Integrand size = 28, antiderivative size = 156 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\frac {2 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}} \] Output:
2/39*a^4*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^6/cos(d*x+c)^(1/2)/(e*s ec(d*x+c))^(1/2)+2/117*a^4*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(3/2)-4/13*I*a* (a+I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^(13/2)-4/117*I*(a^4+I*a^4*tan(d*x+c) )/d/e^2/(e*sec(d*x+c))^(9/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.97 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=-\frac {i a^4 e^{i (c+d x)} \left (31+59 e^{2 i (c+d x)}+37 e^{4 i (c+d x)}+9 e^{6 i (c+d x)}+8 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sqrt {e \sec (c+d x)}}{468 d e^7} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(13/2),x]
Output:
((-1/468*I)*a^4*E^(I*(c + d*x))*(31 + 59*E^((2*I)*(c + d*x)) + 37*E^((4*I) *(c + d*x)) + 9*E^((6*I)*(c + d*x)) + 8*Sqrt[1 + E^((2*I)*(c + d*x))]*Hype rgeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sqrt[e*Sec[c + d*x]])/ (d*e^7)
Time = 0.71 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3977, 3042, 3977, 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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}}dx\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle \frac {a^2 \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{9/2}}dx}{13 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{9/2}}dx}{13 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle \frac {a^2 \left (\frac {5 a^2 \int \frac {1}{(e \sec (c+d x))^{5/2}}dx}{9 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}\right )}{13 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \left (\frac {5 a^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{9 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}\right )}{13 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {a^2 \left (\frac {5 a^2 \left (\frac {3 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}\right )}{13 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \left (\frac {5 a^2 \left (\frac {3 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}\right )}{13 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {a^2 \left (\frac {5 a^2 \left (\frac {3 \int \sqrt {\cos (c+d x)}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}\right )}{13 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \left (\frac {5 a^2 \left (\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}\right )}{13 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {a^2 \left (\frac {5 a^2 \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{9 d (e \sec (c+d x))^{9/2}}\right )}{13 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(13/2),x]
Output:
(((-4*I)/13)*a*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(13/2)) + (a^ 2*((5*a^2*((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) - (((4*I)/9)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(9/2))))/( 13*e^2)
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[(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. 277 vs. \(2 (136 ) = 272\).
Time = 42.07 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.78
| method | result | size |
| default | \(\frac {2 a^{4} \left (\sin \left (d x +c \right ) \left (72 \cos \left (d x +c \right )^{6}+72 \cos \left (d x +c \right )^{5}-16 \cos \left (d x +c \right )^{4}-16 \cos \left (d x +c \right )^{3}+\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right )+3 i \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )-3 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )+i \left (-72 \cos \left (d x +c \right )^{7}-72 \cos \left (d x +c \right )^{6}+52 \cos \left (d x +c \right )^{5}+52 \cos \left (d x +c \right )^{4}\right )\right )}{117 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {e \sec \left (d x +c \right )}\, e^{6}}\) | \(278\) |
| risch | \(-\frac {i \left (9 \,{\mathrm e}^{6 i \left (d x +c \right )}+28 \,{\mathrm e}^{4 i \left (d x +c \right )}+31 \,{\mathrm e}^{2 i \left (d x +c \right )}+24\right ) a^{4} \sqrt {2}}{468 d \,e^{6} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i \operatorname {EllipticE}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) a^{4} \sqrt {2}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{39 d \,e^{6} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\) | \(352\) |
| parts | \(\text {Expression too large to display}\) | \(775\) |
Input:
int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x,method=_RETURNVERBOSE)
Output:
2/117*a^4/d/(cos(d*x+c)+1)/(e*sec(d*x+c))^(1/2)/e^6*(sin(d*x+c)*(72*cos(d* x+c)^6+72*cos(d*x+c)^5-16*cos(d*x+c)^4-16*cos(d*x+c)^3+cos(d*x+c)^2+cos(d* x+c)+3)+3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(co s(d*x+c)+2+sec(d*x+c))*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)-3*I*(1/(cos( d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+2+sec(d*x+c ))*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)+I*(-72*cos(d*x+c)^7-72*cos(d*x+c )^6+52*cos(d*x+c)^5+52*cos(d*x+c)^4))
Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\frac {24 i \, \sqrt {2} a^{4} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (-9 i \, a^{4} e^{\left (7 i \, d x + 7 i \, c\right )} - 37 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 59 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 31 i \, a^{4} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{468 \, d e^{7}} \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x, algorithm="fricas" )
Output:
1/468*(24*I*sqrt(2)*a^4*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse (-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(-9*I*a^4*e^(7*I*d*x + 7*I*c) - 37*I*a ^4*e^(5*I*d*x + 5*I*c) - 59*I*a^4*e^(3*I*d*x + 3*I*c) - 31*I*a^4*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^7)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))**4/(e*sec(d*x+c))**(13/2),x)
Output:
Timed out
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {13}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x, algorithm="maxima" )
Output:
integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(13/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {13}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x, algorithm="giac")
Output:
integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(13/2), x)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{13/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(13/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(13/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx=\frac {\sqrt {e}\, a^{4} \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{7}}d x +\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{7}}d x -4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{7}}d x \right ) i -6 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{7}}d x \right )+4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{7}}d x \right ) i \right )}{e^{7}} \] Input:
int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x)
Output:
(sqrt(e)*a**4*(int(sqrt(sec(c + d*x))/sec(c + d*x)**7,x) + int((sqrt(sec(c + d*x))*tan(c + d*x)**4)/sec(c + d*x)**7,x) - 4*int((sqrt(sec(c + d*x))*t an(c + d*x)**3)/sec(c + d*x)**7,x)*i - 6*int((sqrt(sec(c + d*x))*tan(c + d *x)**2)/sec(c + d*x)**7,x) + 4*int((sqrt(sec(c + d*x))*tan(c + d*x))/sec(c + d*x)**7,x)*i))/e**7