\(\int \sec ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [39]

Optimal result

Integrand size = 24, antiderivative size = 27 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {i (a+i a \tan (c+d x))^4}{4 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 32} \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {i (a+i a \tan (c+d x))^4}{4 a d} \]

[In]

Int[Sec[c + d*x]^2*(a + I*a*Tan[c + d*x])^3,x]

[Out]

((-1/4*I)*(a + I*a*Tan[c + d*x])^4)/(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*} \text {integral}& = -\frac {i \text {Subst}\left (\int (a+x)^3 \, dx,x,i a \tan (c+d x)\right )}{a d} \\ & = -\frac {i (a+i a \tan (c+d x))^4}{4 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 \tan (c+d x) \left (4+6 i \tan (c+d x)-4 \tan ^2(c+d x)-i \tan ^3(c+d x)\right )}{4 d} \]

[In]

Integrate[Sec[c + d*x]^2*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*Tan[c + d*x]*(4 + (6*I)*Tan[c + d*x] - 4*Tan[c + d*x]^2 - I*Tan[c + d*x]^3))/(4*d)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (23 ) = 46\).

Time = 4.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15

method	result	size
risch	\(\frac {4 i a^{3} \left (4 \,{\mathrm e}^{6 i \left (d x +c \right )}+6 \,{\mathrm e}^{4 i \left (d x +c \right )}+4 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}\)	\(58\)
derivativedivides	\(\frac {-\frac {i a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}-\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+\frac {3 i a^{3}}{2 \cos \left (d x +c \right )^{2}}+a^{3} \tan \left (d x +c \right )}{d}\)	\(73\)
default	\(\frac {-\frac {i a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}-\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+\frac {3 i a^{3}}{2 \cos \left (d x +c \right )^{2}}+a^{3} \tan \left (d x +c \right )}{d}\)	\(73\)

[In]

int(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (21) = 42\).

Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 3.74 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {4 \, {\left (-4 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 6 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )}}{d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=- i a^{3} \left (\int i \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (- 3 \tan {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\right )\, dx + \int \tan ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (- 3 i \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

integrate(sec(d*x+c)**2*(a+I*a*tan(d*x+c))**3,x)

[Out]

-I*a**3*(Integral(I*sec(c + d*x)**2, x) + Integral(-3*tan(c + d*x)*sec(c + d*x)**2, x) + Integral(tan(c + d*x)
**3*sec(c + d*x)**2, x) + Integral(-3*I*tan(c + d*x)**2*sec(c + d*x)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{4 \, a d} \]

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/4*I*(I*a*tan(d*x + c) + a)^4/(a*d)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (21) = 42\).

Time = 0.49 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {i \, a^{3} \tan \left (d x + c\right )^{4} + 4 \, a^{3} \tan \left (d x + c\right )^{3} - 6 i \, a^{3} \tan \left (d x + c\right )^{2} - 4 \, a^{3} \tan \left (d x + c\right )}{4 \, d} \]

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.85 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {-\frac {a^3\,{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{4}-a^3\,{\mathrm {tan}\left (c+d\,x\right )}^3+\frac {a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2\,3{}\mathrm {i}}{2}+a^3\,\mathrm {tan}\left (c+d\,x\right )}{d} \]

[In]

int((a + a*tan(c + d*x)*1i)^3/cos(c + d*x)^2,x)

[Out]

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