\(\int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx\) [206]

Optimal result

Integrand size = 28, antiderivative size = 111 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {10 i a^3 \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {10 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}} \] Output:

-10/3*I*a^3*(e*sec(d*x+c))^(1/2)/d/e^2-10/3*a^3*cos(d*x+c)^(1/2)*InverseJa 
cobiAM(1/2*d*x+1/2*c,2^(1/2))*(e*sec(d*x+c))^(1/2)/d/e^2-4/3*I*a*(a+I*a*ta 
n(d*x+c))^2/d/(e*sec(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 1.87 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.11 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 a^3 \sec ^2(c+d x) \left (7 i \cos (c+d x)+5 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (c+d x)-i \sin (c+d x))+3 \sin (c+d x)\right ) (\cos (c+4 d x)+i \sin (c+4 d x))}{3 d (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x))^3} \] Input:

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(3/2),x]
 

Output:

(-2*a^3*Sec[c + d*x]^2*((7*I)*Cos[c + d*x] + 5*Sqrt[Cos[c + d*x]]*Elliptic 
F[(c + d*x)/2, 2]*(Cos[c + d*x] - I*Sin[c + d*x]) + 3*Sin[c + d*x])*(Cos[c 
 + 4*d*x] + I*Sin[c + 4*d*x]))/(3*d*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I*S 
in[d*x])^3)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.97, 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, 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.

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3977

\(\displaystyle -\frac {5 a^2 \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)dx}{3 e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {5 a^2 \int \sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)dx}{3 e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3967

\(\displaystyle -\frac {5 a^2 \left (a \int \sqrt {e \sec (c+d x)}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )}{3 e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {5 a^2 \left (a \int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )}{3 e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle -\frac {5 a^2 \left (a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )}{3 e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {5 a^2 \left (a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )}{3 e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {5 a^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a \sqrt {e \sec (c+d x)}}{d}\right )}{3 e^2}-\frac {4 i a (a+i a \tan (c+d x))^2}{3 d (e \sec (c+d x))^{3/2}}\)

Input:

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(3/2),x]
 

Output:

(-5*a^2*(((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*e^2) - (((4*I)/3)*a*(a + 
I*a*Tan[c + d*x])^2)/(d*(e*Sec[c + d*x])^(3/2))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3120

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]
 

rule 3967

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])
 

rule 3977

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]
 

rule 4258

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]
 

Maple [A] (verified)

Time = 7.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {a^{3} \left (\frac {8 \sin \left (d x +c \right )}{3}+i \left (-\frac {8 \cos \left (d x +c \right )}{3}-2 \sec \left (d x +c \right )\right )+\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (5+5 \sec \left (d x +c \right )\right )}{3}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e}\)	\(115\)
parts	\(\frac {a^{3} \left (-\frac {2 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 (\sec \left (d x +c \right )+1\right )}{3}+\frac {2 \sin \left (d x +c \right )}{3}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e}-\frac {i a^{3} \left (\frac {\sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \ln \left (\frac {4 \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}-2 \cos \left (d x +c \right )+2}{\cos \left (d x +c \right )+1}\right ) \left (-3-3 \sec \left (d x +c \right )\right )}{6}+\frac {\sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \ln \left (\frac {2 \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}-\cos \left (d x +c \right )+1}{\cos \left (d x +c \right )+1}\right ) \left (3+3 \sec \left (d x +c \right )\right )}{6}+\frac {2 \cos \left (d x +c \right )}{3}+2 \sec \left (d x +c \right )\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e}-\frac {2 i a^{3}}{d \left (e \sec \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {3 a^{3} \left (-\frac {2 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 )}{3}-\frac {2 \sin \left (d x +c \right )}{3}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e}\)	\(456\)

Input:

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

a^3/d*(8/3*sin(d*x+c)+I*(-8/3*cos(d*x+c)-2*sec(d*x+c))+2/3*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)*(5+5*sec(d*x+c)))/(e*sec(d*x+c))^(1/2)/e
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.74 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (-5 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (2 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )}}{3 \, d e^{2}} \] Input:

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-2/3*(-5*I*sqrt(2)*a^3*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) 
 + sqrt(2)*(2*I*a^3*e^(2*I*d*x + 2*I*c) + 5*I*a^3)*sqrt(e/(e^(2*I*d*x + 2* 
I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^2)
 

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx=- i a^{3} \left (\int \frac {i}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx\right ) \] Input:

integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(3/2),x)
 

Output:

-I*a**3*(Integral(I/(e*sec(c + d*x))**(3/2), x) + Integral(-3*tan(c + d*x) 
/(e*sec(c + d*x))**(3/2), x) + Integral(tan(c + d*x)**3/(e*sec(c + d*x))** 
(3/2), x) + Integral(-3*I*tan(c + d*x)**2/(e*sec(c + d*x))**(3/2), x))
 

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(3/2), x)
 

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(3/2),x)
 

Output:

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {e}\, a^{3} \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x -\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{2}}d x \right ) i -3 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2}}d x \right )+3 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{2}}d x \right ) i \right )}{e^{2}} \] Input:

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(3/2),x)
 

Output:

(sqrt(e)*a**3*(int(sqrt(sec(c + d*x))/sec(c + d*x)**2,x) - int((sqrt(sec(c 
 + d*x))*tan(c + d*x)**3)/sec(c + d*x)**2,x)*i - 3*int((sqrt(sec(c + d*x)) 
*tan(c + d*x)**2)/sec(c + d*x)**2,x) + 3*int((sqrt(sec(c + d*x))*tan(c + d 
*x))/sec(c + d*x)**2,x)*i))/e**2