\(\int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [382]

Optimal result

Integrand size = 26, antiderivative size = 29 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 i}{5 a d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

2/5*I/a/d/(a+I*a*tan(d*x+c))^(5/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 32} \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 i}{5 a d (a+i a \tan (c+d x))^{5/2}} \]

[In]

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

[Out]

((2*I)/5)/(a*d*(a + I*a*Tan[c + d*x])^(5/2))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int \frac {1}{(a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{a d} \\ & = \frac {2 i}{5 a d (a+i a \tan (c+d x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 i}{5 a d (a+i a \tan (c+d x))^{5/2}} \]

[In]

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

[Out]

((2*I)/5)/(a*d*(a + I*a*Tan[c + d*x])^(5/2))

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {2 i}{5 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\)	\(24\)
default	\(\frac {2 i}{5 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\)	\(24\)

[In]

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

[Out]

2/5*I/a/d/(a+I*a*tan(d*x+c))^(5/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (21) = 42\).

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.48 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{20 \, a^{4} d} \]

[In]

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

[Out]

1/20*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(I*e^(6*I*d*x + 6*I*c) + 3*I*e^(4*I*d*x + 4*I*c) + 3*I*e^(2*I*d
*x + 2*I*c) + I)*e^(-5*I*d*x - 5*I*c)/(a^4*d)

Sympy [F]

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

[In]

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

[Out]

Integral(sec(c + d*x)**2/(I*a*(tan(c + d*x) - I))**(7/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 i}{5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a d} \]

[In]

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

[Out]

2/5*I/((I*a*tan(d*x + c) + a)^(5/2)*a*d)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2{}\mathrm {i}}{5\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \]

[In]

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

[Out]

2i/(5*a*d*(a + a*tan(c + d*x)*1i)^(5/2))