\(\int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [38]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {2 i (a+i a \tan (c+d x))^5}{5 a^2 d}+\frac {i (a+i a \tan (c+d x))^6}{6 a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {i (a+i a \tan (c+d x))^6}{6 a^3 d}-\frac {2 i (a+i a \tan (c+d x))^5}{5 a^2 d} \]

[In]

Int[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((-2*I)/5)*(a + I*a*Tan[c + d*x])^5)/(a^2*d) + ((I/6)*(a + I*a*Tan[c + d*x])^6)/(a^3*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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) (a+x)^4 \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {i \text {Subst}\left (\int \left (2 a (a+x)^4-(a+x)^5\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {2 i (a+i a \tan (c+d x))^5}{5 a^2 d}+\frac {i (a+i a \tan (c+d x))^6}{6 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.62 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 (7-5 i \tan (c+d x)) (-i+\tan (c+d x))^5}{30 d} \]

[In]

Integrate[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*(7 - (5*I)*Tan[c + d*x])*(-I + Tan[c + d*x])^5)/(30*d)

Maple [A] (verified)

Time = 19.53 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.25

method	result	size
risch	\(\frac {32 i a^{3} \left (15 \,{\mathrm e}^{8 i \left (d x +c \right )}+20 \,{\mathrm e}^{6 i \left (d x +c \right )}+15 \,{\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(69\)
derivativedivides	\(\frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {3 i a^{3}}{4 \cos \left (d x +c \right )^{4}}-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(128\)
default	\(\frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {3 i a^{3}}{4 \cos \left (d x +c \right )^{4}}-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(128\)

[In]

int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

32/15*I*a^3*(15*exp(8*I*(d*x+c))+20*exp(6*I*(d*x+c))+15*exp(4*I*(d*x+c))+6*exp(2*I*(d*x+c))+1)/d/(exp(2*I*(d*x
+c))+1)^6

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. 139 vs. \(2 (43) = 86\).

Time = 0.23 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.53 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {32 \, {\left (-15 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 20 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 15 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-32/15*(-15*I*a^3*e^(8*I*d*x + 8*I*c) - 20*I*a^3*e^(6*I*d*x + 6*I*c) - 15*I*a^3*e^(4*I*d*x + 4*I*c) - 6*I*a^3*
e^(2*I*d*x + 2*I*c) - I*a^3)/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) +
20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

\[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=- i a^{3} \left (\int i \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 3 \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int \tan ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 3 i \tan ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

integrate(sec(d*x+c)**4*(a+I*a*tan(d*x+c))**3,x)

[Out]

-I*a**3*(Integral(I*sec(c + d*x)**4, x) + Integral(-3*tan(c + d*x)*sec(c + d*x)**4, x) + Integral(tan(c + d*x)
**3*sec(c + d*x)**4, x) + Integral(-3*I*tan(c + d*x)**2*sec(c + d*x)**4, x))

Maxima [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.49 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {5 i \, a^{3} \tan \left (d x + c\right )^{6} + 18 \, a^{3} \tan \left (d x + c\right )^{5} - 15 i \, a^{3} \tan \left (d x + c\right )^{4} + 20 \, a^{3} \tan \left (d x + c\right )^{3} - 45 i \, a^{3} \tan \left (d x + c\right )^{2} - 30 \, a^{3} \tan \left (d x + c\right )}{30 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/30*(5*I*a^3*tan(d*x + c)^6 + 18*a^3*tan(d*x + c)^5 - 15*I*a^3*tan(d*x + c)^4 + 20*a^3*tan(d*x + c)^3 - 45*I
*a^3*tan(d*x + c)^2 - 30*a^3*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.54 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.49 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {5 i \, a^{3} \tan \left (d x + c\right )^{6} + 18 \, a^{3} \tan \left (d x + c\right )^{5} - 15 i \, a^{3} \tan \left (d x + c\right )^{4} + 20 \, a^{3} \tan \left (d x + c\right )^{3} - 45 i \, a^{3} \tan \left (d x + c\right )^{2} - 30 \, a^{3} \tan \left (d x + c\right )}{30 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/30*(5*I*a^3*tan(d*x + c)^6 + 18*a^3*tan(d*x + c)^5 - 15*I*a^3*tan(d*x + c)^4 + 20*a^3*tan(d*x + c)^3 - 45*I
*a^3*tan(d*x + c)^2 - 30*a^3*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 3.87 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.07 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {a^3\,\sin \left (c+d\,x\right )\,\left (-30\,{\cos \left (c+d\,x\right )}^5-{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,45{}\mathrm {i}+20\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,15{}\mathrm {i}+18\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^5\,5{}\mathrm {i}\right )}{30\,d\,{\cos \left (c+d\,x\right )}^6} \]

[In]

int((a + a*tan(c + d*x)*1i)^3/cos(c + d*x)^4,x)

[Out]

-(a^3*sin(c + d*x)*(18*cos(c + d*x)*sin(c + d*x)^4 - cos(c + d*x)^4*sin(c + d*x)*45i - 30*cos(c + d*x)^5 + sin
(c + d*x)^5*5i - cos(c + d*x)^2*sin(c + d*x)^3*15i + 20*cos(c + d*x)^3*sin(c + d*x)^2))/(30*d*cos(c + d*x)^6)