3.1128 \(\int (d+e x^2)^2 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=124 \[ d^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \log \left (c^2 x^2+1\right )}{30 c^5}-\frac{b e x^2 \left (10 c^2 d-3 e\right )}{30 c^3}-\frac{b e^2 x^4}{20 c} \]

[Out]

-(b*(10*c^2*d - 3*e)*e*x^2)/(30*c^3) - (b*e^2*x^4)/(20*c) + d^2*x*(a + b*ArcTan[c*x]) + (2*d*e*x^3*(a + b*ArcT
an[c*x]))/3 + (e^2*x^5*(a + b*ArcTan[c*x]))/5 - (b*(15*c^4*d^2 - 10*c^2*d*e + 3*e^2)*Log[1 + c^2*x^2])/(30*c^5
)

________________________________________________________________________________________

Rubi [A]  time = 0.159716, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {194, 4912, 1594, 1247, 698} \[ d^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \log \left (c^2 x^2+1\right )}{30 c^5}-\frac{b e x^2 \left (10 c^2 d-3 e\right )}{30 c^3}-\frac{b e^2 x^4}{20 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)^2*(a + b*ArcTan[c*x]),x]

[Out]

-(b*(10*c^2*d - 3*e)*e*x^2)/(30*c^3) - (b*e^2*x^4)/(20*c) + d^2*x*(a + b*ArcTan[c*x]) + (2*d*e*x^3*(a + b*ArcT
an[c*x]))/3 + (e^2*x^5*(a + b*ArcTan[c*x]))/5 - (b*(15*c^4*d^2 - 10*c^2*d*e + 3*e^2)*Log[1 + c^2*x^2])/(30*c^5
)

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 4912

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[u/(1 + c^2*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x]
&& (IntegerQ[q] || ILtQ[q + 1/2, 0])

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \left (d+e x^2\right )^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=d^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac{d^2 x+\frac{2}{3} d e x^3+\frac{e^2 x^5}{5}}{1+c^2 x^2} \, dx\\ &=d^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac{x \left (d^2+\frac{2}{3} d e x^2+\frac{e^2 x^4}{5}\right )}{1+c^2 x^2} \, dx\\ &=d^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{d^2+\frac{2 d e x}{3}+\frac{e^2 x^2}{5}}{1+c^2 x} \, dx,x,x^2\right )\\ &=d^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} (b c) \operatorname{Subst}\left (\int \left (\frac{\left (10 c^2 d-3 e\right ) e}{15 c^4}+\frac{e^2 x}{5 c^2}+\frac{15 c^4 d^2-10 c^2 d e+3 e^2}{15 c^4 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{b \left (10 c^2 d-3 e\right ) e x^2}{30 c^3}-\frac{b e^2 x^4}{20 c}+d^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \log \left (1+c^2 x^2\right )}{30 c^5}\\ \end{align*}

Mathematica [A]  time = 0.0923648, size = 130, normalized size = 1.05 \[ \frac{c^2 x \left (4 a c^3 \left (15 d^2+10 d e x^2+3 e^2 x^4\right )+b e x \left (6 e-c^2 \left (20 d+3 e x^2\right )\right )\right )-2 b \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \log \left (c^2 x^2+1\right )+4 b c^5 x \tan ^{-1}(c x) \left (15 d^2+10 d e x^2+3 e^2 x^4\right )}{60 c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)^2*(a + b*ArcTan[c*x]),x]

[Out]

(c^2*x*(4*a*c^3*(15*d^2 + 10*d*e*x^2 + 3*e^2*x^4) + b*e*x*(6*e - c^2*(20*d + 3*e*x^2))) + 4*b*c^5*x*(15*d^2 +
10*d*e*x^2 + 3*e^2*x^4)*ArcTan[c*x] - 2*b*(15*c^4*d^2 - 10*c^2*d*e + 3*e^2)*Log[1 + c^2*x^2])/(60*c^5)

________________________________________________________________________________________

Maple [A]  time = 0.038, size = 151, normalized size = 1.2 \begin{align*}{\frac{a{x}^{5}{e}^{2}}{5}}+{\frac{2\,a{x}^{3}de}{3}}+a{d}^{2}x+{\frac{b\arctan \left ( cx \right ){x}^{5}{e}^{2}}{5}}+{\frac{2\,b\arctan \left ( cx \right ){x}^{3}de}{3}}+b\arctan \left ( cx \right ){d}^{2}x-{\frac{b{x}^{2}de}{3\,c}}-{\frac{b{e}^{2}{x}^{4}}{20\,c}}+{\frac{b{x}^{2}{e}^{2}}{10\,{c}^{3}}}-{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{2}}{2\,c}}+{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ) ed}{3\,{c}^{3}}}-{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ){e}^{2}}{10\,{c}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^2*(a+b*arctan(c*x)),x)

[Out]

1/5*a*x^5*e^2+2/3*a*x^3*d*e+a*d^2*x+1/5*b*arctan(c*x)*x^5*e^2+2/3*b*arctan(c*x)*x^3*d*e+b*arctan(c*x)*d^2*x-1/
3/c*b*x^2*d*e-1/20*b*e^2*x^4/c+1/10/c^3*b*x^2*e^2-1/2/c*b*ln(c^2*x^2+1)*d^2+1/3/c^3*b*ln(c^2*x^2+1)*e*d-1/10/c
^5*b*ln(c^2*x^2+1)*e^2

________________________________________________________________________________________

Maxima [A]  time = 0.971701, size = 198, normalized size = 1.6 \begin{align*} \frac{1}{5} \, a e^{2} x^{5} + \frac{2}{3} \, a d e x^{3} + \frac{1}{3} \,{\left (2 \, x^{3} \arctan \left (c x\right ) - c{\left (\frac{x^{2}}{c^{2}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d e + \frac{1}{20} \,{\left (4 \, x^{5} \arctan \left (c x\right ) - c{\left (\frac{c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e^{2} + a d^{2} x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{2}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

1/5*a*e^2*x^5 + 2/3*a*d*e*x^3 + 1/3*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*d*e + 1/20*(4*x
^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*e^2 + a*d^2*x + 1/2*(2*c*x*arctan(c*x)
- log(c^2*x^2 + 1))*b*d^2/c

________________________________________________________________________________________

Fricas [A]  time = 1.5331, size = 339, normalized size = 2.73 \begin{align*} \frac{12 \, a c^{5} e^{2} x^{5} + 40 \, a c^{5} d e x^{3} - 3 \, b c^{4} e^{2} x^{4} + 60 \, a c^{5} d^{2} x - 2 \,{\left (10 \, b c^{4} d e - 3 \, b c^{2} e^{2}\right )} x^{2} + 4 \,{\left (3 \, b c^{5} e^{2} x^{5} + 10 \, b c^{5} d e x^{3} + 15 \, b c^{5} d^{2} x\right )} \arctan \left (c x\right ) - 2 \,{\left (15 \, b c^{4} d^{2} - 10 \, b c^{2} d e + 3 \, b e^{2}\right )} \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

1/60*(12*a*c^5*e^2*x^5 + 40*a*c^5*d*e*x^3 - 3*b*c^4*e^2*x^4 + 60*a*c^5*d^2*x - 2*(10*b*c^4*d*e - 3*b*c^2*e^2)*
x^2 + 4*(3*b*c^5*e^2*x^5 + 10*b*c^5*d*e*x^3 + 15*b*c^5*d^2*x)*arctan(c*x) - 2*(15*b*c^4*d^2 - 10*b*c^2*d*e + 3
*b*e^2)*log(c^2*x^2 + 1))/c^5

________________________________________________________________________________________

Sympy [A]  time = 2.32238, size = 194, normalized size = 1.56 \begin{align*} \begin{cases} a d^{2} x + \frac{2 a d e x^{3}}{3} + \frac{a e^{2} x^{5}}{5} + b d^{2} x \operatorname{atan}{\left (c x \right )} + \frac{2 b d e x^{3} \operatorname{atan}{\left (c x \right )}}{3} + \frac{b e^{2} x^{5} \operatorname{atan}{\left (c x \right )}}{5} - \frac{b d^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2 c} - \frac{b d e x^{2}}{3 c} - \frac{b e^{2} x^{4}}{20 c} + \frac{b d e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{3 c^{3}} + \frac{b e^{2} x^{2}}{10 c^{3}} - \frac{b e^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{10 c^{5}} & \text{for}\: c \neq 0 \\a \left (d^{2} x + \frac{2 d e x^{3}}{3} + \frac{e^{2} x^{5}}{5}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**2*(a+b*atan(c*x)),x)

[Out]

Piecewise((a*d**2*x + 2*a*d*e*x**3/3 + a*e**2*x**5/5 + b*d**2*x*atan(c*x) + 2*b*d*e*x**3*atan(c*x)/3 + b*e**2*
x**5*atan(c*x)/5 - b*d**2*log(x**2 + c**(-2))/(2*c) - b*d*e*x**2/(3*c) - b*e**2*x**4/(20*c) + b*d*e*log(x**2 +
c**(-2))/(3*c**3) + b*e**2*x**2/(10*c**3) - b*e**2*log(x**2 + c**(-2))/(10*c**5), Ne(c, 0)), (a*(d**2*x + 2*d
*e*x**3/3 + e**2*x**5/5), True))

________________________________________________________________________________________

Giac [A]  time = 1.13019, size = 231, normalized size = 1.86 \begin{align*} \frac{12 \, b c^{5} x^{5} \arctan \left (c x\right ) e^{2} + 12 \, a c^{5} x^{5} e^{2} + 40 \, b c^{5} d x^{3} \arctan \left (c x\right ) e + 40 \, a c^{5} d x^{3} e + 60 \, b c^{5} d^{2} x \arctan \left (c x\right ) - 3 \, b c^{4} x^{4} e^{2} + 60 \, a c^{5} d^{2} x - 20 \, b c^{4} d x^{2} e - 30 \, b c^{4} d^{2} \log \left (c^{2} x^{2} + 1\right ) + 6 \, b c^{2} x^{2} e^{2} + 20 \, b c^{2} d e \log \left (c^{2} x^{2} + 1\right ) - 6 \, b e^{2} \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

1/60*(12*b*c^5*x^5*arctan(c*x)*e^2 + 12*a*c^5*x^5*e^2 + 40*b*c^5*d*x^3*arctan(c*x)*e + 40*a*c^5*d*x^3*e + 60*b
*c^5*d^2*x*arctan(c*x) - 3*b*c^4*x^4*e^2 + 60*a*c^5*d^2*x - 20*b*c^4*d*x^2*e - 30*b*c^4*d^2*log(c^2*x^2 + 1) +
6*b*c^2*x^2*e^2 + 20*b*c^2*d*e*log(c^2*x^2 + 1) - 6*b*e^2*log(c^2*x^2 + 1))/c^5