\(\int \sec ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [296]

Optimal result

Integrand size = 26, antiderivative size = 29 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 32} \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a d} \]

[In]

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

[Out]

(((-2*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(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)^{3/2} \, dx,x,i a \tan (c+d x)\right )}{a d} \\ & = -\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {2 i \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 a d}\)	\(24\)
default	\(-\frac {2 i \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 a d}\)	\(24\)

[In]

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

[Out]

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

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. 59 vs. \(2 (21) = 42\).

Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.03 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {8 i \, \sqrt {2} a \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (5 i \, d x + 5 i \, c\right )}}{5 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-8/5*I*sqrt(2)*a*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(5*I*d*x + 5*I*c)/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x
+ 2*I*c) + d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{5 \, a d} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 153, normalized size of antiderivative = 5.28 \[ \int \sec ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {4\,a\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,7{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,4{}\mathrm {i}+\cos \left (6\,c+6\,d\,x\right )\,1{}\mathrm {i}-5\,\sin \left (2\,c+2\,d\,x\right )-4\,\sin \left (4\,c+4\,d\,x\right )-\sin \left (6\,c+6\,d\,x\right )+4{}\mathrm {i}\right )}{5\,d\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \]

[In]

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

[Out]

-(4*a*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(2*c + 2*d*x)*7i + c
os(4*c + 4*d*x)*4i + cos(6*c + 6*d*x)*1i - 5*sin(2*c + 2*d*x) - 4*sin(4*c + 4*d*x) - sin(6*c + 6*d*x) + 4i))/(
5*d*(15*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) + cos(6*c + 6*d*x) + 10))