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

Optimal. Leaf size=63 \[ -\frac {a+b \tan ^{-1}(c+d x)}{2 d e^3 (c+d x)^2}-\frac {b}{2 d e^3 (c+d x)}-\frac {b \tan ^{-1}(c+d x)}{2 d e^3} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5043, 12, 4852, 325, 203} \[ -\frac {a+b \tan ^{-1}(c+d x)}{2 d e^3 (c+d x)^2}-\frac {b}{2 d e^3 (c+d x)}-\frac {b \tan ^{-1}(c+d x)}{2 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 51, normalized size = 0.81 \[ -\frac {a+b (c+d x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-(c+d x)^2\right )+b \tan ^{-1}(c+d x)}{2 d e^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a + b*ArcTan[c + d*x] + b*(c + d*x)*Hypergeometric2F1[-1/2, 1, 1/2, -(c + d*x)^2])/(d*e^3*(c + d*x)^2)

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fricas [A]  time = 0.57, size = 70, normalized size = 1.11 \[ -\frac {b d x + b c + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + b\right )} \arctan \left (d x + c\right ) + a}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.04, size = 71, normalized size = 1.13 \[ -\frac {a}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {b \arctan \left (d x +c \right )}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {b}{2 d \,e^{3} \left (d x +c \right )}-\frac {b \arctan \left (d x +c \right )}{2 d \,e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 120, normalized size = 1.90 \[ -\frac {1}{2} \, {\left (d {\left (\frac {1}{d^{3} e^{3} x + c d^{2} e^{3}} + \frac {\arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{2} e^{3}}\right )} + \frac {\arctan \left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} b - \frac {a}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.75, size = 103, normalized size = 1.63 \[ -\frac {\frac {a+b\,c}{d}+b\,x}{2\,c^2\,e^3+4\,c\,d\,e^3\,x+2\,d^2\,e^3\,x^2}-\frac {b\,\mathrm {atan}\left (\frac {b\,c+b\,d\,x}{b}\right )}{2\,d\,e^3}-\frac {b\,\mathrm {atan}\left (c+d\,x\right )}{2\,d^3\,e^3\,\left (x^2+\frac {c^2}{d^2}+\frac {2\,c\,x}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((a + b*c)/d + b*x)/(2*c^2*e^3 + 2*d^2*e^3*x^2 + 4*c*d*e^3*x) - (b*atan((b*c + b*d*x)/b))/(2*d*e^3) - (b*ata
n(c + d*x))/(2*d^3*e^3*(x^2 + c^2/d^2 + (2*c*x)/d))

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sympy [A]  time = 12.39, size = 314, normalized size = 4.98 \[ \begin {cases} - \frac {a}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b c^{2} \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b c d x \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b c}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b d^{2} x^{2} \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b d x}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} & \text {for}\: d \neq 0 \\\frac {x \left (a + b \operatorname {atan}{\relax (c )}\right )}{c^{3} e^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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