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

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, 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 = {3567, 3852} \begin {gather*} \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} \end {gather*}

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 3567

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 3852

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 \text {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*}

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Mathematica [A]
time = 0.06, size = 43, normalized size = 0.93 \begin {gather*} \frac {i a \sec ^4(c+d x)}{4 d}+\frac {a \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \end {gather*}

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 = 0.29, size = 39, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {\frac {i a}{4 \cos \left (d x +c \right )^{4}}-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(39\)
default	\(\frac {\frac {i a}{4 \cos \left (d x +c \right )^{4}}-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(39\)
risch	\(\frac {4 i a \left (6 \,{\mathrm e}^{4 i \left (d x +c \right )}+4 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}\)	\(45\)

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,method=_RETURNVERBOSE)

[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 = 0.28, size = 48, normalized size = 1.04 \begin {gather*} \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 {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (40) = 80\).
time = 0.35, size = 81, normalized size = 1.76 \begin {gather*} -\frac {4 \, {\left (-6 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a\right )}}{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 {gather*}

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]

-4/3*(-6*I*a*e^(4*I*d*x + 4*I*c) - 4*I*a*e^(2*I*d*x + 2*I*c) - 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 = 1.75, size = 48, normalized size = 1.04 \begin {gather*} \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 {gather*}

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 = 0.58, size = 48, normalized size = 1.04 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 3.24, size = 48, normalized size = 1.04 \begin {gather*} \frac {\frac {1{}\mathrm {i}\,a\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}+\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+\frac {1{}\mathrm {i}\,a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+a\,\mathrm {tan}\left (c+d\,x\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)/cos(c + d*x)^4,x)

[Out]

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

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