\(\int \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [282]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.07 (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} \[ \int \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{3 a d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{3 a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {2 i \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 a d}\)	\(24\)
default	\(-\frac {2 i \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 a d}\)	\(24\)

[In]

int(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*I*(a+I*a*tan(d*x+c))^(3/2)/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. 46 vs. \(2 (21) = 42\).

Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {4 i \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (3 i \, d x + 3 i \, c\right )}}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(sec(d*x+c)**2*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{3 \, a d} \]

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2/3*I*(I*a*tan(d*x + c) + a)^(3/2)/(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. 55 vs. \(2 (21) = 42\).

Time = 0.52 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \[ \int \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i \, \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )^{\frac {3}{2}}}{3 \, a d} \]

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.83 \[ \int \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,2{}\mathrm {i}}{3\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

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

[Out]

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