\(\int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx\) [420]

Optimal result

Integrand size = 30, antiderivative size = 121 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i}{5 d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {16 i \sqrt {e \sec (c+d x)}}{15 d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {8 i \sqrt {a+i a \tan (c+d x)}}{15 a d (e \sec (c+d x))^{3/2}} \]

[Out]

2/5*I/d/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2)+16/15*I*(e*sec(d*x+c))^(1/2)/d/e^2/(a+I*a*tan(d*x+c))^(1
/2)-8/15*I*(a+I*a*tan(d*x+c))^(1/2)/a/d/(e*sec(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3583, 3578, 3569} \[ \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {16 i \sqrt {e \sec (c+d x)}}{15 d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {8 i \sqrt {a+i a \tan (c+d x)}}{15 a d (e \sec (c+d x))^{3/2}}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}} \]

[In]

Int[1/((e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((2*I)/5)/(d*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (((16*I)/15)*Sqrt[e*Sec[c + d*x]])/(d*e^2*Sq
rt[a + I*a*Tan[c + d*x]]) - (((8*I)/15)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*(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}{5 d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {4 \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{5 a} \\ & = \frac {2 i}{5 d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}-\frac {8 i \sqrt {a+i a \tan (c+d x)}}{15 a d (e \sec (c+d x))^{3/2}}+\frac {8 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 e^2} \\ & = \frac {2 i}{5 d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {16 i \sqrt {e \sec (c+d x)}}{15 d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {8 i \sqrt {a+i a \tan (c+d x)}}{15 a d (e \sec (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i \sec ^2(c+d x) (-15+\cos (2 (c+d x))+4 i \sin (2 (c+d x)))}{15 d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[1/((e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((-1/15*I)*Sec[c + d*x]^2*(-15 + Cos[2*(c + d*x)] + (4*I)*Sin[2*(c + d*x)]))/(d*(e*Sec[c + d*x])^(3/2)*Sqrt[a
+ I*a*Tan[c + d*x]])

Maple [A] (verified)

Time = 10.35 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.50

method	result	size
default	\(-\frac {2 \left (i \cos \left (d x +c \right )-4 \sin \left (d x +c \right )-8 i \sec \left (d x +c \right )\right )}{15 d \sqrt {e \sec \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, e}\)	\(61\)

[In]

int(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/15/d/(e*sec(d*x+c))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)/e*(I*cos(d*x+c)-4*sin(d*x+c)-8*I*sec(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-5 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 25 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 33 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-\frac {5}{2} i \, d x - \frac {5}{2} i \, c\right )}}{30 \, a d e^{2}} \]

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/30*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^(6*I*d*x + 6*I*c) + 25*I*e^(4
*I*d*x + 4*I*c) + 33*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-5/2*I*d*x - 5/2*I*c)/(a*d*e^2)

Sympy [F]

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

[In]

integrate(1/(e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(1/((e*sec(c + d*x))**(3/2)*sqrt(I*a*(tan(c + d*x) - I))), x)

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {3 i \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 i \, \cos \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 30 i \, \cos \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 3 \, \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 30 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )}{30 \, \sqrt {a} d e^{\frac {3}{2}}} \]

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/30*(3*I*cos(5/2*d*x + 5/2*c) - 5*I*cos(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*I*cos(1
/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 3*sin(5/2*d*x + 5/2*c) + 5*sin(3/5*arctan2(sin(5/2*d
*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))/(sqrt(a
)*d*e^(3/2))

Giac [F]

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

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(3/2)*sqrt(I*a*tan(d*x + c) + a)), x)

Mupad [B] (verification not implemented)

Time = 4.56 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (4\,\sin \left (2\,c+2\,d\,x\right )-\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+15{}\mathrm {i}\right )}{15\,d\,e^2\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]

[In]

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

[Out]

((e/cos(c + d*x))^(1/2)*(4*sin(2*c + 2*d*x) - cos(2*c + 2*d*x)*1i + 15i))/(15*d*e^2*((a*(cos(2*c + 2*d*x) + si
n(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2))