Integrand size = 28, antiderivative size = 124 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{9/2}} \, dx=\frac {2 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d e^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}-\frac {4 i \left (a^3+i a^3 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}} \]
2/15*a^3*(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))/d/e^4/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-2/9*I*(a+ I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^(9/2)-4/15*I*(a^3+I*a^3*tan(d*x+c))/d/e ^2/(e*sec(d*x+c))^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.95 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.95 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{9/2}} \, dx=-\frac {a^3 e^{-2 i (c+d x)} \left (11+16 e^{2 i (c+d x)}+5 e^{4 i (c+d x)}+4 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) (-i+\tan (c+d x))^3}{90 d e^2 (e \sec (c+d x))^{5/2}} \]
-1/90*(a^3*(11 + 16*E^((2*I)*(c + d*x)) + 5*E^((4*I)*(c + d*x)) + 4*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d* x))])*(-I + Tan[c + d*x])^3)/(d*e^2*E^((2*I)*(c + d*x))*(e*Sec[c + d*x])^( 5/2))
Time = 0.61 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3978, 3042, 3977, 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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3978 |
| \(\displaystyle \frac {a \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{5/2}}dx}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{5/2}}dx}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle \frac {a \left (\frac {a^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{5 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{5 d (e \sec (c+d x))^{5/2}}\right )}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \left (\frac {a^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{5 d (e \sec (c+d x))^{5/2}}\right )}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {a \left (\frac {a^2 \int \sqrt {\cos (c+d x)}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{5 d (e \sec (c+d x))^{5/2}}\right )}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \left (\frac {a^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{5 d (e \sec (c+d x))^{5/2}}\right )}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {a \left (\frac {2 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{5 d (e \sec (c+d x))^{5/2}}\right )}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}\) |
(((-2*I)/9)*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(9/2)) + (a*((2* a^2*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (((4*I)/5)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(5/2)) ))/(3*e^2)
3.3.9.3.1 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*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m)) Int[(d*Sec[e + f*x]) ^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & & LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/( a*f*m)), x] + Simp[a*((m + n)/(m*d^2)) Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b ^2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 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. 338 vs. \(2 (132 ) = 264\).
Time = 14.16 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.73
| method | result | size |
| risch | \(-\frac {i \left (5 \,{\mathrm e}^{4 i \left (d x +c \right )}+11 \,{\mathrm e}^{2 i \left (d x +c \right )}+12\right ) a^{3} \sqrt {2}}{90 d \,e^{4} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i E\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) a^{3} \sqrt {2}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{15 d \,e^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\) | \(339\) |
| default | \(\frac {2 i a^{3} \left (-20 i \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-20 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-20 \left (\cos ^{5}\left (d x +c \right )\right )+3 \cos \left (d x +c \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}}\, E\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-3 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}}\, \cos \left (d x +c \right )-i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-20 \left (\cos ^{4}\left (d x +c \right )\right )+6 \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}}\, E\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-6 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}}-i \cos \left (d x +c \right ) \sin \left (d x +c \right )+9 \left (\cos ^{3}\left (d x +c \right )\right )+3 \sec \left (d x +c \right ) \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}}-3 \sec \left (d x +c \right ) \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}}-3 i \sin \left (d x +c \right )+9 \left (\cos ^{2}\left (d x +c \right )\right )\right )}{45 e^{4} d \left (\cos \left (d x +c \right )+1\right ) \sqrt {e \sec \left (d x +c \right )}}\) | \(497\) |
| parts | \(\text {Expression too large to display}\) | \(976\) |
-1/90*I*(5*exp(I*(d*x+c))^4+11*exp(I*(d*x+c))^2+12)/d*a^3*2^(1/2)/e^4/(e*e xp(I*(d*x+c))/(exp(I*(d*x+c))^2+1))^(1/2)-1/15*I/d*(-2*(e*exp(I*(d*x+c))^2 +e)/e/(exp(I*(d*x+c))*(e*exp(I*(d*x+c))^2+e))^(1/2)+I*(-I*(exp(I*(d*x+c))+ I))^(1/2)*2^(1/2)*(I*(exp(I*(d*x+c))-I))^(1/2)*(I*exp(I*(d*x+c)))^(1/2)/(e *exp(I*(d*x+c))^3+e*exp(I*(d*x+c)))^(1/2)*(-2*I*EllipticE((-I*(exp(I*(d*x+ c))+I))^(1/2),1/2*2^(1/2))+I*EllipticF((-I*(exp(I*(d*x+c))+I))^(1/2),1/2*2 ^(1/2))))*a^3*2^(1/2)/e^4/(exp(I*(d*x+c))^2+1)/(e*exp(I*(d*x+c))/(exp(I*(d *x+c))^2+1))^(1/2)*(e*exp(I*(d*x+c))*(exp(I*(d*x+c))^2+1))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.87 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{9/2}} \, dx=\frac {12 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (-5 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 16 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 11 i \, a^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{90 \, d e^{5}} \]
1/90*(12*I*sqrt(2)*a^3*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse( -4, 0, e^(I*d*x + I*c))) + sqrt(2)*(-5*I*a^3*e^(5*I*d*x + 5*I*c) - 16*I*a^ 3*e^(3*I*d*x + 3*I*c) - 11*I*a^3*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I *c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^5)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{9/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{9/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{9/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{9/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{9/2}} \,d x \]