3.8.28 \(\int \frac {(a+i a \tan (c+d x))^2}{\sqrt {\cot (c+d x)}} \, dx\) [728]

Optimal. Leaf size=71 \[ \frac {4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a^2}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 i a^2}{d \sqrt {\cot (c+d x)}} \]

[Out]

4*(-1)^(1/4)*a^2*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d-2/3*a^2/d/cot(d*x+c)^(3/2)+4*I*a^2/d/cot(d*x+c)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3754, 3623, 3610, 3614, 214} \begin {gather*} -\frac {2 a^2}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 i a^2}{d \sqrt {\cot (c+d x)}}+\frac {4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1)^(1/4)*a^2*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (2*a^2)/(3*d*Cot[c + d*x]^(3/2)) + ((4*I)*a^2)/(
d*Sqrt[Cot[c + d*x]])

Rule 214

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

Rule 3610

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

Rule 3614

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

Rule 3623

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

Rule 3754

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^2}{\sqrt {\cot (c+d x)}} \, dx &=\int \frac {(i a+a \cot (c+d x))^2}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\int \frac {2 i a^2+2 a^2 \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 i a^2}{d \sqrt {\cot (c+d x)}}+\int \frac {2 a^2-2 i a^2 \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=-\frac {2 a^2}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 i a^2}{d \sqrt {\cot (c+d x)}}+\frac {\left (8 a^4\right ) \text {Subst}\left (\int \frac {1}{-2 a^2-2 i a^2 x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a^2}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 i a^2}{d \sqrt {\cot (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.43, size = 90, normalized size = 1.27 \begin {gather*} -\frac {2 a^2 \left (1-6 i \cot (c+d x)+6 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) \cot ^2(c+d x) \sqrt {i \tan (c+d x)}\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(1 - (6*I)*Cot[c + d*x] + 6*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Cot[c
+ d*x]^2*Sqrt[I*Tan[c + d*x]]))/(3*d*Cot[c + d*x]^(3/2))

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 26.90, size = 485, normalized size = 6.83

method	result	size
default	\(\frac {a^{2} \left (\cos \left (d x +c \right )+1\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (6 i \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-6 i \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \EllipticF \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )+6 \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+6 i \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )-6 i \cos \left (d x +c \right ) \sqrt {2}-\sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+\sqrt {2}\, \sin \left (d x +c \right )\right ) \sqrt {2}}{3 d \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} \sqrt {\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}}\)	\(485\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2/d*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))*(6*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/
sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*sin(d*x+c)*EllipticPi((-(cos(d*x+c)
-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-6*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+
sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*sin(d*x+c)*EllipticF((-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+6*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos
(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+
c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)+6*I*2^(1/2)*cos(d*x+c)^2-6*I*2^(1/2)*cos(d*
x+c)-2^(1/2)*cos(d*x+c)*sin(d*x+c)+2^(1/2)*sin(d*x+c))/cos(d*x+c)/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(1/2)*2
^(1/2)

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (57) = 114\).
time = 0.49, size = 146, normalized size = 2.06 \begin {gather*} -\frac {3 \, {\left (-\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \left (i + 1\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \left (i + 1\right ) \, \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{2} + 4 \, {\left (a^{2} - \frac {6 i \, a^{2}}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(-(2*I - 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) - (2*I - 2)*sqrt(2)*arctan(-1
/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + (I + 1)*sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c)
+ 1) - (I + 1)*sqrt(2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^2 + 4*(a^2 - 6*I*a^2/tan(d*x
+ c))*tan(d*x + c)^(3/2))/d

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (57) = 114\).
time = 0.68, size = 340, normalized size = 4.79 \begin {gather*} -\frac {3 \, \sqrt {\frac {16 i \, a^{4}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {16 i \, a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) - 3 \, \sqrt {\frac {16 i \, a^{4}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {16 i \, a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) - 8 \, {\left (7 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 \, a^{2}\right )} \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}}}{12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*sqrt(16*I*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/2*(4*I*a^2*e^(2*I*d*x
+ 2*I*c) + sqrt(16*I*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*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)/a^2) - 3*sqrt(16*I*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d
)*log(1/2*(4*I*a^2*e^(2*I*d*x + 2*I*c) - sqrt(16*I*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*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)/a^2) - 8*(7*a^2*e^(4*I*d*x + 4*I*c) - 2*a^2*e^(2*I
*d*x + 2*I*c) - 5*a^2)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - a^{2} \left (\int \frac {\tan ^{2}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {2 i \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \left (- \frac {1}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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