3.2.18 \(\int \frac {(a+b x)^2 \cot ^{-1}(a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx\) [118]

Optimal. Leaf size=187 \[ \frac {\sqrt {1+(a+b x)^2}}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \cot ^{-1}(a+b x)}{2 b}+\frac {i \cot ^{-1}(a+b x) \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}+\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}-\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b} \]

[Out]

I*arccot(b*x+a)*arctan((1+I*(b*x+a))^(1/2)/(1-I*(b*x+a))^(1/2))/b+1/2*I*polylog(2,-I*(1+I*(b*x+a))^(1/2)/(1-I*
(b*x+a))^(1/2))/b-1/2*I*polylog(2,I*(1+I*(b*x+a))^(1/2)/(1-I*(b*x+a))^(1/2))/b+1/2*(1+(b*x+a)^2)^(1/2)/b+1/2*(
b*x+a)*arccot(b*x+a)*(1+(b*x+a)^2)^(1/2)/b

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Rubi [A]
time = 0.15, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {5166, 5073, 267, 5007} \begin {gather*} \frac {i \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \cot ^{-1}(a+b x)}{b}+\frac {i \text {Li}_2\left (-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b}-\frac {i \text {Li}_2\left (\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b}+\frac {\sqrt {(a+b x)^2+1}}{2 b}+\frac {(a+b x) \sqrt {(a+b x)^2+1} \cot ^{-1}(a+b x)}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 + (a + b*x)^2]/(2*b) + ((a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcCot[a + b*x])/(2*b) + (I*ArcCot[a + b*x]*Arc
Tan[Sqrt[1 + I*(a + b*x)]/Sqrt[1 - I*(a + b*x)]])/b + ((I/2)*PolyLog[2, ((-I)*Sqrt[1 + I*(a + b*x)])/Sqrt[1 -
I*(a + b*x)]])/b - ((I/2)*PolyLog[2, (I*Sqrt[1 + I*(a + b*x)])/Sqrt[1 - I*(a + b*x)]])/b

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 5007

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[-2*I*(a + b*ArcCot[c*x])*(
ArcTan[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]]/(c*Sqrt[d])), x] + (-Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 + I*c*x]/Sqrt[1
- I*c*x])]/(c*Sqrt[d])), x] + Simp[I*b*(PolyLog[2, I*(Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x])]/(c*Sqrt[d])), x]) /;
FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]

Rule 5073

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

Rule 5166

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_
)^2)^(q_.), x_Symbol] :> Dist[1/d, Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(C/d^2 + (C/d^2)*x^2)^q*(a + b*ArcCot
[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, p, q}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0]
&& EqQ[2*c*C - B*d, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x)^2 \cot ^{-1}(a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\text {Subst}\left (\int \frac {x^2 \cot ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sqrt {1+(a+b x)^2} \cot ^{-1}(a+b x)}{2 b}+\frac {\text {Subst}\left (\int \frac {x}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b}-\frac {\text {Subst}\left (\int \frac {\cot ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=\frac {\sqrt {1+(a+b x)^2}}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \cot ^{-1}(a+b x)}{2 b}+\frac {i \cot ^{-1}(a+b x) \tan ^{-1}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}+\frac {i \text {Li}_2\left (-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}-\frac {i \text {Li}_2\left (\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}\\ \end {align*}

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Mathematica [A]
time = 1.05, size = 202, normalized size = 1.08 \begin {gather*} -\frac {\sqrt {(a+b x)^2 \left (1+\frac {1}{(a+b x)^2}\right )} \left (-2 \cot \left (\frac {1}{2} \cot ^{-1}(a+b x)\right )-\cot ^{-1}(a+b x) \csc ^2\left (\frac {1}{2} \cot ^{-1}(a+b x)\right )-4 \cot ^{-1}(a+b x) \log \left (1-e^{i \cot ^{-1}(a+b x)}\right )+4 \cot ^{-1}(a+b x) \log \left (1+e^{i \cot ^{-1}(a+b x)}\right )-4 i \text {PolyLog}\left (2,-e^{i \cot ^{-1}(a+b x)}\right )+4 i \text {PolyLog}\left (2,e^{i \cot ^{-1}(a+b x)}\right )+\cot ^{-1}(a+b x) \sec ^2\left (\frac {1}{2} \cot ^{-1}(a+b x)\right )-2 \tan \left (\frac {1}{2} \cot ^{-1}(a+b x)\right )\right )}{8 b (a+b x) \sqrt {1+\frac {1}{(a+b x)^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(Sqrt[(a + b*x)^2*(1 + (a + b*x)^(-2))]*(-2*Cot[ArcCot[a + b*x]/2] - ArcCot[a + b*x]*Csc[ArcCot[a + b*x]/
2]^2 - 4*ArcCot[a + b*x]*Log[1 - E^(I*ArcCot[a + b*x])] + 4*ArcCot[a + b*x]*Log[1 + E^(I*ArcCot[a + b*x])] - (
4*I)*PolyLog[2, -E^(I*ArcCot[a + b*x])] + (4*I)*PolyLog[2, E^(I*ArcCot[a + b*x])] + ArcCot[a + b*x]*Sec[ArcCot
[a + b*x]/2]^2 - 2*Tan[ArcCot[a + b*x]/2]))/(b*(a + b*x)*Sqrt[1 + (a + b*x)^(-2)])

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Maple [A]
time = 0.69, size = 160, normalized size = 0.86

method	result	size
default	\(\frac {\left (\mathrm {arccot}\left (b x +a \right ) b x +\mathrm {arccot}\left (b x +a \right ) a +1\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b}-\frac {i \left (i \mathrm {arccot}\left (b x +a \right ) \ln \left (1-\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \mathrm {arccot}\left (b x +a \right ) \ln \left (\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}+1\right )-\polylog \left (2, -\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+\polylog \left (2, \frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )\right )}{2 b}\)	\(160\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*arccot(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(arccot(b*x+a)*b*x+arccot(b*x+a)*a+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)/b-1/2*I*(I*arccot(b*x+a)*ln(1-(I+a+b*x
)/(1+(b*x+a)^2)^(1/2))-I*arccot(b*x+a)*ln((I+a+b*x)/(1+(b*x+a)^2)^(1/2)+1)-polylog(2,-(I+a+b*x)/(1+(b*x+a)^2)^
(1/2))+polylog(2,(I+a+b*x)/(1+(b*x+a)^2)^(1/2)))/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{2} \operatorname {acot}{\left (a + b x \right )}}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**2*acot(a + b*x)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1), 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((b*x+a)^2*arccot(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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