3.215 \(\int \frac{1}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=120 \[ -\frac{3 \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}+\frac{1}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

((1/2 + I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) + 1/(d*Sqrt
[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (3*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.268913, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3559, 3598, 12, 3544, 205} \[ -\frac{3 \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}+\frac{1}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((1/2 + I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) + 1/(d*Sqrt
[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (3*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*Sqrt[Tan[c + d*x]])

Rule 3559

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

Rule 3598

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3544

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{1}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\left (\frac{3 a}{2}-i a \tan (c+d x)\right ) \sqrt{a+i a \tan (c+d x)}}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac{1}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{3 \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}+\frac{2 \int \frac{i a^2 \sqrt{a+i a \tan (c+d x)}}{4 \sqrt{\tan (c+d x)}} \, dx}{a^3}\\ &=\frac{1}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{3 \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}+\frac{i \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{2 a}\\ &=\frac{1}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{3 \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}+\frac{a \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}+\frac{1}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{3 \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.4928, size = 160, normalized size = 1.33 \[ \frac{i \sqrt{\tan (c+d x)} \left (\sqrt{-1+e^{2 i (c+d x)}} \left (1-5 e^{2 i (c+d x)}\right )+e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt{2} d \left (-1+e^{2 i (c+d x)}\right )^{3/2} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

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

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Maple [B]  time = 0.071, size = 395, normalized size = 3.3 \begin{align*} -{\frac{1}{4\,ad \left ( -\tan \left ( dx+c \right ) +i \right ) ^{2}}\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 2\,i\sqrt{2}\ln \left ( -{\frac{1}{\tan \left ( dx+c \right ) +i} \left ( -2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+ia-3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{2}a-\sqrt{2}\ln \left ( -{\frac{1}{\tan \left ( dx+c \right ) +i} \left ( -2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+ia-3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{3}a-20\,i\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\tan \left ( dx+c \right ) +\sqrt{2}\ln \left ( -{\frac{1}{\tan \left ( dx+c \right ) +i} \left ( -2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+ia-3\,a\tan \left ( dx+c \right ) \right ) } \right ) \tan \left ( dx+c \right ) a+12\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}-8\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia} \right ){\frac{1}{\sqrt{\tan \left ( dx+c \right ) }}}{\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{-ia}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*(a*(1+I*tan(d*x+c)))^(1/2)*(2*I*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a-20*I*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)*(-I*a)^(1/2)*tan(d*x+c)+2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a
*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a+12*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-8
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/a/tan(d*x+c)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/
(-tan(d*x+c)+I)^2/(-I*a)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(I*a*tan(d*x + c) + a)*tan(d*x + c)^(3/2)), x)

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Fricas [B]  time = 3.08007, size = 1108, normalized size = 9.23 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-10 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (i \, d x + i \, c\right )} -{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{2 i}{a d^{2}}} \log \left (\frac{1}{4} \,{\left (i \, a d \sqrt{\frac{2 i}{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}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{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 ) +{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{2 i}{a d^{2}}} \log \left (\frac{1}{4} \,{\left (-i \, a d \sqrt{\frac{2 i}{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}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{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 )}{4 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \tan ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(3/2),x)

[Out]

Integral(1/(sqrt(a*(I*tan(c + d*x) + 1))*tan(c + d*x)**(3/2)), x)

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Giac [A]  time = 1.27339, size = 147, normalized size = 1.22 \begin{align*} -\frac{2 \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} a^{2} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{-\left (i + 1\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} + \left (4 i + 4\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a - \left (5 i + 5\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} + \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2),x, algorithm="giac")

[Out]

-2*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*a^2*log(sqrt(I*a*tan(d*x + c) + a))/(-(I + 1)*(I*a*tan(d*x + c) +
a)^4 + (4*I + 4)*(I*a*tan(d*x + c) + a)^3*a - (5*I + 5)*(I*a*tan(d*x + c) + a)^2*a^2 + (2*I + 2)*(I*a*tan(d*x
+ c) + a)*a^3)