3.96 \(\int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=79 \[ \frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.176079, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3554, 3480, 206, 3599, 63, 208} \[ \frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3554

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(2
*a^2)/(a*c - b*d), Int[Sqrt[a + b*Tan[e + f*x]], x], x] - Dist[(2*b*c*d + a*(c^2 - d^2))/(a*(c^2 + d^2)), Int[
((a - b*Tan[e + f*x])*Sqrt[a + b*Tan[e + f*x]])/(c + d*Tan[e + f*x]), 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 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])

Rule 3599

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [B]  time = 1.062, size = 201, normalized size = 2.54 \[ \frac{e^{-3 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \left (\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (\sqrt{2} \left (\log \left (1-e^{i (c+d x)}\right )-\log \left (1+e^{i (c+d x)}\right )+\log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-\log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}+e^{i (c+d x)}+1\right )\right )+4 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{\sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^(3/2)*(4*ArcSinh[E^(I*(c
+ d*x))] + Sqrt[2]*(Log[1 - E^(I*(c + d*x))] - Log[1 + E^(I*(c + d*x))] + Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sq
rt[1 + E^((2*I)*(c + d*x))]] - Log[1 + E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]])))/(Sqrt[2]*d*
E^((3*I)*(c + d*x)))

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Maple [B]  time = 0.255, size = 231, normalized size = 2.9 \begin{align*} -{\frac{a\sin \left ( dx+c \right ) }{d \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( 2\,i\sqrt{2}\arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) +i\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) -2\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) -\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*a*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(2*I*2^(1/2)*arctan
(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+I*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2*2^(1/2)*
arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))-ln(((-2*cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c)))*sin(d*x+c)/(I*sin(d*x+c)+cos(d*x+c)-1)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.39063, size = 967, normalized size = 12.24 \begin{align*} \sqrt{2} \sqrt{\frac{a^{3}}{d^{2}}} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - \sqrt{2} \sqrt{\frac{a^{3}}{d^{2}}} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - \sqrt{\frac{a^{3}}{d^{2}}} \log \left (\frac{{\left (\sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + 2 \, \sqrt{\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) + \sqrt{\frac{a^{3}}{d^{2}}} \log \left (\frac{{\left (\sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 2 \, \sqrt{\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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