\(\int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx\) [398]

Optimal result

Integrand size = 30, antiderivative size = 122 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx=\frac {8 i a}{15 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{15 d e^2 \sqrt {e \sec (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 122, 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 = {3578, 3583, 3569} \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx=\frac {8 i a}{15 d e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{15 d e^2 \sqrt {e \sec (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}} \]

[In]

Int[Sqrt[a + I*a*Tan[c + d*x]]/(e*Sec[c + d*x])^(5/2),x]

[Out]

(((8*I)/15)*a)/(d*e^2*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/5)*Sqrt[a + I*a*Tan[c + d*x]]
)/(d*(e*Sec[c + d*x])^(5/2)) - (((16*I)/15)*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^2*Sqrt[e*Sec[c + d*x]])

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)}}{5 d (e \sec (c+d x))^{5/2}}+\frac {(4 a) \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{5 e^2} \\ & = \frac {8 i a}{15 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}+\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx}{15 e^2} \\ & = \frac {8 i a}{15 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{15 d e^2 \sqrt {e \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx=\frac {i (-15+\cos (2 (c+d x))-4 i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{15 d e^2 \sqrt {e \sec (c+d x)}} \]

[In]

Integrate[Sqrt[a + I*a*Tan[c + d*x]]/(e*Sec[c + d*x])^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 10.52 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.51

method	result	size
default	\(\frac {2 \left (i \left (\cos ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right ) \cos \left (d x +c \right )-8 i\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{15 d \sqrt {e \sec \left (d x +c \right )}\, e^{2}}\)	\(62\)
risch	\(-\frac {i \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (30-2 \cos \left (2 d x +2 c \right )+8 i \sin \left (2 d x +2 c \right )\right )}{30 e^{2} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(87\)

[In]

int((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 33 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 25 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac {3}{2} i \, d x - \frac {3}{2} i \, c\right )}}{30 \, d e^{3}} \]

[In]

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

Sympy [F]

\[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((a+I*a*tan(d*x+c))**(1/2)/(e*sec(d*x+c))**(5/2),x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))/(e*sec(c + d*x))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a} {\left (5 i \, \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 i \, \cos \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) - 30 i \, \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sin \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 30 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )}}{30 \, d e^{\frac {5}{2}}} \]

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sqrt {i \, a \tan \left (d x + c\right ) + a}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)/(e*sec(d*x + c))^(5/2), x)

Mupad [B] (verification not implemented)

Time = 5.80 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (4\,\sin \left (c+d\,x\right )+4\,\sin \left (3\,c+3\,d\,x\right )-\cos \left (c+d\,x\right )\,29{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}\right )}{30\,d\,e^3} \]

[In]

int((a + a*tan(c + d*x)*1i)^(1/2)/(e/cos(c + d*x))^(5/2),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)*(4*sin
(c + d*x) - cos(c + d*x)*29i + cos(3*c + 3*d*x)*1i + 4*sin(3*c + 3*d*x)))/(30*d*e^3)