3.1126 \(\int x^2 (d+e x^2)^2 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=161 \[ \frac{1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} d e x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{7} e^2 x^7 \left (a+b \tan ^{-1}(c x)\right )-\frac{b x^2 \left (35 c^4 d^2-42 c^2 d e+15 e^2\right )}{210 c^5}+\frac{b \left (35 c^4 d^2-42 c^2 d e+15 e^2\right ) \log \left (c^2 x^2+1\right )}{210 c^7}-\frac{b e x^4 \left (14 c^2 d-5 e\right )}{140 c^3}-\frac{b e^2 x^6}{42 c} \]

[Out]

-(b*(35*c^4*d^2 - 42*c^2*d*e + 15*e^2)*x^2)/(210*c^5) - (b*(14*c^2*d - 5*e)*e*x^4)/(140*c^3) - (b*e^2*x^6)/(42
*c) + (d^2*x^3*(a + b*ArcTan[c*x]))/3 + (2*d*e*x^5*(a + b*ArcTan[c*x]))/5 + (e^2*x^7*(a + b*ArcTan[c*x]))/7 +
(b*(35*c^4*d^2 - 42*c^2*d*e + 15*e^2)*Log[1 + c^2*x^2])/(210*c^7)

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Rubi [A]  time = 0.245806, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {270, 4976, 12, 1251, 771} \[ \frac{1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} d e x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{7} e^2 x^7 \left (a+b \tan ^{-1}(c x)\right )-\frac{b x^2 \left (35 c^4 d^2-42 c^2 d e+15 e^2\right )}{210 c^5}+\frac{b \left (35 c^4 d^2-42 c^2 d e+15 e^2\right ) \log \left (c^2 x^2+1\right )}{210 c^7}-\frac{b e x^4 \left (14 c^2 d-5 e\right )}{140 c^3}-\frac{b e^2 x^6}{42 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(35*c^4*d^2 - 42*c^2*d*e + 15*e^2)*x^2)/(210*c^5) - (b*(14*c^2*d - 5*e)*e*x^4)/(140*c^3) - (b*e^2*x^6)/(42
*c) + (d^2*x^3*(a + b*ArcTan[c*x]))/3 + (2*d*e*x^5*(a + b*ArcTan[c*x]))/5 + (e^2*x^7*(a + b*ArcTan[c*x]))/7 +
(b*(35*c^4*d^2 - 42*c^2*d*e + 15*e^2)*Log[1 + c^2*x^2])/(210*c^7)

Rule 270

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

Rule 4976

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(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] &&
IntegerQ[(m - 1)/2]

Rule 771

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

Rubi steps

\begin{align*} \int x^2 \left (d+e x^2\right )^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} d e x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{7} e^2 x^7 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac{x^3 \left (35 d^2+42 d e x^2+15 e^2 x^4\right )}{105 \left (1+c^2 x^2\right )} \, dx\\ &=\frac{1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} d e x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{7} e^2 x^7 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{105} (b c) \int \frac{x^3 \left (35 d^2+42 d e x^2+15 e^2 x^4\right )}{1+c^2 x^2} \, dx\\ &=\frac{1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} d e x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{7} e^2 x^7 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{210} (b c) \operatorname{Subst}\left (\int \frac{x \left (35 d^2+42 d e x+15 e^2 x^2\right )}{1+c^2 x} \, dx,x,x^2\right )\\ &=\frac{1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} d e x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{7} e^2 x^7 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{210} (b c) \operatorname{Subst}\left (\int \left (\frac{35 c^4 d^2-42 c^2 d e+15 e^2}{c^6}+\frac{3 \left (14 c^2 d-5 e\right ) e x}{c^4}+\frac{15 e^2 x^2}{c^2}+\frac{-35 c^4 d^2+42 c^2 d e-15 e^2}{c^6 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{b \left (35 c^4 d^2-42 c^2 d e+15 e^2\right ) x^2}{210 c^5}-\frac{b \left (14 c^2 d-5 e\right ) e x^4}{140 c^3}-\frac{b e^2 x^6}{42 c}+\frac{1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} d e x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{7} e^2 x^7 \left (a+b \tan ^{-1}(c x)\right )+\frac{b \left (35 c^4 d^2-42 c^2 d e+15 e^2\right ) \log \left (1+c^2 x^2\right )}{210 c^7}\\ \end{align*}

Mathematica [A]  time = 0.117087, size = 162, normalized size = 1.01 \[ \frac{c^2 x^2 \left (4 a c^5 x \left (35 d^2+42 d e x^2+15 e^2 x^4\right )-2 b c^4 \left (35 d^2+21 d e x^2+5 e^2 x^4\right )+3 b c^2 e \left (28 d+5 e x^2\right )-30 b e^2\right )+2 b \left (35 c^4 d^2-42 c^2 d e+15 e^2\right ) \log \left (c^2 x^2+1\right )+4 b c^7 x^3 \tan ^{-1}(c x) \left (35 d^2+42 d e x^2+15 e^2 x^4\right )}{420 c^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x^2*(-30*b*e^2 + 3*b*c^2*e*(28*d + 5*e*x^2) - 2*b*c^4*(35*d^2 + 21*d*e*x^2 + 5*e^2*x^4) + 4*a*c^5*x*(35*d
^2 + 42*d*e*x^2 + 15*e^2*x^4)) + 4*b*c^7*x^3*(35*d^2 + 42*d*e*x^2 + 15*e^2*x^4)*ArcTan[c*x] + 2*b*(35*c^4*d^2
- 42*c^2*d*e + 15*e^2)*Log[1 + c^2*x^2])/(420*c^7)

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Maple [A]  time = 0.036, size = 192, normalized size = 1.2 \begin{align*}{\frac{a{e}^{2}{x}^{7}}{7}}+{\frac{2\,aed{x}^{5}}{5}}+{\frac{a{d}^{2}{x}^{3}}{3}}+{\frac{b\arctan \left ( cx \right ){e}^{2}{x}^{7}}{7}}+{\frac{2\,b\arctan \left ( cx \right ) ed{x}^{5}}{5}}+{\frac{b\arctan \left ( cx \right ){d}^{2}{x}^{3}}{3}}-{\frac{b{d}^{2}{x}^{2}}{6\,c}}-{\frac{bde{x}^{4}}{10\,c}}-{\frac{b{e}^{2}{x}^{6}}{42\,c}}+{\frac{b{x}^{2}de}{5\,{c}^{3}}}+{\frac{b{x}^{4}{e}^{2}}{28\,{c}^{3}}}-{\frac{b{x}^{2}{e}^{2}}{14\,{c}^{5}}}+{\frac{b{d}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{6\,{c}^{3}}}-{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ) ed}{5\,{c}^{5}}}+{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ){e}^{2}}{14\,{c}^{7}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*e^2*x^7+2/5*a*e*d*x^5+1/3*a*d^2*x^3+1/7*b*arctan(c*x)*e^2*x^7+2/5*b*arctan(c*x)*e*d*x^5+1/3*b*arctan(c*x
)*d^2*x^3-1/6*b*d^2*x^2/c-1/10/c*b*e*d*x^4-1/42*b*e^2*x^6/c+1/5/c^3*b*x^2*d*e+1/28/c^3*b*x^4*e^2-1/14/c^5*b*x^
2*e^2+1/6*b*d^2*ln(c^2*x^2+1)/c^3-1/5/c^5*b*ln(c^2*x^2+1)*e*d+1/14/c^7*b*ln(c^2*x^2+1)*e^2

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Maxima [A]  time = 0.972984, size = 244, normalized size = 1.52 \begin{align*} \frac{1}{7} \, a e^{2} x^{7} + \frac{2}{5} \, a d e x^{5} + \frac{1}{3} \, a d^{2} x^{3} + \frac{1}{6} \,{\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^{2} + \frac{1}{10} \,{\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 d e + \frac{1}{84} \,{\left (12 \, x^{7} \arctan \left (c x\right ) - c{\left (\frac{2 \, c^{4} x^{6} - 3 \, c^{2} x^{4} + 6 \, x^{2}}{c^{6}} - \frac{6 \, \log \left (c^{2} x^{2} + 1\right )}{c^{8}}\right )}\right )} b e^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*e^2*x^7 + 2/5*a*d*e*x^5 + 1/3*a*d^2*x^3 + 1/6*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b
*d^2 + 1/10*(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*d*e + 1/84*(12*x^7*arct
an(c*x) - c*((2*c^4*x^6 - 3*c^2*x^4 + 6*x^2)/c^6 - 6*log(c^2*x^2 + 1)/c^8))*b*e^2

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Fricas [A]  time = 1.68684, size = 424, normalized size = 2.63 \begin{align*} \frac{60 \, a c^{7} e^{2} x^{7} + 168 \, a c^{7} d e x^{5} - 10 \, b c^{6} e^{2} x^{6} + 140 \, a c^{7} d^{2} x^{3} - 3 \,{\left (14 \, b c^{6} d e - 5 \, b c^{4} e^{2}\right )} x^{4} - 2 \,{\left (35 \, b c^{6} d^{2} - 42 \, b c^{4} d e + 15 \, b c^{2} e^{2}\right )} x^{2} + 4 \,{\left (15 \, b c^{7} e^{2} x^{7} + 42 \, b c^{7} d e x^{5} + 35 \, b c^{7} d^{2} x^{3}\right )} \arctan \left (c x\right ) + 2 \,{\left (35 \, b c^{4} d^{2} - 42 \, b c^{2} d e + 15 \, b e^{2}\right )} \log \left (c^{2} x^{2} + 1\right )}{420 \, c^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(60*a*c^7*e^2*x^7 + 168*a*c^7*d*e*x^5 - 10*b*c^6*e^2*x^6 + 140*a*c^7*d^2*x^3 - 3*(14*b*c^6*d*e - 5*b*c^4
*e^2)*x^4 - 2*(35*b*c^6*d^2 - 42*b*c^4*d*e + 15*b*c^2*e^2)*x^2 + 4*(15*b*c^7*e^2*x^7 + 42*b*c^7*d*e*x^5 + 35*b
*c^7*d^2*x^3)*arctan(c*x) + 2*(35*b*c^4*d^2 - 42*b*c^2*d*e + 15*b*e^2)*log(c^2*x^2 + 1))/c^7

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Sympy [A]  time = 4.09677, size = 245, normalized size = 1.52 \begin{align*} \begin{cases} \frac{a d^{2} x^{3}}{3} + \frac{2 a d e x^{5}}{5} + \frac{a e^{2} x^{7}}{7} + \frac{b d^{2} x^{3} \operatorname{atan}{\left (c x \right )}}{3} + \frac{2 b d e x^{5} \operatorname{atan}{\left (c x \right )}}{5} + \frac{b e^{2} x^{7} \operatorname{atan}{\left (c x \right )}}{7} - \frac{b d^{2} x^{2}}{6 c} - \frac{b d e x^{4}}{10 c} - \frac{b e^{2} x^{6}}{42 c} + \frac{b d^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{6 c^{3}} + \frac{b d e x^{2}}{5 c^{3}} + \frac{b e^{2} x^{4}}{28 c^{3}} - \frac{b d e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{5 c^{5}} - \frac{b e^{2} x^{2}}{14 c^{5}} + \frac{b e^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{14 c^{7}} & \text{for}\: c \neq 0 \\a \left (\frac{d^{2} x^{3}}{3} + \frac{2 d e x^{5}}{5} + \frac{e^{2} x^{7}}{7}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d**2*x**3/3 + 2*a*d*e*x**5/5 + a*e**2*x**7/7 + b*d**2*x**3*atan(c*x)/3 + 2*b*d*e*x**5*atan(c*x)/5
+ b*e**2*x**7*atan(c*x)/7 - b*d**2*x**2/(6*c) - b*d*e*x**4/(10*c) - b*e**2*x**6/(42*c) + b*d**2*log(x**2 + c*
*(-2))/(6*c**3) + b*d*e*x**2/(5*c**3) + b*e**2*x**4/(28*c**3) - b*d*e*log(x**2 + c**(-2))/(5*c**5) - b*e**2*x*
*2/(14*c**5) + b*e**2*log(x**2 + c**(-2))/(14*c**7), Ne(c, 0)), (a*(d**2*x**3/3 + 2*d*e*x**5/5 + e**2*x**7/7),
True))

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Giac [A]  time = 1.1001, size = 284, normalized size = 1.76 \begin{align*} \frac{60 \, b c^{7} x^{7} \arctan \left (c x\right ) e^{2} + 60 \, a c^{7} x^{7} e^{2} + 168 \, b c^{7} d x^{5} \arctan \left (c x\right ) e + 168 \, a c^{7} d x^{5} e + 140 \, b c^{7} d^{2} x^{3} \arctan \left (c x\right ) - 10 \, b c^{6} x^{6} e^{2} + 140 \, a c^{7} d^{2} x^{3} - 42 \, b c^{6} d x^{4} e - 70 \, b c^{6} d^{2} x^{2} + 15 \, b c^{4} x^{4} e^{2} + 84 \, b c^{4} d x^{2} e + 70 \, b c^{4} d^{2} \log \left (c^{2} x^{2} + 1\right ) - 30 \, b c^{2} x^{2} e^{2} - 84 \, b c^{2} d e \log \left (c^{2} x^{2} + 1\right ) + 30 \, b e^{2} \log \left (c^{2} x^{2} + 1\right )}{420 \, c^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(60*b*c^7*x^7*arctan(c*x)*e^2 + 60*a*c^7*x^7*e^2 + 168*b*c^7*d*x^5*arctan(c*x)*e + 168*a*c^7*d*x^5*e + 1
40*b*c^7*d^2*x^3*arctan(c*x) - 10*b*c^6*x^6*e^2 + 140*a*c^7*d^2*x^3 - 42*b*c^6*d*x^4*e - 70*b*c^6*d^2*x^2 + 15
*b*c^4*x^4*e^2 + 84*b*c^4*d*x^2*e + 70*b*c^4*d^2*log(c^2*x^2 + 1) - 30*b*c^2*x^2*e^2 - 84*b*c^2*d*e*log(c^2*x^
2 + 1) + 30*b*e^2*log(c^2*x^2 + 1))/c^7