Integrand size = 28, antiderivative size = 192 \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {154 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {154 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}-\frac {154 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac {44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )} \]
-154/15*e^5*(e*sec(d*x+c))^(5/2)*sin(d*x+c)/a^4/d+154/5*e^8*(cos(1/2*d*x+1 /2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^ 4/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-154/5*e^7*sin(d*x+c)*(e*sec(d*x+ c))^(1/2)/a^4/d+4*I*e^2*(e*sec(d*x+c))^(11/2)/a/d/(a+I*a*tan(d*x+c))^3+44/ 3*I*e^4*(e*sec(d*x+c))^(7/2)/d/(a^4+I*a^4*tan(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.17 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.65 \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {i e^5 (e \sec (c+d x))^{5/2} \left (-1133 \cos (c+d x)+77 e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )-3 (117 \cos (3 (c+d x))+33 i \sin (c+d x)+37 i \sin (3 (c+d x)))\right )}{30 a^4 d} \]
((-1/30*I)*e^5*(e*Sec[c + d*x])^(5/2)*(-1133*Cos[c + d*x] + (77*(1 + E^((2 *I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x) )])/E^(I*(c + d*x)) - 3*(117*Cos[3*(c + d*x)] + (33*I)*Sin[c + d*x] + (37* I)*Sin[3*(c + d*x)])))/(a^4*d)
Time = 0.88 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3981, 3042, 3981, 3042, 4255, 3042, 4255, 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.
| \(\displaystyle \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4}dx\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \int \frac {(e \sec (c+d x))^{11/2}}{(i \tan (c+d x) a+a)^2}dx}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \int \frac {(e \sec (c+d x))^{11/2}}{(i \tan (c+d x) a+a)^2}dx}{a^2}\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \left (\frac {7 e^2 \int (e \sec (c+d x))^{7/2}dx}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \left (\frac {7 e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \left (\frac {7 e^2 \left (\frac {3}{5} e^2 \int (e \sec (c+d x))^{3/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \left (\frac {7 e^2 \left (\frac {3}{5} e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \left (\frac {7 e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \left (\frac {7 e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \left (\frac {7 e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \left (\frac {7 e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac {11 e^2 \left (\frac {7 e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )}{3 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{a^2}\) |
((4*I)*e^2*(e*Sec[c + d*x])^(11/2))/(a*d*(a + I*a*Tan[c + d*x])^3) - (11*e ^2*((7*e^2*((2*e*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d) + (3*e^2*((-2* e^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*e*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/d))/5))/(3*a^2) - (((4*I)/3)*e^ 2*(e*Sec[c + d*x])^(7/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))))/a^2
3.3.55.3.1 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[(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. 465 vs. \(2 (194 ) = 388\).
Time = 11.66 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.43
| method | result | size |
| default | \(\frac {2 \sqrt {e \sec \left (d x +c \right )}\, e^{7} \left (-231 i \left (\cos ^{2}\left (d x +c \right )\right ) E\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+231 i \left (\cos ^{2}\left (d x +c \right )\right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-462 i \cos \left (d x +c \right ) E\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+462 i \cos \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-231 i E\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+231 i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+120 i \left (\cos ^{2}\left (d x +c \right )\right )+120 \sin \left (d x +c \right ) \cos \left (d x +c \right )+120 i \cos \left (d x +c \right )-111 \sin \left (d x +c \right )+20 i+3 \tan \left (d x +c \right )+20 i \sec \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 a^{4} d \left (\cos \left (d x +c \right )+1\right )}\) | \(466\) |
2/15/a^4/d*(e*sec(d*x+c))^(1/2)*e^7/(cos(d*x+c)+1)*(-231*I*cos(d*x+c)^2*El lipticE(I*(-csc(d*x+c)+cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1 /(cos(d*x+c)+1))^(1/2)+231*I*cos(d*x+c)^2*EllipticF(I*(-csc(d*x+c)+cot(d*x +c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)-462*I*c os(d*x+c)*EllipticE(I*(-csc(d*x+c)+cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+ 1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)+462*I*cos(d*x+c)*EllipticF(I*(-csc(d*x+ c)+cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/ 2)-231*I*EllipticE(I*(-csc(d*x+c)+cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)+231*I*EllipticF(I*(-csc(d*x+c)+cot(d*x+c )),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)+120*I*cos (d*x+c)^2+120*sin(d*x+c)*cos(d*x+c)+120*I*cos(d*x+c)-111*sin(d*x+c)+20*I+3 *tan(d*x+c)+20*I*sec(d*x+c)+3*sec(d*x+c)*tan(d*x+c))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.01 \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (-231 i \, e^{7} e^{\left (6 i \, d x + 6 i \, c\right )} - 616 i \, e^{7} e^{\left (4 i \, d x + 4 i \, c\right )} - 517 i \, e^{7} e^{\left (2 i \, d x + 2 i \, c\right )} - 120 i \, e^{7}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 231 \, \sqrt {2} {\left (-i \, e^{7} e^{\left (5 i \, d x + 5 i \, c\right )} - 2 i \, e^{7} e^{\left (3 i \, d x + 3 i \, c\right )} - i \, e^{7} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{15 \, {\left (a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )}} \]
-2/15*(sqrt(2)*(-231*I*e^7*e^(6*I*d*x + 6*I*c) - 616*I*e^7*e^(4*I*d*x + 4* I*c) - 517*I*e^7*e^(2*I*d*x + 2*I*c) - 120*I*e^7)*sqrt(e/(e^(2*I*d*x + 2*I *c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 231*sqrt(2)*(-I*e^7*e^(5*I*d*x + 5*I*c ) - 2*I*e^7*e^(3*I*d*x + 3*I*c) - I*e^7*e^(I*d*x + I*c))*sqrt(e)*weierstra ssZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))/(a^4*d*e^(5*I* d*x + 5*I*c) + 2*a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))
Timed out. \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{15/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \]