3.5 \(\int \frac{a+b \tan ^{-1}(c+d x)}{(c e+d e x)^2} \, dx\)
Optimal. Leaf size=61 \[ -\frac{a+b \tan ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \log (c+d x)}{d e^2}-\frac{b \log \left ((c+d x)^2+1\right )}{2 d e^2} \]
[Out]
-((a + b*ArcTan[c + d*x])/(d*e^2*(c + d*x))) + (b*Log[c + d*x])/(d*e^2) - (b*Log[1 + (c + d*x)^2])/(2*d*e^2)
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Rubi [A] time = 0.0472608, antiderivative size = 61, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used =
{5043, 12, 4852, 266, 36, 29, 31} \[ -\frac{a+b \tan ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \log (c+d x)}{d e^2}-\frac{b \log \left ((c+d x)^2+1\right )}{2 d e^2} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTan[c + d*x])/(c*e + d*e*x)^2,x]
[Out]
-((a + b*ArcTan[c + d*x])/(d*e^2*(c + d*x))) + (b*Log[c + d*x])/(d*e^2) - (b*Log[1 + (c + d*x)^2])/(2*d*e^2)
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 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]
Rubi steps
\begin{align*} \int \frac{a+b \tan ^{-1}(c+d x)}{(c e+d e x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{a+b \tan ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \operatorname{Subst}\left (\int \frac{1}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{a+b \tan ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \operatorname{Subst}\left (\int \frac{1}{x (1+x)} \, dx,x,(c+d x)^2\right )}{2 d e^2}\\ &=-\frac{a+b \tan ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,(c+d x)^2\right )}{2 d e^2}-\frac{b \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,(c+d x)^2\right )}{2 d e^2}\\ &=-\frac{a+b \tan ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \log (c+d x)}{d e^2}-\frac{b \log \left (1+(c+d x)^2\right )}{2 d e^2}\\ \end{align*}
Mathematica [A] time = 0.0181161, size = 50, normalized size = 0.82 \[ \frac{-\frac{a+b \tan ^{-1}(c+d x)}{c+d x}+b \log (c+d x)-\frac{1}{2} b \log \left ((c+d x)^2+1\right )}{d e^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcTan[c + d*x])/(c*e + d*e*x)^2,x]
[Out]
(-((a + b*ArcTan[c + d*x])/(c + d*x)) + b*Log[c + d*x] - (b*Log[1 + (c + d*x)^2])/2)/(d*e^2)
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Maple [A] time = 0.041, size = 73, normalized size = 1.2 \begin{align*} -{\frac{a}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{b\arctan \left ( dx+c \right ) }{d{e}^{2} \left ( dx+c \right ) }}-{\frac{b\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{2\,d{e}^{2}}}+{\frac{b\ln \left ( dx+c \right ) }{d{e}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctan(d*x+c))/(d*e*x+c*e)^2,x)
[Out]
-1/d*a/e^2/(d*x+c)-1/d*b/e^2/(d*x+c)*arctan(d*x+c)-1/2*b*ln(1+(d*x+c)^2)/d/e^2+b*ln(d*x+c)/d/e^2
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Maxima [A] time = 0.988298, size = 124, normalized size = 2.03 \begin{align*} -\frac{1}{2} \,{\left (d{\left (\frac{\log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{2} e^{2}} - \frac{2 \, \log \left (d x + c\right )}{d^{2} e^{2}}\right )} + \frac{2 \, \arctan \left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}}\right )} b - \frac{a}{d^{2} e^{2} x + c d e^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(d*x+c))/(d*e*x+c*e)^2,x, algorithm="maxima")
[Out]
-1/2*(d*(log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*e^2) - 2*log(d*x + c)/(d^2*e^2)) + 2*arctan(d*x + c)/(d^2*e^2*x
+ c*d*e^2))*b - a/(d^2*e^2*x + c*d*e^2)
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Fricas [A] time = 1.67866, size = 184, normalized size = 3.02 \begin{align*} -\frac{2 \, b \arctan \left (d x + c\right ) +{\left (b d x + b c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) - 2 \,{\left (b d x + b c\right )} \log \left (d x + c\right ) + 2 \, a}{2 \,{\left (d^{2} e^{2} x + c d e^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(d*x+c))/(d*e*x+c*e)^2,x, algorithm="fricas")
[Out]
-1/2*(2*b*arctan(d*x + c) + (b*d*x + b*c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - 2*(b*d*x + b*c)*log(d*x + c) + 2*
a)/(d^2*e^2*x + c*d*e^2)
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Sympy [A] time = 125.41, size = 2814, normalized size = 46.13 \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))/(d*e*x+c*e)**2,x)
[Out]
Piecewise((x*(a + b*atan(c))/(c**2*e**2), Eq(d, 0)), (-9*a*d**2*x**2/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2
*x**2 - 2*sqrt(3)*I*d*e**2) + 6*sqrt(3)*I*a*d*x/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)*I*d
*e**2) - 6*a/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)*I*d*e**2) + 9*b*d**3*x**3*log(x - sqrt
(3)*I/(3*d))/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)*I*d*e**2) - 9*b*d**3*x**3*log(x - I/d
- sqrt(3)*I/(3*d))/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)*I*d*e**2) + 9*I*b*d**3*x**3*atan
(d*x - sqrt(3)*I/3)/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)*I*d*e**2) - 9*sqrt(3)*I*b*d**2*
x**2*log(x - sqrt(3)*I/(3*d))/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)*I*d*e**2) + 9*sqrt(3)
*I*b*d**2*x**2*log(x - I/d - sqrt(3)*I/(3*d))/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)*I*d*e
**2) - 9*b*d**2*x**2*atan(d*x - sqrt(3)*I/3)/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)*I*d*e*
*2) + 9*sqrt(3)*b*d**2*x**2*atan(d*x - sqrt(3)*I/3)/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)
*I*d*e**2) + 6*sqrt(3)*I*b*d*x*atan(d*x - sqrt(3)*I/3)/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt
(3)*I*d*e**2) - 2*sqrt(3)*I*b*log(x - sqrt(3)*I/(3*d))/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt
(3)*I*d*e**2) + 2*sqrt(3)*I*b*log(x - I/d - sqrt(3)*I/(3*d))/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 -
2*sqrt(3)*I*d*e**2) - 6*b*atan(d*x - sqrt(3)*I/3)/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)*I
*d*e**2) + 2*sqrt(3)*b*atan(d*x - sqrt(3)*I/3)/(9*d**4*e**2*x**3 - 9*sqrt(3)*I*d**3*e**2*x**2 - 2*sqrt(3)*I*d*
e**2), Eq(c, -sqrt(3)*I/3)), (-3*sqrt(3)*I*a*d**3*x**3/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt
(3)*I*d*e**2) - 6*sqrt(3)*I*a*d*x/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2) - 4*a/(
9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2) + 9*b*d**3*x**3*log(x + sqrt(3)*I/(3*d))/(
9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2) - 9*b*d**3*x**3*log(x - I/d + sqrt(3)*I/(3
*d))/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2) + 9*I*b*d**3*x**3*atan(d*x + sqrt(3)
*I/3)/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2) + 9*sqrt(3)*I*b*d**2*x**2*log(x + s
qrt(3)*I/(3*d))/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2) - 9*sqrt(3)*I*b*d**2*x**2
*log(x - I/d + sqrt(3)*I/(3*d))/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2) - 9*sqrt(
3)*b*d**2*x**2*atan(d*x + sqrt(3)*I/3)/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2) -
9*b*d**2*x**2*atan(d*x + sqrt(3)*I/3)/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2) - 6
*sqrt(3)*I*b*d*x*atan(d*x + sqrt(3)*I/3)/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2)
+ 2*sqrt(3)*I*b*log(x + sqrt(3)*I/(3*d))/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2)
- 2*sqrt(3)*I*b*log(x - I/d + sqrt(3)*I/(3*d))/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*
e**2) - 6*b*atan(d*x + sqrt(3)*I/3)/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2) - 2*s
qrt(3)*b*atan(d*x + sqrt(3)*I/3)/(9*d**4*e**2*x**3 + 9*sqrt(3)*I*d**3*e**2*x**2 + 2*sqrt(3)*I*d*e**2), Eq(c, s
qrt(3)*I/3)), (zoo*a*x, Eq(c, -d*x)), (-2*a*c**2/(6*c**3*d*e**2 + 6*c**2*d**2*e**2*x + 2*c*d*e**2 + 2*d**2*e**
2*x) + 4*a*c*d*x/(6*c**3*d*e**2 + 6*c**2*d**2*e**2*x + 2*c*d*e**2 + 2*d**2*e**2*x) - 2*a/(6*c**3*d*e**2 + 6*c*
*2*d**2*e**2*x + 2*c*d*e**2 + 2*d**2*e**2*x) + 6*b*c**3*log(c/d + x)/(6*c**3*d*e**2 + 6*c**2*d**2*e**2*x + 2*c
*d*e**2 + 2*d**2*e**2*x) - 3*b*c**3*log(c**2/d**2 + 2*c*x/d + x**2 + d**(-2))/(6*c**3*d*e**2 + 6*c**2*d**2*e**
2*x + 2*c*d*e**2 + 2*d**2*e**2*x) + 6*b*c**2*d*x*log(c/d + x)/(6*c**3*d*e**2 + 6*c**2*d**2*e**2*x + 2*c*d*e**2
+ 2*d**2*e**2*x) - 3*b*c**2*d*x*log(c**2/d**2 + 2*c*x/d + x**2 + d**(-2))/(6*c**3*d*e**2 + 6*c**2*d**2*e**2*x
+ 2*c*d*e**2 + 2*d**2*e**2*x) - 6*b*c**2*atan(c + d*x)/(6*c**3*d*e**2 + 6*c**2*d**2*e**2*x + 2*c*d*e**2 + 2*d
**2*e**2*x) + 2*b*c*log(c/d + x)/(6*c**3*d*e**2 + 6*c**2*d**2*e**2*x + 2*c*d*e**2 + 2*d**2*e**2*x) - b*c*log(c
**2/d**2 + 2*c*x/d + x**2 + d**(-2))/(6*c**3*d*e**2 + 6*c**2*d**2*e**2*x + 2*c*d*e**2 + 2*d**2*e**2*x) + 2*b*d
*x*log(c/d + x)/(6*c**3*d*e**2 + 6*c**2*d**2*e**2*x + 2*c*d*e**2 + 2*d**2*e**2*x) - b*d*x*log(c**2/d**2 + 2*c*
x/d + x**2 + d**(-2))/(6*c**3*d*e**2 + 6*c**2*d**2*e**2*x + 2*c*d*e**2 + 2*d**2*e**2*x) - 2*b*atan(c + d*x)/(6
*c**3*d*e**2 + 6*c**2*d**2*e**2*x + 2*c*d*e**2 + 2*d**2*e**2*x), True))
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Giac [A] time = 1.08692, size = 100, normalized size = 1.64 \begin{align*} -\frac{1}{2} \, b{\left (\frac{e^{\left (-2\right )} \log \left (\frac{e^{2}}{{\left (d x e + c e\right )}^{2}} + 1\right )}{d} + \frac{2 \, \arctan \left (d x + c\right ) e^{\left (-1\right )}}{{\left (d x e + c e\right )} d}\right )} - \frac{a e^{\left (-1\right )}}{{\left (d x e + c e\right )} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(d*x+c))/(d*e*x+c*e)^2,x, algorithm="giac")
[Out]
-1/2*b*(e^(-2)*log(e^2/(d*x*e + c*e)^2 + 1)/d + 2*arctan(d*x + c)*e^(-1)/((d*x*e + c*e)*d)) - a*e^(-1)/((d*x*e
+ c*e)*d)