3.8.22 \(\int \frac {a+i a \tan (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\) [722]

Optimal. Leaf size=47 \[ \frac {2 \sqrt [4]{-1} a \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i a}{d \sqrt {\cot (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3754, 3610, 3614, 214} \begin {gather*} \frac {2 \sqrt [4]{-1} a \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i a}{d \sqrt {\cot (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.77, size = 71, normalized size = 1.51 \begin {gather*} \frac {a \left (2 i-\frac {2 i \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {i \tan (c+d x)}}\right )}{d \sqrt {\cot (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 26.53, size = 420, normalized size = 8.94

method	result	size
default	\(\frac {a \left (\cos \left (d x +c \right )+1\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (-i \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticF \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )+i \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c \right )+\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c \right )+i \cos \left (d x +c \right ) \sqrt {2}-i \sqrt {2}\right ) \sqrt {2}}{d \sin \left (d x +c \right )^{4} \sqrt {\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}}\)	\(420\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/d*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))*(-I*sin(d*x+c)*((-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)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(
d*x+c))^(1/2),1/2*2^(1/2))+I*sin(d*x+c)*((-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)*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)-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))*sin(d*x+c)+I*cos(d*x+c)*2^(1/2)-I*2^(1/2))/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. 127 vs. \(2 (37) = 74\).
time = 0.49, size = 127, normalized size = 2.70 \begin {gather*} -\frac {{\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 - 8 i \, a \sqrt {\tan \left (d x + c\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((-(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 - 8*I*a*sqrt(tan(d*x + c)))/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. 278 vs. \(2 (37) = 74\).
time = 0.67, size = 278, normalized size = 5.91 \begin {gather*} -\frac {{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {4 i \, a^{2}}{d^{2}}} \log \left (\frac {{\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {4 i \, a^{2}}{d^{2}}} \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}} + 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {4 i \, a^{2}}{d^{2}}} \log \left (-\frac {{\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {4 i \, a^{2}}{d^{2}}} \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}} - 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - 8 \, {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} - a\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}}}{4 \, {\left (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))/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/4*((d*e^(2*I*d*x + 2*I*c) + d)*sqrt(4*I*a^2/d^2)*log(((d*e^(2*I*d*x + 2*I*c) - d)*sqrt(4*I*a^2/d^2)*sqrt((I
*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) + 2*I*a*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a) - (d
*e^(2*I*d*x + 2*I*c) + d)*sqrt(4*I*a^2/d^2)*log(-((d*e^(2*I*d*x + 2*I*c) - d)*sqrt(4*I*a^2/d^2)*sqrt((I*e^(2*I
*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) - 2*I*a*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a) - 8*(a*e^(2
*I*d*x + 2*I*c) - a)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(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*} i a \left (\int \left (- \frac {i}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \frac {\tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 5.64, size = 102, normalized size = 2.17 \begin {gather*} \frac {a\,2{}\mathrm {i}}{d\,\sqrt {\mathrm {cot}\left (c+d\,x\right )}}-\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}}\right )}{d}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {cot}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d}-\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}}\right )}{d}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {cot}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*2i)/(d*cot(c + d*x)^(1/2)) - ((-1)^(1/4)*a*atan((-1)^(1/4)/cot(c + d*x)^(1/2)))/d + ((-1)^(1/4)*a*atan((-1)
^(1/4)*cot(c + d*x)^(1/2))*1i)/d - ((-1)^(1/4)*a*atanh((-1)^(1/4)/cot(c + d*x)^(1/2)))/d + ((-1)^(1/4)*a*atanh
((-1)^(1/4)*cot(c + d*x)^(1/2))*1i)/d

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