Integrand size = 28, antiderivative size = 146 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {10 i a^4 \sqrt {e \sec (c+d x)}}{d e^2}-\frac {10 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{d e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}-\frac {2 i \sqrt {e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}{d e^2} \]
-10*I*a^4*(e*sec(d*x+c))^(1/2)/d/e^2-10*a^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c os(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*( e*sec(d*x+c))^(1/2)/d/e^2-4/3*I*a*(a+I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^(3 /2)-2*I*(e*sec(d*x+c))^(1/2)*(a^4+I*a^4*tan(d*x+c))/d/e^2
Time = 2.97 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.89 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}} \, dx=\frac {a^4 \sec ^3(c+d x) \left (21+19 \cos (2 (c+d x))-30 i \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (c+d x)-i \sin (c+d x))-11 i \sin (2 (c+d x))\right ) (-i \cos (c+5 d x)+\sin (c+5 d x))}{3 d (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x))^4} \]
(a^4*Sec[c + d*x]^3*(21 + 19*Cos[2*(c + d*x)] - (30*I)*Cos[c + d*x]^(3/2)* EllipticF[(c + d*x)/2, 2]*(Cos[c + d*x] - I*Sin[c + d*x]) - (11*I)*Sin[2*( c + d*x)])*((-I)*Cos[c + 5*d*x] + Sin[c + 5*d*x]))/(3*d*(e*Sec[c + d*x])^( 3/2)*(Cos[d*x] + I*Sin[d*x])^4)
Time = 0.68 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.03, 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, 3979, 3042, 3967, 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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle -\frac {3 a^2 \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^2dx}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 a^2 \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^2dx}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle -\frac {3 a^2 \left (\frac {5}{3} a \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 a^2 \left (\frac {5}{3} a \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle -\frac {3 a^2 \left (\frac {5}{3} a \left (a \int \sqrt {e \sec (c+d x)}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 a^2 \left (\frac {5}{3} a \left (a \int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle -\frac {3 a^2 \left (\frac {5}{3} a \left (a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 a^2 \left (\frac {5}{3} a \left (a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {3 a^2 \left (\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {5}{3} a \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )\right )}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{3 d (e \sec (c+d x))^{3/2}}\) |
(((-4*I)/3)*a*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(3/2)) - (3*a^ 2*((5*a*(((2*I)*a*Sqrt[e*Sec[c + d*x]])/d + (2*a*Sqrt[Cos[c + d*x]]*Ellipt icF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/d))/3 + (((2*I)/3)*Sqrt[e*Sec[c + d*x]]*(a^2 + I*a^2*Tan[c + d*x]))/d))/e^2
3.3.16.3.1 Defintions of rubi rules used
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_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
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 - 1)/(f*(m + n - 1))), x] + Simp[a*((m + 2*n - 2)/(m + n - 1)) Int[(d*Se c[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && NeQ[m + n - 1, 0] && 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]
Time = 17.23 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.23
| method | result | size |
| default | \(-\frac {2 a^{4} \left (15 i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+15 i \sec \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+8 i \cos \left (d x +c \right )-8 \sin \left (d x +c \right )+12 i \sec \left (d x +c \right )-\sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{3 e d \sqrt {e \sec \left (d x +c \right )}}\) | \(179\) |
| parts | \(\text {Expression too large to display}\) | \(1303\) |
-2/3*a^4/e/d/(e*sec(d*x+c))^(1/2)*(15*I*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)+15*I*sec( d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*Elliptic F(I*(-csc(d*x+c)+cot(d*x+c)),I)+8*I*cos(d*x+c)-8*sin(d*x+c)+12*I*sec(d*x+c )-sec(d*x+c)*tan(d*x+c))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (4 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 21 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i \, a^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 15 \, \sqrt {2} {\left (-i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{3 \, {\left (d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )}} \]
-2/3*(sqrt(2)*(4*I*a^4*e^(4*I*d*x + 4*I*c) + 21*I*a^4*e^(2*I*d*x + 2*I*c) + 15*I*a^4)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 15 *sqrt(2)*(-I*a^4*e^(2*I*d*x + 2*I*c) - I*a^4)*sqrt(e)*weierstrassPInverse( -4, 0, e^(I*d*x + I*c)))/(d*e^2*e^(2*I*d*x + 2*I*c) + d*e^2)
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}} \, dx=a^{4} \left (\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {6 \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx + \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {4 i \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {4 i \tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx\right ) \]
a**4*(Integral((e*sec(c + d*x))**(-3/2), x) + Integral(-6*tan(c + d*x)**2/ (e*sec(c + d*x))**(3/2), x) + Integral(tan(c + d*x)**4/(e*sec(c + d*x))**( 3/2), x) + Integral(4*I*tan(c + d*x)/(e*sec(c + d*x))**(3/2), x) + Integra l(-4*I*tan(c + d*x)**3/(e*sec(c + d*x))**(3/2), x))
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]