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

Optimal. Leaf size=46 \[ \frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{i a \sec ^4(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0342313, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3486, 3767} \[ \frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{i a \sec ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0422621, size = 43, normalized size = 0.93 \[ \frac{a \left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{i a \sec ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.08, size = 39, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{{\frac{i}{4}}a}{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-a \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)),x)

[Out]

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

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Maxima [A]  time = 1.06162, size = 65, normalized size = 1.41 \begin{align*} \frac{3 i \, a \tan \left (d x + c\right )^{4} + 4 \, a \tan \left (d x + c\right )^{3} + 6 i \, a \tan \left (d x + c\right )^{2} + 12 \, a \tan \left (d x + c\right )}{12 \, 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)),x, algorithm="maxima")

[Out]

1/12*(3*I*a*tan(d*x + c)^4 + 4*a*tan(d*x + c)^3 + 6*I*a*tan(d*x + c)^2 + 12*a*tan(d*x + c))/d

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Fricas [B]  time = 1.10474, size = 239, normalized size = 5.2 \begin{align*} \frac{24 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, a}{3 \,{\left (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\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)),x, algorithm="fricas")

[Out]

1/3*(24*I*a*e^(4*I*d*x + 4*I*c) + 16*I*a*e^(2*I*d*x + 2*I*c) + 4*I*a)/(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 [A]  time = 8.66531, size = 48, normalized size = 1.04 \begin{align*} \begin{cases} \frac{a \left (\frac{\tan ^{3}{\left (c + d x \right )}}{3} + \tan{\left (c + d x \right )}\right ) + \frac{i a \sec ^{4}{\left (c + d x \right )}}{4}}{d} & \text{for}\: d \neq 0 \\x \left (i a \tan{\left (c \right )} + a\right ) \sec ^{4}{\left (c \right )} & \text{otherwise} \end{cases} \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)),x)

[Out]

Piecewise(((a*(tan(c + d*x)**3/3 + tan(c + d*x)) + I*a*sec(c + d*x)**4/4)/d, Ne(d, 0)), (x*(I*a*tan(c) + a)*se
c(c)**4, True))

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Giac [A]  time = 1.12926, size = 65, normalized size = 1.41 \begin{align*} -\frac{-3 i \, a \tan \left (d x + c\right )^{4} - 4 \, a \tan \left (d x + c\right )^{3} - 6 i \, a \tan \left (d x + c\right )^{2} - 12 \, a \tan \left (d x + c\right )}{12 \, 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)),x, algorithm="giac")

[Out]

-1/12*(-3*I*a*tan(d*x + c)^4 - 4*a*tan(d*x + c)^3 - 6*I*a*tan(d*x + c)^2 - 12*a*tan(d*x + c))/d