3.39 \(\int \sec ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=27 \[ -\frac{i (a+i a \tan (c+d x))^4}{4 a d} \]

[Out]

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

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Rubi [A]  time = 0.0365777, 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} \[ -\frac{i (a+i a \tan (c+d x))^4}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 0.488852, size = 84, normalized size = 3.11 \[ \frac{a^3 \sec (c) \sec ^4(c+d x) (2 \sin (c+2 d x)-2 \sin (3 c+2 d x)+\sin (3 c+4 d x)+2 i \cos (c+2 d x)+2 i \cos (3 c+2 d x)-3 \sin (c)+3 i \cos (c))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.057, size = 73, normalized size = 2.7 \begin{align*}{\frac{1}{d} \left ({\frac{-{\frac{i}{4}}{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{\frac{3\,i}{2}}{a}^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{a}^{3}\tan \left ( dx+c \right ) \right ) } \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))^3,x)

[Out]

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

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Maxima [A]  time = 1.11796, size = 28, normalized size = 1.04 \begin{align*} -\frac{i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{4 \, 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))^3,x, algorithm="maxima")

[Out]

-1/4*I*(I*a*tan(d*x + c) + a)^4/(a*d)

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Fricas [B]  time = 1.19093, size = 284, normalized size = 10.52 \begin{align*} \frac{16 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 24 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, a^{3}}{d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 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))^3,x, algorithm="fricas")

[Out]

(16*I*a^3*e^(6*I*d*x + 6*I*c) + 24*I*a^3*e^(4*I*d*x + 4*I*c) + 16*I*a^3*e^(2*I*d*x + 2*I*c) + 4*I*a^3)/(d*e^(8
*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int - 3 \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 i \tan{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int - i \tan ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sec ^{2}{\left (c + d x \right )}\, dx\right ) \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))**3,x)

[Out]

a**3*(Integral(-3*tan(c + d*x)**2*sec(c + d*x)**2, x) + Integral(3*I*tan(c + d*x)*sec(c + d*x)**2, x) + Integr
al(-I*tan(c + d*x)**3*sec(c + d*x)**2, x) + Integral(sec(c + d*x)**2, x))

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Giac [B]  time = 1.22505, size = 76, normalized size = 2.81 \begin{align*} -\frac{i \, a^{3} \tan \left (d x + c\right )^{4} + 4 \, a^{3} \tan \left (d x + c\right )^{3} - 6 i \, a^{3} \tan \left (d x + c\right )^{2} - 4 \, a^{3} \tan \left (d x + c\right )}{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))^3,x, algorithm="giac")

[Out]

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