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

Optimal. Leaf size=29 \[ -\frac{2 i (a+i a \tan (c+d x))^{3/2}}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.0556067, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3487, 32} \[ -\frac{2 i (a+i a \tan (c+d x))^{3/2}}{3 a d} \]

Antiderivative was successfully verified.

[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 3487

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]

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]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx &=-\frac{i \operatorname{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]  time = 0.122623, size = 34, normalized size = 1.17 \[ \frac{2 (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 24, normalized size = 0.8 \begin{align*}{\frac{-{\frac{2\,i}{3}}}{ad} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.971031, size = 28, normalized size = 0.97 \begin{align*} -\frac{2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{3 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 2.27043, size = 132, normalized size = 4.55 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(I*tan(c + d*x) + 1))*sec(c + d*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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