3.1.22 \(\int \sec ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [22]

Optimal. Leaf size=27 \[ -\frac {i (a+i a \tan (c+d x))^3}{3 a d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 32} \begin {gather*} -\frac {i (a+i a \tan (c+d x))^3}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/3*I)*(a + I*a*Tan[c + d*x])^3)/(a*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3568

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]

Rubi steps

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(68\) vs. \(2(27)=54\).
time = 0.32, size = 68, normalized size = 2.52 \begin {gather*} \frac {a^2 \sec (c) \sec ^3(c+d x) (3 i \cos (d x)+3 i \cos (2 c+d x)+3 \sin (d x)-3 \sin (2 c+d x)+2 \sin (2 c+3 d x))}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[c]*Sec[c + d*x]^3*((3*I)*Cos[d*x] + (3*I)*Cos[2*c + d*x] + 3*Sin[d*x] - 3*Sin[2*c + d*x] + 2*Sin[2*c
+ 3*d*x]))/(6*d)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (23 ) = 46\).
time = 0.27, size = 51, normalized size = 1.89

method	result	size
risch	\(\frac {8 i a^{2} \left (3 \,{\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(47\)
derivativedivides	\(\frac {-\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {i a^{2}}{\cos \left (d x +c \right )^{2}}+a^{2} \tan \left (d x +c \right )}{d}\)	\(51\)
default	\(\frac {-\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {i a^{2}}{\cos \left (d x +c \right )^{2}}+a^{2} \tan \left (d x +c \right )}{d}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 21, normalized size = 0.78 \begin {gather*} -\frac {i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{3 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I*(I*a*tan(d*x + c) + a)^3/(a*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. 75 vs. \(2 (21) = 42\).
time = 0.33, size = 75, normalized size = 2.78 \begin {gather*} -\frac {8 \, {\left (-3 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2}\right )}}{3 \, {\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/3*(-3*I*a^2*e^(4*I*d*x + 4*I*c) - 3*I*a^2*e^(2*I*d*x + 2*I*c) - I*a^2)/(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - a^{2} \left (\int \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (- 2 i \tan {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\right )\, dx + \int \left (- \sec ^{2}{\left (c + d x \right )}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.52, size = 42, normalized size = 1.56 \begin {gather*} -\frac {a^{2} \tan \left (d x + c\right )^{3} - 3 i \, a^{2} \tan \left (d x + c\right )^{2} - 3 \, a^{2} \tan \left (d x + c\right )}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.22, size = 35, normalized size = 1.30 \begin {gather*} \frac {a^2\,\mathrm {tan}\left (c+d\,x\right )\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+3\right )}{3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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