Integrand size = 21, antiderivative size = 63 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {b}{2 d e^3 (c+d x)}-\frac {b \arctan (c+d x)}{2 d e^3}-\frac {a+b \arctan (c+d x)}{2 d e^3 (c+d x)^2} \] Output:
-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
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 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {a+b \arctan (c+d x)+b (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-(c+d x)^2\right )}{2 d e^3 (c+d x)^2} \] Input:
Integrate[(a + b*ArcTan[c + d*x])/(c*e + d*e*x)^3,x]
Output:
-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)
Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5566, 27, 5361, 264, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx\) |
\(\Big \downarrow \) 5566 |
\(\displaystyle \frac {\int \frac {a+b \arctan (c+d x)}{e^3 (c+d x)^3}d(c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a+b \arctan (c+d x)}{(c+d x)^3}d(c+d x)}{d e^3}\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {\frac {1}{2} b \int \frac {1}{(c+d x)^2 \left ((c+d x)^2+1\right )}d(c+d x)-\frac {a+b \arctan (c+d x)}{2 (c+d x)^2}}{d e^3}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {\frac {1}{2} b \left (-\int \frac {1}{(c+d x)^2+1}d(c+d x)-\frac {1}{c+d x}\right )-\frac {a+b \arctan (c+d x)}{2 (c+d x)^2}}{d e^3}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {1}{2} b \left (-\arctan (c+d x)-\frac {1}{c+d x}\right )-\frac {a+b \arctan (c+d x)}{2 (c+d x)^2}}{d e^3}\) |
Input:
Int[(a + b*ArcTan[c + d*x])/(c*e + d*e*x)^3,x]
Output:
((b*(-(c + d*x)^(-1) - ArcTan[c + d*x]))/2 - (a + b*ArcTan[c + d*x])/(2*(c + d*x)^2))/(d*e^3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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] - Simp[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]
Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[(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]
Time = 0.40 (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 | \(\frac {-4 b \,d^{4} \arctan \left (d x +c \right ) x^{2} c -8 b \,c^{2} \arctan \left (d x +c \right ) x \,d^{3}+b \,d^{4} x^{2}-4 \arctan \left (d x +c \right ) b \,c^{3} d^{2}-2 x b c \,d^{3}-4 b \arctan \left (d x +c \right ) c \,d^{2}-3 b \,c^{2} d^{2}-4 a c \,d^{2}}{8 \left (d x +c \right )^{2} e^{3} c \,d^{3}}\) | \(112\) |
orering | \(-\frac {2 \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}+d x +c \right ) \left (a +b \arctan \left (d x +c \right )\right )}{d \left (d e x +c e \right )^{3}}-\frac {\left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) \left (d x +c \right )^{2} \left (\frac {b d}{\left (1+\left (d x +c \right )^{2}\right ) \left (d e x +c e \right )^{3}}-\frac {3 \left (a +b \arctan \left (d x +c \right )\right ) d e}{\left (d e x +c e \right )^{4}}\right )}{2 d^{2}}\) | \(136\) |
risch | \(\frac {i b \ln \left (1+i \left (d x +c \right )\right )}{4 d \,e^{3} \left (d x +c \right )^{2}}-\frac {i \ln \left (-d x -c -i\right ) b \,d^{2} x^{2}-i \ln \left (-d x -c +i\right ) b \,d^{2} x^{2}+2 i \ln \left (-d x -c -i\right ) b c d x -2 i \ln \left (-d x -c +i\right ) b c d x +i \ln \left (-d x -c -i\right ) b \,c^{2}-i \ln \left (-d x -c +i\right ) b \,c^{2}+i b \ln \left (1-i \left (d x +c \right )\right )+2 b d x +2 b c +2 a}{4 e^{3} \left (d x +c \right )^{2} d}\) | \(187\) |
Input:
int((a+b*arctan(d*x+c))/(d*e*x+c*e)^3,x,method=_RETURNVERBOSE)
Output:
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)))
Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.11 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {b d x + b c + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + b\right )} \arctan \left (d x + c\right ) + a}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \] Input:
integrate((a+b*arctan(d*x+c))/(d*e*x+c*e)^3,x, algorithm="fricas")
Output:
-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)
Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (54) = 108\).
Time = 3.75 (sec) , antiderivative size = 314, normalized size of antiderivative = 4.98 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=\begin {cases} - \frac {a}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b c^{2} \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b c d x \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b c}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b d^{2} x^{2} \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b d x}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b \operatorname {atan}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} & \text {for}\: d \neq 0 \\\frac {x \left (a + b \operatorname {atan}{\left (c \right )}\right )}{c^{3} e^{3}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*atan(d*x+c))/(d*e*x+c*e)**3,x)
Output:
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*a tan(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*at an(c))/(c**3*e**3), True))
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (57) = 114\).
Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.90 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {1}{2} \, {\left (d {\left (\frac {1}{d^{3} e^{3} x + c d^{2} e^{3}} + \frac {\arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{2} e^{3}}\right )} + \frac {\arctan \left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} b - \frac {a}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \] Input:
integrate((a+b*arctan(d*x+c))/(d*e*x+c*e)^3,x, algorithm="maxima")
Output:
-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)
Result contains complex when optimal does not.
Time = 0.15 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.48 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=\frac {b d^{2} x^{2} \log \left (i \, d x + i \, c + 1\right ) - b d^{2} x^{2} \log \left (-i \, d x - i \, c + 1\right ) + 2 \, b c d x \log \left (i \, d x + i \, c + 1\right ) - 2 \, b c d x \log \left (-i \, d x - i \, c + 1\right ) + b c^{2} \log \left (i \, d x + i \, c + 1\right ) - b c^{2} \log \left (-i \, d x - i \, c + 1\right ) + 2 i \, b d x + 2 i \, b c + 2 i \, b \arctan \left (d x + c\right ) + 2 i \, a}{-4 i \, d^{3} e^{3} x^{2} - 8 i \, c d^{2} e^{3} x - 4 i \, c^{2} d e^{3}} \] Input:
integrate((a+b*arctan(d*x+c))/(d*e*x+c*e)^3,x, algorithm="giac")
Output:
(b*d^2*x^2*log(I*d*x + I*c + 1) - b*d^2*x^2*log(-I*d*x - I*c + 1) + 2*b*c* d*x*log(I*d*x + I*c + 1) - 2*b*c*d*x*log(-I*d*x - I*c + 1) + b*c^2*log(I*d *x + I*c + 1) - b*c^2*log(-I*d*x - I*c + 1) + 2*I*b*d*x + 2*I*b*c + 2*I*b* arctan(d*x + c) + 2*I*a)/(-4*I*d^3*e^3*x^2 - 8*I*c*d^2*e^3*x - 4*I*c^2*d*e ^3)
Time = 1.16 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.63 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=-\frac {\frac {a+b\,c}{d}+b\,x}{2\,c^2\,e^3+4\,c\,d\,e^3\,x+2\,d^2\,e^3\,x^2}-\frac {b\,\mathrm {atan}\left (\frac {b\,c+b\,d\,x}{b}\right )}{2\,d\,e^3}-\frac {b\,\mathrm {atan}\left (c+d\,x\right )}{2\,d^3\,e^3\,\left (x^2+\frac {c^2}{d^2}+\frac {2\,c\,x}{d}\right )} \] Input:
int((a + b*atan(c + d*x))/(c*e + d*e*x)^3,x)
Output:
- ((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*atan(c + d*x))/(2*d^3*e^3*(x^2 + c^2/d^2 + (2*c*x)/d))
Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.59 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^3} \, dx=\frac {-2 \mathit {atan} \left (d x +c \right ) b \,c^{3}-4 \mathit {atan} \left (d x +c \right ) b \,c^{2} d x -2 \mathit {atan} \left (d x +c \right ) b c \,d^{2} x^{2}-2 \mathit {atan} \left (d x +c \right ) b c -2 a c -b \,c^{2}+b \,d^{2} x^{2}}{4 c d \,e^{3} \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:
int((a+b*atan(d*x+c))/(d*e*x+c*e)^3,x)
Output:
( - 2*atan(c + d*x)*b*c**3 - 4*atan(c + d*x)*b*c**2*d*x - 2*atan(c + d*x)* b*c*d**2*x**2 - 2*atan(c + d*x)*b*c - 2*a*c - b*c**2 + b*d**2*x**2)/(4*c*d *e**3*(c**2 + 2*c*d*x + d**2*x**2))