3.91 \(\int \cot ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=145 \[ \frac{7 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{\cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}-\frac{i \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d} \]

[Out]

(7*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c +
d*x]]/(Sqrt[2]*Sqrt[a])])/d - ((I/4)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d - (Cot[c + d*x]^2*Sqrt[a + I*a
*Tan[c + d*x]])/(2*d)

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Rubi [A]  time = 0.437621, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3561, 3598, 3600, 3480, 206, 3599, 63, 208} \[ \frac{7 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{\cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}-\frac{i \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(7*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c +
d*x]]/(Sqrt[2]*Sqrt[a])])/d - ((I/4)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d - (Cot[c + d*x]^2*Sqrt[a + I*a
*Tan[c + d*x]])/(2*d)

Rule 3561

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(d*(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*(c^2 + d^2)*
(n + 1)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - a*c*(n + 1) + a*d*(m + n + 1)*T
an[e + f*x], 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] && LtQ[n, -1] && (IntegerQ[n] || 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 3600

Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(A*b + a*B)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m, x], x] - Dist[(B*c
- A*d)/(b*c + a*d), Int[((a + b*Tan[e + f*x])^m*(a - b*Tan[e + f*x]))/(c + d*Tan[e + f*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] && NeQ[A*b + a*B, 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 ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx &=-\frac{\cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{\int \cot ^2(c+d x) \left (\frac{i a}{2}-\frac{3}{2} a \tan (c+d x)\right ) \sqrt{a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac{i \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{\int \cot (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{7 a^2}{4}-\frac{1}{4} i a^2 \tan (c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac{i \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}-i \int \sqrt{a+i a \tan (c+d x)} \, dx-\frac{7 \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)} \, dx}{8 a}\\ &=-\frac{i \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}-\frac{(7 a) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 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{i \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{(7 i) \operatorname{Subst}\left (\int \frac{1}{i-\frac{i x^2}{a}} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{4 d}\\ &=\frac{7 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 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{i \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}\\ \end{align*}

Mathematica [A]  time = 1.88593, size = 144, normalized size = 0.99 \[ \frac{\sqrt{a+i a \tan (c+d x)} \left (-4 \csc ^2(c+d x)+2 i \csc (c) \sin (d x) \csc (c+d x)+e^{-i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (7 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{1+e^{2 i (c+d x)}}}\right )-8 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )-2 i \cot (c)+4\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((4 + (Sqrt[1 + E^((2*I)*(c + d*x))]*(-8*ArcSinh[E^(I*(c + d*x))] + 7*Sqrt[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x))
)/Sqrt[1 + E^((2*I)*(c + d*x))]]))/E^(I*(c + d*x)) - (2*I)*Cot[c] - 4*Csc[c + d*x]^2 + (2*I)*Csc[c]*Csc[c + d*
x]*Sin[d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(8*d)

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Maple [B]  time = 0.417, size = 904, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(8*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(
1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-8*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(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))-2*I*cos(d*x+c)^3+2*I*cos(d*x+c
)-7*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*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))+16*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(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))-14*I*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ar
ctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-6*I*cos(d*x+c)^2-16*I*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+8*I*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-6*cos(d*x+c)^3*sin(d*x+c)+14*
cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*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))-8*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(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))+6*I*cos(d*x+c)^4+7*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(1/2))+4*cos(d*x+c)^2*sin(d*x+c)+7*I*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2*cos(d*x+c)*sin(d*x+c)-7*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
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)))/(I*sin(d*x+c)+cos(d*x+c)-1)/sin
(d*x+c)^3

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.39702, size = 1461, normalized size = 10.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(3*e^(4*I*d*x + 4*I*c) + 4*e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x
+ I*c) - 4*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/d^2)*log((sqrt(2)*d*sqrt(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
^(-2*I*d*x - 2*I*c)) + 4*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/d^2)*log(-(sqrt(
2)*d*sqrt(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^(-2*I*d*x - 2*I*c)) + 7*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/d^2)*log(
(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) + 2*d*sqrt(a/d^2)*e^(2*I*
d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) - 7*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/d^2)*log(
(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) - 2*d*sqrt(a/d^2)*e^(2*I*
d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)))/(d*e^(4*I*d*x + 4*I*c) - 2*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 )} \cot ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(I*tan(c + d*x) + 1))*cot(c + d*x)**3, 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} \cot \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(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)^3, x)