3.13.41 \(\int x^{-3-2 p} (d+e x^2)^p (a+b \text {ArcTan}(c x)) \, dx\) [1241]

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

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {270, 5096, 12, 525, 524} \begin {gather*} -\frac {x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} (a+b \text {ArcTan}(c x))}{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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(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)])
/((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 12

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

Rule 270

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 5096

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^{-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*}

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Mathematica [A]
time = 0.45, size = 166, normalized size = 1.29 \begin {gather*} -\frac {x^{-2 (1+p)} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \left (b \left (c^2 d-e\right ) x F_1\left (-\frac {1}{2}-p;-p,1;\frac {1}{2}-p;-\frac {e x^2}{d},-c^2 x^2\right )+c (1+2 p) \left (d+e x^2\right ) \left (1+\frac {e x^2}{d}\right )^p (a+b \text {ArcTan}(c x))+b e x \, _2F_1\left (-\frac {1}{2}-p,-p;\frac {1}{2}-p;-\frac {e x^2}{d}\right )\right )}{2 c d (1+p) (1+2 p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((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)]))/(c*d*(1 + p)*(1 + 2*p)*x^(2*(1 + p))*(1 + (e*x^2)/d)^p)

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Maple [F]
time = 1.28, size = 0, normalized size = 0.00 \[\int x^{-3-2 p} \left (e \,x^{2}+d \right )^{p} \left (a +b \arctan \left (c x \right )\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*(x^2*e + d)^p*x^(-2*p - 3), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 6193 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^p}{x^{2\,p+3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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