3.21 \(\int \sec ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=55 \[ \frac{i (a+i a \tan (c+d x))^5}{5 a^3 d}-\frac{i (a+i a \tan (c+d x))^4}{2 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0431053, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3487, 43} \[ \frac{i (a+i a \tan (c+d x))^5}{5 a^3 d}-\frac{i (a+i a \tan (c+d x))^4}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*(a + I*a*Tan[c + d*x])^4)/(a^2*d) + ((I/5)*(a + I*a*Tan[c + d*x])^5)/(a^3*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 43

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])

Rubi steps

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

Mathematica [A]  time = 0.373512, size = 77, normalized size = 1.4 \[ \frac{a^2 \sec (c) \sec ^5(c+d x) (-5 \sin (2 c+d x)+5 \sin (2 c+3 d x)+\sin (4 c+5 d x)+5 i \cos (2 c+d x)+5 \sin (d x)+5 i \cos (d x))}{20 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.055, size = 85, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ( -{a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{{\frac{i}{2}}{a}^{2}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{a}^{2} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-a^2*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)+1/2*I*a^2/cos(d*x+c)^4-a^2*(-2/3-1/3*
sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.12444, size = 76, normalized size = 1.38 \begin{align*} -\frac{6 \, a^{2} \tan \left (d x + c\right )^{5} - 15 i \, a^{2} \tan \left (d x + c\right )^{4} - 30 i \, a^{2} \tan \left (d x + c\right )^{2} - 30 \, a^{2} \tan \left (d x + c\right )}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(6*a^2*tan(d*x + c)^5 - 15*I*a^2*tan(d*x + c)^4 - 30*I*a^2*tan(d*x + c)^2 - 30*a^2*tan(d*x + c))/d

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Fricas [B]  time = 1.18037, size = 329, normalized size = 5.98 \begin{align*} \frac{80 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 80 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{2}}{5 \,{\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, 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)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/5*(80*I*a^2*e^(6*I*d*x + 6*I*c) + 80*I*a^2*e^(4*I*d*x + 4*I*c) + 40*I*a^2*e^(2*I*d*x + 2*I*c) + 8*I*a^2)/(d*
e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*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^{2} \left (\int - \tan ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 i \tan{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \sec ^{4}{\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)**4*(a+I*a*tan(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.1673, size = 76, normalized size = 1.38 \begin{align*} -\frac{2 \, a^{2} \tan \left (d x + c\right )^{5} - 5 i \, a^{2} \tan \left (d x + c\right )^{4} - 10 i \, a^{2} \tan \left (d x + c\right )^{2} - 10 \, a^{2} \tan \left (d x + c\right )}{10 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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