Integrand size = 28, antiderivative size = 114 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx=\frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 \sin (c+d x)}{45 a d e (e \sec (c+d x))^{3/2}}+\frac {2 i}{9 d (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \] Output:
14/15*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d/e^2/cos(d*x+c)^(1/2)/(e*se c(d*x+c))^(1/2)+14/45*sin(d*x+c)/a/d/e/(e*sec(d*x+c))^(3/2)+2/9*I/d/(e*sec (d*x+c))^(5/2)/(a+I*a*tan(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.90 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.18 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx=\frac {\left (106+104 \cos (2 (c+d x))-2 \cos (4 (c+d x))-56 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+70 i \sin (2 (c+d x))-7 i \sin (4 (c+d x))\right ) (i+\tan (c+d x))}{180 a d e^2 \sqrt {e \sec (c+d x)}} \] Input:
Integrate[1/((e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])),x]
Output:
((106 + 104*Cos[2*(c + d*x)] - 2*Cos[4*(c + d*x)] - 56*E^((2*I)*(c + d*x)) *Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)* (c + d*x))] + (70*I)*Sin[2*(c + d*x)] - (7*I)*Sin[4*(c + d*x)])*(I + Tan[c + d*x]))/(180*a*d*e^2*Sqrt[e*Sec[c + d*x]])
Time = 0.52 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3983, 3042, 4256, 3042, 4258, 3042, 3119}
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.
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {7 \int \frac {1}{(e \sec (c+d x))^{5/2}}dx}{9 a}+\frac {2 i}{9 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {7 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{9 a}+\frac {2 i}{9 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle \frac {7 \left (\frac {3 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a}+\frac {2 i}{9 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {7 \left (\frac {3 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a}+\frac {2 i}{9 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {7 \left (\frac {3 \int \sqrt {\cos (c+d x)}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a}+\frac {2 i}{9 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {7 \left (\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a}+\frac {2 i}{9 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {7 \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a}+\frac {2 i}{9 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{5/2}}\) |
Input:
Int[1/((e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])),x]
Output:
(7*((6*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*Sin[c + d*x])/(5*d*e*(e*Sec[c + d*x])^(3/2))))/(9*a) + ((2* I)/9)/(d*(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x]))
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ (b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n)) Int[(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x ] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2* n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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. 443 vs. \(2 (100 ) = 200\).
Time = 4.20 (sec) , antiderivative size = 444, normalized size of antiderivative = 3.89
method | result | size |
default | \(\frac {-\frac {14}{15}+\frac {14 i \tan \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )}{15}+\frac {14 \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )}{15}+\frac {14 i \tan \left (d x +c \right ) \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )}{15}+\frac {14 \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )}{15}+\frac {14 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right )}{45}+\frac {4 \cos \left (d x +c \right )^{4}}{45}+\frac {4 \cos \left (d x +c \right )^{3}}{45}+\frac {28 \cos \left (d x +c \right )^{2}}{45}-\frac {14 \cos \left (d x +c \right )}{45}}{a d \left (-\cos \left (d x +c \right ) \sin \left (d x +c \right )-\sin \left (d x +c \right )+i \cos \left (d x +c \right )^{2}+i \cos \left (d x +c \right )\right ) \sqrt {e \sec \left (d x +c \right )}\, e^{2}}\) | \(444\) |
Input:
int(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
2/45/a/d/(-cos(d*x+c)*sin(d*x+c)-sin(d*x+c)+I*cos(d*x+c)^2+I*cos(d*x+c))/( e*sec(d*x+c))^(1/2)*(-21+21*I*tan(d*x+c)*(cos(d*x+c)^2+2*cos(d*x+c)+1)*(co s(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d *x+c)-csc(d*x+c)),I)+21*(cos(d*x+c)^2+2*cos(d*x+c)+1)*(cos(d*x+c)/(cos(d*x +c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)) ,I)+21*I*tan(d*x+c)*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(cos(d*x+c)/(cos(d*x+c) +1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I) +21*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(c os(d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)+7*I*cos(d*x+c)* sin(d*x+c)*(cos(d*x+c)^2+cos(d*x+c)+3)+2*cos(d*x+c)^4+2*cos(d*x+c)^3+14*co s(d*x+c)^2-7*cos(d*x+c))/e^2
Time = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.13 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-9 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 174 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 212 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 34 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 336 i \, \sqrt {2} \sqrt {e} e^{\left (5 i \, d x + 5 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{360 \, a d e^{3}} \] Input:
integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
1/360*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-9*I*e^(8*I*d*x + 8*I*c) + 174*I*e^(6*I*d*x + 6*I*c) + 212*I*e^(4*I*d*x + 4*I*c) + 34*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(1/2*I*d*x + 1/2*I*c) + 336*I*sqrt(2)*sqrt(e)*e^(5*I*d* x + 5*I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I* c))))*e^(-5*I*d*x - 5*I*c)/(a*d*e^3)
\[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx=- \frac {i \int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan {\left (c + d x \right )} - i \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx}{a} \] Input:
integrate(1/(e*sec(d*x+c))**(5/2)/(a+I*a*tan(d*x+c)),x)
Output:
-I*Integral(1/((e*sec(c + d*x))**(5/2)*tan(c + d*x) - I*(e*sec(c + d*x))** (5/2)), x)/a
Exception generated. \[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
\[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}} \,d x } \] Input:
integrate(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
integrate(1/((e*sec(d*x + c))^(5/2)*(I*a*tan(d*x + c) + a)), x)
Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx=\int \frac {1}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \] Input:
int(1/((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)),x)
Output:
int(1/((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)), x)
\[ \int \frac {1}{(e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3} \tan \left (d x +c \right ) i +\sec \left (d x +c \right )^{3}}d x \right )}{a \,e^{3}} \] Input:
int(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c)),x)
Output:
(sqrt(e)*int(sqrt(sec(c + d*x))/(sec(c + d*x)**3*tan(c + d*x)*i + sec(c + d*x)**3),x))/(a*e**3)