\(\int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx\) [208]

Optimal result

Integrand size = 28, antiderivative size = 124 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {2 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d e^4}-\frac {2 i (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}-\frac {4 i \left (a^3+i a^3 \tan (c+d x)\right )}{21 d e^2 (e \sec (c+d x))^{3/2}} \] Output:

-2/21*a^3*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^4-2/7*I*(a+I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^(7/2)-4/21* 
I*(a^3+I*a^3*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 1.78 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {a^3 \sqrt {e \sec (c+d x)} \left (5 i+5 i \cos (2 (c+d x))+2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (2 (c+d x))-i \sin (2 (c+d x)))-\sin (2 (c+d x))\right ) (\cos (2 c+5 d x)+i \sin (2 c+5 d x))}{21 d e^4 (\cos (d x)+i \sin (d x))^3} \] Input:

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(7/2),x]
 

Output:

-1/21*(a^3*Sqrt[e*Sec[c + d*x]]*(5*I + (5*I)*Cos[2*(c + d*x)] + 2*Sqrt[Cos 
[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x) 
]) - Sin[2*(c + d*x)])*(Cos[2*c + 5*d*x] + I*Sin[2*c + 5*d*x]))/(d*e^4*(Co 
s[d*x] + I*Sin[d*x])^3)
 

Rubi [A] (verified)

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, 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.

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {a \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3977

\(\displaystyle \frac {a \left (-\frac {a^2 \int \sqrt {e \sec (c+d x)}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (-\frac {a^2 \int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {a \left (-\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (-\frac {a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {a \left (-\frac {2 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{3 d (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}\)

Input:

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(7/2),x]
 

Output:

(((-2*I)/7)*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(7/2)) + (a*((-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)
 

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 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 3978

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]
 

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 = 11.74 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.04

method	result	size
default	\(-\frac {2 a^{3} \left (\sin \left (d x +c \right ) \left (-12 \cos \left (d x +c \right )^{2}+1\right )+i \left (12 \cos \left (d x +c \right )^{3}-7 \cos \left (d x +c \right )\right )+i \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (-1-\sec \left (d x +c \right )\right )\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\)	\(129\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+2\right ) a^{3} \sqrt {2}}{21 d \,e^{3} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {2 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a^{3} \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{21 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(247\)
parts	\(-\frac {2 a^{3} \left (\sin \left (d x +c \right ) \left (-3 \cos \left (d x +c \right )^{2}-5\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (5+5 \sec \left (d x +c \right )\right )\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}-\frac {i a^{3} \left (\frac {2 \cos \left (d x +c \right )^{3}}{7}-\frac {2 \cos \left (d x +c \right )}{3}\right )}{d \,e^{3} \sqrt {e \sec \left (d x +c \right )}}-\frac {6 i a^{3}}{7 d \left (e \sec \left (d x +c \right )\right )^{\frac {7}{2}}}+\frac {2 a^{3} \left (\sin \left (d x +c \right ) \left (3 \cos \left (d x +c \right )^{2}-2\right )+i \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (2+2 \sec \left (d x +c \right )\right )\right )}{7 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\)	\(274\)

Input:

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-2/21*a^3/d/(e*sec(d*x+c))^(1/2)/e^3*(sin(d*x+c)*(-12*cos(d*x+c)^2+1)+I*(1 
2*cos(d*x+c)^3-7*cos(d*x+c))+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)*(-1-sec(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=\frac {2 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-3 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 5 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{21 \, d e^{4}} \] Input:

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

1/21*(2*I*sqrt(2)*a^3*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) 
+ sqrt(2)*(-3*I*a^3*e^(4*I*d*x + 4*I*c) - 5*I*a^3*e^(2*I*d*x + 2*I*c) - 2* 
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^4)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=- i a^{3} \left (\int \frac {i}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\right )\, dx\right ) \] Input:

integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(7/2),x)
 

Output:

-I*a**3*(Integral(I/(e*sec(c + d*x))**(7/2), x) + Integral(-3*tan(c + d*x) 
/(e*sec(c + d*x))**(7/2), x) + Integral(tan(c + d*x)**3/(e*sec(c + d*x))** 
(7/2), x) + Integral(-3*I*tan(c + d*x)**2/(e*sec(c + d*x))**(7/2), x))
 

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(7/2), x)
 

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \] Input:

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(7/2),x)
 

Output:

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(7/2), x)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {e}\, a^{3} \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4}}d x -\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{4}}d x \right ) i -3 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{4}}d x \right )+3 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{4}}d x \right ) i \right )}{e^{4}} \] Input:

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(7/2),x)
 

Output:

(sqrt(e)*a**3*(int(sqrt(sec(c + d*x))/sec(c + d*x)**4,x) - int((sqrt(sec(c 
 + d*x))*tan(c + d*x)**3)/sec(c + d*x)**4,x)*i - 3*int((sqrt(sec(c + d*x)) 
*tan(c + d*x)**2)/sec(c + d*x)**4,x) + 3*int((sqrt(sec(c + d*x))*tan(c + d 
*x))/sec(c + d*x)**4,x)*i))/e**4