\(\int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx\) [399]

Optimal result

Integrand size = 30, antiderivative size = 164 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {12 i a}{35 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i a \sqrt {e \sec (c+d x)}}{35 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{35 d e^2 (e \sec (c+d x))^{3/2}} \]

[Out]

12/35*I*a/d/e^2/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2)+32/35*I*a*(e*sec(d*x+c))^(1/2)/d/e^4/(a+I*a*tan(
d*x+c))^(1/2)-2/7*I*(a+I*a*tan(d*x+c))^(1/2)/d/(e*sec(d*x+c))^(7/2)-16/35*I*(a+I*a*tan(d*x+c))^(1/2)/d/e^2/(e*
sec(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3578, 3583, 3569} \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {32 i a \sqrt {e \sec (c+d x)}}{35 d e^4 \sqrt {a+i a \tan (c+d x)}}+\frac {12 i a}{35 d e^2 \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{35 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}} \]

[In]

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

[Out]

(((12*I)/35)*a)/(d*e^2*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (((32*I)/35)*a*Sqrt[e*Sec[c + d*x]
])/(d*e^4*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/7)*Sqrt[a + I*a*Tan[c + d*x]])/(d*(e*Sec[c + d*x])^(7/2)) - ((
(16*I)/35)*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^2*(e*Sec[c + d*x])^(3/2))

Rule 3569

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] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

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 3583

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(b*f*(m + 2*n))), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
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] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}+\frac {(6 a) \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx}{7 e^2} \\ & = \frac {12 i a}{35 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}+\frac {24 \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{35 e^2} \\ & = \frac {12 i a}{35 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{35 d e^2 (e \sec (c+d x))^{3/2}}+\frac {(16 a) \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{35 e^4} \\ & = \frac {12 i a}{35 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i a \sqrt {e \sec (c+d x)}}{35 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{35 d e^2 (e \sec (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {(35 i \cos (c+d x)+i \cos (3 (c+d x))+70 \sin (c+d x)+6 \sin (3 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{70 d e^3 \sqrt {e \sec (c+d x)}} \]

[In]

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

[Out]

(((35*I)*Cos[c + d*x] + I*Cos[3*(c + d*x)] + 70*Sin[c + d*x] + 6*Sin[3*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])
/(70*d*e^3*Sqrt[e*Sec[c + d*x]])

Maple [A] (verified)

Time = 10.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.49

method	result	size
default	\(-\frac {2 i \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (6 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-\left (\cos ^{3}\left (d x +c \right )\right )+16 i \sin \left (d x +c \right )-8 \cos \left (d x +c \right )\right )}{35 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\)	\(80\)

[In]

int((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/35*I/d*(a*(1+I*tan(d*x+c)))^(1/2)*(6*I*cos(d*x+c)^2*sin(d*x+c)-cos(d*x+c)^3+16*I*sin(d*x+c)-8*cos(d*x+c))/(
e*sec(d*x+c))^(1/2)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.59 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-5 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 40 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 112 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (-\frac {5}{2} i \, d x - \frac {5}{2} i \, c\right )}}{140 \, d e^{4}} \]

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/140*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-5*I*e^(8*I*d*x + 8*I*c) - 40*I*e^(
6*I*d*x + 6*I*c) + 70*I*e^(4*I*d*x + 4*I*c) + 112*I*e^(2*I*d*x + 2*I*c) + 7*I)*e^(-5/2*I*d*x - 5/2*I*c)/(d*e^4
)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(d*x+c))**(1/2)/(e*sec(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {a} {\left (7 i \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 i \, \cos \left (\frac {7}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) - 35 i \, \cos \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 105 i \, \cos \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 7 \, \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sin \left (\frac {7}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 105 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )}}{140 \, d e^{\frac {7}{2}}} \]

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

1/140*sqrt(a)*(7*I*cos(5/2*d*x + 5/2*c) - 5*I*cos(7/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 3
5*I*cos(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 105*I*cos(1/5*arctan2(sin(5/2*d*x + 5/2*c),
cos(5/2*d*x + 5/2*c))) + 7*sin(5/2*d*x + 5/2*c) + 5*sin(7/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c
))) + 35*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 105*sin(1/5*arctan2(sin(5/2*d*x + 5/2*
c), cos(5/2*d*x + 5/2*c))))/(d*e^(7/2))

Giac [F]

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

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.78 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,36{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}+76\,\sin \left (2\,c+2\,d\,x\right )+6\,\sin \left (4\,c+4\,d\,x\right )+35{}\mathrm {i}\right )}{140\,d\,e^4} \]

[In]

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

[Out]

((e/cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(2
*c + 2*d*x)*36i + cos(4*c + 4*d*x)*1i + 76*sin(2*c + 2*d*x) + 6*sin(4*c + 4*d*x) + 35i))/(140*d*e^4)