\(\int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx\) [62]

Optimal result

Integrand size = 24, antiderivative size = 27 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i (a+i a \tan (c+d x))^6}{6 a d} \]

[Out]

-1/6*I*(a+I*a*tan(d*x+c))^6/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} \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i (a+i a \tan (c+d x))^6}{6 a d} \]

[In]

Int[Sec[c + d*x]^2*(a + I*a*Tan[c + d*x])^5,x]

[Out]

((-1/6*I)*(a + I*a*Tan[c + d*x])^6)/(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)^5 \, dx,x,i a \tan (c+d x)\right )}{a d} \\ & = -\frac {i (a+i a \tan (c+d x))^6}{6 a d} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(72\) vs. \(2(27)=54\).

Time = 0.34 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.67 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5 \tan (c+d x) \left (6+15 i \tan (c+d x)-20 \tan ^2(c+d x)-15 i \tan ^3(c+d x)+6 \tan ^4(c+d x)+i \tan ^5(c+d x)\right )}{6 d} \]

[In]

Integrate[Sec[c + d*x]^2*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*Tan[c + d*x]*(6 + (15*I)*Tan[c + d*x] - 20*Tan[c + d*x]^2 - (15*I)*Tan[c + d*x]^3 + 6*Tan[c + d*x]^4 + I*
Tan[c + d*x]^5))/(6*d)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (23 ) = 46\).

Time = 20.70 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.96

method	result	size
risch	\(\frac {32 i a^{5} \left (6 \,{\mathrm e}^{10 i \left (d x +c \right )}+15 \,{\mathrm e}^{8 i \left (d x +c \right )}+20 \,{\mathrm e}^{6 i \left (d x +c \right )}+15 \,{\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(80\)
derivativedivides	\(\frac {\frac {i a^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{6 \cos \left (d x +c \right )^{6}}+\frac {a^{5} \left (\sin ^{5}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{5}}-\frac {5 i a^{5} \left (\sin ^{4}\left (d x +c \right )\right )}{2 \cos \left (d x +c \right )^{4}}-\frac {10 a^{5} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {5 i a^{5}}{2 \cos \left (d x +c \right )^{2}}+a^{5} \tan \left (d x +c \right )}{d}\)	\(115\)
default	\(\frac {\frac {i a^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{6 \cos \left (d x +c \right )^{6}}+\frac {a^{5} \left (\sin ^{5}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{5}}-\frac {5 i a^{5} \left (\sin ^{4}\left (d x +c \right )\right )}{2 \cos \left (d x +c \right )^{4}}-\frac {10 a^{5} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {5 i a^{5}}{2 \cos \left (d x +c \right )^{2}}+a^{5} \tan \left (d x +c \right )}{d}\)	\(115\)

[In]

int(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

32/3*I*a^5*(6*exp(10*I*(d*x+c))+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. 153 vs. \(2 (21) = 42\).

Time = 0.23 (sec) , antiderivative size = 153, normalized size of antiderivative = 5.67 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {32 \, {\left (-6 i \, a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} - 15 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} - 20 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} - 15 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{5}\right )}}{3 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

-32/3*(-6*I*a^5*e^(10*I*d*x + 10*I*c) - 15*I*a^5*e^(8*I*d*x + 8*I*c) - 20*I*a^5*e^(6*I*d*x + 6*I*c) - 15*I*a^5
*e^(4*I*d*x + 4*I*c) - 6*I*a^5*e^(2*I*d*x + 2*I*c) - I*a^5)/(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]

\[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=i a^{5} \left (\int \left (- i \sec ^{2}{\left (c + d x \right )}\right )\, dx + \int 5 \tan {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (- 10 \tan ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\right )\, dx + \int \tan ^{5}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 10 i \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (- 5 i \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

integrate(sec(d*x+c)**2*(a+I*a*tan(d*x+c))**5,x)

[Out]

I*a**5*(Integral(-I*sec(c + d*x)**2, x) + Integral(5*tan(c + d*x)*sec(c + d*x)**2, x) + Integral(-10*tan(c + d
*x)**3*sec(c + d*x)**2, x) + Integral(tan(c + d*x)**5*sec(c + d*x)**2, x) + Integral(10*I*tan(c + d*x)**2*sec(
c + d*x)**2, x) + Integral(-5*I*tan(c + d*x)**4*sec(c + d*x)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6}}{6 \, a d} \]

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/6*I*(I*a*tan(d*x + c) + a)^6/(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. 82 vs. \(2 (21) = 42\).

Time = 0.75 (sec) , antiderivative size = 82, normalized size of antiderivative = 3.04 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {-i \, a^{5} \tan \left (d x + c\right )^{6} - 6 \, a^{5} \tan \left (d x + c\right )^{5} + 15 i \, a^{5} \tan \left (d x + c\right )^{4} + 20 \, a^{5} \tan \left (d x + c\right )^{3} - 15 i \, a^{5} \tan \left (d x + c\right )^{2} - 6 \, a^{5} \tan \left (d x + c\right )}{6 \, d} \]

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.75 (sec) , antiderivative size = 114, normalized size of antiderivative = 4.22 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5\,\sin \left (c+d\,x\right )\,\left (6\,{\cos \left (c+d\,x\right )}^5+{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,15{}\mathrm {i}-20\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,15{}\mathrm {i}+6\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^5\,1{}\mathrm {i}\right )}{6\,d\,{\cos \left (c+d\,x\right )}^6} \]

[In]

int((a + a*tan(c + d*x)*1i)^5/cos(c + d*x)^2,x)

[Out]

(a^5*sin(c + d*x)*(6*cos(c + d*x)*sin(c + d*x)^4 + cos(c + d*x)^4*sin(c + d*x)*15i + 6*cos(c + d*x)^5 + sin(c
+ d*x)^5*1i - cos(c + d*x)^2*sin(c + d*x)^3*15i - 20*cos(c + d*x)^3*sin(c + d*x)^2))/(6*d*cos(c + d*x)^6)