\(\int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [381]

Optimal result

Integrand size = 26, antiderivative size = 57 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {4 i}{3 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac {2 i}{a^3 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-2*I/a^3/d/(a+I*a*tan(d*x+c))^(1/2)+4/3*I/a^2/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 45} \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {4 i}{3 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac {2 i}{a^3 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Int[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((4*I)/3)/(a^2*d*(a + I*a*Tan[c + d*x])^(3/2)) - (2*I)/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 {a-x}{(a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {i \text {Subst}\left (\int \left (\frac {2 a}{(a+x)^{5/2}}-\frac {1}{(a+x)^{3/2}}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = \frac {4 i}{3 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac {2 i}{a^3 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {i \left (-\frac {4 a}{3 (a+i a \tan (c+d x))^{3/2}}+\frac {2}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^3 d} \]

[In]

Integrate[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-I)*((-4*a)/(3*(a + I*a*Tan[c + d*x])^(3/2)) + 2/Sqrt[a + I*a*Tan[c + d*x]]))/(a^3*d)

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {1}{\sqrt {a +i a \tan \left (d x +c \right )}}+\frac {2 a}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{3}}\)	\(44\)
default	\(\frac {2 i \left (-\frac {1}{\sqrt {a +i a \tan \left (d x +c \right )}}+\frac {2 a}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{3}}\)	\(44\)

[In]

int(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2*I/d/a^3*(-1/(a+I*a*tan(d*x+c))^(1/2)+2/3*a/(a+I*a*tan(d*x+c))^(3/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-2 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{3 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**4/(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Integral(sec(c + d*x)**4/(I*a*(tan(c + d*x) - I))**(7/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.56 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {2 i \, {\left (3 i \, a \tan \left (d x + c\right ) + a\right )}}{3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} d} \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-2/3*I*(3*I*a*tan(d*x + c) + a)/((I*a*tan(d*x + c) + a)^(3/2)*a^3*d)

Giac [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(I*a*tan(d*x + c) + a)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \]

[In]

int(1/(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(7/2)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(7/2)), x)