\(\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx\) [218]

Optimal result

Integrand size = 28, antiderivative size = 125 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\frac {10 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d e^4}-\frac {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}+\frac {20 i \left (a^4+i a^4 \tan (c+d x)\right )}{21 d e^2 (e \sec (c+d x))^{3/2}} \]

[Out]

10/21*a^4*(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^4-4/7*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)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3577, 3856, 2720} \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\frac {10 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d e^4}+\frac {20 i \left (a^4+i a^4 \tan (c+d x)\right )}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(7/2),x]

[Out]

(10*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(21*d*e^4) - (((4*I)/7)*a*(a + I*a*
Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(7/2)) + (((20*I)/21)*(a^4 + I*a^4*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 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 {4 i a (a+i a \tan (c+d x))^3}{7 d (e \sec (c+d x))^{7/2}}-\frac {\left (5 a^2\right ) \int \frac {(a+i a \tan (c+d x))^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}}+\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}}+\frac {\left (5 a^4\right ) \int \sqrt {e \sec (c+d x)} \, dx}{21 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}}+\frac {\left (5 a^4 \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 {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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.71 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\frac {2 a^4 \sqrt {e \sec (c+d x)} \left (2 i+2 i \cos (2 (c+d x))+5 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (2 (c+d x))-i \sin (2 (c+d x)))+8 \sin (2 (c+d x))\right ) (\cos (2 (c+3 d x))+i \sin (2 (c+3 d x)))}{21 d e^4 (\cos (d x)+i \sin (d x))^4} \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(7/2),x]

[Out]

(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]*(Co
s[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)

Maple [A] (verified)

Time = 14.52 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.46

method	result	size
default	\(\frac {2 a^{4} \left (5 i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-24 i \left (\cos ^{3}\left (d x +c \right )\right )+5 i \sec \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+24 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+28 i \cos \left (d x +c \right )-16 \sin \left (d x +c \right )\right )}{21 e^{3} d \sqrt {e \sec \left (d x +c \right )}}\)	\(183\)
risch	\(-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-5\right ) a^{4} \sqrt {2}}{21 d \,e^{3} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}+\frac {10 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a^{4} \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{21 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(247\)
parts	\(-\frac {2 a^{4} \left (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 a^{4} \left (-4 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}}+\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 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 \sin \left (d x +c \right )\right )}{7 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}-\frac {8 i a^{4} \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 {8 i a^{4}}{7 d \left (e \sec \left (d x +c \right )\right )^{\frac {7}{2}}}+\frac {4 a^{4} \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}}\)	\(547\)

[In]

int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/21*a^4/e^3/d/(e*sec(d*x+c))^(1/2)*(5*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)
),I)*(1/(cos(d*x+c)+1))^(1/2)-24*I*cos(d*x+c)^3+5*I*sec(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(
-csc(d*x+c)+cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+24*cos(d*x+c)^2*sin(d*x+c)+28*I*cos(d*x+c)-16*sin(d*x+c))

Fricas [C] (verification not implemented)

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

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

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-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
)

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=a^{4} \left (\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx + \int \left (- \frac {6 \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\right )\, dx + \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {4 i \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx + \int \left (- \frac {4 i \tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\right )\, dx\right ) \]

[In]

integrate((a+I*a*tan(d*x+c))**4/(e*sec(d*x+c))**(7/2),x)

[Out]

a**4*(Integral((e*sec(c + d*x))**(-7/2), x) + Integral(-6*tan(c + d*x)**2/(e*sec(c + d*x))**(7/2), x) + Integr
al(tan(c + d*x)**4/(e*sec(c + d*x))**(7/2), x) + Integral(4*I*tan(c + d*x)/(e*sec(c + d*x))**(7/2), x) + Integ
ral(-4*I*tan(c + d*x)**3/(e*sec(c + d*x))**(7/2), x))

Maxima [F]

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

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(7/2), x)

Giac [F]

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

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]

[In]

int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(7/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(7/2), x)