\displaystyle \int \frac {1}{(3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}dx\) |
| \(\Big \downarrow \) 3595 |
| \(\displaystyle \frac {3}{40} \int \frac {1}{(4 \cos (d+e x)+3 \sin (d+e x)+5)^{3/2}}dx-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{40} \int \frac {1}{(4 \cos (d+e x)+3 \sin (d+e x)+5)^{3/2}}dx-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}\) |
| \(\Big \downarrow \) 3595 |
| \(\displaystyle \frac {3}{40} \left (\frac {1}{20} \int \frac {1}{\sqrt {4 \cos (d+e x)+3 \sin (d+e x)+5}}dx-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}}\right )-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{40} \left (\frac {1}{20} \int \frac {1}{\sqrt {4 \cos (d+e x)+3 \sin (d+e x)+5}}dx-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}}\right )-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}\) |
| \(\Big \downarrow \) 3594 |
| \(\displaystyle \frac {3}{40} \left (\frac {1}{20} \int \frac {1}{\sqrt {5 \cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )+5}}dx-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}}\right )-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{40} \left (\frac {1}{20} \int \frac {1}{\sqrt {5 \sin \left (d+e x-\arctan \left (\frac {3}{4}\right )+\frac {\pi }{2}\right )+5}}dx-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}}\right )-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {3}{40} \left (-\frac {\int \frac {1}{10-\frac {5 \sin ^2\left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )+1}}d\left (-\frac {\sqrt {5} \sin \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )}{\sqrt {\cos \left (d+e x-\arctan \left (\frac {3}{4}\right )\right )+1}}\right )}{10 e}-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}}\right )-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3}{40} \left (\frac {\text {arctanh}\left (\frac {\sin \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )}{\sqrt {2} \sqrt {\cos \left (-\arctan \left (\frac {3}{4}\right )+d+e x\right )+1}}\right )}{10 \sqrt {10} e}-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}}\right )-\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{5/2}}\) |
Input:
Int[(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(-5/2),x]
Output:
-1/20*(3*Cos[d + e*x] - 4*Sin[d + e*x])/(e*(5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(5/2)) + (3*(ArcTanh[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[1 + C os[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
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])
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]
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]
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]
Time = 0.36 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.34
| method | result | size |
| default | \(-\frac {\left (3 \sqrt {10}\, \operatorname {arctanh}\left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\, \sqrt {10}}{10}\right ) \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )^{2}+6 \sqrt {10}\, \operatorname {arctanh}\left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\, \sqrt {10}}{10}\right ) \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+3 \sqrt {10}\, \operatorname {arctanh}\left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\, \sqrt {10}}{10}\right )+6 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\, \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+14 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}\right ) \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+5}}{4000 \left (1+\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )\right ) \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e}\) | \(190\) |
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)*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+arc tan(4/3))/(5+5*sin(e*x+d+arctan(4/3)))^(1/2)/e
Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (123) = 246\).
Time = 0.09 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.40 \[ \int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\frac {3 \, {\left (3 \, \sqrt {10} \cos \left (e x + d\right )^{3} - 111 \, \sqrt {10} \cos \left (e x + d\right )^{2} - {\left (79 \, \sqrt {10} \cos \left (e x + d\right )^{2} + 202 \, \sqrt {10} \cos \left (e x + d\right ) + 124 \, \sqrt {10}\right )} \sin \left (e x + d\right ) - 246 \, \sqrt {10} \cos \left (e x + d\right ) - 132 \, \sqrt {10}\right )} \log \left (-\frac {9 \, \cos \left (e x + d\right )^{2} + {\left (13 \, \cos \left (e x + d\right ) - 6\right )} \sin \left (e x + d\right ) + 2 \, {\left (\sqrt {10} \cos \left (e x + d\right ) - 3 \, \sqrt {10} \sin \left (e x + d\right ) + \sqrt {10}\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5} - 33 \, \cos \left (e x + d\right ) - 42}{9 \, \cos \left (e x + d\right )^{2} + {\left (13 \, \cos \left (e x + d\right ) + 14\right )} \sin \left (e x + d\right ) + 27 \, \cos \left (e x + d\right ) + 18}\right ) + 20 \, {\left (39 \, \cos \left (e x + d\right )^{2} - 3 \, {\left (9 \, \cos \left (e x + d\right ) + 32\right )} \sin \left (e x + d\right ) + 47 \, \cos \left (e x + d\right ) + 8\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5}}{8000 \, {\left (3 \, e \cos \left (e x + d\right )^{3} - 111 \, e \cos \left (e x + d\right )^{2} - 246 \, e \cos \left (e x + d\right ) - {\left (79 \, e \cos \left (e x + d\right )^{2} + 202 \, e \cos \left (e x + d\right ) + 124 \, e\right )} \sin \left (e x + d\right ) - 132 \, e\right )}} \] Input:
integrate(1/(5+4*cos(e*x+d)+3*sin(e*x+d))^(5/2),x, algorithm="fricas")
Output:
1/8000*(3*(3*sqrt(10)*cos(e*x + d)^3 - 111*sqrt(10)*cos(e*x + d)^2 - (79*s qrt(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)
\[ \int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\int \frac {1}{\left (3 \sin {\left (d + e x \right )} + 4 \cos {\left (d + e x \right )} + 5\right )^{\frac {5}{2}}}\, dx \] 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)
\[ \int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\int { \frac {1}{{\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5\right )}^{\frac {5}{2}}} \,d x } \] 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)
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
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)
\[ \int \frac {1}{(5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx=\int \frac {\sqrt {4 \cos \left (e x +d \right )+3 \sin \left (e x +d \right )+5}}{64 \cos \left (e x +d \right )^{3}+144 \cos \left (e x +d \right )^{2} \sin \left (e x +d \right )+240 \cos \left (e x +d \right )^{2}+108 \cos \left (e x +d \right ) \sin \left (e x +d \right )^{2}+360 \cos \left (e x +d \right ) \sin \left (e x +d \right )+300 \cos \left (e x +d \right )+27 \sin \left (e x +d \right )^{3}+135 \sin \left (e x +d \right )^{2}+225 \sin \left (e x +d \right )+125}d x \] 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)