Integrand size = 28, antiderivative size = 92 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\frac {2 \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{3/2}}+\frac {4 i \cos ^2(c+d x)}{5 d (e \cos (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]
2/5*cos(d*x+c)^(3/2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip ticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d/(e*cos(d*x+c))^(3/2)+4/5*I*cos(d*x+ c)^2/d/(e*cos(d*x+c))^(3/2)/(a^2+I*a^2*tan(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 3.55 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\frac {2 \cos ^2(c+d x) \left (1+2^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {3}{4},\frac {1}{2} (1+i \tan (c+d x))\right ) \sqrt [4]{1-i \tan (c+d x)} (1+i \tan (c+d x))-i \tan (c+d x)\right )}{5 a^2 d (e \cos (c+d x))^{3/2} (-i+\tan (c+d x))} \]
(2*Cos[c + d*x]^2*(1 + 2^(3/4)*Hypergeometric2F1[-1/4, 1/4, 3/4, (1 + I*Ta n[c + d*x])/2]*(1 - I*Tan[c + d*x])^(1/4)*(1 + I*Tan[c + d*x]) - I*Tan[c + d*x]))/(5*a^2*d*(e*Cos[c + d*x])^(3/2)*(-I + Tan[c + d*x]))
Time = 0.56 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3998, 3042, 3981, 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.
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \cos (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \cos (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3998 |
\(\displaystyle \frac {\int \frac {(e \sec (c+d x))^{3/2}}{(i \tan (c+d x) a+a)^2}dx}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(e \sec (c+d x))^{3/2}}{(i \tan (c+d x) a+a)^2}dx}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3981 |
\(\displaystyle \frac {\frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{5 a^2}+\frac {4 i e^2}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 a^2}+\frac {4 i e^2}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {\frac {e^2 \int \sqrt {\cos (c+d x)}dx}{5 a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {e^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\) |
((2*e^2*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[ c + d*x]]) + (((4*I)/5)*e^2)/(d*Sqrt[e*Sec[c + d*x]]*(a^2 + I*a^2*Tan[c + d*x])))/((e*Cos[c + d*x])^(3/2)*(e*Sec[c + d*x])^(3/2))
3.7.68.3.1 Defintions of rubi rules used
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[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*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 = 4.18 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.24
method | result | size |
default | \(\frac {-\frac {32 i \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {32 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+\frac {48 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {32 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {24 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}}{5}+\frac {4 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}}{e \,a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(206\) |
2/5/e/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(-16*I*si n(1/2*d*x+1/2*c)^7+16*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+24*I*sin(1/2 *d*x+1/2*c)^5-16*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-12*I*sin(1/2*d*x+ 1/2*c)^3+4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+EllipticE(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)^(1/ 2)+2*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 = 94, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (\sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (-2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - i \, \sqrt {2} \sqrt {e} e^{\left (2 i \, d x + 2 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{5 \, a^{2} d e^{2}} \]
-2/5*(sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(-2*I*e^(2*I*d*x + 2*I*c) - I)*e^(-1/2*I*d*x - 1/2*I*c) - I*sqrt(2)*sqrt(e)*e^(2*I*d*x + 2*I*c)*weie rstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))*e^(-2*I*d *x - 2*I*c)/(a^2*d*e^2)
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]