3.67 \(\int \frac{(a+b x)^2 \tan ^{-1}(a+b x)}{\sqrt{1+a^2+2 a b x+b^2 x^2}} \, dx\)

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

[Out]

-Sqrt[1 + (a + b*x)^2]/(2*b) + ((a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcTan[a + b*x])/(2*b) + (I*ArcTan[a + b*x]*Ar
cTan[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

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Rubi [A]  time = 0.215784, antiderivative size = 187, normalized size of antiderivative = 1., 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 = {5057, 4952, 261, 4886} \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{2 b}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{2 b}-\frac{\sqrt{(a+b x)^2+1}}{2 b}+\frac{i \tan ^{-1}\left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right ) \tan ^{-1}(a+b x)}{b}+\frac{(a+b x) \sqrt{(a+b x)^2+1} \tan ^{-1}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*ArcTan[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]*ArcTan[a + b*x])/(2*b) + (I*ArcTan[a + b*x]*Ar
cTan[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 5057

Int[((a_.) + ArcTan[(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*x^2)/d^2)^q*(a + b*ArcTan
[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]

Rule 4952

Int[(((a_.) + ArcTan[(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*ArcTan[c*x])^p)/(c^2*d*m), x] + (-Dist[(b*f*p)/(c*m), Int[((f*x)^(m -
1)*(a + b*ArcTan[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*ArcTan[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] && Gt
Q[m, 1]

Rule 261

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 4886

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

Rubi steps

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

Mathematica [A]  time = 0.657008, size = 145, normalized size = 0.78 \[ \frac{-i \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a+b x)}\right )+i \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a+b x)}\right )-\sqrt{(a+b x)^2+1}+(a+b x) \sqrt{(a+b x)^2+1} \tan ^{-1}(a+b x)+\tan ^{-1}(a+b x) \left (-\log \left (1-i e^{i \tan ^{-1}(a+b x)}\right )\right )+\tan ^{-1}(a+b x) \log \left (1+i e^{i \tan ^{-1}(a+b x)}\right )}{2 b} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.619, size = 187, normalized size = 1. \begin{align*}{\frac{\arctan \left ( bx+a \right ) xb+\arctan \left ( bx+a \right ) a-1}{2\,b}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{\arctan \left ( bx+a \right ) }{2\,b}\ln \left ( 1+{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) }-{\frac{\arctan \left ( bx+a \right ) }{2\,b}\ln \left ( 1-{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) }-{\frac{{\frac{i}{2}}}{b}{\it dilog} \left ( 1+{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) }+{\frac{{\frac{i}{2}}}{b}{\it dilog} \left ( 1-{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(arctan(b*x+a)*x*b+arctan(b*x+a)*a-1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)/b+1/2/b*arctan(b*x+a)*ln(1+I*(1+I*(b*x
+a))/(1+(b*x+a)^2)^(1/2))-1/2/b*arctan(b*x+a)*ln(1-I*(1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))-1/2*I/b*dilog(1+I*(1+I
*(b*x+a))/(1+(b*x+a)^2)^(1/2))+1/2*I/b*dilog(1-I*(1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{2} \operatorname{atan}{\left (a + b x \right )}}{\sqrt{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((b*x+a)**2*atan(b*x+a)/(b**2*x**2+2*a*b*x+a**2+1)**(1/2),x)

[Out]

Integral((a + b*x)**2*atan(a + b*x)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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