\(\int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx\) [406]

Optimal result

Integrand size = 30, antiderivative size = 125 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {16 i a^2 \sqrt {e \sec (c+d x)}}{21 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {8 i a \sqrt {a+i a \tan (c+d x)}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{7 d (e \sec (c+d x))^{7/2}} \]

[Out]

16/21*I*a^2*(e*sec(d*x+c))^(1/2)/d/e^4/(a+I*a*tan(d*x+c))^(1/2)-8/21*I*a*(a+I*a*tan(d*x+c))^(1/2)/d/e^2/(e*sec
(d*x+c))^(3/2)-2/7*I*(a+I*a*tan(d*x+c))^(3/2)/d/(e*sec(d*x+c))^(7/2)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3578, 3569} \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {16 i a^2 \sqrt {e \sec (c+d x)}}{21 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {8 i a \sqrt {a+i a \tan (c+d x)}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{7 d (e \sec (c+d x))^{7/2}} \]

[In]

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

[Out]

(((16*I)/21)*a^2*Sqrt[e*Sec[c + d*x]])/(d*e^4*Sqrt[a + I*a*Tan[c + d*x]]) - (((8*I)/21)*a*Sqrt[a + I*a*Tan[c +
d*x]])/(d*e^2*(e*Sec[c + d*x])^(3/2)) - (((2*I)/7)*(a + I*a*Tan[c + d*x])^(3/2))/(d*(e*Sec[c + d*x])^(7/2))

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i (a+i a \tan (c+d x))^{3/2}}{7 d (e \sec (c+d x))^{7/2}}+\frac {(4 a) \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{7 e^2} \\ & = -\frac {8 i a \sqrt {a+i a \tan (c+d x)}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{7 d (e \sec (c+d x))^{7/2}}+\frac {\left (8 a^2\right ) \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{21 e^4} \\ & = \frac {16 i a^2 \sqrt {e \sec (c+d x)}}{21 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {8 i a \sqrt {a+i a \tan (c+d x)}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{7 d (e \sec (c+d x))^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.45 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {a (\cos (d x)-i \sin (d x)) (-7 i+9 i \cos (2 (c+d x))+12 \sin (2 (c+d x))) (\cos (c+2 d x)+i \sin (c+2 d x)) \sqrt {a+i a \tan (c+d x)}}{21 d e^3 \sqrt {e \sec (c+d x)}} \]

[In]

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

[Out]

(a*(Cos[d*x] - I*Sin[d*x])*(-7*I + (9*I)*Cos[2*(c + d*x)] + 12*Sin[2*(c + d*x)])*(Cos[c + 2*d*x] + I*Sin[c + 2
*d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(21*d*e^3*Sqrt[e*Sec[c + d*x]])

Maple [A] (verified)

Time = 9.70 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.62

method	result	size
default	\(-\frac {2 \cos \left (d x +c \right ) \left (\tan \left (d x +c \right )-i\right ) a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-12 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+9 \left (\cos ^{2}\left (d x +c \right )\right )-8\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\)	\(77\)
risch	\(-\frac {i a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (3 \,{\mathrm e}^{3 i \left (d x +c \right )}-7 \cos \left (d x +c \right )+35 i \sin \left (d x +c \right )\right )}{42 e^{3} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(92\)

[In]

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

[Out]

-2/21/d*cos(d*x+c)*(tan(d*x+c)-I)*a*(a*(1+I*tan(d*x+c)))^(1/2)*(-12*I*sin(d*x+c)*cos(d*x+c)+9*cos(d*x+c)^2-8)/
(e*sec(d*x+c))^(1/2)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {{\left (-3 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 17 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 7 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 21 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{42 \, d e^{4}} \]

[In]

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

[Out]

1/42*(-3*I*a*e^(6*I*d*x + 6*I*c) - 17*I*a*e^(4*I*d*x + 4*I*c) + 7*I*a*e^(2*I*d*x + 2*I*c) + 21*I*a)*sqrt(a/(e^
(2*I*d*x + 2*I*c) + 1))*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(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.67 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {{\left (-3 i \, a \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 14 i \, a \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 21 i \, a \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 14 \, a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 21 \, a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{42 \, d e^{\frac {7}{2}}} \]

[In]

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

[Out]

1/42*(-3*I*a*cos(7/2*d*x + 7/2*c) - 14*I*a*cos(3/2*d*x + 3/2*c) + 21*I*a*cos(1/2*d*x + 1/2*c) + 3*a*sin(7/2*d*
x + 7/2*c) + 14*a*sin(3/2*d*x + 3/2*c) + 21*a*sin(1/2*d*x + 1/2*c))*sqrt(a)/(d*e^(7/2))

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.85 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {a\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,4{}\mathrm {i}-\cos \left (4\,c+4\,d\,x\right )\,3{}\mathrm {i}+38\,\sin \left (2\,c+2\,d\,x\right )+3\,\sin \left (4\,c+4\,d\,x\right )+7{}\mathrm {i}\right )}{84\,d\,e^4} \]

[In]

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

[Out]

(a*(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)*4i - cos(4*c + 4*d*x)*3i + 38*sin(2*c + 2*d*x) + 3*sin(4*c + 4*d*x) + 7i))/(84*d*e^4)