\(\int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [150]

Optimal result

Integrand size = 24, antiderivative size = 27 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i (a-i a \tan (c+d x))^5}{5 a^9 d} \]

[Out]

1/5*I*(a-I*a*tan(d*x+c))^5/a^9/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 32} \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i (a-i a \tan (c+d x))^5}{5 a^9 d} \]

[In]

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I/5)*(a - I*a*Tan[c + d*x])^5)/(a^9*d)

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 (a-x)^4 \, dx,x,i a \tan (c+d x)\right )}{a^9 d} \\ & = \frac {i (a-i a \tan (c+d x))^5}{5 a^9 d} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(58\) vs. \(2(27)=54\).

Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\tan (c+d x) \left (5-10 i \tan (c+d x)-10 \tan ^2(c+d x)+5 i \tan ^3(c+d x)+\tan ^4(c+d x)\right )}{5 a^4 d} \]

[In]

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Tan[c + d*x]*(5 - (10*I)*Tan[c + d*x] - 10*Tan[c + d*x]^2 + (5*I)*Tan[c + d*x]^3 + Tan[c + d*x]^4))/(5*a^4*d)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {\left (\tan \left (d x +c \right )+i\right )^{5}}{5 a^{4} d}\)	\(20\)
default	\(\frac {\left (\tan \left (d x +c \right )+i\right )^{5}}{5 a^{4} d}\)	\(20\)
risch	\(\frac {32 i}{5 d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\)	\(23\)

[In]

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/5/a^4/d*(tan(d*x+c)+I)^5

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. 84 vs. \(2 (21) = 42\).

Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.11 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {32 i}{5 \, {\left (a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \]

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

32/5*I/(a^4*d*e^(10*I*d*x + 10*I*c) + 5*a^4*d*e^(8*I*d*x + 8*I*c) + 10*a^4*d*e^(6*I*d*x + 6*I*c) + 10*a^4*d*e^
(4*I*d*x + 4*I*c) + 5*a^4*d*e^(2*I*d*x + 2*I*c) + a^4*d)

Sympy [F]

\[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{10}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate(sec(d*x+c)**10/(a+I*a*tan(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)**10/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x) + 1),
x)/a**4

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (21) = 42\).

Time = 0.35 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\tan \left (d x + c\right )^{5} + 5 i \, \tan \left (d x + c\right )^{4} - 10 \, \tan \left (d x + c\right )^{3} - 10 i \, \tan \left (d x + c\right )^{2} + 5 \, \tan \left (d x + c\right )}{5 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/5*(tan(d*x + c)^5 + 5*I*tan(d*x + c)^4 - 10*tan(d*x + c)^3 - 10*I*tan(d*x + c)^2 + 5*tan(d*x + c))/(a^4*d)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (21) = 42\).

Time = 0.75 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\tan \left (d x + c\right )^{5} + 5 i \, \tan \left (d x + c\right )^{4} - 10 \, \tan \left (d x + c\right )^{3} - 10 i \, \tan \left (d x + c\right )^{2} + 5 \, \tan \left (d x + c\right )}{5 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/5*(tan(d*x + c)^5 + 5*I*tan(d*x + c)^4 - 10*tan(d*x + c)^3 - 10*I*tan(d*x + c)^2 + 5*tan(d*x + c))/(a^4*d)

Mupad [B] (verification not implemented)

Time = 4.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.44 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sin \left (c+d\,x\right )\,\left (5\,{\cos \left (c+d\,x\right )}^4-{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )\,10{}\mathrm {i}-10\,{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^2+\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^3\,5{}\mathrm {i}+{\sin \left (c+d\,x\right )}^4\right )}{5\,a^4\,d\,{\cos \left (c+d\,x\right )}^5} \]

[In]

int(1/(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)^4),x)

[Out]

(sin(c + d*x)*(cos(c + d*x)*sin(c + d*x)^3*5i - cos(c + d*x)^3*sin(c + d*x)*10i + 5*cos(c + d*x)^4 + sin(c + d
*x)^4 - 10*cos(c + d*x)^2*sin(c + d*x)^2))/(5*a^4*d*cos(c + d*x)^5)