Integrand size = 21, antiderivative size = 75 \[ \int \frac {1}{(2 \cos (c+d x)+3 \sin (c+d x))^{5/2}} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\arctan \left (\frac {3}{2}\right )\right ),2\right )}{39 \sqrt [4]{13} d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{39 d (2 \cos (c+d x)+3 \sin (c+d x))^{3/2}} \] Output:
2/507*13^(3/4)*InverseJacobiAM(1/2*c+1/2*d*x-1/2*arctan(3/2),2^(1/2))/d-2/ 39*(3*cos(d*x+c)-2*sin(d*x+c))/d/(2*cos(d*x+c)+3*sin(d*x+c))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.62 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.09 \[ \int \frac {1}{(2 \cos (c+d x)+3 \sin (c+d x))^{5/2}} \, dx=\frac {-78 \cos (c+d x)+52 \sin (c+d x)+\sqrt {2} 13^{3/4} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (c+d x+\arctan \left (\frac {2}{3}\right )\right )\right ) \sec \left (c+d x+\arctan \left (\frac {2}{3}\right )\right ) (2 \cos (c+d x)+3 \sin (c+d x))^{3/2} \sqrt {1+\sin \left (c+d x+\arctan \left (\frac {2}{3}\right )\right )} \sqrt {-1+\cos \left (2 \left (c+d x+\arctan \left (\frac {2}{3}\right )\right )\right )+2 \sin \left (c+d x+\arctan \left (\frac {2}{3}\right )\right )}}{507 d (2 \cos (c+d x)+3 \sin (c+d x))^{3/2}} \] Input:
Integrate[(2*Cos[c + d*x] + 3*Sin[c + d*x])^(-5/2),x]
Output:
(-78*Cos[c + d*x] + 52*Sin[c + d*x] + Sqrt[2]*13^(3/4)*HypergeometricPFQ[{ 1/4, 1/2}, {5/4}, Sin[c + d*x + ArcTan[2/3]]^2]*Sec[c + d*x + ArcTan[2/3]] *(2*Cos[c + d*x] + 3*Sin[c + d*x])^(3/2)*Sqrt[1 + Sin[c + d*x + ArcTan[2/3 ]]]*Sqrt[-1 + Cos[2*(c + d*x + ArcTan[2/3])] + 2*Sin[c + d*x + ArcTan[2/3] ]])/(507*d*(2*Cos[c + d*x] + 3*Sin[c + d*x])^(3/2))
Time = 0.34 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3555, 3042, 3556, 3042, 3120}
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.
| \(\displaystyle \int \frac {1}{(3 \sin (c+d x)+2 \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(3 \sin (c+d x)+2 \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3555 |
| \(\displaystyle \frac {1}{39} \int \frac {1}{\sqrt {2 \cos (c+d x)+3 \sin (c+d x)}}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{39 d (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{39} \int \frac {1}{\sqrt {2 \cos (c+d x)+3 \sin (c+d x)}}dx-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{39 d (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3556 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {\cos \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )}}dx}{39 \sqrt [4]{13}}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{39 d (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {\sin \left (c+d x-\arctan \left (\frac {3}{2}\right )+\frac {\pi }{2}\right )}}dx}{39 \sqrt [4]{13}}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{39 d (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\arctan \left (\frac {3}{2}\right )\right ),2\right )}{39 \sqrt [4]{13} d}-\frac {2 (3 \cos (c+d x)-2 \sin (c+d x))}{39 d (3 \sin (c+d x)+2 \cos (c+d x))^{3/2}}\) |
Input:
Int[(2*Cos[c + d*x] + 3*Sin[c + d*x])^(-5/2),x]
Output:
(2*EllipticF[(c + d*x - ArcTan[3/2])/2, 2])/(39*13^(1/4)*d) - (2*(3*Cos[c + d*x] - 2*Sin[c + d*x]))/(39*d*(2*Cos[c + d*x] + 3*Sin[c + d*x])^(3/2))
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(b*Cos[c + d*x] - a*Sin[c + d*x])*((a*Cos[c + d*x] + b*Sin [c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[(n + 2)/((n + 1)*(a^ 2 + b^2)) Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1] && NeQ[n, -2]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(a^2 + b^2)^(n/2) Int[Cos[c + d*x - ArcTan[a, b]]^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && !(GeQ[n, 1] || LeQ[n, -1]) && GtQ[a^2 + b^2, 0]
Time = 0.24 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.57
| method | result | size |
| default | \(\frac {\sqrt {\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+1}\, \sqrt {-2 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+2}\, \sqrt {-\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )+1}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )-2 \cos \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )^{2}}{39 \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right ) \cos \left (d x +c +\arctan \left (\frac {2}{3}\right )\right ) \sqrt {\sqrt {13}\, \sin \left (d x +c +\arctan \left (\frac {2}{3}\right )\right )}\, d}\) | \(118\) |
Input:
int(1/(2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/39/sin(d*x+c+arctan(2/3))*((sin(d*x+c+arctan(2/3))+1)^(1/2)*(-2*sin(d*x+ c+arctan(2/3))+2)^(1/2)*(-sin(d*x+c+arctan(2/3)))^(1/2)*EllipticF((sin(d*x +c+arctan(2/3))+1)^(1/2),1/2*2^(1/2))*sin(d*x+c+arctan(2/3))-2*cos(d*x+c+a rctan(2/3))^2)/cos(d*x+c+arctan(2/3))/(13^(1/2)*sin(d*x+c+arctan(2/3)))^(1 /2)/d
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.25 \[ \int \frac {1}{(2 \cos (c+d x)+3 \sin (c+d x))^{5/2}} \, dx=\frac {2 \, {\left (\sqrt {\frac {3}{2} i + 1} {\left (\left (10 i + 15\right ) \, \cos \left (d x + c\right )^{2} - \left (24 i + 36\right ) \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 18 i - 27\right )} {\rm weierstrassPInverse}\left (\frac {48}{13} i + \frac {20}{13}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + \sqrt {-\frac {3}{2} i + 1} {\left (-\left (10 i - 15\right ) \, \cos \left (d x + c\right )^{2} + \left (24 i - 36\right ) \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 18 i - 27\right )} {\rm weierstrassPInverse}\left (-\frac {48}{13} i + \frac {20}{13}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 13 \, {\left (3 \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right )\right )} \sqrt {2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )}\right )}}{507 \, {\left (5 \, d \cos \left (d x + c\right )^{2} - 12 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 9 \, d\right )}} \] Input:
integrate(1/(2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
2/507*(sqrt(3/2*I + 1)*((10*I + 15)*cos(d*x + c)^2 - (24*I + 36)*cos(d*x + c)*sin(d*x + c) - 18*I - 27)*weierstrassPInverse(48/13*I + 20/13, 0, cos( d*x + c) - I*sin(d*x + c)) + sqrt(-3/2*I + 1)*(-(10*I - 15)*cos(d*x + c)^2 + (24*I - 36)*cos(d*x + c)*sin(d*x + c) + 18*I - 27)*weierstrassPInverse( -48/13*I + 20/13, 0, cos(d*x + c) + I*sin(d*x + c)) + 13*(3*cos(d*x + c) - 2*sin(d*x + c))*sqrt(2*cos(d*x + c) + 3*sin(d*x + c)))/(5*d*cos(d*x + c)^ 2 - 12*d*cos(d*x + c)*sin(d*x + c) - 9*d)
Timed out. \[ \int \frac {1}{(2 \cos (c+d x)+3 \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(2*cos(d*x+c)+3*sin(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \frac {1}{(2 \cos (c+d x)+3 \sin (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((2*cos(d*x + c) + 3*sin(d*x + c))^(-5/2), x)
\[ \int \frac {1}{(2 \cos (c+d x)+3 \sin (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate((2*cos(d*x + c) + 3*sin(d*x + c))^(-5/2), x)
Timed out. \[ \int \frac {1}{(2 \cos (c+d x)+3 \sin (c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (2\,\cos \left (c+d\,x\right )+3\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int(1/(2*cos(c + d*x) + 3*sin(c + d*x))^(5/2),x)
Output:
int(1/(2*cos(c + d*x) + 3*sin(c + d*x))^(5/2), x)
\[ \int \frac {1}{(2 \cos (c+d x)+3 \sin (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {2 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )}}{8 \cos \left (d x +c \right )^{3}+36 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+54 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+27 \sin \left (d x +c \right )^{3}}d x \] Input:
int(1/(2*cos(d*x+c)+3*sin(d*x+c))^(5/2),x)
Output:
int(sqrt(2*cos(c + d*x) + 3*sin(c + d*x))/(8*cos(c + d*x)**3 + 36*cos(c + d*x)**2*sin(c + d*x) + 54*cos(c + d*x)*sin(c + d*x)**2 + 27*sin(c + d*x)** 3),x)