Integrand size = 28, antiderivative size = 163 \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {42 e^6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {42 e^5 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {28 i e^4 (e \sec (c+d x))^{3/2}}{5 d \left (a^4+i a^4 \tan (c+d x)\right )} \]
-42/5*e^6*(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)+42/5*e^5 *sin(d*x+c)*(e*sec(d*x+c))^(1/2)/a^4/d+4/5*I*e^2*(e*sec(d*x+c))^(7/2)/a/d/ (a+I*a*tan(d*x+c))^3-28/5*I*e^4*(e*sec(d*x+c))^(3/2)/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 = 1.59 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.65 \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {2 i e^5 e^{-3 i (c+d x)} \left (-2-7 e^{2 i (c+d x)}+21 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )\right ) \sqrt {e \sec (c+d x)}}{5 a^4 d} \]
(((-2*I)/5)*e^5*(-2 - 7*E^((2*I)*(c + d*x)) + 21*E^((2*I)*(c + d*x))*Sqrt[ 1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])*Sqrt[e*Sec[c + d*x]])/(a^4*d*E^((3*I)*(c + d*x)))
Time = 0.72 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3981, 3042, 3981, 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.
| \(\displaystyle \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4}dx\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {7 e^2 \int \frac {(e \sec (c+d x))^{7/2}}{(i \tan (c+d x) a+a)^2}dx}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {7 e^2 \int \frac {(e \sec (c+d x))^{7/2}}{(i \tan (c+d x) a+a)^2}dx}{5 a^2}\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {7 e^2 \left (-\frac {3 e^2 \int (e \sec (c+d x))^{3/2}dx}{a^2}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {7 e^2 \left (-\frac {3 e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx}{a^2}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {7 e^2 \left (-\frac {3 e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx\right )}{a^2}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {7 e^2 \left (-\frac {3 e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{a^2}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {7 e^2 \left (-\frac {3 e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{a^2}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {7 e^2 \left (-\frac {3 e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{a^2}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {4 i e^2 (e \sec (c+d x))^{7/2}}{5 a d (a+i a \tan (c+d x))^3}-\frac {7 e^2 \left (-\frac {3 e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{a^2}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{5 a^2}\) |
(((4*I)/5)*e^2*(e*Sec[c + d*x])^(7/2))/(a*d*(a + I*a*Tan[c + d*x])^3) - (7 *e^2*((-3*e^2*((-2*e^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sq rt[e*Sec[c + d*x]]) + (2*e*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/d))/a^2 + (( 4*I)*e^2*(e*Sec[c + d*x])^(3/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))))/(5*a^2)
3.3.57.3.1 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. 481 vs. \(2 (167 ) = 334\).
Time = 9.68 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.96
| method | result | size |
| default | \(-\frac {2 i \sqrt {e \sec \left (d x +c \right )}\, \left (21 \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}}-21 \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}}+8 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+42 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, 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}}\, \cos \left (d x +c \right )-42 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\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}}\, \cos \left (d x +c \right )-8 \left (\cos ^{4}\left (d x +c \right )\right )+8 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+21 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}}-21 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\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}}-8 \left (\cos ^{3}\left (d x +c \right )\right )-16 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+20 \left (\cos ^{2}\left (d x +c \right )\right )+5 i \sin \left (d x +c \right )+20 \cos \left (d x +c \right )\right ) e^{5}}{5 a^{4} d \left (\cos \left (d x +c \right )+1\right )}\) | \(482\) |
-2/5*I/a^4/d*(e*sec(d*x+c))^(1/2)*(21*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 )-21*cos(d*x+c)^2*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)+8*I*sin(d*x+c)*cos(d*x+c)^3+42* (cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)),I)* (1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)-42*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*E llipticE(I*(-csc(d*x+c)+cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c) -8*cos(d*x+c)^4+8*I*sin(d*x+c)*cos(d*x+c)^2+21*EllipticF(I*(-csc(d*x+c)+co t(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-21 *(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-csc(d*x+c)+cot(d*x+c)),I) *(1/(cos(d*x+c)+1))^(1/2)-8*cos(d*x+c)^3-16*I*sin(d*x+c)*cos(d*x+c)+20*cos (d*x+c)^2+5*I*sin(d*x+c)+20*cos(d*x+c))*e^5/(cos(d*x+c)+1)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.70 \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {2 \, {\left (21 i \, \sqrt {2} e^{\frac {11}{2}} e^{\left (3 i \, d x + 3 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (21 i \, e^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 14 i \, e^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, e^{5}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{5 \, a^{4} d} \]
-2/5*(21*I*sqrt(2)*e^(11/2)*e^(3*I*d*x + 3*I*c)*weierstrassZeta(-4, 0, wei erstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(21*I*e^5*e^(4*I*d*x + 4*I*c) + 14*I*e^5*e^(2*I*d*x + 2*I*c) - 2*I*e^5)*sqrt(e/(e^(2*I*d*x + 2*I *c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-3*I*d*x - 3*I*c)/(a^4*d)
Timed out. \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^4} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{11/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \]