3.89 \(\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=78 \[ \frac {\sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3562, 3480, 206, 3599, 63, 208} \[ \frac {\sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

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

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 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 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]

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 3562

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

Rubi steps

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

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Mathematica [F]  time = 0.64, size = 0, normalized size = 0.00 \[ \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

Integrate[Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]], x]

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fricas [B]  time = 0.45, size = 336, normalized size = 4.31 \[ \frac {1}{2} \, \sqrt {2} \sqrt {\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \frac {1}{2} \, \sqrt {2} \sqrt {\frac {a}{d^{2}}} \log \left (-4 \, {\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \frac {1}{2} \, \sqrt {\frac {a}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \frac {1}{2} \, \sqrt {\frac {a}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*sqrt(a/d^2)*log(4*((d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(a/d^2) + a*e
^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 1/2*sqrt(2)*sqrt(a/d^2)*log(-4*((d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I
*d*x + 2*I*c) + 1))*sqrt(a/d^2) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 1/2*sqrt(a/d^2)*log(16*(3*a^2*e^(2*I*
d*x + 2*I*c) + 2*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr
t(a/d^2) + a^2)*e^(-2*I*d*x - 2*I*c)) + 1/2*sqrt(a/d^2)*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) - 2*sqrt(2)*(a*d*e^(
3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(a/d^2) + a^2)*e^(-2*I*d*x - 2*I
*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)*cot(d*x + c), x)

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maple [B]  time = 1.43, size = 230, normalized size = 2.95 \[ -\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (i \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )+i \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-\sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )-\ln \left (-\frac {-\sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )\right ) \sin \left (d x +c \right )}{d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(I*2^(1/2)*arctan(1/2
*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+I*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-2^(1/2)*arctan
h(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))-ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)+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 [A]  time = 0.59, size = 107, normalized size = 1.37 \[ -\frac {\sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 2 \, \sqrt {a} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.17, size = 61, normalized size = 0.78 \[ -\frac {2\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a}}\right )}{d}+\frac {\sqrt {2}\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*a^(1/2)*atanh((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^(1/2))))/d - (2*a^(1/2)*atanh((a + a*tan(c
+ d*x)*1i)^(1/2)/a^(1/2)))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \cot {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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