3.1241 \(\int x^{-3-2 p} (d+e x^2)^p (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=129 \[ -\frac{x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d (p+1)}-\frac{b c x^{-2 p-1} \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} F_1\left (\frac{1}{2} (-2 p-1);1,-p-1;\frac{1}{2} (1-2 p);-c^2 x^2,-\frac{e x^2}{d}\right )}{2 \left (2 p^2+3 p+1\right )} \]

[Out]

-(b*c*x^(-1 - 2*p)*(d + e*x^2)^p*AppellF1[(-1 - 2*p)/2, 1, -1 - p, (1 - 2*p)/2, -(c^2*x^2), -((e*x^2)/d)])/(2*
(1 + 3*p + 2*p^2)*(1 + (e*x^2)/d)^p) - ((d + e*x^2)^(1 + p)*(a + b*ArcTan[c*x]))/(2*d*(1 + p)*x^(2*(1 + p)))

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Rubi [A]  time = 0.166492, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {264, 4976, 12, 511, 510} \[ -\frac{x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d (p+1)}-\frac{b c x^{-2 p-1} \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} F_1\left (\frac{1}{2} (-2 p-1);1,-p-1;\frac{1}{2} (1-2 p);-c^2 x^2,-\frac{e x^2}{d}\right )}{2 \left (2 p^2+3 p+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^(-3 - 2*p)*(d + e*x^2)^p*(a + b*ArcTan[c*x]),x]

[Out]

-(b*c*x^(-1 - 2*p)*(d + e*x^2)^p*AppellF1[(-1 - 2*p)/2, 1, -1 - p, (1 - 2*p)/2, -(c^2*x^2), -((e*x^2)/d)])/(2*
(1 + 3*p + 2*p^2)*(1 + (e*x^2)/d)^p) - ((d + e*x^2)^(1 + p)*(a + b*ArcTan[c*x]))/(2*d*(1 + p)*x^(2*(1 + p)))

Rule 264

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

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 511

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

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int x^{-3-2 p} \left (d+e x^2\right )^p \left (a+b \tan ^{-1}(c x)\right ) \, dx &=-\frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (1+p)}-(b c) \int -\frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p}}{2 d (1+p) \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (1+p)}+\frac{(b c) \int \frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p}}{1+c^2 x^2} \, dx}{2 d (1+p)}\\ &=-\frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (1+p)}+\frac{\left (b c \left (d+e x^2\right )^p \left (1+\frac{e x^2}{d}\right )^{-p}\right ) \int \frac{x^{-2 (1+p)} \left (1+\frac{e x^2}{d}\right )^{1+p}}{1+c^2 x^2} \, dx}{2 (1+p)}\\ &=-\frac{b c x^{-1-2 p} \left (d+e x^2\right )^p \left (1+\frac{e x^2}{d}\right )^{-p} F_1\left (\frac{1}{2} (-1-2 p);1,-1-p;\frac{1}{2} (1-2 p);-c^2 x^2,-\frac{e x^2}{d}\right )}{2 \left (1+3 p+2 p^2\right )}-\frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.434221, size = 166, normalized size = 1.29 \[ -\frac{x^{-2 (p+1)} \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} \left (b e x \text{Hypergeometric2F1}\left (-p-\frac{1}{2},-p,\frac{1}{2}-p,-\frac{e x^2}{d}\right )+c (2 p+1) \left (d+e x^2\right ) \left (\frac{e x^2}{d}+1\right )^p \left (a+b \tan ^{-1}(c x)\right )+b x \left (c^2 d-e\right ) F_1\left (-p-\frac{1}{2};-p,1;\frac{1}{2}-p;-\frac{e x^2}{d},-c^2 x^2\right )\right )}{2 c d (p+1) (2 p+1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^(-3 - 2*p)*(d + e*x^2)^p*(a + b*ArcTan[c*x]),x]

[Out]

-((d + e*x^2)^p*(b*(c^2*d - e)*x*AppellF1[-1/2 - p, -p, 1, 1/2 - p, -((e*x^2)/d), -(c^2*x^2)] + c*(1 + 2*p)*(d
+ e*x^2)*(1 + (e*x^2)/d)^p*(a + b*ArcTan[c*x]) + b*e*x*Hypergeometric2F1[-1/2 - p, -p, 1/2 - p, -((e*x^2)/d)]
))/(2*c*d*(1 + p)*(1 + 2*p)*x^(2*(1 + p))*(1 + (e*x^2)/d)^p)

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Maple [F]  time = 0.886, size = 0, normalized size = 0. \begin{align*} \int{x}^{-3-2\,p} \left ( e{x}^{2}+d \right ) ^{p} \left ( a+b\arctan \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-3-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x)

[Out]

int(x^(-3-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{\arctan \left (c x\right ) e^{\left (p \log \left (e x^{2} + d\right ) - 2 \, p \log \left (x\right )\right )}}{x^{3}}\,{d x} - \frac{{\left (e x^{2} + d\right )} a e^{\left (p \log \left (e x^{2} + d\right ) - 2 \, p \log \left (x\right )\right )}}{2 \, d{\left (p + 1\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-3-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \arctan \left (c x\right ) + a\right )}{\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-3-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

integral((b*arctan(c*x) + a)*(e*x^2 + d)^p*x^(-2*p - 3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-3-2*p)*(e*x**2+d)**p*(a+b*atan(c*x)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arctan \left (c x\right ) + a\right )}{\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-3-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)*(e*x^2 + d)^p*x^(-2*p - 3), x)