\(\int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx\) [6]

Optimal result

Integrand size = 21, antiderivative size = 63 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {b}{2 d e^3 (c+d x)}-\frac {b \arctan (c+d x)}{2 d e^3}-\frac {a+b \arctan (c+d x)}{2 d e^3 (c+d x)^2} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5151, 12, 4946, 331, 209} \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {a+b \arctan (c+d x)}{2 d e^3 (c+d x)^2}-\frac {b \arctan (c+d x)}{2 d e^3}-\frac {b}{2 d e^3 (c+d x)} \]

[In]

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

[Out]

-1/2*b/(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 209

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

Rule 331

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 4946

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

Rule 5151

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {a+b \arctan (c+d x)+b (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-(c+d x)^2\right )}{2 d e^3 (c+d x)^2} \]

[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)

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {-\frac {a}{2 e^{3} \left (d x +c \right )^{2}}+\frac {b \left (-\frac {\arctan \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {1}{2 \left (d x +c \right )}-\frac {\arctan \left (d x +c \right )}{2}\right )}{e^{3}}}{d}\)	\(57\)
default	\(\frac {-\frac {a}{2 e^{3} \left (d x +c \right )^{2}}+\frac {b \left (-\frac {\arctan \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {1}{2 \left (d x +c \right )}-\frac {\arctan \left (d x +c \right )}{2}\right )}{e^{3}}}{d}\)	\(57\)
parts	\(-\frac {a}{2 e^{3} \left (d x +c \right )^{2} d}+\frac {b \left (-\frac {\arctan \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {1}{2 \left (d x +c \right )}-\frac {\arctan \left (d x +c \right )}{2}\right )}{e^{3} d}\)	\(59\)
parallelrisch	\(\frac {-4 b \,d^{4} \arctan \left (d x +c \right ) x^{2} c -8 b \,c^{2} \arctan \left (d x +c \right ) x \,d^{3}+b \,d^{4} x^{2}-4 \arctan \left (d x +c \right ) b \,c^{3} d^{2}-2 x b c \,d^{3}-4 b \arctan \left (d x +c \right ) c \,d^{2}-3 b \,c^{2} d^{2}-4 a c \,d^{2}}{8 \left (d x +c \right )^{2} e^{3} c \,d^{3}}\)	\(112\)
risch	\(\frac {i b \ln \left (1+i \left (d x +c \right )\right )}{4 d \,e^{3} \left (d x +c \right )^{2}}-\frac {-i \ln \left (-d x -c +i\right ) b \,d^{2} x^{2}+i \ln \left (-d x -c -i\right ) b \,d^{2} x^{2}-2 i \ln \left (-d x -c +i\right ) b c d x +2 i \ln \left (-d x -c -i\right ) b c d x -i \ln \left (-d x -c +i\right ) b \,c^{2}+i \ln \left (-d x -c -i\right ) b \,c^{2}+i b \ln \left (1-i \left (d x +c \right )\right )+2 b d x +2 b c +2 a}{4 e^{3} \left (d x +c \right )^{2} d}\)	\(187\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.11 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {b d x + b c + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + b\right )} \arctan \left (d x + c\right ) + a}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (54) = 108\).

Time = 15.21 (sec) , antiderivative size = 314, normalized size of antiderivative = 4.98 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=\begin {cases} - \frac {a}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b c^{2} \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b c d x \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b c}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b d^{2} x^{2} \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b d x}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} & \text {for}\: d \neq 0 \\\frac {x \left (a + b \operatorname {atan}{\left (c \right )}\right )}{c^{3} e^{3}} & \text {otherwise} \end {cases} \]

[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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (57) = 114\).

Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.90 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {1}{2} \, {\left (d {\left (\frac {1}{d^{3} e^{3} x + c d^{2} e^{3}} + \frac {\arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{2} e^{3}}\right )} + \frac {\arctan \left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} b - \frac {a}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]

[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)

Giac [F]

\[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=\int { \frac {b \arctan \left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{3}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.63 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {\frac {a+b\,c}{d}+b\,x}{2\,c^2\,e^3+4\,c\,d\,e^3\,x+2\,d^2\,e^3\,x^2}-\frac {b\,\mathrm {atan}\left (\frac {b\,c+b\,d\,x}{b}\right )}{2\,d\,e^3}-\frac {b\,\mathrm {atan}\left (c+d\,x\right )}{2\,d^3\,e^3\,\left (x^2+\frac {c^2}{d^2}+\frac {2\,c\,x}{d}\right )} \]

[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))