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

Optimal. Leaf size=122 \[ -\frac{b x^{m+2} \left (\frac{c^2 d}{m+1}-\frac{e}{m+3}\right ) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-c^2 x^2\right )}{c (m+2)}+\frac{d x^{m+1} \left (a+b \tan ^{-1}(c x)\right )}{m+1}+\frac{e x^{m+3} \left (a+b \tan ^{-1}(c x)\right )}{m+3}-\frac{b e x^{m+2}}{c \left (m^2+5 m+6\right )} \]

[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))

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Rubi [A]  time = 0.126968, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, 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 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 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 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])

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*}

Mathematica [A]  time = 0.172725, size = 119, normalized size = 0.98 \[ x^{m+1} \left (\frac{\frac{\left (d (m+3)+e (m+1) x^2\right ) \left (a+b \tan ^{-1}(c x)\right )}{m+1}-\frac{b c e x^3 \text{Hypergeometric2F1}\left (1,\frac{m+4}{2},\frac{m+6}{2},-c^2 x^2\right )}{m+4}}{m+3}-\frac{b c d x \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-c^2 x^2\right )}{m^2+3 m+2}\right ) \]

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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Maple [F]  time = 0.678, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( e{x}^{2}+d \right ) \left ( a+b\arctan \left ( cx \right ) \right ) \, dx \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \left (a + b \operatorname{atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}{\left (b \arctan \left (c x\right ) + a\right )} x^{m}\,{d x} \end{align*}

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]

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