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

Optimal. Leaf size=27 \[ -\frac {i (a+i a \tan (c+d x))^2}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3486, 3767, 8} \[ \frac {a \tan (c+d x)}{d}+\frac {i a \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.11 \[ \frac {a \tan (c+d x)}{d}+\frac {i a \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.40, size = 44, normalized size = 1.63 \[ \frac {4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*I*a*e^(2*I*d*x + 2*I*c) + 2*I*a)/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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giac [A]  time = 0.76, size = 26, normalized size = 0.96 \[ -\frac {-i \, a \tan \left (d x + c\right )^{2} - 2 \, a \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.39, size = 26, normalized size = 0.96 \[ \frac {\frac {i a}{2 \cos \left (d x +c \right )^{2}}+a \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/2*I*a/cos(d*x+c)^2+a*tan(d*x+c))

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maxima [A]  time = 0.46, size = 21, normalized size = 0.78 \[ -\frac {i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{2 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*(I*a*tan(d*x + c) + a)^2/(a*d)

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mupad [B]  time = 3.21, size = 23, normalized size = 0.85 \[ \frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (2+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)/cos(c + d*x)^2,x)

[Out]

(a*tan(c + d*x)*(tan(c + d*x)*1i + 2))/(2*d)

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sympy [A]  time = 2.36, size = 37, normalized size = 1.37 \[ \begin {cases} \frac {\frac {i a \tan ^{2}{\left (c + d x \right )}}{2} + a \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (i a \tan {\relax (c )} + a\right ) \sec ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((I*a*tan(c + d*x)**2/2 + a*tan(c + d*x))/d, Ne(d, 0)), (x*(I*a*tan(c) + a)*sec(c)**2, True))

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