3.13.43 \(\int x^{-5-2 p} (d+e x^2)^p (a+b \text {ArcTan}(c x)) \, dx\) [1243]

Optimal. Leaf size=285 \[ -\frac {b \left (e+c^2 d (1+p)\right ) x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} F_1\left (\frac {1}{2} (-3-2 p);1,-1-p;\frac {1}{2} (-1-2 p);-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c d (1+p) (2+p) (3+2 p)}+\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \text {ArcTan}(c x))}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} (a+b \text {ArcTan}(c x))}{2 d (2+p)}+\frac {b e x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \, _2F_1\left (\frac {1}{2} (-3-2 p),-1-p;\frac {1}{2} (-1-2 p);-\frac {e x^2}{d}\right )}{2 c d \left (6+13 p+9 p^2+2 p^3\right )} \]

[Out]

-1/2*b*(e+c^2*d*(1+p))*x^(-3-2*p)*(e*x^2+d)^p*AppellF1(-3/2-p,1,-1-p,-1/2-p,-c^2*x^2,-e*x^2/d)/c/d/(3+2*p)/(p^
2+3*p+2)/((1+e*x^2/d)^p)+1/2*e*(e*x^2+d)^(1+p)*(a+b*arctan(c*x))/d^2/(1+p)/(2+p)/(x^(2+2*p))-1/2*(e*x^2+d)^(1+
p)*(a+b*arctan(c*x))/d/(2+p)/(x^(4+2*p))+1/2*b*e*x^(-3-2*p)*(e*x^2+d)^p*hypergeom([-1-p, -3/2-p],[-1/2-p],-e*x
^2/d)/c/d/(2*p^3+9*p^2+13*p+6)/((1+e*x^2/d)^p)

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Rubi [A]
time = 0.27, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {277, 270, 5096, 12, 598, 372, 371, 525, 524} \begin {gather*} \frac {e x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} (a+b \text {ArcTan}(c x))}{2 d^2 (p+1) (p+2)}-\frac {x^{-2 (p+2)} \left (d+e x^2\right )^{p+1} (a+b \text {ArcTan}(c x))}{2 d (p+2)}-\frac {b x^{-2 p-3} \left (c^2 d (p+1)+e\right ) \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} F_1\left (\frac {1}{2} (-2 p-3);1,-p-1;\frac {1}{2} (-2 p-1);-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c d (p+1) (p+2) (2 p+3)}+\frac {b e x^{-2 p-3} \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} (-2 p-3),-p-1;\frac {1}{2} (-2 p-1);-\frac {e x^2}{d}\right )}{2 c d \left (2 p^3+9 p^2+13 p+6\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-5 - 2*p)*(d + e*x^2)^p*(a + b*ArcTan[c*x]),x]

[Out]

-1/2*(b*(e + c^2*d*(1 + p))*x^(-3 - 2*p)*(d + e*x^2)^p*AppellF1[(-3 - 2*p)/2, 1, -1 - p, (-1 - 2*p)/2, -(c^2*x
^2), -((e*x^2)/d)])/(c*d*(1 + p)*(2 + p)*(3 + 2*p)*(1 + (e*x^2)/d)^p) + (e*(d + e*x^2)^(1 + p)*(a + b*ArcTan[c
*x]))/(2*d^2*(1 + p)*(2 + p)*x^(2*(1 + p))) - ((d + e*x^2)^(1 + p)*(a + b*ArcTan[c*x]))/(2*d*(2 + p)*x^(2*(2 +
p))) + (b*e*x^(-3 - 2*p)*(d + e*x^2)^p*Hypergeometric2F1[(-3 - 2*p)/2, -1 - p, (-1 - 2*p)/2, -((e*x^2)/d)])/(
2*c*d*(6 + 13*p + 9*p^2 + 2*p^3)*(1 + (e*x^2)/d)^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 277

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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

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 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 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^{-5-2 p} \left (d+e x^2\right )^p \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}-(b c) \int \frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (-d-d p+e x^2\right )}{2 d^2 (1+p) (2+p) \left (1+c^2 x^2\right )} \, dx\\ &=\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}-\frac {(b c) \int \frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (-d-d p+e x^2\right )}{1+c^2 x^2} \, dx}{2 d^2 (1+p) (2+p)}\\ &=\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}-\frac {(b c) \int \left (\frac {e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p}}{c^2}+\frac {\left (-e+c^2 (-d-d p)\right ) x^{-2 (2+p)} \left (d+e x^2\right )^{1+p}}{c^2 \left (1+c^2 x^2\right )}\right ) \, dx}{2 d^2 (1+p) (2+p)}\\ &=\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}-\frac {(b e) \int x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \, dx}{2 c d^2 (1+p) (2+p)}+\frac {\left (b \left (e+c^2 d (1+p)\right )\right ) \int \frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p}}{1+c^2 x^2} \, dx}{2 c d^2 (1+p) (2+p)}\\ &=\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}-\frac {\left (b e \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p}\right ) \int x^{-2 (2+p)} \left (1+\frac {e x^2}{d}\right )^{1+p} \, dx}{2 c d (1+p) (2+p)}+\frac {\left (b \left (e+c^2 d (1+p)\right ) \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p}\right ) \int \frac {x^{-2 (2+p)} \left (1+\frac {e x^2}{d}\right )^{1+p}}{1+c^2 x^2} \, dx}{2 c d (1+p) (2+p)}\\ &=-\frac {b \left (e+c^2 d (1+p)\right ) x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} F_1\left (\frac {1}{2} (-3-2 p);1,-1-p;\frac {1}{2} (-1-2 p);-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c d (1+p) (2+p) (3+2 p)}+\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}+\frac {b e x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \, _2F_1\left (\frac {1}{2} (-3-2 p),-1-p;\frac {1}{2} (-1-2 p);-\frac {e x^2}{d}\right )}{2 c d \left (6+13 p+9 p^2+2 p^3\right )}\\ \end {align*}

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Mathematica [F]
time = 3.33, size = 0, normalized size = 0.00 \begin {gather*} \int x^{-5-2 p} \left (d+e x^2\right )^p (a+b \text {ArcTan}(c x)) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[x^(-5 - 2*p)*(d + e*x^2)^p*(a + b*ArcTan[c*x]),x]

[Out]

Integrate[x^(-5 - 2*p)*(d + e*x^2)^p*(a + b*ArcTan[c*x]), x]

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Maple [F]
time = 1.30, size = 0, normalized size = 0.00 \[\int x^{-5-2 p} \left (e \,x^{2}+d \right )^{p} \left (a +b \arctan \left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-5-2*p)*(e*x^2+d)^p*(a+b*arctan(c*x)),x)

[Out]

int(x^(-5-2*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^(-5-2*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^5, x) + 1/2*(x^4*e^2 - d*p*x^2*e - d^2*(p + 1))*a*
e^(p*log(x^2*e + d) - 2*p*log(x))/((p^2 + 3*p + 2)*d^2*x^4)

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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^(-5-2*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 - 5), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-5-2*p)*(e*x**2+d)**p*(a+b*atan(c*x)),x)

[Out]

Timed out

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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^(-5-2*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 - 5), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^p}{x^{2\,p+5}} \,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 + 5),x)

[Out]

int(((a + b*atan(c*x))*(d + e*x^2)^p)/x^(2*p + 5), x)

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