Integrand size = 26, antiderivative size = 96 \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\frac {2 i a}{5 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d (e \cos (c+d x))^{5/2}}+\frac {2 a \cos (c+d x) \sin (c+d x)}{3 d (e \cos (c+d x))^{5/2}} \] Output:
2/5*I*a/d/(e*cos(d*x+c))^(5/2)+2/3*a*cos(d*x+c)^(5/2)*InverseJacobiAM(1/2* d*x+1/2*c,2^(1/2))/d/(e*cos(d*x+c))^(5/2)+2/3*a*cos(d*x+c)*sin(d*x+c)/d/(e *cos(d*x+c))^(5/2)
Time = 0.99 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.59 \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\frac {a \left (6 i+10 \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+5 \sin (2 (c+d x))\right )}{15 d (e \cos (c+d x))^{5/2}} \] Input:
Integrate[(a + I*a*Tan[c + d*x])/(e*Cos[c + d*x])^(5/2),x]
Output:
(a*(6*I + 10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d *x)]))/(15*d*(e*Cos[c + d*x])^(5/2))
Time = 0.64 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3998, 3042, 3967, 3042, 4255, 3042, 4258, 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.
| \(\displaystyle \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3998 |
| \(\displaystyle \frac {\int (e \sec (c+d x))^{5/2} (i \tan (c+d x) a+a)dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (e \sec (c+d x))^{5/2} (i \tan (c+d x) a+a)dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle \frac {a \int (e \sec (c+d x))^{5/2}dx+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {a \left (\frac {1}{3} e^2 \int \sqrt {e \sec (c+d x)}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \left (\frac {1}{3} e^2 \int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {a \left (\frac {1}{3} e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \left (\frac {1}{3} e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {a \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])/(e*Cos[c + d*x])^(5/2),x]
Output:
((((2*I)/5)*a*(e*Sec[c + d*x])^(5/2))/d + a*((2*e^2*Sqrt[Cos[c + d*x]]*Ell ipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d) + (2*e*(e*Sec[c + d*x]) ^(3/2)*Sin[c + d*x])/(3*d)))/((e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2 ))
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[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m Int[( a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m , n}, x] && !IntegerQ[m]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (82 ) = 164\).
Time = 2.94 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.71
| method | result | size |
| parts | \(-\frac {2 a \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 e^{2} \sqrt {-e \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\, d}+\frac {2 i a}{5 d \left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}}}\) | \(260\) |
| default | \(-\frac {2 \left (20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+20 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{15 \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, e^{2} d}\) | \(283\) |
Input:
int((a+I*a*tan(d*x+c))/(e*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*a*(-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/ 2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-2*sin(1/2*d*x+1/ 2*c)^2*cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))/e^2*(e*(2*cos(1/2*d*x +1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1 /2*d*x+1/2*c)^2))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/(e*( 2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/d+2/5*I*a/d/(e*cos(d*x+c))^(5/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (81) = 162\).
Time = 0.08 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.88 \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {4 \, {\left (\sqrt {\frac {1}{2}} {\left (5 i \, a e^{\left (5 i \, d x + 5 i \, c\right )} - 12 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 5 i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 5 \, \sqrt {\frac {1}{2}} {\left (i \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{15 \, {\left (d e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3}\right )}} \] Input:
integrate((a+I*a*tan(d*x+c))/(e*cos(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-4/15*(sqrt(1/2)*(5*I*a*e^(5*I*d*x + 5*I*c) - 12*I*a*e^(3*I*d*x + 3*I*c) - 5*I*a*e^(I*d*x + I*c))*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*e^(-1/2*I*d*x - 1/ 2*I*c) + 5*sqrt(1/2)*(I*a*e^(6*I*d*x + 6*I*c) + 3*I*a*e^(4*I*d*x + 4*I*c) + 3*I*a*e^(2*I*d*x + 2*I*c) + I*a)*sqrt(e)*weierstrassPInverse(-4, 0, e^(I *d*x + I*c)))/(d*e^3*e^(6*I*d*x + 6*I*c) + 3*d*e^3*e^(4*I*d*x + 4*I*c) + 3 *d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3)
Timed out. \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))/(e*cos(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\int { \frac {i \, a \tan \left (d x + c\right ) + a}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))/(e*cos(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((I*a*tan(d*x + c) + a)/(e*cos(d*x + c))^(5/2), x)
\[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\int { \frac {i \, a \tan \left (d x + c\right ) + a}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))/(e*cos(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate((I*a*tan(d*x + c) + a)/(e*cos(d*x + c))^(5/2), x)
Timed out. \[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\int \frac {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)/(e*cos(c + d*x))^(5/2),x)
Output:
int((a + a*tan(c + d*x)*1i)/(e*cos(c + d*x))^(5/2), x)
\[ \int \frac {a+i a \tan (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {e}\, a \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \tan \left (d x +c \right )}{\cos \left (d x +c \right )^{3}}d x \right ) i \right )}{e^{3}} \] Input:
int((a+I*a*tan(d*x+c))/(e*cos(d*x+c))^(5/2),x)
Output:
(sqrt(e)*a*(int(sqrt(cos(c + d*x))/cos(c + d*x)**3,x) + int((sqrt(cos(c + d*x))*tan(c + d*x))/cos(c + d*x)**3,x)*i))/e**3