\(\int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {(1+a^2) c+2 a b c x+b^2 c x^2}} \, dx\) [69]

Optimal result

Integrand size = 40, antiderivative size = 281 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=-\frac {\sqrt {c+c (a+b x)^2}}{2 b c}+\frac {(a+b x) \sqrt {c+c (a+b x)^2} \arctan (a+b x)}{2 b c}+\frac {i \sqrt {1+(a+b x)^2} \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}-\frac {i \sqrt {1+(a+b x)^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c+c (a+b x)^2}}+\frac {i \sqrt {1+(a+b x)^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c+c (a+b x)^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\frac {\sqrt {1+a^2+2 a b x+b^2 x^2} \left (-\sqrt {1+(a+b x)^2}+(a+b x) \sqrt {1+(a+b x)^2} \arctan (a+b x)-\arctan (a+b x) \log \left (1-i e^{i \arctan (a+b x)}\right )+\arctan (a+b x) \log \left (1+i e^{i \arctan (a+b x)}\right )-i \operatorname {PolyLog}\left (2,-i e^{i \arctan (a+b x)}\right )+i \operatorname {PolyLog}\left (2,i e^{i \arctan (a+b x)}\right )\right )}{2 b \sqrt {c \left (1+a^2+2 a b x+b^2 x^2\right )}} \] Input:

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

Output:

(Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(-Sqrt[1 + (a + b*x)^2] + (a + b*x)*Sqr 
t[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*Sqrt[c*(1 + a^2 + 2*a*b*x + b^2*x^2)])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5580, 5487, 241, 5425, 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.

\(\displaystyle \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {\left (a^2+1\right ) c+2 a b c x+b^2 c x^2}} \, dx\)

\(\Big \downarrow \) 5580

\(\displaystyle \frac {\int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt {c (a+b x)^2+c}}d(a+b x)}{b}\)

\(\Big \downarrow \) 5487

\(\displaystyle \frac {-\frac {1}{2} \int \frac {\arctan (a+b x)}{\sqrt {c (a+b x)^2+c}}d(a+b x)-\frac {1}{2} \int \frac {a+b x}{\sqrt {c (a+b x)^2+c}}d(a+b x)+\frac {(a+b x) \arctan (a+b x) \sqrt {c (a+b x)^2+c}}{2 c}}{b}\)

\(\Big \downarrow \) 241

\(\displaystyle \frac {-\frac {1}{2} \int \frac {\arctan (a+b x)}{\sqrt {c (a+b x)^2+c}}d(a+b x)+\frac {(a+b x) \arctan (a+b x) \sqrt {c (a+b x)^2+c}}{2 c}-\frac {\sqrt {c (a+b x)^2+c}}{2 c}}{b}\)

\(\Big \downarrow \) 5425

\(\displaystyle \frac {-\frac {\sqrt {(a+b x)^2+1} \int \frac {\arctan (a+b x)}{\sqrt {(a+b x)^2+1}}d(a+b x)}{2 \sqrt {c (a+b x)^2+c}}+\frac {(a+b x) \arctan (a+b x) \sqrt {c (a+b x)^2+c}}{2 c}-\frac {\sqrt {c (a+b x)^2+c}}{2 c}}{b}\)

\(\Big \downarrow \) 5421

\(\displaystyle \frac {-\frac {\sqrt {(a+b x)^2+1} \left (-2 i \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )+i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )-i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )\right )}{2 \sqrt {c (a+b x)^2+c}}+\frac {(a+b x) \arctan (a+b x) \sqrt {c (a+b x)^2+c}}{2 c}-\frac {\sqrt {c (a+b x)^2+c}}{2 c}}{b}\)

Input:

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

Output:

(-1/2*Sqrt[c + c*(a + b*x)^2]/c + ((a + b*x)*Sqrt[c + c*(a + b*x)^2]*ArcTa 
n[a + b*x])/(2*c) - (Sqrt[1 + (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*Sqrt[c + c*(a + b*x)^2]))/b
 

Defintions of rubi rules used

rule 241

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]
 

rule 5421

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]
 

rule 5425

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]   Int[(a + b*ArcTan[c*x])^ 
p/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] & 
& IGtQ[p, 0] &&  !GtQ[d, 0]
 

rule 5487

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]
 

rule 5580

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]
 

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {\left (\arctan \left (b x +a \right ) b x +a \arctan \left (b x +a \right )-1\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{2 b c}+\frac {\left (\arctan \left (b x +a \right ) \ln \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-\arctan \left (b x +a \right ) \ln \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b c}\)	\(222\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((b*x+a)^2*atan(b*x+a)/((a^2+1)*c+2*a*b*c*x+b^2*c*x^2)^(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((ata 
n(a + b*x)*x**2)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1),x)*b**2 + 2*int((ata 
n(a + b*x)*x)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1),x)*a*b)/sqrt(c)