Integrand size = 28, antiderivative size = 126 \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\frac {10 \cos ^{\frac {9}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d (e \cos (c+d x))^{9/2}}+\frac {10 \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (e \cos (c+d x))^{9/2}}-\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]
10/3*cos(d*x+c)^(9/2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli pticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d/(e*cos(d*x+c))^(9/2)+10/3*cos(d*x+ c)^3*sin(d*x+c)/a^2/d/(e*cos(d*x+c))^(9/2)-4*I*cos(d*x+c)^2/d/(e*cos(d*x+c ))^(9/2)/(a^2+I*a^2*tan(d*x+c))
Time = 1.57 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\frac {2 \left (-6 i \cos (c+d x)+5 \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\sin (c+d x)\right )}{3 a^2 d e^3 (e \cos (c+d x))^{3/2}} \]
(2*((-6*I)*Cos[c + d*x] + 5*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] - Sin[c + d*x]))/(3*a^2*d*e^3*(e*Cos[c + d*x])^(3/2))
Time = 0.76 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3998, 3042, 3981, 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.
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \cos (c+d x))^{9/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \cos (c+d x))^{9/2}}dx\) |
\(\Big \downarrow \) 3998 |
\(\displaystyle \frac {\int \frac {(e \sec (c+d x))^{9/2}}{(i \tan (c+d x) a+a)^2}dx}{(e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(e \sec (c+d x))^{9/2}}{(i \tan (c+d x) a+a)^2}dx}{(e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 3981 |
\(\displaystyle \frac {\frac {5 e^2 \int (e \sec (c+d x))^{5/2}dx}{a^2}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5 e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx}{a^2}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {5 e^2 \left (\frac {1}{3} e^2 \int \sqrt {e \sec (c+d x)}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )}{a^2}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5 e^2 \left (\frac {1}{3} e^2 \int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )}{a^2}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {\frac {5 e^2 \left (\frac {1}{3} e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )}{a^2}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5 e^2 \left (\frac {1}{3} e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )}{a^2}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {5 e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d}\right )}{a^2}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}}{(e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}}\) |
((5*e^2*((2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(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)))/a^2 - ( (4*I)*e^2*(e*Sec[c + d*x])^(5/2))/(d*(a^2 + I*a^2*Tan[c + d*x])))/((e*Cos[ c + d*x])^(9/2)*(e*Sec[c + d*x])^(9/2))
3.7.71.3.1 Defintions of rubi rules used
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 _)])^(n_), x_Symbol] :> Simp[2*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Simp[d^2*((m - 2)/(b^2*(m + 2*n))) Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[ {a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (IntegersQ[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
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]
Time = 4.56 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.65
method | result | size |
default | \(\frac {-\frac {20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-8 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {10 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}}{3}+4 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{4} d}\) | \(208\) |
2/3/(2*sin(1/2*d*x+1/2*c)^2-1)/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2* c)^2*e+e)^(1/2)/e^4*(-10*(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)^2-12 *I*sin(1/2*d*x+1/2*c)^3+2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+5*(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))*(sin(1/2* d*x+1/2*c)^2)^(1/2)+6*I*sin(1/2*d*x+1/2*c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (5 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 7 i \, e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 5 \, {\left (i \, \sqrt {2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{3 \, {\left (a^{2} d e^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} d e^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{5}\right )}} \]
-2/3*(2*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(5*I*e^(3*I*d*x + 3*I*c) + 7*I*e^(I*d*x + I*c))*e^(-1/2*I*d*x - 1/2*I*c) + 5*(I*sqrt(2)*e^(4*I*d*x + 4*I*c) + 2*I*sqrt(2)*e^(2*I*d*x + 2*I*c) + I*sqrt(2))*sqrt(e)*weierstra ssPInverse(-4, 0, e^(I*d*x + I*c)))/(a^2*d*e^5*e^(4*I*d*x + 4*I*c) + 2*a^2 *d*e^5*e^(2*I*d*x + 2*I*c) + a^2*d*e^5)
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]