Integrand size = 35, antiderivative size = 187 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\sqrt {1+(a+b x)^2}}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \arctan (a+b x)}{2 b}+\frac {i \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b} \] Output:
-1/2*(1+(b*x+a)^2)^(1/2)/b+1/2*(b*x+a)*(1+(b*x+a)^2)^(1/2)*arctan(b*x+a)/b +I*arctan(b*x+a)*arctan((1+I*(b*x+a))^(1/2)/(1-I*(b*x+a))^(1/2))/b-1/2*I*p olylog(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
Time = 0.52 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\frac {-\sqrt {1+(a+b x)^2}+(a+b x) \sqrt {1+(a+b x)^2} \arctan (a+b x)-\arctan (a+b x) \log \left (1-i e^{i \arctan (a+b x)}\right )+\arctan (a+b x) \log \left (1+i e^{i \arctan (a+b x)}\right )-i \operatorname {PolyLog}\left (2,-i e^{i \arctan (a+b x)}\right )+i \operatorname {PolyLog}\left (2,i e^{i \arctan (a+b x)}\right )}{2 b} \] Input:
Integrate[((a + b*x)^2*ArcTan[a + b*x])/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2], x]
Output:
(-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*ArcTan[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)
Time = 0.44 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {5580, 5487, 241, 5421}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2+1}} \, dx\) |
\(\Big \downarrow \) 5580 |
\(\displaystyle \frac {\int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {(a+b x)^2+1}}d(a+b x)}{b}\) |
\(\Big \downarrow \) 5487 |
\(\displaystyle \frac {-\frac {1}{2} \int \frac {\arctan (a+b x)}{\sqrt {(a+b x)^2+1}}d(a+b x)-\frac {1}{2} \int \frac {a+b x}{\sqrt {(a+b x)^2+1}}d(a+b x)+\frac {1}{2} (a+b x) \sqrt {(a+b x)^2+1} \arctan (a+b x)}{b}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {-\frac {1}{2} \int \frac {\arctan (a+b x)}{\sqrt {(a+b x)^2+1}}d(a+b x)+\frac {1}{2} (a+b x) \sqrt {(a+b x)^2+1} \arctan (a+b x)-\frac {1}{2} \sqrt {(a+b x)^2+1}}{b}\) |
\(\Big \downarrow \) 5421 |
\(\displaystyle \frac {\frac {1}{2} \left (2 i \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )-i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )+i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )\right )+\frac {1}{2} (a+b x) \sqrt {(a+b x)^2+1} \arctan (a+b x)-\frac {1}{2} \sqrt {(a+b x)^2+1}}{b}\) |
Input:
Int[((a + b*x)^2*ArcTan[a + b*x])/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2],x]
Output:
(-1/2*Sqrt[1 + (a + b*x)^2] + ((a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcTan[a + b*x])/2 + ((2*I)*ArcTan[a + b*x]*ArcTan[Sqrt[1 + I*(a + b*x)]/Sqrt[1 - I*( a + b*x)]] - I*PolyLog[2, ((-I)*Sqrt[1 + I*(a + b*x)])/Sqrt[1 - I*(a + b*x )]] + I*PolyLog[2, (I*Sqrt[1 + I*(a + b*x)])/Sqrt[1 - I*(a + b*x)]])/2)/b
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
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]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]
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] + (-Simp[b*f*(p/(c*m)) Int[(f*x)^(m - 1)*(( a + b*ArcTan[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] - Simp[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] && GtQ[m, 1]
Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/d Subs t[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]
Time = 1.06 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {\left (\arctan \left (b x +a \right ) b x +a \arctan \left (b x +a \right )-1\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b}+\frac {\arctan \left (b x +a \right ) \ln \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-\arctan \left (b x +a \right ) \ln \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{2 b}\) | \(180\) |
Input:
int((b*x+a)^2*arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x,method=_RETURN VERBOSE)
Output:
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
\[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} \,d x } \] Input:
integrate((b*x+a)^2*arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorit hm="fricas")
Output:
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)
Timed out. \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\text {Timed out} \] Input:
integrate((b*x+a)**2*atan(b*x+a)/(b**2*x**2+2*a*b*x+a**2+1)**(1/2),x)
Output:
Timed out
\[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} \,d x } \] Input:
integrate((b*x+a)^2*arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorit hm="maxima")
Output:
integrate((b*x + a)^2*arctan(b*x + a)/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1), x )
\[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} \,d x } \] Input:
integrate((b*x+a)^2*arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorit hm="giac")
Output:
integrate((b*x + a)^2*arctan(b*x + a)/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1), x )
Timed out. \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\mathrm {atan}\left (a+b\,x\right )\,{\left (a+b\,x\right )}^2}{\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}} \,d x \] Input:
int((atan(a + b*x)*(a + b*x)^2)/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2),x)
Output:
int((atan(a + b*x)*(a + b*x)^2)/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2), x)
\[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\left (\int \frac {\mathit {atan} \left (b x +a \right )}{\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}d x \right ) a^{2}+\left (\int \frac {\mathit {atan} \left (b x +a \right ) x^{2}}{\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}d x \right ) b^{2}+2 \left (\int \frac {\mathit {atan} \left (b x +a \right ) x}{\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}d x \right ) a b \] Input:
int((b*x+a)^2*atan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x)
Output:
int(atan(a + b*x)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1),x)*a**2 + int((atan (a + b*x)*x**2)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1),x)*b**2 + 2*int((atan (a + b*x)*x)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1),x)*a*b