3.116 \(\int \frac{\cot ^{-1}(a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=22 \[ \text{Unintegrable}\left (\frac{\cot ^{-1}(a+b x)}{\sqrt [3]{(a+b x)^2+1}},x\right ) \]

[Out]

Unintegrable[ArcCot[a + b*x]/(1 + (a + b*x)^2)^(1/3), x]

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Rubi [A]  time = 0.0400934, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\cot ^{-1}(a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx \]

Verification is Not applicable to the result.

[In]

Int[ArcCot[a + b*x]/(1 + a^2 + 2*a*b*x + b^2*x^2)^(1/3),x]

[Out]

Defer[Subst][Defer[Int][ArcCot[x]/(1 + x^2)^(1/3), x], x, a + b*x]/b

Rubi steps

\begin{align*} \int \frac{\cot ^{-1}(a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cot ^{-1}(x)}{\sqrt [3]{1+x^2}} \, dx,x,a+b x\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.37931, size = 177, normalized size = 8.05 \[ \frac{6 \text{Gamma}\left (\frac{11}{6}\right ) \text{Gamma}\left (\frac{7}{3}\right ) \left (4 (a+b x) \, _2F_1\left (1,\frac{4}{3};\frac{11}{6};\frac{1}{a^2+2 b x a+b^2 x^2+1}\right ) \cot ^{-1}(a+b x)+5 \left (a^2+2 a b x+b^2 x^2+1\right ) \left (2 (a+b x) \cot ^{-1}(a+b x)-3\right )\right )-5 \sqrt [3]{2} \sqrt{\pi } \text{Gamma}\left (\frac{5}{3}\right ) \text{HypergeometricPFQ}\left (\left \{1,\frac{4}{3},\frac{4}{3}\right \},\left \{\frac{11}{6},\frac{7}{3}\right \},\frac{1}{a^2+2 a b x+b^2 x^2+1}\right )}{20 b \text{Gamma}\left (\frac{11}{6}\right ) \text{Gamma}\left (\frac{7}{3}\right ) \left (a^2+2 a b x+b^2 x^2+1\right )^{4/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCot[a + b*x]/(1 + a^2 + 2*a*b*x + b^2*x^2)^(1/3),x]

[Out]

(6*Gamma[11/6]*Gamma[7/3]*(5*(1 + a^2 + 2*a*b*x + b^2*x^2)*(-3 + 2*(a + b*x)*ArcCot[a + b*x]) + 4*(a + b*x)*Ar
cCot[a + b*x]*Hypergeometric2F1[1, 4/3, 11/6, (1 + a^2 + 2*a*b*x + b^2*x^2)^(-1)]) - 5*2^(1/3)*Sqrt[Pi]*Gamma[
5/3]*HypergeometricPFQ[{1, 4/3, 4/3}, {11/6, 7/3}, (1 + a^2 + 2*a*b*x + b^2*x^2)^(-1)])/(20*b*(1 + a^2 + 2*a*b
*x + b^2*x^2)^(4/3)*Gamma[11/6]*Gamma[7/3])

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Maple [A]  time = 1.135, size = 0, normalized size = 0. \begin{align*} \int{{\rm arccot} \left (bx+a\right ){\frac{1}{\sqrt [3]{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/3),x)

[Out]

int(arccot(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/3),x)

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arccot}\left (b x + a\right )}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/3),x, algorithm="maxima")

[Out]

integrate(arccot(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 + 1)^(1/3), x)

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arccot}\left (b x + a\right )}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{1}{3}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/3),x, algorithm="fricas")

[Out]

integral(arccot(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 + 1)^(1/3), x)

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Sympy [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acot}{\left (a + b x \right )}}{\sqrt [3]{a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(b*x+a)/(b**2*x**2+2*a*b*x+a**2+1)**(1/3),x)

[Out]

Integral(acot(a + b*x)/(a**2 + 2*a*b*x + b**2*x**2 + 1)**(1/3), x)

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arccot}\left (b x + a\right )}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/3),x, algorithm="giac")

[Out]

integrate(arccot(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 + 1)^(1/3), x)