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

Optimal. Leaf size=216 \[ \frac{i \sqrt{(a+b x)^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b \sqrt{c (a+b x)^2+c}}-\frac{i \sqrt{(a+b x)^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b \sqrt{c (a+b x)^2+c}}-\frac{2 i \sqrt{(a+b x)^2+1} \tan ^{-1}(a+b x) \tan ^{-1}\left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b \sqrt{c (a+b x)^2+c}} \]

[Out]

((-2*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*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*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])

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Rubi [A]  time = 0.162288, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {5055, 4890, 4886} \[ \frac{i \sqrt{(a+b x)^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b \sqrt{c (a+b x)^2+c}}-\frac{i \sqrt{(a+b x)^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b \sqrt{c (a+b x)^2+c}}-\frac{2 i \sqrt{(a+b x)^2+1} \tan ^{-1}(a+b x) \tan ^{-1}\left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b \sqrt{c (a+b x)^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*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*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*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])

Rule 5055

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(q_.), x_Symbol] :> Di
st[1/d, Subst[Int[(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, A, B,
C, p, q}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rule 4890

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

Mathematica [A]  time = 0.0636544, size = 125, normalized size = 0.58 \[ \frac{\sqrt{(a+b x)^2+1} \left (i \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a+b x)}\right )-i \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a+b x)}\right )+\tan ^{-1}(a+b x) \left (\log \left (1-i e^{i \tan ^{-1}(a+b x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a+b x)}\right )\right )\right )}{b \sqrt{c \left ((a+b x)^2+1\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 + (a + b*x)^2]*(ArcTan[a + b*x]*(Log[1 - I*E^(I*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])]))/(b*Sqrt[c*(1 + (a + b*x)^
2)])

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Maple [A]  time = 0.439, size = 176, normalized size = 0.8 \begin{align*}{\frac{i}{bc} \left ( i\arctan \left ( bx+a \right ) \ln \left ( 1+{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) -i\arctan \left ( bx+a \right ) \ln \left ( 1-{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) +{\it dilog} \left ( 1+{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) -{\it dilog} \left ( 1-{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) \right ) \sqrt{c \left ( -i+a+bx \right ) \left ( i+a+bx \right ) }{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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(-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(arctan(b*x+a)/((a^2+1)*c+2*a*b*c*x+c*x^2*b^2)^(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{\arctan \left (b x + a\right )}{\sqrt{b^{2} c x^{2} + 2 \, a b c x +{\left (a^{2} + 1\right )} c}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(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., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atan}{\left (a + b x \right )}}{\sqrt{c \left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(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., size = 0, normalized size = 0. \begin{align*} \int \frac{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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