3.62 \(\int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx\)

Optimal. Leaf size=27 \[ -\frac{i (a+i a \tan (c+d x))^6}{6 a d} \]

[Out]

((-I/6)*(a + I*a*Tan[c + d*x])^6)/(a*d)

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Rubi [A]  time = 0.0361545, 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} \[ -\frac{i (a+i a \tan (c+d x))^6}{6 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + I*a*Tan[c + d*x])^5,x]

[Out]

((-I/6)*(a + I*a*Tan[c + d*x])^6)/(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))^5 \, dx &=-\frac{i \operatorname{Subst}\left (\int (a+x)^5 \, dx,x,i a \tan (c+d x)\right )}{a d}\\ &=-\frac{i (a+i a \tan (c+d x))^6}{6 a d}\\ \end{align*}

Mathematica [B]  time = 1.45354, size = 134, normalized size = 4.96 \[ \frac{a^5 \sec (c) \sec ^6(c+d x) (15 \sin (c+2 d x)-15 \sin (3 c+2 d x)+6 \sin (3 c+4 d x)-6 \sin (5 c+4 d x)+2 \sin (5 c+6 d x)+15 i \cos (c+2 d x)+15 i \cos (3 c+2 d x)+6 i \cos (3 c+4 d x)+6 i \cos (5 c+4 d x)-20 \sin (c)+20 i \cos (c))}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*Sec[c]*Sec[c + d*x]^6*((20*I)*Cos[c] + (15*I)*Cos[c + 2*d*x] + (15*I)*Cos[3*c + 2*d*x] + (6*I)*Cos[3*c +
4*d*x] + (6*I)*Cos[5*c + 4*d*x] - 20*Sin[c] + 15*Sin[c + 2*d*x] - 15*Sin[3*c + 2*d*x] + 6*Sin[3*c + 4*d*x] - 6
*Sin[5*c + 4*d*x] + 2*Sin[5*c + 6*d*x]))/(12*d)

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Maple [B]  time = 0.076, size = 115, normalized size = 4.3 \begin{align*}{\frac{1}{d} \left ({\frac{{\frac{i}{6}}{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{\frac{{\frac{5\,i}{2}}{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{10\,{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{\frac{5\,i}{2}}{a}^{5}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{a}^{5}\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))^5,x)

[Out]

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

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Maxima [A]  time = 1.1303, size = 28, normalized size = 1.04 \begin{align*} -\frac{i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6}}{6 \, 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))^5,x, algorithm="maxima")

[Out]

-1/6*I*(I*a*tan(d*x + c) + a)^6/(a*d)

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Fricas [B]  time = 1.10871, size = 463, normalized size = 17.15 \begin{align*} \frac{192 i \, a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 480 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 640 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 480 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 192 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + 32 i \, a^{5}}{3 \,{\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\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))^5,x, algorithm="fricas")

[Out]

1/3*(192*I*a^5*e^(10*I*d*x + 10*I*c) + 480*I*a^5*e^(8*I*d*x + 8*I*c) + 640*I*a^5*e^(6*I*d*x + 6*I*c) + 480*I*a
^5*e^(4*I*d*x + 4*I*c) + 192*I*a^5*e^(2*I*d*x + 2*I*c) + 32*I*a^5)/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x
+ 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*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^{5} \left (\int - 10 \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 5 \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 5 i \tan{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int - 10 i \tan ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int i \tan ^{5}{\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))**5,x)

[Out]

a**5*(Integral(-10*tan(c + d*x)**2*sec(c + d*x)**2, x) + Integral(5*tan(c + d*x)**4*sec(c + d*x)**2, x) + Inte
gral(5*I*tan(c + d*x)*sec(c + d*x)**2, x) + Integral(-10*I*tan(c + d*x)**3*sec(c + d*x)**2, x) + Integral(I*ta
n(c + d*x)**5*sec(c + d*x)**2, x) + Integral(sec(c + d*x)**2, x))

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Giac [B]  time = 1.58311, size = 111, normalized size = 4.11 \begin{align*} -\frac{-i \, a^{5} \tan \left (d x + c\right )^{6} - 6 \, a^{5} \tan \left (d x + c\right )^{5} + 15 i \, a^{5} \tan \left (d x + c\right )^{4} + 20 \, a^{5} \tan \left (d x + c\right )^{3} - 15 i \, a^{5} \tan \left (d x + c\right )^{2} - 6 \, a^{5} \tan \left (d x + c\right )}{6 \, 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))^5,x, algorithm="giac")

[Out]

-1/6*(-I*a^5*tan(d*x + c)^6 - 6*a^5*tan(d*x + c)^5 + 15*I*a^5*tan(d*x + c)^4 + 20*a^5*tan(d*x + c)^3 - 15*I*a^
5*tan(d*x + c)^2 - 6*a^5*tan(d*x + c))/d