3.243 \(\int \frac {\coth ^{-1}(1-i d+d \tan (a+b x))}{x} \, dx\)

Optimal. Leaf size=23 \[ \text {Int}\left (\frac {\coth ^{-1}(d \tan (a+b x)-i d+1)}{x},x\right ) \]

[Out]

CannotIntegrate(arccoth(1-I*d+d*tan(b*x+a))/x,x)

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Rubi [A]  time = 0.09, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\coth ^{-1}(1-i d+d \tan (a+b x))}{x} \, dx \]

Verification is Not applicable to the result.

[In]

Int[ArcCoth[1 - I*d + d*Tan[a + b*x]]/x,x]

[Out]

Defer[Int][ArcCoth[1 - I*d + d*Tan[a + b*x]]/x, x]

Rubi steps

\begin {align*} \int \frac {\coth ^{-1}(1-i d+d \tan (a+b x))}{x} \, dx &=\int \frac {\coth ^{-1}(1-i d+d \tan (a+b x))}{x} \, dx\\ \end {align*}

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Mathematica [A]  time = 0.81, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{-1}(1-i d+d \tan (a+b x))}{x} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[ArcCoth[1 - I*d + d*Tan[a + b*x]]/x,x]

[Out]

Integrate[ArcCoth[1 - I*d + d*Tan[a + b*x]]/x, x]

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fricas [A]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (\frac {{\left ({\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right )}{2 \, x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1-I*d+d*tan(b*x+a))/x,x, algorithm="fricas")

[Out]

integral(1/2*log(((d + I)*e^(2*I*b*x + 2*I*a) + I)*e^(-2*I*b*x - 2*I*a)/d)/x, x)

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1-I*d+d*tan(b*x+a))/x,x, algorithm="giac")

[Out]

integrate(arccoth(d*tan(b*x + a) - I*d + 1)/x, x)

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maple [A]  time = 1.82, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {arccoth}\left (1-i d +d \tan \left (b x +a \right )\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(1-I*d+d*tan(b*x+a))/x,x)

[Out]

int(arccoth(1-I*d+d*tan(b*x+a))/x,x)

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ -i \, b x + \frac {1}{4} \, {\left (-i \, \pi - 4 i \, a - 2 \, \log \relax (d)\right )} \log \relax (x) - \frac {1}{2} i \, \int \frac {\arctan \left (-d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), -d \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right ) - 1\right )}{x}\,{d x} + \frac {1}{4} \, \int \frac {\log \left ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, d \sin \left (2 \, b x + 2 \, a\right ) + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1-I*d+d*tan(b*x+a))/x,x, algorithm="maxima")

[Out]

-I*b*x + 1/4*(-I*pi - 4*I*a - 2*log(d))*log(x) - 1/2*I*integrate(arctan2(-d*cos(2*b*x + 2*a) + sin(2*b*x + 2*a
), -d*sin(2*b*x + 2*a) - cos(2*b*x + 2*a) - 1)/x, x) + 1/4*integrate(log((d^2 + 1)*cos(2*b*x + 2*a)^2 + (d^2 +
1)*sin(2*b*x + 2*a)^2 + 2*d*sin(2*b*x + 2*a) + 2*cos(2*b*x + 2*a) + 1)/x, x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {\mathrm {acoth}\left (d\,\mathrm {tan}\left (a+b\,x\right )+1-d\,1{}\mathrm {i}\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(d*tan(a + b*x) - d*1i + 1)/x,x)

[Out]

int(acoth(d*tan(a + b*x) - d*1i + 1)/x, x)

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acoth}{\left (d \tan {\left (a + b x \right )} - i d + 1 \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(1-I*d+d*tan(b*x+a))/x,x)

[Out]

Integral(acoth(d*tan(a + b*x) - I*d + 1)/x, x)

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