3.1.68 \(\int \frac {(a+b x)^2 \text {ArcTan}(a+b x)}{\sqrt {(1+a^2) c+2 a b c x+b^2 c x^2}} \, dx\) [68]

Optimal. Leaf size=281 \[ -\frac {\sqrt {c+c (a+b x)^2}}{2 b c}+\frac {(a+b x) \sqrt {c+c (a+b x)^2} \text {ArcTan}(a+b x)}{2 b c}+\frac {i \sqrt {1+(a+b x)^2} \text {ArcTan}(a+b x) \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}-\frac {i \sqrt {1+(a+b x)^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c+c (a+b x)^2}}+\frac {i \sqrt {1+(a+b x)^2} \text {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c+c (a+b x)^2}} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5165, 5072, 267, 5010, 5006} \begin {gather*} \frac {i \sqrt {(a+b x)^2+1} \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \text {ArcTan}(a+b x)}{b \sqrt {c (a+b x)^2+c}}+\frac {(a+b x) \text {ArcTan}(a+b x) \sqrt {c (a+b x)^2+c}}{2 b c}-\frac {i \sqrt {(a+b x)^2+1} \text {Li}_2\left (-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c (a+b x)^2+c}}+\frac {i \sqrt {(a+b x)^2+1} \text {Li}_2\left (\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c (a+b x)^2+c}}-\frac {\sqrt {c (a+b x)^2+c}}{2 b c} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

-1/2*Sqrt[c + c*(a + b*x)^2]/(b*c) + ((a + b*x)*Sqrt[c + c*(a + b*x)^2]*ArcTan[a + b*x])/(2*b*c) + (I*Sqrt[1 +
(a + b*x)^2]*ArcTan[a + b*x]*ArcTan[Sqrt[1 + I*(a + b*x)]/Sqrt[1 - I*(a + b*x)]])/(b*Sqrt[c + c*(a + b*x)^2])
- ((I/2)*Sqrt[1 + (a + b*x)^2]*PolyLog[2, ((-I)*Sqrt[1 + I*(a + b*x)])/Sqrt[1 - I*(a + b*x)]])/(b*Sqrt[c + c*
(a + b*x)^2]) + ((I/2)*Sqrt[1 + (a + b*x)^2]*PolyLog[2, (I*Sqrt[1 + I*(a + b*x)])/Sqrt[1 - I*(a + b*x)]])/(b*S
qrt[c + c*(a + b*x)^2])

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 5010

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/Sq
rt[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 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*} \int \frac {(a+b x)^2 \tan ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx &=\frac {\text {Subst}\left (\int \frac {x^2 \tan ^{-1}(x)}{\sqrt {c+c x^2}} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sqrt {c+c (a+b x)^2} \tan ^{-1}(a+b x)}{2 b c}-\frac {\text {Subst}\left (\int \frac {x}{\sqrt {c+c x^2}} \, dx,x,a+b x\right )}{2 b}-\frac {\text {Subst}\left (\int \frac {\tan ^{-1}(x)}{\sqrt {c+c x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=-\frac {\sqrt {c+c (a+b x)^2}}{2 b c}+\frac {(a+b x) \sqrt {c+c (a+b x)^2} \tan ^{-1}(a+b x)}{2 b c}-\frac {\sqrt {1+(a+b x)^2} \text {Subst}\left (\int \frac {\tan ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b \sqrt {c+c (a+b x)^2}}\\ &=-\frac {\sqrt {c+c (a+b x)^2}}{2 b c}+\frac {(a+b x) \sqrt {c+c (a+b x)^2} \tan ^{-1}(a+b x)}{2 b c}+\frac {i \sqrt {1+(a+b x)^2} \tan ^{-1}(a+b x) \tan ^{-1}\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} \text {Li}_2\left (-\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} \text {Li}_2\left (\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c+c (a+b x)^2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 189, normalized size = 0.67 \begin {gather*} \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} \text {ArcTan}(a+b x)-\text {ArcTan}(a+b x) \log \left (1-i e^{i \text {ArcTan}(a+b x)}\right )+\text {ArcTan}(a+b x) \log \left (1+i e^{i \text {ArcTan}(a+b x)}\right )-i \text {PolyLog}\left (2,-i e^{i \text {ArcTan}(a+b x)}\right )+i \text {PolyLog}\left (2,i e^{i \text {ArcTan}(a+b x)}\right )\right )}{2 b \sqrt {c \left (1+a^2+2 a b x+b^2 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

(Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(-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*PolyL
og[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)])

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Maple [A]
time = 0.48, size = 222, normalized size = 0.79

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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

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

[Out]

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)

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

[Out]

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)

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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 {atan}{\left (a + b x \right )}}{\sqrt {c \left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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)

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