\(\int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx\) [67]

Optimal result

Integrand size = 35, antiderivative size = 187 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\sqrt {1+(a+b x)^2}}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \arctan (a+b x)}{2 b}+\frac {i \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {5165, 5072, 267, 5006} \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\frac {i \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \arctan (a+b x)}{b}+\frac {(a+b x) \sqrt {(a+b x)^2+1} \arctan (a+b x)}{2 b}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b}-\frac {\sqrt {(a+b x)^2+1}}{2 b} \]

[In]

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

[Out]

-1/2*Sqrt[1 + (a + b*x)^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 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 5006

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]

Rule 5072

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 5165

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/d^2)*x^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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^2 \arctan (x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b} \\ & = \frac {(a+b x) \sqrt {1+(a+b x)^2} \arctan (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 {\arctan (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} \arctan (a+b x)}{2 b}+\frac {i \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\frac {-\sqrt {1+(a+b x)^2}+(a+b x) \sqrt {1+(a+b x)^2} \arctan (a+b x)-\arctan (a+b x) \log \left (1-i e^{i \arctan (a+b x)}\right )+\arctan (a+b x) \log \left (1+i e^{i \arctan (a+b x)}\right )-i \operatorname {PolyLog}\left (2,-i e^{i \arctan (a+b x)}\right )+i \operatorname {PolyLog}\left (2,i e^{i \arctan (a+b x)}\right )}{2 b} \]

[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)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {\left (\arctan \left (b x +a \right ) b x +a \arctan \left (b x +a \right )-1\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b}+\frac {\arctan \left (b x +a \right ) \ln \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-\arctan \left (b x +a \right ) \ln \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{2 b}\)	\(180\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\text {Timed out} \]

[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]

Timed out

Maxima [F]

\[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} \,d x } \]

[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]

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

Giac [F]

\[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} \,d x } \]

[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]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\mathrm {atan}\left (a+b\,x\right )\,{\left (a+b\,x\right )}^2}{\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}} \,d x \]

[In]

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

[Out]

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