Integrand size = 28, antiderivative size = 125 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\frac {10 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d e^4}-\frac {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}+\frac {20 i \left (a^4+i a^4 \tan (c+d x)\right )}{21 d e^2 (e \sec (c+d x))^{3/2}} \] Output:
10/21*a^4*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(e*sec(d *x+c))^(1/2)/d/e^4-4/7*I*a*(a+I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^(7/2)+20/ 21*I*(a^4+I*a^4*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(3/2)
Time = 2.55 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\frac {2 a^4 \sqrt {e \sec (c+d x)} \left (2 i+2 i \cos (2 (c+d x))+5 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (2 (c+d x))-i \sin (2 (c+d x)))+8 \sin (2 (c+d x))\right ) (\cos (2 (c+3 d x))+i \sin (2 (c+3 d x)))}{21 d e^4 (\cos (d x)+i \sin (d x))^4} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(7/2),x]
Output:
(2*a^4*Sqrt[e*Sec[c + d*x]]*(2*I + (2*I)*Cos[2*(c + d*x)] + 5*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)]) + 8*Sin[2*(c + d*x)])*(Cos[2*(c + 3*d*x)] + I*Sin[2*(c + 3*d*x)]))/(21*d*e^ 4*(Cos[d*x] + I*Sin[d*x])^4)
Time = 0.57 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3977, 3042, 3977, 3042, 4258, 3042, 3120}
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.
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}}dx\) |
\(\Big \downarrow \) 3977 |
\(\displaystyle -\frac {5 a^2 \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {5 a^2 \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3977 |
\(\displaystyle -\frac {5 a^2 \left (-\frac {a^2 \int \sqrt {e \sec (c+d x)}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {5 a^2 \left (-\frac {a^2 \int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle -\frac {5 a^2 \left (-\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {5 a^2 \left (-\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {5 a^2 \left (-\frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(7/2),x]
Output:
(((-4*I)/7)*a*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(7/2)) - (5*a^ 2*((-2*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x ]])/(3*d*e^2) - (((4*I)/3)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(e*Sec[c + d*x]) ^(3/2))))/(7*e^2)
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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[(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]
Time = 17.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03
method | result | size |
default | \(\frac {2 a^{4} \left (\sin \left (d x +c \right ) \left (24 \cos \left (d x +c \right )^{2}-16\right )+i \left (-24 \cos \left (d x +c \right )^{3}+28 \cos \left (d x +c \right )\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (5+5 \sec \left (d x +c \right )\right )\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\) | \(129\) |
risch | \(-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-5\right ) a^{4} \sqrt {2}}{21 d \,e^{3} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}+\frac {10 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a^{4} \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{21 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\) | \(247\) |
parts | \(-\frac {2 a^{4} \left (\sin \left (d x +c \right ) \left (-3 \cos \left (d x +c \right )^{2}-5\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (5+5 \sec \left (d x +c \right )\right )\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}-\frac {2 a^{4} \left (\sin \left (d x +c \right ) \left (-\cos \left (d x +c \right )^{2}+3\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (4+4 \sec \left (d x +c \right )\right )\right )}{7 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}-\frac {4 i a^{4} \left (\frac {2 \cos \left (d x +c \right )^{3}}{7}-\frac {2 \cos \left (d x +c \right )}{3}\right )}{d \,e^{3} \sqrt {e \sec \left (d x +c \right )}}-\frac {8 i a^{4}}{7 d \left (e \sec \left (d x +c \right )\right )^{\frac {7}{2}}}+\frac {4 a^{4} \left (\sin \left (d x +c \right ) \left (3 \cos \left (d x +c \right )^{2}-2\right )+i \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (2+2 \sec \left (d x +c \right )\right )\right )}{7 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\) | \(380\) |
Input:
int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
2/21*a^4/d/(e*sec(d*x+c))^(1/2)/e^3*(sin(d*x+c)*(24*cos(d*x+c)^2-16)+I*(-2 4*cos(d*x+c)^3+28*cos(d*x+c))+I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos( d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*(5+5*sec(d*x+c)))
Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} a^{4} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (3 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, a^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )}}{21 \, d e^{4}} \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(7/2),x, algorithm="fricas")
Output:
-2/21*(5*I*sqrt(2)*a^4*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(3*I*a^4*e^(4*I*d*x + 4*I*c) - 2*I*a^4*e^(2*I*d*x + 2*I*c) - 5* I*a^4)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^4)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))**4/(e*sec(d*x+c))**(7/2),x)
Output:
Timed out
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(7/2),x, algorithm="maxima")
Output:
integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(7/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(7/2),x, algorithm="giac")
Output:
integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(7/2), x)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(7/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(7/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {e}\, a^{4} \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4}}d x +\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{4}}d x -4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{4}}d x \right ) i -6 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{4}}d x \right )+4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{4}}d x \right ) i \right )}{e^{4}} \] Input:
int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(7/2),x)
Output:
(sqrt(e)*a**4*(int(sqrt(sec(c + d*x))/sec(c + d*x)**4,x) + int((sqrt(sec(c + d*x))*tan(c + d*x)**4)/sec(c + d*x)**4,x) - 4*int((sqrt(sec(c + d*x))*t an(c + d*x)**3)/sec(c + d*x)**4,x)*i - 6*int((sqrt(sec(c + d*x))*tan(c + d *x)**2)/sec(c + d*x)**4,x) + 4*int((sqrt(sec(c + d*x))*tan(c + d*x))/sec(c + d*x)**4,x)*i))/e**4