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

Optimal. Leaf size=54 \[ \frac {i a \sec ^3(c+d x)}{3 d}+\frac {a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

1/2*a*arctanh(sin(d*x+c))/d+1/3*I*a*sec(d*x+c)^3/d+1/2*a*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, 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, 3768, 3770} \[ \frac {i a \sec ^3(c+d x)}{3 d}+\frac {a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTanh[Sin[c + d*x]])/(2*d) + ((I/3)*a*Sec[c + d*x]^3)/d + (a*Sec[c + d*x]*Tan[c + d*x])/(2*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 54, normalized size = 1.00 \[ \frac {i a \sec ^3(c+d x)}{3 d}+\frac {a \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTanh[Sin[c + d*x]])/(2*d) + ((I/3)*a*Sec[c + d*x]^3)/d + (a*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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fricas [B]  time = 0.59, size = 180, normalized size = 3.33 \[ \frac {-6 i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 16 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} + 6 i \, a e^{\left (i \, d x + i \, c\right )} + 3 \, {\left (a e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, {\left (a e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{6 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(-6*I*a*e^(5*I*d*x + 5*I*c) + 16*I*a*e^(3*I*d*x + 3*I*c) + 6*I*a*e^(I*d*x + I*c) + 3*(a*e^(6*I*d*x + 6*I*c
) + 3*a*e^(4*I*d*x + 4*I*c) + 3*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(I*d*x + I*c) + I) - 3*(a*e^(6*I*d*x + 6*I*c)
+ 3*a*e^(4*I*d*x + 4*I*c) + 3*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(I*d*x + I*c) - I))/(d*e^(6*I*d*x + 6*I*c) + 3
*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

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giac [B]  time = 1.42, size = 97, normalized size = 1.80 \[ \frac {3 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 3 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, {\left (3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 i \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.42, size = 55, normalized size = 1.02 \[ \frac {i a}{3 d \cos \left (d x +c \right )^{3}}+\frac {a \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.45, size = 61, normalized size = 1.13 \[ -\frac {3 \, a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {4 i \, a}{\cos \left (d x + c\right )^{3}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(3*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*I*a/cos(d
*x + c)^3)/d

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mupad [B]  time = 5.13, size = 107, normalized size = 1.98 \[ \frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2{}\mathrm {i}\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,2{}\mathrm {i}}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)/cos(c + d*x)^3,x)

[Out]

(a*atanh(tan(c/2 + (d*x)/2)))/d - ((a*2i)/3 + a*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^4*2i - a*tan(c/2 + (
d*x)/2)^5)/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \left (- i \sec ^{3}{\left (c + d x \right )}\right )\, dx + \int \tan {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+I*a*tan(d*x+c)),x)

[Out]

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

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