3.1124 \(\int \frac {(d+e x^2) (a+b \tan ^{-1}(c x))}{x^7} \, dx\)

Optimal. Leaf size=105 \[ -\frac {d \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}+\frac {b c \left (2 c^2 d-3 e\right )}{36 x^3}-\frac {1}{12} b c^4 \left (2 c^2 d-3 e\right ) \tan ^{-1}(c x)-\frac {b c^3 \left (2 c^2 d-3 e\right )}{12 x}-\frac {b c d}{30 x^5} \]

[Out]

-1/30*b*c*d/x^5+1/36*b*c*(2*c^2*d-3*e)/x^3-1/12*b*c^3*(2*c^2*d-3*e)/x-1/12*b*c^4*(2*c^2*d-3*e)*arctan(c*x)-1/6
*d*(a+b*arctan(c*x))/x^6-1/4*e*(a+b*arctan(c*x))/x^4

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Rubi [A]  time = 0.11, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {14, 4976, 12, 453, 325, 203} \[ -\frac {d \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}+\frac {b c \left (2 c^2 d-3 e\right )}{36 x^3}-\frac {b c^3 \left (2 c^2 d-3 e\right )}{12 x}-\frac {1}{12} b c^4 \left (2 c^2 d-3 e\right ) \tan ^{-1}(c x)-\frac {b c d}{30 x^5} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcTan[c*x]))/x^7,x]

[Out]

-(b*c*d)/(30*x^5) + (b*c*(2*c^2*d - 3*e))/(36*x^3) - (b*c^3*(2*c^2*d - 3*e))/(12*x) - (b*c^4*(2*c^2*d - 3*e)*A
rcTan[c*x])/12 - (d*(a + b*ArcTan[c*x]))/(6*x^6) - (e*(a + b*ArcTan[c*x]))/(4*x^4)

Rule 12

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

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 325

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] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 453

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

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right ) \left (a+b \tan ^{-1}(c x)\right )}{x^7} \, dx &=-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-(b c) \int \frac {-2 d-3 e x^2}{12 x^6 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {1}{12} (b c) \int \frac {-2 d-3 e x^2}{x^6 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b c d}{30 x^5}-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {1}{12} \left (b c \left (2 c^2 d-3 e\right )\right ) \int \frac {1}{x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b c d}{30 x^5}+\frac {b c \left (2 c^2 d-3 e\right )}{36 x^3}-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}+\frac {1}{12} \left (b c^3 \left (2 c^2 d-3 e\right )\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b c d}{30 x^5}+\frac {b c \left (2 c^2 d-3 e\right )}{36 x^3}-\frac {b c^3 \left (2 c^2 d-3 e\right )}{12 x}-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {1}{12} \left (b c^5 \left (2 c^2 d-3 e\right )\right ) \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {b c d}{30 x^5}+\frac {b c \left (2 c^2 d-3 e\right )}{36 x^3}-\frac {b c^3 \left (2 c^2 d-3 e\right )}{12 x}-\frac {1}{12} b c^4 \left (2 c^2 d-3 e\right ) \tan ^{-1}(c x)-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 97, normalized size = 0.92 \[ -\frac {a d}{6 x^6}-\frac {a e}{4 x^4}-\frac {b c d \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-c^2 x^2\right )}{30 x^5}-\frac {b c e \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-c^2 x^2\right )}{12 x^3}-\frac {b d \tan ^{-1}(c x)}{6 x^6}-\frac {b e \tan ^{-1}(c x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*ArcTan[c*x]))/x^7,x]

[Out]

-1/6*(a*d)/x^6 - (a*e)/(4*x^4) - (b*d*ArcTan[c*x])/(6*x^6) - (b*e*ArcTan[c*x])/(4*x^4) - (b*c*d*Hypergeometric
2F1[-5/2, 1, -3/2, -(c^2*x^2)])/(30*x^5) - (b*c*e*Hypergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)])/(12*x^3)

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fricas [A]  time = 0.41, size = 98, normalized size = 0.93 \[ -\frac {15 \, {\left (2 \, b c^{5} d - 3 \, b c^{3} e\right )} x^{5} + 6 \, b c d x + 45 \, a e x^{2} - 5 \, {\left (2 \, b c^{3} d - 3 \, b c e\right )} x^{3} + 30 \, a d + 15 \, {\left ({\left (2 \, b c^{6} d - 3 \, b c^{4} e\right )} x^{6} + 3 \, b e x^{2} + 2 \, b d\right )} \arctan \left (c x\right )}{180 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^7,x, algorithm="fricas")

[Out]

-1/180*(15*(2*b*c^5*d - 3*b*c^3*e)*x^5 + 6*b*c*d*x + 45*a*e*x^2 - 5*(2*b*c^3*d - 3*b*c*e)*x^3 + 30*a*d + 15*((
2*b*c^6*d - 3*b*c^4*e)*x^6 + 3*b*e*x^2 + 2*b*d)*arctan(c*x))/x^6

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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((e*x^2+d)*(a+b*arctan(c*x))/x^7,x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.05, size = 106, normalized size = 1.01 \[ -\frac {a e}{4 x^{4}}-\frac {a d}{6 x^{6}}-\frac {b \arctan \left (c x \right ) e}{4 x^{4}}-\frac {b \arctan \left (c x \right ) d}{6 x^{6}}-\frac {c^{5} b d}{6 x}+\frac {b \,c^{3} e}{4 x}+\frac {c^{3} b d}{18 x^{3}}-\frac {c b e}{12 x^{3}}-\frac {b c d}{30 x^{5}}-\frac {c^{6} b \arctan \left (c x \right ) d}{6}+\frac {b \,c^{4} e \arctan \left (c x \right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arctan(c*x))/x^7,x)

[Out]

-1/4*a*e/x^4-1/6*a*d/x^6-1/4*b*arctan(c*x)*e/x^4-1/6*b*arctan(c*x)*d/x^6-1/6*c^5*b*d/x+1/4*b*c^3*e/x+1/18*c^3*
b*d/x^3-1/12*c*b*e/x^3-1/30*b*c*d/x^5-1/6*c^6*b*arctan(c*x)*d+1/4*b*c^4*e*arctan(c*x)

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maxima [A]  time = 0.42, size = 103, normalized size = 0.98 \[ -\frac {1}{90} \, {\left ({\left (15 \, c^{5} \arctan \left (c x\right ) + \frac {15 \, c^{4} x^{4} - 5 \, c^{2} x^{2} + 3}{x^{5}}\right )} c + \frac {15 \, \arctan \left (c x\right )}{x^{6}}\right )} b d + \frac {1}{12} \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} b e - \frac {a e}{4 \, x^{4}} - \frac {a d}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arctan(c*x))/x^7,x, algorithm="maxima")

[Out]

-1/90*((15*c^5*arctan(c*x) + (15*c^4*x^4 - 5*c^2*x^2 + 3)/x^5)*c + 15*arctan(c*x)/x^6)*b*d + 1/12*((3*c^3*arct
an(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*e - 1/4*a*e/x^4 - 1/6*a*d/x^6

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mupad [B]  time = 0.59, size = 130, normalized size = 1.24 \[ \frac {b\,c^4\,\mathrm {atan}\left (\frac {b\,c^2\,x\,\left (3\,e-2\,c^2\,d\right )}{3\,b\,c\,e-2\,b\,c^3\,d}\right )\,\left (3\,e-2\,c^2\,d\right )}{12}-\frac {\mathrm {atan}\left (c\,x\right )\,\left (\frac {b\,e\,x^2}{4}+\frac {b\,d}{6}\right )}{x^6}-\frac {x^3\,\left (b\,c\,e-\frac {2\,b\,c^3\,d}{3}\right )+2\,a\,d-c^2\,x^5\,\left (3\,b\,c\,e-2\,b\,c^3\,d\right )+3\,a\,e\,x^2+\frac {2\,b\,c\,d\,x}{5}}{12\,x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + e*x^2))/x^7,x)

[Out]

(b*c^4*atan((b*c^2*x*(3*e - 2*c^2*d))/(3*b*c*e - 2*b*c^3*d))*(3*e - 2*c^2*d))/12 - (atan(c*x)*((b*d)/6 + (b*e*
x^2)/4))/x^6 - (x^3*(b*c*e - (2*b*c^3*d)/3) + 2*a*d - c^2*x^5*(3*b*c*e - 2*b*c^3*d) + 3*a*e*x^2 + (2*b*c*d*x)/
5)/(12*x^6)

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sympy [A]  time = 1.60, size = 122, normalized size = 1.16 \[ - \frac {a d}{6 x^{6}} - \frac {a e}{4 x^{4}} - \frac {b c^{6} d \operatorname {atan}{\left (c x \right )}}{6} - \frac {b c^{5} d}{6 x} + \frac {b c^{4} e \operatorname {atan}{\left (c x \right )}}{4} + \frac {b c^{3} d}{18 x^{3}} + \frac {b c^{3} e}{4 x} - \frac {b c d}{30 x^{5}} - \frac {b c e}{12 x^{3}} - \frac {b d \operatorname {atan}{\left (c x \right )}}{6 x^{6}} - \frac {b e \operatorname {atan}{\left (c x \right )}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*atan(c*x))/x**7,x)

[Out]

-a*d/(6*x**6) - a*e/(4*x**4) - b*c**6*d*atan(c*x)/6 - b*c**5*d/(6*x) + b*c**4*e*atan(c*x)/4 + b*c**3*d/(18*x**
3) + b*c**3*e/(4*x) - b*c*d/(30*x**5) - b*c*e/(12*x**3) - b*d*atan(c*x)/(6*x**6) - b*e*atan(c*x)/(4*x**4)

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