3.154 \(\int \frac{\sec ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=27 \[ \frac{i}{3 a d (a+i a \tan (c+d x))^3} \]

[Out]

(I/3)/(a*d*(a + I*a*Tan[c + d*x])^3)

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Rubi [A]  time = 0.039359, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3487, 32} \[ \frac{i}{3 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I/3)/(a*d*(a + I*a*Tan[c + d*x])^3)

Rule 3487

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]

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]

Rubi steps

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

Mathematica [B]  time = 0.143224, size = 56, normalized size = 2.07 \[ \frac{i \sec ^4(c+d x) (2 i \sin (2 (c+d x))+4 \cos (2 (c+d x))+3)}{24 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/24)*Sec[c + d*x]^4*(3 + 4*Cos[2*(c + d*x)] + (2*I)*Sin[2*(c + d*x)]))/(a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]  time = 0.039, size = 24, normalized size = 0.9 \begin{align*}{\frac{{\frac{i}{3}}}{ad \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.963541, size = 28, normalized size = 1.04 \begin{align*} \frac{i}{3 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*I/((I*a*tan(d*x + c) + a)^3*a*d)

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Fricas [A]  time = 2.38951, size = 123, normalized size = 4.56 \begin{align*} \frac{{\left (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 (-6 i \, d x - 6 i \, c\right )}}{24 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*I*e^(4*I*d*x + 4*I*c) + 3*I*e^(2*I*d*x + 2*I*c) + I)*e^(-6*I*d*x - 6*I*c)/(a^4*d)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.19635, size = 115, normalized size = 4.26 \begin{align*} -\frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{3 \, a^{4} d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*tan(1/2*d*x + 1/2*c)^5 - 6*I*tan(1/2*d*x + 1/2*c)^4 - 10*tan(1/2*d*x + 1/2*c)^3 + 6*I*tan(1/2*d*x + 1/
2*c)^2 + 3*tan(1/2*d*x + 1/2*c))/(a^4*d*(tan(1/2*d*x + 1/2*c) - I)^6)