3.1230 \(\int x^m (d+e x^2) (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=122 \[ \frac {d x^{m+1} \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac {e x^{m+3} \left (a+b \tan ^{-1}(c x)\right )}{m+3}-\frac {b x^{m+2} \left (\frac {c^2 d}{m+1}-\frac {e}{m+3}\right ) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{c (m+2)}-\frac {b e x^{m+2}}{c \left (m^2+5 m+6\right )} \]

[Out]

-b*e*x^(2+m)/c/(m^2+5*m+6)+d*x^(1+m)*(a+b*arctan(c*x))/(1+m)+e*x^(3+m)*(a+b*arctan(c*x))/(3+m)-b*(c^2*d/(1+m)-
e/(3+m))*x^(2+m)*hypergeom([1, 1+1/2*m],[2+1/2*m],-c^2*x^2)/c/(2+m)

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Rubi [A]  time = 0.13, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {14, 4976, 459, 364} \[ \frac {d x^{m+1} \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac {e x^{m+3} \left (a+b \tan ^{-1}(c x)\right )}{m+3}-\frac {b x^{m+2} \left (\frac {c^2 d}{m+1}-\frac {e}{m+3}\right ) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{c (m+2)}-\frac {b e x^{m+2}}{c \left (m^2+5 m+6\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*e*x^(2 + m))/(c*(6 + 5*m + m^2))) + (d*x^(1 + m)*(a + b*ArcTan[c*x]))/(1 + m) + (e*x^(3 + m)*(a + b*ArcTa
n[c*x]))/(3 + m) - (b*((c^2*d)/(1 + m) - e/(3 + m))*x^(2 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(c^2
*x^2)])/(c*(2 + m))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 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]))

Rubi steps

\begin {align*} \int x^m \left (d+e x^2\right ) \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {d x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac {e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}-(b c) \int \frac {x^{1+m} \left (\frac {d}{1+m}+\frac {e x^2}{3+m}\right )}{1+c^2 x^2} \, dx\\ &=-\frac {b e x^{2+m}}{c \left (6+5 m+m^2\right )}+\frac {d x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac {e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}+\left (b c \left (-\frac {d}{1+m}+\frac {e}{c^2 (3+m)}\right )\right ) \int \frac {x^{1+m}}{1+c^2 x^2} \, dx\\ &=-\frac {b e x^{2+m}}{c \left (6+5 m+m^2\right )}+\frac {d x^{1+m} \left (a+b \tan ^{-1}(c x)\right )}{1+m}+\frac {e x^{3+m} \left (a+b \tan ^{-1}(c x)\right )}{3+m}-\frac {b c \left (\frac {d}{1+m}-\frac {e}{c^2 (3+m)}\right ) x^{2+m} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-c^2 x^2\right )}{2+m}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 119, normalized size = 0.98 \[ x^{m+1} \left (\frac {\frac {\left (d (m+3)+e (m+1) x^2\right ) \left (a+b \tan ^{-1}(c x)\right )}{m+1}-\frac {b c e x^3 \, _2F_1\left (1,\frac {m+4}{2};\frac {m+6}{2};-c^2 x^2\right )}{m+4}}{m+3}-\frac {b c d x \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{m^2+3 m+2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^(1 + m)*(-((b*c*d*x*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(c^2*x^2)])/(2 + 3*m + m^2)) + (((d*(3 + m)
+ e*(1 + m)*x^2)*(a + b*ArcTan[c*x]))/(1 + m) - (b*c*e*x^3*Hypergeometric2F1[1, (4 + m)/2, (6 + m)/2, -(c^2*x^
2)])/(4 + m))/(3 + m))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a e x^{2} + a d + {\left (b e x^{2} + b d\right )} \arctan \left (c x\right )\right )} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [F]  time = 1.23, size = 0, normalized size = 0.00 \[ \int x^{m} \left (e \,x^{2}+d \right ) \left (a +b \arctan \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a e x^{m + 3}}{m + 3} + \frac {a d x^{m + 1}}{m + 1} + \frac {{\left ({\left (b e m + b e\right )} x^{3} + {\left (b d m + 3 \, b d\right )} x\right )} x^{m} \arctan \left (c x\right ) - {\left (m^{2} + 4 \, m + 3\right )} \int \frac {{\left ({\left (b c e m + b c e\right )} x^{3} + {\left (b c d m + 3 \, b c d\right )} x\right )} x^{m}}{{\left (c^{2} m^{2} + 4 \, c^{2} m + 3 \, c^{2}\right )} x^{2} + m^{2} + 4 \, m + 3}\,{d x}}{m^{2} + 4 \, m + 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e*x^(m + 3)/(m + 3) + a*d*x^(m + 1)/(m + 1) + (((b*e*m + b*e)*x^3 + (b*d*m + 3*b*d)*x)*x^m*arctan(c*x) - (m^
2 + 4*m + 3)*integrate(((b*c*e*m + b*c*e)*x^3 + (b*c*d*m + 3*b*c*d)*x)*x^m/((c^2*m^2 + 4*c^2*m + 3*c^2)*x^2 +
m^2 + 4*m + 3), x))/(m^2 + 4*m + 3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^m\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,\left (e\,x^2+d\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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