Integrand size = 23, antiderivative size = 100 \[ \int (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=-2 a e x-2 b e x \arctan (c x)+\frac {e (a+b \arctan (c x))^2}{b c}+\frac {b e \log \left (1+c^2 x^2\right )}{c}+x (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {b \left (d+e \log \left (1+c^2 x^2\right )\right )^2}{4 c e} \] Output:
-2*a*e*x-2*b*e*x*arctan(c*x)+e*(a+b*arctan(c*x))^2/b/c+b*e*ln(c^2*x^2+1)/c +x*(a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1))-1/4*b*(d+e*ln(c^2*x^2+1))^2/c/e
Time = 0.02 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.38 \[ \int (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=a d x-2 a e x+\frac {2 a e \arctan (c x)}{c}+b d x \arctan (c x)-2 b e x \arctan (c x)+\frac {b e \arctan (c x)^2}{c}-\frac {b d \log \left (1+c^2 x^2\right )}{2 c}+\frac {b e \log \left (1+c^2 x^2\right )}{c}+a e x \log \left (1+c^2 x^2\right )+b e x \arctan (c x) \log \left (1+c^2 x^2\right )-\frac {b e \log ^2\left (1+c^2 x^2\right )}{4 c} \] Input:
Integrate[(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]),x]
Output:
a*d*x - 2*a*e*x + (2*a*e*ArcTan[c*x])/c + b*d*x*ArcTan[c*x] - 2*b*e*x*ArcT an[c*x] + (b*e*ArcTan[c*x]^2)/c - (b*d*Log[1 + c^2*x^2])/(2*c) + (b*e*Log[ 1 + c^2*x^2])/c + a*e*x*Log[1 + c^2*x^2] + b*e*x*ArcTan[c*x]*Log[1 + c^2*x ^2] - (b*e*Log[1 + c^2*x^2]^2)/(4*c)
Time = 0.78 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5544, 2925, 2837, 2738, 5451, 2009, 5419}
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.
\(\displaystyle \int (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right ) \, dx\) |
\(\Big \downarrow \) 5544 |
\(\displaystyle -2 c^2 e \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx-b c \int \frac {x \left (d+e \log \left (c^2 x^2+1\right )\right )}{c^2 x^2+1}dx+x (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )\) |
\(\Big \downarrow \) 2925 |
\(\displaystyle -2 c^2 e \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx-\frac {1}{2} b c \int \frac {d+e \log \left (c^2 x^2+1\right )}{c^2 x^2+1}dx^2+x (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )\) |
\(\Big \downarrow \) 2837 |
\(\displaystyle -2 c^2 e \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx-\frac {b \int \frac {d+e \log \left (c^2 x^2+1\right )}{x^2}d\left (c^2 x^2+1\right )}{2 c}+x (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )\) |
\(\Big \downarrow \) 2738 |
\(\displaystyle -2 c^2 e \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx+x (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {b \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{4 c e}\) |
\(\Big \downarrow \) 5451 |
\(\displaystyle -2 c^2 e \left (\frac {\int (a+b \arctan (c x))dx}{c^2}-\frac {\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx}{c^2}\right )+x (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {b \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{4 c e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 c^2 e \left (\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx}{c^2}\right )+x (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )-\frac {b \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{4 c e}\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle x (a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )-2 c^2 e \left (\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}\right )-\frac {b \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{4 c e}\) |
Input:
Int[(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]),x]
Output:
x*(a + b*ArcTan[c*x])*(d + e*Log[1 + c^2*x^2]) - (b*(d + e*Log[1 + c^2*x^2 ])^2)/(4*c*e) - 2*c^2*e*(-1/2*(a + b*ArcTan[c*x])^2/(b*c^3) + (a*x + b*x*A rcTan[c*x] - (b*Log[1 + c^2*x^2])/(2*c))/c^2)
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Lo g[c*x^n])^2/(2*b*n), x] /; FreeQ[{a, b, c, n}, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. )*(x_))^(q_.), x_Symbol] :> Simp[1/e Subst[Int[(f*(x/d))^q*(a + b*Log[c*x ^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x] && EqQ[e*f - d*g, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Si mplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && Integer Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 ] || IGtQ[q, 0])
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo l] :> Simp[(a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e _.)*(x_)^2), x_Symbol] :> Simp[f^2/e Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x] )^p, x], x] - Simp[d*(f^2/e) Int[(f*x)^(m - 2)*((a + b*ArcTan[c*x])^p/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*( e_.)), x_Symbol] :> Simp[x*(d + e*Log[f + g*x^2])*(a + b*ArcTan[c*x]), x] + (-Simp[b*c Int[x*((d + e*Log[f + g*x^2])/(1 + c^2*x^2)), x], x] - Simp[2 *e*g Int[x^2*((a + b*ArcTan[c*x])/(f + g*x^2)), x], x]) /; FreeQ[{a, b, c , d, e, f, g}, x]
Time = 1.79 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.37
method | result | size |
parallelrisch | \(\frac {4 e b \ln \left (c^{2} x^{2}+1\right ) \arctan \left (c x \right ) x c +4 b \arctan \left (c x \right ) x c d -8 e b \arctan \left (c x \right ) x c +4 a e \ln \left (c^{2} x^{2}+1\right ) x c +4 a c d x -8 x a c e +4 e b \arctan \left (c x \right )^{2}-e b \ln \left (c^{2} x^{2}+1\right )^{2}+8 a e \arctan \left (c x \right )-2 \ln \left (c^{2} x^{2}+1\right ) b d +4 \ln \left (c^{2} x^{2}+1\right ) b e}{4 c}\) | \(137\) |
default | \(\text {Expression too large to display}\) | \(2516\) |
parts | \(\text {Expression too large to display}\) | \(2516\) |
risch | \(\text {Expression too large to display}\) | \(22111\) |
Input:
int((a+b*arctan(c*x))*(d+e*ln(c^2*x^2+1)),x,method=_RETURNVERBOSE)
Output:
1/4*(4*e*b*ln(c^2*x^2+1)*arctan(c*x)*x*c+4*b*arctan(c*x)*x*c*d-8*e*b*arcta n(c*x)*x*c+4*a*e*ln(c^2*x^2+1)*x*c+4*a*c*d*x-8*x*a*c*e+4*e*b*arctan(c*x)^2 -e*b*ln(c^2*x^2+1)^2+8*a*e*arctan(c*x)-2*ln(c^2*x^2+1)*b*d+4*ln(c^2*x^2+1) *b*e)/c
Time = 0.10 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.05 \[ \int (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {4 \, b e \arctan \left (c x\right )^{2} - b e \log \left (c^{2} x^{2} + 1\right )^{2} + 4 \, {\left (a c d - 2 \, a c e\right )} x + 4 \, {\left (2 \, a e + {\left (b c d - 2 \, b c e\right )} x\right )} \arctan \left (c x\right ) + 2 \, {\left (2 \, b c e x \arctan \left (c x\right ) + 2 \, a c e x - b d + 2 \, b e\right )} \log \left (c^{2} x^{2} + 1\right )}{4 \, c} \] Input:
integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="fricas")
Output:
1/4*(4*b*e*arctan(c*x)^2 - b*e*log(c^2*x^2 + 1)^2 + 4*(a*c*d - 2*a*c*e)*x + 4*(2*a*e + (b*c*d - 2*b*c*e)*x)*arctan(c*x) + 2*(2*b*c*e*x*arctan(c*x) + 2*a*c*e*x - b*d + 2*b*e)*log(c^2*x^2 + 1))/c
Time = 0.36 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.48 \[ \int (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\begin {cases} a d x + a e x \log {\left (c^{2} x^{2} + 1 \right )} - 2 a e x + \frac {2 a e \operatorname {atan}{\left (c x \right )}}{c} + b d x \operatorname {atan}{\left (c x \right )} + b e x \log {\left (c^{2} x^{2} + 1 \right )} \operatorname {atan}{\left (c x \right )} - 2 b e x \operatorname {atan}{\left (c x \right )} - \frac {b d \log {\left (c^{2} x^{2} + 1 \right )}}{2 c} - \frac {b e \log {\left (c^{2} x^{2} + 1 \right )}^{2}}{4 c} + \frac {b e \log {\left (c^{2} x^{2} + 1 \right )}}{c} + \frac {b e \operatorname {atan}^{2}{\left (c x \right )}}{c} & \text {for}\: c \neq 0 \\a d x & \text {otherwise} \end {cases} \] Input:
integrate((a+b*atan(c*x))*(d+e*ln(c**2*x**2+1)),x)
Output:
Piecewise((a*d*x + a*e*x*log(c**2*x**2 + 1) - 2*a*e*x + 2*a*e*atan(c*x)/c + b*d*x*atan(c*x) + b*e*x*log(c**2*x**2 + 1)*atan(c*x) - 2*b*e*x*atan(c*x) - b*d*log(c**2*x**2 + 1)/(2*c) - b*e*log(c**2*x**2 + 1)**2/(4*c) + b*e*lo g(c**2*x**2 + 1)/c + b*e*atan(c*x)**2/c, Ne(c, 0)), (a*d*x, True))
Time = 0.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.53 \[ \int (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=-{\left (2 \, c^{2} {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )} - x \log \left (c^{2} x^{2} + 1\right )\right )} b e \arctan \left (c x\right ) - {\left (2 \, c^{2} {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )} - x \log \left (c^{2} x^{2} + 1\right )\right )} a e + a d x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d}{2 \, c} - \frac {{\left (4 \, \arctan \left (c x\right )^{2} + \log \left (c^{2} x^{2} + 1\right )^{2} - 4 \, \log \left (c^{2} x^{2} + 1\right )\right )} b e}{4 \, c} \] Input:
integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="maxima")
Output:
-(2*c^2*(x/c^2 - arctan(c*x)/c^3) - x*log(c^2*x^2 + 1))*b*e*arctan(c*x) - (2*c^2*(x/c^2 - arctan(c*x)/c^3) - x*log(c^2*x^2 + 1))*a*e + a*d*x + 1/2*( 2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d/c - 1/4*(4*arctan(c*x)^2 + log(c ^2*x^2 + 1)^2 - 4*log(c^2*x^2 + 1))*b*e/c
Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (98) = 196\).
Time = 0.20 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.61 \[ \int (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {2 \, \pi b c e x \log \left (c^{2} x^{2} + 1\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (x\right ) + 2 \, \pi b c d x \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (x\right ) - 4 \, \pi b c e x \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (x\right ) - 4 \, b c e x \arctan \left (\frac {1}{c x}\right ) \log \left (c^{2} x^{2} + 1\right ) - 6 \, \pi ^{2} b e \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (x\right ) - 4 \, \pi b e \arctan \left (\frac {1}{c x}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (x\right ) - 4 \, b c d x \arctan \left (\frac {1}{c x}\right ) + 8 \, b c e x \arctan \left (\frac {1}{c x}\right ) + 4 \, a c e x \log \left (c^{2} x^{2} + 1\right ) - 8 \, \pi a e \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (x\right ) + 2 \, \pi ^{2} b e + 4 \, a c d x - 8 \, a c e x + 4 \, \pi b e \arctan \left (c x\right ) + 4 \, \pi b e \arctan \left (\frac {1}{c x}\right ) + 4 \, b e \arctan \left (\frac {1}{c x}\right )^{2} - b e \log \left (c^{2} x^{2} + 1\right )^{2} + 8 \, a e \arctan \left (c x\right ) - 2 \, b d \log \left (c^{2} x^{2} + 1\right ) + 4 \, b e \log \left (c^{2} x^{2} + 1\right )}{4 \, c} \] Input:
integrate((a+b*arctan(c*x))*(d+e*log(c^2*x^2+1)),x, algorithm="giac")
Output:
1/4*(2*pi*b*c*e*x*log(c^2*x^2 + 1)*sgn(c)*sgn(x) + 2*pi*b*c*d*x*sgn(c)*sgn (x) - 4*pi*b*c*e*x*sgn(c)*sgn(x) - 4*b*c*e*x*arctan(1/(c*x))*log(c^2*x^2 + 1) - 6*pi^2*b*e*sgn(c)*sgn(x) - 4*pi*b*e*arctan(1/(c*x))*sgn(c)*sgn(x) - 4*b*c*d*x*arctan(1/(c*x)) + 8*b*c*e*x*arctan(1/(c*x)) + 4*a*c*e*x*log(c^2* x^2 + 1) - 8*pi*a*e*sgn(c)*sgn(x) + 2*pi^2*b*e + 4*a*c*d*x - 8*a*c*e*x + 4 *pi*b*e*arctan(c*x) + 4*pi*b*e*arctan(1/(c*x)) + 4*b*e*arctan(1/(c*x))^2 - b*e*log(c^2*x^2 + 1)^2 + 8*a*e*arctan(c*x) - 2*b*d*log(c^2*x^2 + 1) + 4*b *e*log(c^2*x^2 + 1))/c
Time = 1.12 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.34 \[ \int (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=a\,d\,x-2\,a\,e\,x-\frac {b\,e\,{\ln \left (c^2\,x^2+1\right )}^2}{4\,c}+b\,d\,x\,\mathrm {atan}\left (c\,x\right )-2\,b\,e\,x\,\mathrm {atan}\left (c\,x\right )+a\,e\,x\,\ln \left (c^2\,x^2+1\right )+\frac {2\,a\,e\,\mathrm {atan}\left (c\,x\right )}{c}-\frac {b\,d\,\ln \left (c^2\,x^2+1\right )}{2\,c}+\frac {b\,e\,\ln \left (c^2\,x^2+1\right )}{c}+\frac {b\,e\,{\mathrm {atan}\left (c\,x\right )}^2}{c}+b\,e\,x\,\mathrm {atan}\left (c\,x\right )\,\ln \left (c^2\,x^2+1\right ) \] Input:
int((a + b*atan(c*x))*(d + e*log(c^2*x^2 + 1)),x)
Output:
a*d*x - 2*a*e*x - (b*e*log(c^2*x^2 + 1)^2)/(4*c) + b*d*x*atan(c*x) - 2*b*e *x*atan(c*x) + a*e*x*log(c^2*x^2 + 1) + (2*a*e*atan(c*x))/c - (b*d*log(c^2 *x^2 + 1))/(2*c) + (b*e*log(c^2*x^2 + 1))/c + (b*e*atan(c*x)^2)/c + b*e*x* atan(c*x)*log(c^2*x^2 + 1)
Time = 0.24 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.36 \[ \int (a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right ) \, dx=\frac {4 \mathit {atan} \left (c x \right )^{2} b e +4 \mathit {atan} \left (c x \right ) \mathrm {log}\left (c^{2} x^{2}+1\right ) b c e x +8 \mathit {atan} \left (c x \right ) a e +4 \mathit {atan} \left (c x \right ) b c d x -8 \mathit {atan} \left (c x \right ) b c e x -\mathrm {log}\left (c^{2} x^{2}+1\right )^{2} b e +4 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) a c e x -2 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b d +4 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b e +4 a c d x -8 a c e x}{4 c} \] Input:
int((a+b*atan(c*x))*(d+e*log(c^2*x^2+1)),x)
Output:
(4*atan(c*x)**2*b*e + 4*atan(c*x)*log(c**2*x**2 + 1)*b*c*e*x + 8*atan(c*x) *a*e + 4*atan(c*x)*b*c*d*x - 8*atan(c*x)*b*c*e*x - log(c**2*x**2 + 1)**2*b *e + 4*log(c**2*x**2 + 1)*a*c*e*x - 2*log(c**2*x**2 + 1)*b*d + 4*log(c**2* x**2 + 1)*b*e + 4*a*c*d*x - 8*a*c*e*x)/(4*c)