3.12 \(\int \frac{(a+b \tan ^{-1}(c+d x))^2}{(c e+d e x)^3} \, dx\)

Optimal. Leaf size=117 \[ -\frac{b \left (a+b \tan ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3}+\frac{b^2 \log (c+d x)}{d e^3}-\frac{b^2 \log \left ((c+d x)^2+1\right )}{2 d e^3} \]

[Out]

-((b*(a + b*ArcTan[c + d*x]))/(d*e^3*(c + d*x))) - (a + b*ArcTan[c + d*x])^2/(2*d*e^3) - (a + b*ArcTan[c + d*x
])^2/(2*d*e^3*(c + d*x)^2) + (b^2*Log[c + d*x])/(d*e^3) - (b^2*Log[1 + (c + d*x)^2])/(2*d*e^3)

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Rubi [A]  time = 0.152844, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5043, 12, 4852, 4918, 266, 36, 29, 31, 4884} \[ -\frac{b \left (a+b \tan ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3}+\frac{b^2 \log (c+d x)}{d e^3}-\frac{b^2 \log \left ((c+d x)^2+1\right )}{2 d e^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^2/(c*e + d*e*x)^3,x]

[Out]

-((b*(a + b*ArcTan[c + d*x]))/(d*e^3*(c + d*x))) - (a + b*ArcTan[c + d*x])^2/(2*d*e^3) - (a + b*ArcTan[c + d*x
])^2/(2*d*e^3*(c + d*x)^2) + (b^2*Log[c + d*x])/(d*e^3) - (b^2*Log[1 + (c + d*x)^2])/(2*d*e^3)

Rule 5043

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((f*x)/d)^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0
] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTan[c*x])^p)/(d + e*
x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 4884

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{(c e+d e x)^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{x^3} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac{b \operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac{b \operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^3}-\frac{b \operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x (1+x)} \, dx,x,(c+d x)^2\right )}{2 d e^3}\\ &=-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,(c+d x)^2\right )}{2 d e^3}-\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,(c+d x)^2\right )}{2 d e^3}\\ &=-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac{b^2 \log (c+d x)}{d e^3}-\frac{b^2 \log \left (1+(c+d x)^2\right )}{2 d e^3}\\ \end{align*}

Mathematica [A]  time = 0.108117, size = 194, normalized size = 1.66 \[ -\frac{a^2+2 b \tan ^{-1}(c+d x) \left (a \left (c^2+2 c d x+d^2 x^2+1\right )+b (c+d x)\right )+2 a b c+2 a b d x+b^2 c^2 \log \left (c^2+2 c d x+d^2 x^2+1\right )+b^2 d^2 x^2 \log \left (c^2+2 c d x+d^2 x^2+1\right )+2 b^2 c d x \log \left (c^2+2 c d x+d^2 x^2+1\right )+b^2 \left (c^2+2 c d x+d^2 x^2+1\right ) \tan ^{-1}(c+d x)^2-2 b^2 (c+d x)^2 \log (c+d x)}{2 d e^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c + d*x])^2/(c*e + d*e*x)^3,x]

[Out]

-(a^2 + 2*a*b*c + 2*a*b*d*x + 2*b*(b*(c + d*x) + a*(1 + c^2 + 2*c*d*x + d^2*x^2))*ArcTan[c + d*x] + b^2*(1 + c
^2 + 2*c*d*x + d^2*x^2)*ArcTan[c + d*x]^2 - 2*b^2*(c + d*x)^2*Log[c + d*x] + b^2*c^2*Log[1 + c^2 + 2*c*d*x + d
^2*x^2] + 2*b^2*c*d*x*Log[1 + c^2 + 2*c*d*x + d^2*x^2] + b^2*d^2*x^2*Log[1 + c^2 + 2*c*d*x + d^2*x^2])/(2*d*e^
3*(c + d*x)^2)

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Maple [A]  time = 0.053, size = 182, normalized size = 1.6 \begin{align*} -{\frac{{a}^{2}}{2\,d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{2} \left ( \arctan \left ( dx+c \right ) \right ) ^{2}}{2\,d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{2} \left ( \arctan \left ( dx+c \right ) \right ) ^{2}}{2\,d{e}^{3}}}-{\frac{{b}^{2}\arctan \left ( dx+c \right ) }{d{e}^{3} \left ( dx+c \right ) }}-{\frac{{b}^{2}\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{2\,d{e}^{3}}}+{\frac{{b}^{2}\ln \left ( dx+c \right ) }{d{e}^{3}}}-{\frac{ab\arctan \left ( dx+c \right ) }{d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{ab\arctan \left ( dx+c \right ) }{d{e}^{3}}}-{\frac{ab}{d{e}^{3} \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))^2/(d*e*x+c*e)^3,x)

[Out]

-1/2/d*a^2/e^3/(d*x+c)^2-1/2/d*b^2/e^3/(d*x+c)^2*arctan(d*x+c)^2-1/2/d*b^2/e^3*arctan(d*x+c)^2-1/d*b^2/e^3*arc
tan(d*x+c)/(d*x+c)-1/2*b^2*ln(1+(d*x+c)^2)/d/e^3+b^2*ln(d*x+c)/d/e^3-1/d*a*b/e^3/(d*x+c)^2*arctan(d*x+c)-1/d*a
*b/e^3*arctan(d*x+c)-1/d*a*b/e^3/(d*x+c)

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Maxima [B]  time = 1.62124, size = 362, normalized size = 3.09 \begin{align*} -{\left (d{\left (\frac{1}{d^{3} e^{3} x + c d^{2} e^{3}} + \frac{\arctan \left (\frac{d^{2} x + c d}{d}\right )}{d^{2} e^{3}}\right )} + \frac{\arctan \left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} a b - \frac{1}{2} \,{\left (2 \, d{\left (\frac{1}{d^{3} e^{3} x + c d^{2} e^{3}} + \frac{\arctan \left (\frac{d^{2} x + c d}{d}\right )}{d^{2} e^{3}}\right )} \arctan \left (d x + c\right ) - \frac{\arctan \left (d x + c\right )^{2} - \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + 2 \, \log \left (d x + c\right )}{d e^{3}}\right )} b^{2} - \frac{b^{2} \arctan \left (d x + c\right )^{2}}{2 \,{\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} - \frac{a^{2}}{2 \,{\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^2/(d*e*x+c*e)^3,x, algorithm="maxima")

[Out]

-(d*(1/(d^3*e^3*x + c*d^2*e^3) + arctan((d^2*x + c*d)/d)/(d^2*e^3)) + arctan(d*x + c)/(d^3*e^3*x^2 + 2*c*d^2*e
^3*x + c^2*d*e^3))*a*b - 1/2*(2*d*(1/(d^3*e^3*x + c*d^2*e^3) + arctan((d^2*x + c*d)/d)/(d^2*e^3))*arctan(d*x +
c) - (arctan(d*x + c)^2 - log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 2*log(d*x + c))/(d*e^3))*b^2 - 1/2*b^2*arctan(d*
x + c)^2/(d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3) - 1/2*a^2/(d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3)

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Fricas [A]  time = 1.87855, size = 470, normalized size = 4.02 \begin{align*} -\frac{2 \, a b d x + 2 \, a b c +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + b^{2}\right )} \arctan \left (d x + c\right )^{2} + a^{2} + 2 \,{\left (a b d^{2} x^{2} + a b c^{2} + b^{2} c +{\left (2 \, a b c + b^{2}\right )} d x + a b\right )} \arctan \left (d x + c\right ) +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) - 2 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (d x + c\right )}{2 \,{\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^2/(d*e*x+c*e)^3,x, algorithm="fricas")

[Out]

-1/2*(2*a*b*d*x + 2*a*b*c + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 + b^2)*arctan(d*x + c)^2 + a^2 + 2*(a*b*d^2*x
^2 + a*b*c^2 + b^2*c + (2*a*b*c + b^2)*d*x + a*b)*arctan(d*x + c) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(
d^2*x^2 + 2*c*d*x + c^2 + 1) - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(d*x + c))/(d^3*e^3*x^2 + 2*c*d^2*e^
3*x + c^2*d*e^3)

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Sympy [A]  time = 6.51563, size = 986, normalized size = 8.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(d*x+c))**2/(d*e*x+c*e)**3,x)

[Out]

Piecewise((-a**2/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*a*b*c**2*atan(c + d*x)/(2*c**2*d*e**
3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 4*a*b*c*d*x*atan(c + d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*
e**3*x**2) - 2*a*b*c/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*a*b*d**2*x**2*atan(c + d*x)/(2*c
**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*a*b*d*x/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x*
*2) - 2*a*b*atan(c + d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + 2*b**2*c**2*log(c/d + x)/(2*c
**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - b**2*c**2*log(c**2/d**2 + 2*c*x/d + x**2 + d**(-2))/(2*c**2
*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - b**2*c**2*atan(c + d*x)**2/(2*c**2*d*e**3 + 4*c*d**2*e**3*x +
2*d**3*e**3*x**2) + 4*b**2*c*d*x*log(c/d + x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*b**2*c*
d*x*log(c**2/d**2 + 2*c*x/d + x**2 + d**(-2))/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*b**2*c*
d*x*atan(c + d*x)**2/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*b**2*c*atan(c + d*x)/(2*c**2*d*e
**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + 2*b**2*d**2*x**2*log(c/d + x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2
*d**3*e**3*x**2) - b**2*d**2*x**2*log(c**2/d**2 + 2*c*x/d + x**2 + d**(-2))/(2*c**2*d*e**3 + 4*c*d**2*e**3*x +
2*d**3*e**3*x**2) - b**2*d**2*x**2*atan(c + d*x)**2/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*
b**2*d*x*atan(c + d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - b**2*atan(c + d*x)**2/(2*c**2*d*
e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2), Ne(d, 0)), (x*(a + b*atan(c))**2/(c**3*e**3), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^2/(d*e*x+c*e)^3,x, algorithm="giac")

[Out]

integrate((b*arctan(d*x + c) + a)^2/(d*e*x + c*e)^3, x)