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

Optimal. Leaf size=98 \[ -\frac{2 i \sqrt{a+i a \tan (c+d x)}}{a d}-\frac{i}{d \sqrt{a+i a \tan (c+d x)}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d} \]

[Out]

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

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Rubi [A]  time = 0.0957389, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3543, 3479, 3480, 206} \[ -\frac{2 i \sqrt{a+i a \tan (c+d x)}}{a d}-\frac{i}{d \sqrt{a+i a \tan (c+d x)}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^2(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx &=-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{a d}-\int \frac{1}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=-\frac{i}{d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{a d}-\frac{\int \sqrt{a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac{i}{d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{i \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d}-\frac{i}{d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{a d}\\ \end{align*}

Mathematica [A]  time = 0.733642, size = 113, normalized size = 1.15 \[ -\frac{i \left (\sqrt{1+e^{2 i (c+d x)}} \left (1+5 e^{2 i (c+d x)}\right )-e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{d \left (1+e^{2 i (c+d x)}\right )^{3/2} \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(1 + 5*E^((2*I)*(c + d*x))) - E^(I*(c + d*x))*(1 + E^((2*I)*(c + d*x)))*A
rcSinh[E^(I*(c + d*x))]))/(d*(1 + E^((2*I)*(c + d*x)))^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.044, size = 73, normalized size = 0.7 \begin{align*}{\frac{-2\,i}{ad} \left ( \sqrt{a+ia\tan \left ( dx+c \right ) }+{\frac{a}{2}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{\sqrt{2}}{4}\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/d/a*((a+I*a*tan(d*x+c))^(1/2)+1/2/(a+I*a*tan(d*x+c))^(1/2)*a-1/4*a^(1/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d
*x+c))^(1/2)*2^(1/2)/a^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.15209, size = 738, normalized size = 7.53 \begin{align*} \frac{{\left (i \, \sqrt{2} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left ({\left (\sqrt{2} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - i \, \sqrt{2} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-{\left (\sqrt{2} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(I*sqrt(2)*a*d*sqrt(1/(a*d^2))*e^(2*I*d*x + 2*I*c)*log((sqrt(2)*a*d*sqrt(1/(a*d^2))*e^(2*I*d*x + 2*I*c) +
sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - I*sqr
t(2)*a*d*sqrt(1/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(-(sqrt(2)*a*d*sqrt(1/(a*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*sqrt(a
/(e^(2*I*d*x + 2*I*c) + 1))*(-10*I*e^(2*I*d*x + 2*I*c) - 2*I)*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*I*c)/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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