Integrand size = 28, antiderivative size = 154 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {42 (e \cos (c+d x))^{5/2} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{65 a^2 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \cos (c+d x) (e \cos (c+d x))^{5/2} \sin (c+d x)}{13 a^2 d}+\frac {14 (e \cos (c+d x))^{5/2} \tan (c+d x)}{65 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{5/2}}{13 d \left (a^2+i a^2 \tan (c+d x)\right )} \]
42/65*(e*cos(d*x+c))^(5/2)*(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^2/d/cos(d*x+c)^(5/2)+2/13*cos(d*x +c)*(e*cos(d*x+c))^(5/2)*sin(d*x+c)/a^2/d+14/65*(e*cos(d*x+c))^(5/2)*tan(d *x+c)/a^2/d+4/13*I*cos(d*x+c)^2*(e*cos(d*x+c))^(5/2)/d/(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 = 6.00 (sec) , antiderivative size = 464, normalized size of antiderivative = 3.01 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {(e \cos (c+d x))^{5/2} \sec ^5(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-21 \cos (c) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c)))-42 i \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (c) \sin (d x+\arctan (\tan (c)))+21 i (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}+\frac {21}{2} (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \cot (c) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}-\frac {1}{8} \cos (c+d x) \csc (c) \sqrt {\sec ^2(c)} (\cos (2 d x)-i \sin (2 d x)) (178 \cos (c+2 d x)+158 \cos (3 c+2 d x)-9 \cos (3 c+4 d x)+9 \cos (5 c+4 d x)-88 i \sin (c)+208 i \sin (c+2 d x)+128 i \sin (3 c+2 d x)-4 i \sin (3 c+4 d x)+4 i \sin (5 c+4 d x)) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}+21 \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (c) \sin (d x+\arctan (\tan (c))) \tan (c)-\frac {21}{2} (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sqrt {\sin ^2(d x+\arctan (\tan (c)))} \tan (c)\right )}{65 d \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))} (a+i a \tan (c+d x))^2} \]
((e*Cos[c + d*x])^(5/2)*Sec[c + d*x]^5*(Cos[d*x] + I*Sin[d*x])^2*(-21*Cos[ c]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin [d*x + ArcTan[Tan[c]]] - (42*I)*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos [d*x + ArcTan[Tan[c]]]^2]*Sin[c]*Sin[d*x + ArcTan[Tan[c]]] + (21*I)*(3*Cos [c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2] + (21*(3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Cot[c]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])/2 - (Cos[c + d *x]*Csc[c]*Sqrt[Sec[c]^2]*(Cos[2*d*x] - I*Sin[2*d*x])*(178*Cos[c + 2*d*x] + 158*Cos[3*c + 2*d*x] - 9*Cos[3*c + 4*d*x] + 9*Cos[5*c + 4*d*x] - (88*I)* Sin[c] + (208*I)*Sin[c + 2*d*x] + (128*I)*Sin[3*c + 2*d*x] - (4*I)*Sin[3*c + 4*d*x] + (4*I)*Sin[5*c + 4*d*x])*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])/8 + 21*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Si n[c]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c] - (21*(3*Cos[c - d*x - ArcTan[Tan[c] ]] + Cos[c + d*x + ArcTan[Tan[c]]])*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]*Tan[ c])/2))/(65*d*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]*(a + I*a*Ta n[c + d*x])^2)
Time = 0.84 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3998, 3042, 3981, 3042, 4256, 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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3998 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \int \frac {1}{(e \sec (c+d x))^{5/2} (i \tan (c+d x) a+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \int \frac {1}{(e \sec (c+d x))^{5/2} (i \tan (c+d x) a+a)^2}dx\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \int \frac {1}{(e \sec (c+d x))^{9/2}}dx}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}}dx}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \int \frac {1}{(e \sec (c+d x))^{5/2}}dx}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \sqrt {\cos (c+d x)}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\) |
(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)*((9*e^2*((2*Sin[c + d*x])/(9 *d*e*(e*Sec[c + d*x])^(7/2)) + (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*e^2)))/(13*a^2) + (((4*I)/13)*e^2)/(d*(e*Sec[c + d*x ])^(9/2)*(a^2 + I*a^2*Tan[c + d*x])))
3.7.64.3.1 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[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. 350 vs. \(2 (160 ) = 320\).
Time = 8.42 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.28
| method | result | size |
| default | \(\frac {2 e^{3} \left (840 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1280 \left (\sin ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-140 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3840 \left (\sin ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-6720 i \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4960 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-1280 i \left (\sin ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3520 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2800 i \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1496 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+10 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )-376 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5600 i \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+44 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+21 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}+4480 i \left (\sin ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{65 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(351\) |
2/65/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^3*(840*I *sin(1/2*d*x+1/2*c)^5+1280*sin(1/2*d*x+1/2*c)^14*cos(1/2*d*x+1/2*c)-140*I* sin(1/2*d*x+1/2*c)^3-3840*sin(1/2*d*x+1/2*c)^12*cos(1/2*d*x+1/2*c)-6720*I* sin(1/2*d*x+1/2*c)^11+4960*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-1280*I *sin(1/2*d*x+1/2*c)^15-3520*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-2800*I *sin(1/2*d*x+1/2*c)^7+1496*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+10*I*si n(1/2*d*x+1/2*c)-376*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5600*I*sin(1/ 2*d*x+1/2*c)^9+44*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+21*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)+4480*I*sin(1/2*d*x+1/2*c)^13)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.91 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (336 i \, \sqrt {2} e^{\frac {5}{2}} e^{\left (6 i \, d x + 6 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {\frac {1}{2}} {\left (-13 i \, e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 386 i \, e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 88 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 30 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{2}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{520 \, a^{2} d} \]
1/520*(336*I*sqrt(2)*e^(5/2)*e^(6*I*d*x + 6*I*c)*weierstrassZeta(-4, 0, we ierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(1/2)*(-13*I*e^2*e^(8*I*d *x + 8*I*c) + 386*I*e^2*e^(6*I*d*x + 6*I*c) + 88*I*e^2*e^(4*I*d*x + 4*I*c) + 30*I*e^2*e^(2*I*d*x + 2*I*c) + 5*I*e^2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e) *e^(-1/2*I*d*x - 1/2*I*c))*e^(-6*I*d*x - 6*I*c)/(a^2*d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]