Integrand size = 21, antiderivative size = 67 \[ \int (c e+d e x)^2 (a+b \arctan (c+d x)) \, dx=-\frac {b e^2 (c+d x)^2}{6 d}+\frac {e^2 (c+d x)^3 (a+b \arctan (c+d x))}{3 d}+\frac {b e^2 \log \left (1+(c+d x)^2\right )}{6 d} \] Output:
-1/6*b*e^2*(d*x+c)^2/d+1/3*e^2*(d*x+c)^3*(a+b*arctan(d*x+c))/d+1/6*b*e^2*l n(1+(d*x+c)^2)/d
Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.81 \[ \int (c e+d e x)^2 (a+b \arctan (c+d x)) \, dx=\frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))-\frac {1}{6} b \left ((c+d x)^2-\log \left (1+(c+d x)^2\right )\right )\right )}{d} \] Input:
Integrate[(c*e + d*e*x)^2*(a + b*ArcTan[c + d*x]),x]
Output:
(e^2*(((c + d*x)^3*(a + b*ArcTan[c + d*x]))/3 - (b*((c + d*x)^2 - Log[1 + (c + d*x)^2]))/6))/d
Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5566, 27, 5361, 243, 49, 2009}
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.
\(\displaystyle \int (c e+d e x)^2 (a+b \arctan (c+d x)) \, dx\) |
\(\Big \downarrow \) 5566 |
\(\displaystyle \frac {\int e^2 (c+d x)^2 (a+b \arctan (c+d x))d(c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^2 \int (c+d x)^2 (a+b \arctan (c+d x))d(c+d x)}{d}\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))-\frac {1}{3} b \int \frac {(c+d x)^3}{(c+d x)^2+1}d(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))-\frac {1}{6} b \int \frac {(c+d x)^2}{(c+d x)^2+1}d(c+d x)^2\right )}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))-\frac {1}{6} b \int \left (1+\frac {1}{-c-d x-1}\right )d(c+d x)^2\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \arctan (c+d x))-\frac {1}{6} b \left ((c+d x)^2-\log (c+d x+1)\right )\right )}{d}\) |
Input:
Int[(c*e + d*e*x)^2*(a + b*ArcTan[c + d*x]),x]
Output:
(e^2*(((c + d*x)^3*(a + b*ArcTan[c + d*x]))/3 - (b*((c + d*x)^2 - Log[1 + c + d*x]))/6))/d
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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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 = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {\frac {e^{2} a \left (d x +c \right )^{3}}{3}+e^{2} b \left (\frac {\left (d x +c \right )^{3} \arctan \left (d x +c \right )}{3}-\frac {\left (d x +c \right )^{2}}{6}+\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{6}\right )}{d}\) | \(61\) |
default | \(\frac {\frac {e^{2} a \left (d x +c \right )^{3}}{3}+e^{2} b \left (\frac {\left (d x +c \right )^{3} \arctan \left (d x +c \right )}{3}-\frac {\left (d x +c \right )^{2}}{6}+\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{6}\right )}{d}\) | \(61\) |
parts | \(\frac {e^{2} a \left (d x +c \right )^{3}}{3 d}+\frac {e^{2} b \left (\frac {\left (d x +c \right )^{3} \arctan \left (d x +c \right )}{3}-\frac {\left (d x +c \right )^{2}}{6}+\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{6}\right )}{d}\) | \(63\) |
parallelrisch | \(\frac {2 d^{4} e^{2} b \,x^{3} \arctan \left (d x +c \right )+2 x^{3} a \,d^{4} e^{2}+6 d^{3} e^{2} c b \,x^{2} \arctan \left (d x +c \right )+6 x^{2} a c \,d^{3} e^{2}+6 b \,c^{2} e^{2} x \arctan \left (d x +c \right ) d^{2}-x^{2} b \,d^{3} e^{2}+6 x a \,c^{2} d^{2} e^{2}+2 c^{3} b \,e^{2} \arctan \left (d x +c \right ) d -2 x b c \,d^{2} e^{2}-18 a \,c^{3} d \,e^{2}+5 b \,c^{2} d \,e^{2}+e^{2} b \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) d -6 d \,e^{2} c a +b d \,e^{2}}{6 d^{2}}\) | \(198\) |
risch | \(-\frac {i e^{2} \left (d x +c \right )^{3} b \ln \left (1+i \left (d x +c \right )\right )}{6 d}+\frac {i e^{2} d^{2} b \,x^{3} \ln \left (1-i \left (d x +c \right )\right )}{6}+\frac {i e^{2} d b c \,x^{2} \ln \left (1-i \left (d x +c \right )\right )}{2}+\frac {i e^{2} b \,c^{2} x \ln \left (1-i \left (d x +c \right )\right )}{2}+\frac {e^{2} d^{2} a \,x^{3}}{3}+\frac {i e^{2} b \,c^{3} \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{12 d}+\frac {e^{2} b \,c^{3} \arctan \left (d x +c \right )}{6 d}+e^{2} d a c \,x^{2}+e^{2} a \,c^{2} x -\frac {e^{2} d b \,x^{2}}{6}-\frac {e^{2} b c x}{3}+\frac {e^{2} b \ln \left (-d^{2} x^{2}-2 c d x -c^{2}-1\right )}{6 d}\) | \(227\) |
Input:
int((d*e*x+c*e)^2*(a+b*arctan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/3*e^2*a*(d*x+c)^3+e^2*b*(1/3*(d*x+c)^3*arctan(d*x+c)-1/6*(d*x+c)^2+ 1/6*ln(1+(d*x+c)^2)))
Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (61) = 122\).
Time = 0.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.93 \[ \int (c e+d e x)^2 (a+b \arctan (c+d x)) \, dx=\frac {2 \, a d^{3} e^{2} x^{3} + {\left (6 \, a c - b\right )} d^{2} e^{2} x^{2} + 2 \, {\left (3 \, a c^{2} - b c\right )} d e^{2} x + b e^{2} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + 2 \, {\left (b d^{3} e^{2} x^{3} + 3 \, b c d^{2} e^{2} x^{2} + 3 \, b c^{2} d e^{2} x + b c^{3} e^{2}\right )} \arctan \left (d x + c\right )}{6 \, d} \] Input:
integrate((d*e*x+c*e)^2*(a+b*arctan(d*x+c)),x, algorithm="fricas")
Output:
1/6*(2*a*d^3*e^2*x^3 + (6*a*c - b)*d^2*e^2*x^2 + 2*(3*a*c^2 - b*c)*d*e^2*x + b*e^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 2*(b*d^3*e^2*x^3 + 3*b*c*d^2*e ^2*x^2 + 3*b*c^2*d*e^2*x + b*c^3*e^2)*arctan(d*x + c))/d
Result contains complex when optimal does not.
Time = 1.17 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.72 \[ \int (c e+d e x)^2 (a+b \arctan (c+d x)) \, dx=\begin {cases} a c^{2} e^{2} x + a c d e^{2} x^{2} + \frac {a d^{2} e^{2} x^{3}}{3} + \frac {b c^{3} e^{2} \operatorname {atan}{\left (c + d x \right )}}{3 d} + b c^{2} e^{2} x \operatorname {atan}{\left (c + d x \right )} + b c d e^{2} x^{2} \operatorname {atan}{\left (c + d x \right )} - \frac {b c e^{2} x}{3} + \frac {b d^{2} e^{2} x^{3} \operatorname {atan}{\left (c + d x \right )}}{3} - \frac {b d e^{2} x^{2}}{6} + \frac {b e^{2} \log {\left (\frac {c}{d} + x - \frac {i}{d} \right )}}{3 d} - \frac {i b e^{2} \operatorname {atan}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\c^{2} e^{2} x \left (a + b \operatorname {atan}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((d*e*x+c*e)**2*(a+b*atan(d*x+c)),x)
Output:
Piecewise((a*c**2*e**2*x + a*c*d*e**2*x**2 + a*d**2*e**2*x**3/3 + b*c**3*e **2*atan(c + d*x)/(3*d) + b*c**2*e**2*x*atan(c + d*x) + b*c*d*e**2*x**2*at an(c + d*x) - b*c*e**2*x/3 + b*d**2*e**2*x**3*atan(c + d*x)/3 - b*d*e**2*x **2/6 + b*e**2*log(c/d + x - I/d)/(3*d) - I*b*e**2*atan(c + d*x)/(3*d), Ne (d, 0)), (c**2*e**2*x*(a + b*atan(c)), True))
Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (61) = 122\).
Time = 0.12 (sec) , antiderivative size = 238, normalized size of antiderivative = 3.55 \[ \int (c e+d e x)^2 (a+b \arctan (c+d x)) \, dx=\frac {1}{3} \, a d^{2} e^{2} x^{3} + a c d e^{2} x^{2} + {\left (x^{2} \arctan \left (d x + c\right ) - d {\left (\frac {x}{d^{2}} + \frac {{\left (c^{2} - 1\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{3}} - \frac {c \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{3}}\right )}\right )} b c d e^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (d x + c\right ) - d {\left (\frac {d x^{2} - 4 \, c x}{d^{3}} - \frac {2 \, {\left (c^{3} - 3 \, c\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{4}} + \frac {{\left (3 \, c^{2} - 1\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{4}}\right )}\right )} b d^{2} e^{2} + a c^{2} e^{2} x + \frac {{\left (2 \, {\left (d x + c\right )} \arctan \left (d x + c\right ) - \log \left ({\left (d x + c\right )}^{2} + 1\right )\right )} b c^{2} e^{2}}{2 \, d} \] Input:
integrate((d*e*x+c*e)^2*(a+b*arctan(d*x+c)),x, algorithm="maxima")
Output:
1/3*a*d^2*e^2*x^3 + a*c*d*e^2*x^2 + (x^2*arctan(d*x + c) - d*(x/d^2 + (c^2 - 1)*arctan((d^2*x + c*d)/d)/d^3 - c*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^3 ))*b*c*d*e^2 + 1/6*(2*x^3*arctan(d*x + c) - d*((d*x^2 - 4*c*x)/d^3 - 2*(c^ 3 - 3*c)*arctan((d^2*x + c*d)/d)/d^4 + (3*c^2 - 1)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^4))*b*d^2*e^2 + a*c^2*e^2*x + 1/2*(2*(d*x + c)*arctan(d*x + c) - log((d*x + c)^2 + 1))*b*c^2*e^2/d
Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (61) = 122\).
Time = 0.19 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.96 \[ \int (c e+d e x)^2 (a+b \arctan (c+d x)) \, dx=\frac {2 \, b d^{3} e^{2} x^{3} \arctan \left (d x + c\right ) + 2 \, a d^{3} e^{2} x^{3} + 6 \, b c d^{2} e^{2} x^{2} \arctan \left (d x + c\right ) + 6 \, a c d^{2} e^{2} x^{2} + 6 \, b c^{2} d e^{2} x \arctan \left (d x + c\right ) - \pi b c^{3} e^{2} \mathrm {sgn}\left (d x + c\right ) + \pi b c^{3} e^{2} \mathrm {sgn}\left (-d x - c\right ) + 6 \, a c^{2} d e^{2} x - b d^{2} e^{2} x^{2} + b c^{3} e^{2} \arctan \left (d x + c\right ) - b c^{3} e^{2} \arctan \left (-d x - c\right ) - 2 \, b c d e^{2} x + b e^{2} \log \left ({\left (d x + c\right )}^{2} + 1\right )}{6 \, d} \] Input:
integrate((d*e*x+c*e)^2*(a+b*arctan(d*x+c)),x, algorithm="giac")
Output:
1/6*(2*b*d^3*e^2*x^3*arctan(d*x + c) + 2*a*d^3*e^2*x^3 + 6*b*c*d^2*e^2*x^2 *arctan(d*x + c) + 6*a*c*d^2*e^2*x^2 + 6*b*c^2*d*e^2*x*arctan(d*x + c) - p i*b*c^3*e^2*sgn(d*x + c) + pi*b*c^3*e^2*sgn(-d*x - c) + 6*a*c^2*d*e^2*x - b*d^2*e^2*x^2 + b*c^3*e^2*arctan(d*x + c) - b*c^3*e^2*arctan(-d*x - c) - 2 *b*c*d*e^2*x + b*e^2*log((d*x + c)^2 + 1))/d
Time = 1.66 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.15 \[ \int (c e+d e x)^2 (a+b \arctan (c+d x)) \, dx=\frac {a\,d^2\,e^2\,x^3}{3}-\frac {b\,c\,e^2\,x}{3}+\frac {b\,e^2\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2+1\right )}{6\,d}+a\,c^2\,e^2\,x-\frac {b\,d\,e^2\,x^2}{6}+b\,c^2\,e^2\,x\,\mathrm {atan}\left (c+d\,x\right )+a\,c\,d\,e^2\,x^2+\frac {b\,c^3\,e^2\,\mathrm {atan}\left (c+d\,x\right )}{3\,d}+\frac {b\,d^2\,e^2\,x^3\,\mathrm {atan}\left (c+d\,x\right )}{3}+b\,c\,d\,e^2\,x^2\,\mathrm {atan}\left (c+d\,x\right ) \] Input:
int((c*e + d*e*x)^2*(a + b*atan(c + d*x)),x)
Output:
(a*d^2*e^2*x^3)/3 - (b*c*e^2*x)/3 + (b*e^2*log(c^2 + d^2*x^2 + 2*c*d*x + 1 ))/(6*d) + a*c^2*e^2*x - (b*d*e^2*x^2)/6 + b*c^2*e^2*x*atan(c + d*x) + a*c *d*e^2*x^2 + (b*c^3*e^2*atan(c + d*x))/(3*d) + (b*d^2*e^2*x^3*atan(c + d*x ))/3 + b*c*d*e^2*x^2*atan(c + d*x)
Time = 0.19 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.91 \[ \int (c e+d e x)^2 (a+b \arctan (c+d x)) \, dx=\frac {e^{2} \left (2 \mathit {atan} \left (d x +c \right ) b \,c^{3}+6 \mathit {atan} \left (d x +c \right ) b \,c^{2} d x +6 \mathit {atan} \left (d x +c \right ) b c \,d^{2} x^{2}+2 \mathit {atan} \left (d x +c \right ) b \,d^{3} x^{3}+\mathrm {log}\left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) b +6 a \,c^{2} d x +6 a c \,d^{2} x^{2}+2 a \,d^{3} x^{3}-2 b c d x -b \,d^{2} x^{2}\right )}{6 d} \] Input:
int((d*e*x+c*e)^2*(a+b*atan(d*x+c)),x)
Output:
(e**2*(2*atan(c + d*x)*b*c**3 + 6*atan(c + d*x)*b*c**2*d*x + 6*atan(c + d* x)*b*c*d**2*x**2 + 2*atan(c + d*x)*b*d**3*x**3 + log(c**2 + 2*c*d*x + d**2 *x**2 + 1)*b + 6*a*c**2*d*x + 6*a*c*d**2*x**2 + 2*a*d**3*x**3 - 2*b*c*d*x - b*d**2*x**2))/(6*d)