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

Optimal result

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

[Out]

-2/21*a^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(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^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)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3578, 3577, 3856, 2720} \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {2 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d e^4}-\frac {4 i \left (a^3+i a^3 \tan (c+d x)\right )}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}} \]

[In]

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

[Out]

(-2*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(21*d*e^4) - (((2*I)/7)*(a + I*a*Ta
n[c + d*x])^3)/(d*(e*Sec[c + d*x])^(7/2)) - (((4*I)/21)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(3
/2))

Rule 2720

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 3577

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] - Dist[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]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] + Dist[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 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}+\frac {a \int \frac {(a+i a \tan (c+d x))^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}}-\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}}-\frac {a^3 \int \sqrt {e \sec (c+d x)} \, dx}{21 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}}-\frac {\left (a^3 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 e^4} \\ & = -\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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.87 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {a^3 \sqrt {e \sec (c+d x)} \left (5 i+5 i \cos (2 (c+d x))+2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (2 (c+d x))-i \sin (2 (c+d x)))-\sin (2 (c+d x))\right ) (\cos (2 c+5 d x)+i \sin (2 c+5 d x))}{21 d e^4 (\cos (d x)+i \sin (d x))^3} \]

[In]

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

[Out]

-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*(
Cos[d*x] + I*Sin[d*x])^3)

Maple [A] (verified)

Time = 13.45 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.46

method	result	size
default	\(-\frac {2 a^{3} \left (12 i \left (\cos ^{3}\left (d x +c \right )\right )+i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-12 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+i \sec \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-7 i \cos \left (d x +c \right )+\sin \left (d x +c \right )\right )}{21 e^{3} d \sqrt {e \sec \left (d x +c \right )}}\)	\(181\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+2\right ) a^{3} \sqrt {2}}{21 d \,e^{3} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {2 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a^{3} \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{21 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(247\)
parts	\(-\frac {2 a^{3} \left (5 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+5 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-5 \sin \left (d x +c \right )\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}-\frac {2 i a^{3} \left (3 \left (\cos ^{3}\left (d x +c \right )\right )-7 \cos \left (d x +c \right )\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}-\frac {6 i a^{3}}{7 d \left (e \sec \left (d x +c \right )\right )^{\frac {7}{2}}}+\frac {2 a^{3} \left (2 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+2 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2 \sin \left (d x +c \right )\right )}{7 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\)	\(386\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{7/2}} \, dx=\frac {2 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-3 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 5 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{21 \, d e^{4}} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

-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) + Integ
ral(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]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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