3.1140 \(\int (d+e x^2)^3 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=188 \[ d^3 x \left (a+b \tan ^{-1}(c x)\right )+d^2 e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{5} d e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{7} e^3 x^7 \left (a+b \tan ^{-1}(c x)\right )-\frac {b e^2 x^4 \left (21 c^2 d-5 e\right )}{140 c^3}-\frac {b e x^2 \left (35 c^4 d^2-21 c^2 d e+5 e^2\right )}{70 c^5}-\frac {b \left (35 c^6 d^3-35 c^4 d^2 e+21 c^2 d e^2-5 e^3\right ) \log \left (c^2 x^2+1\right )}{70 c^7}-\frac {b e^3 x^6}{42 c} \]

[Out]

-1/70*b*e*(35*c^4*d^2-21*c^2*d*e+5*e^2)*x^2/c^5-1/140*b*(21*c^2*d-5*e)*e^2*x^4/c^3-1/42*b*e^3*x^6/c+d^3*x*(a+b
*arctan(c*x))+d^2*e*x^3*(a+b*arctan(c*x))+3/5*d*e^2*x^5*(a+b*arctan(c*x))+1/7*e^3*x^7*(a+b*arctan(c*x))-1/70*b
*(35*c^6*d^3-35*c^4*d^2*e+21*c^2*d*e^2-5*e^3)*ln(c^2*x^2+1)/c^7

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Rubi [A]  time = 0.15, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {194, 4912, 1810, 260} \[ d^2 e x^3 \left (a+b \tan ^{-1}(c x)\right )+d^3 x \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{5} d e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{7} e^3 x^7 \left (a+b \tan ^{-1}(c x)\right )-\frac {b e x^2 \left (35 c^4 d^2-21 c^2 d e+5 e^2\right )}{70 c^5}-\frac {b \left (-35 c^4 d^2 e+35 c^6 d^3+21 c^2 d e^2-5 e^3\right ) \log \left (c^2 x^2+1\right )}{70 c^7}-\frac {b e^2 x^4 \left (21 c^2 d-5 e\right )}{140 c^3}-\frac {b e^3 x^6}{42 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*e*(35*c^4*d^2 - 21*c^2*d*e + 5*e^2)*x^2)/(70*c^5) - (b*(21*c^2*d - 5*e)*e^2*x^4)/(140*c^3) - (b*e^3*x^6)/(
42*c) + d^3*x*(a + b*ArcTan[c*x]) + d^2*e*x^3*(a + b*ArcTan[c*x]) + (3*d*e^2*x^5*(a + b*ArcTan[c*x]))/5 + (e^3
*x^7*(a + b*ArcTan[c*x]))/7 - (b*(35*c^6*d^3 - 35*c^4*d^2*e + 21*c^2*d*e^2 - 5*e^3)*Log[1 + c^2*x^2])/(70*c^7)

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 260

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

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 4912

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[u/(1 + c^2*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x]
&& (IntegerQ[q] || ILtQ[q + 1/2, 0])

Rubi steps

\begin {align*} \int \left (d+e x^2\right )^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=d^3 x \left (a+b \tan ^{-1}(c x)\right )+d^2 e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{5} d e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{7} e^3 x^7 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac {d^3 x+d^2 e x^3+\frac {3}{5} d e^2 x^5+\frac {e^3 x^7}{7}}{1+c^2 x^2} \, dx\\ &=d^3 x \left (a+b \tan ^{-1}(c x)\right )+d^2 e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{5} d e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{7} e^3 x^7 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \left (\frac {e \left (35 c^4 d^2-21 c^2 d e+5 e^2\right ) x}{35 c^6}+\frac {\left (21 c^2 d-5 e\right ) e^2 x^3}{35 c^4}+\frac {e^3 x^5}{7 c^2}+\frac {\left (35 c^6 d^3-35 c^4 d^2 e+21 c^2 d e^2-5 e^3\right ) x}{35 c^6 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac {b e \left (35 c^4 d^2-21 c^2 d e+5 e^2\right ) x^2}{70 c^5}-\frac {b \left (21 c^2 d-5 e\right ) e^2 x^4}{140 c^3}-\frac {b e^3 x^6}{42 c}+d^3 x \left (a+b \tan ^{-1}(c x)\right )+d^2 e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{5} d e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{7} e^3 x^7 \left (a+b \tan ^{-1}(c x)\right )-\frac {\left (b \left (35 c^6 d^3-35 c^4 d^2 e+21 c^2 d e^2-5 e^3\right )\right ) \int \frac {x}{1+c^2 x^2} \, dx}{35 c^5}\\ &=-\frac {b e \left (35 c^4 d^2-21 c^2 d e+5 e^2\right ) x^2}{70 c^5}-\frac {b \left (21 c^2 d-5 e\right ) e^2 x^4}{140 c^3}-\frac {b e^3 x^6}{42 c}+d^3 x \left (a+b \tan ^{-1}(c x)\right )+d^2 e x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{5} d e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{7} e^3 x^7 \left (a+b \tan ^{-1}(c x)\right )-\frac {b \left (35 c^6 d^3-35 c^4 d^2 e+21 c^2 d e^2-5 e^3\right ) \log \left (1+c^2 x^2\right )}{70 c^7}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 192, normalized size = 1.02 \[ \frac {c^2 x \left (12 a c^5 \left (35 d^3+35 d^2 e x^2+21 d e^2 x^4+5 e^3 x^6\right )-b e x \left (c^4 \left (210 d^2+63 d e x^2+10 e^2 x^4\right )-3 c^2 e \left (42 d+5 e x^2\right )+30 e^2\right )\right )+12 b c^7 x \tan ^{-1}(c x) \left (35 d^3+35 d^2 e x^2+21 d e^2 x^4+5 e^3 x^6\right )-6 b \left (35 c^6 d^3-35 c^4 d^2 e+21 c^2 d e^2-5 e^3\right ) \log \left (c^2 x^2+1\right )}{420 c^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x*(12*a*c^5*(35*d^3 + 35*d^2*e*x^2 + 21*d*e^2*x^4 + 5*e^3*x^6) - b*e*x*(30*e^2 - 3*c^2*e*(42*d + 5*e*x^2)
+ c^4*(210*d^2 + 63*d*e*x^2 + 10*e^2*x^4))) + 12*b*c^7*x*(35*d^3 + 35*d^2*e*x^2 + 21*d*e^2*x^4 + 5*e^3*x^6)*A
rcTan[c*x] - 6*b*(35*c^6*d^3 - 35*c^4*d^2*e + 21*c^2*d*e^2 - 5*e^3)*Log[1 + c^2*x^2])/(420*c^7)

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fricas [A]  time = 0.51, size = 229, normalized size = 1.22 \[ \frac {60 \, a c^{7} e^{3} x^{7} + 252 \, a c^{7} d e^{2} x^{5} - 10 \, b c^{6} e^{3} x^{6} + 420 \, a c^{7} d^{2} e x^{3} + 420 \, a c^{7} d^{3} x - 3 \, {\left (21 \, b c^{6} d e^{2} - 5 \, b c^{4} e^{3}\right )} x^{4} - 6 \, {\left (35 \, b c^{6} d^{2} e - 21 \, b c^{4} d e^{2} + 5 \, b c^{2} e^{3}\right )} x^{2} + 12 \, {\left (5 \, b c^{7} e^{3} x^{7} + 21 \, b c^{7} d e^{2} x^{5} + 35 \, b c^{7} d^{2} e x^{3} + 35 \, b c^{7} d^{3} x\right )} \arctan \left (c x\right ) - 6 \, {\left (35 \, b c^{6} d^{3} - 35 \, b c^{4} d^{2} e + 21 \, b c^{2} d e^{2} - 5 \, b e^{3}\right )} \log \left (c^{2} x^{2} + 1\right )}{420 \, c^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(60*a*c^7*e^3*x^7 + 252*a*c^7*d*e^2*x^5 - 10*b*c^6*e^3*x^6 + 420*a*c^7*d^2*e*x^3 + 420*a*c^7*d^3*x - 3*(
21*b*c^6*d*e^2 - 5*b*c^4*e^3)*x^4 - 6*(35*b*c^6*d^2*e - 21*b*c^4*d*e^2 + 5*b*c^2*e^3)*x^2 + 12*(5*b*c^7*e^3*x^
7 + 21*b*c^7*d*e^2*x^5 + 35*b*c^7*d^2*e*x^3 + 35*b*c^7*d^3*x)*arctan(c*x) - 6*(35*b*c^6*d^3 - 35*b*c^4*d^2*e +
21*b*c^2*d*e^2 - 5*b*e^3)*log(c^2*x^2 + 1))/c^7

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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)^3*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.04, size = 239, normalized size = 1.27 \[ \frac {a \,x^{7} e^{3}}{7}+\frac {3 a \,x^{5} d \,e^{2}}{5}+a \,x^{3} d^{2} e +a \,d^{3} x +\frac {b \arctan \left (c x \right ) x^{7} e^{3}}{7}+\frac {3 b \arctan \left (c x \right ) x^{5} d \,e^{2}}{5}+b \arctan \left (c x \right ) x^{3} d^{2} e +b \arctan \left (c x \right ) d^{3} x -\frac {b \,x^{2} d^{2} e}{2 c}-\frac {3 b \,x^{4} d \,e^{2}}{20 c}-\frac {b \,e^{3} x^{6}}{42 c}+\frac {3 b \,x^{2} d \,e^{2}}{10 c^{3}}+\frac {b \,e^{3} x^{4}}{28 c^{3}}-\frac {b \,x^{2} e^{3}}{14 c^{5}}-\frac {b \ln \left (c^{2} x^{2}+1\right ) d^{3}}{2 c}+\frac {b \ln \left (c^{2} x^{2}+1\right ) d^{2} e}{2 c^{3}}-\frac {3 b \ln \left (c^{2} x^{2}+1\right ) d \,e^{2}}{10 c^{5}}+\frac {b \ln \left (c^{2} x^{2}+1\right ) e^{3}}{14 c^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*x^7*e^3+3/5*a*x^5*d*e^2+a*x^3*d^2*e+a*d^3*x+1/7*b*arctan(c*x)*x^7*e^3+3/5*b*arctan(c*x)*x^5*d*e^2+b*arct
an(c*x)*x^3*d^2*e+b*arctan(c*x)*d^3*x-1/2/c*b*x^2*d^2*e-3/20/c*b*x^4*d*e^2-1/42*b*e^3*x^6/c+3/10/c^3*b*x^2*d*e
^2+1/28/c^3*b*e^3*x^4-1/14/c^5*b*x^2*e^3-1/2/c*b*ln(c^2*x^2+1)*d^3+1/2/c^3*b*ln(c^2*x^2+1)*d^2*e-3/10/c^5*b*ln
(c^2*x^2+1)*d*e^2+1/14/c^7*b*ln(c^2*x^2+1)*e^3

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maxima [A]  time = 0.33, size = 222, normalized size = 1.18 \[ \frac {1}{7} \, a e^{3} x^{7} + \frac {3}{5} \, a d e^{2} x^{5} + a d^{2} e x^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d^{2} e + \frac {3}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b d e^{2} + \frac {1}{84} \, {\left (12 \, x^{7} \arctan \left (c x\right ) - c {\left (\frac {2 \, c^{4} x^{6} - 3 \, c^{2} x^{4} + 6 \, x^{2}}{c^{6}} - \frac {6 \, \log \left (c^{2} x^{2} + 1\right )}{c^{8}}\right )}\right )} b e^{3} + a d^{3} x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{3}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*e^3*x^7 + 3/5*a*d*e^2*x^5 + a*d^2*e*x^3 + 1/2*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b
*d^2*e + 3/20*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*d*e^2 + 1/84*(12*x^7*
arctan(c*x) - c*((2*c^4*x^6 - 3*c^2*x^4 + 6*x^2)/c^6 - 6*log(c^2*x^2 + 1)/c^8))*b*e^3 + a*d^3*x + 1/2*(2*c*x*a
rctan(c*x) - log(c^2*x^2 + 1))*b*d^3/c

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mupad [B]  time = 0.43, size = 238, normalized size = 1.27 \[ \frac {a\,e^3\,x^7}{7}+a\,d^3\,x-\frac {b\,d^3\,\ln \left (c^2\,x^2+1\right )}{2\,c}+\frac {b\,e^3\,\ln \left (c^2\,x^2+1\right )}{14\,c^7}-\frac {b\,e^3\,x^6}{42\,c}+\frac {b\,e^3\,x^4}{28\,c^3}-\frac {b\,e^3\,x^2}{14\,c^5}+b\,d^3\,x\,\mathrm {atan}\left (c\,x\right )+a\,d^2\,e\,x^3+\frac {3\,a\,d\,e^2\,x^5}{5}+\frac {b\,e^3\,x^7\,\mathrm {atan}\left (c\,x\right )}{7}+b\,d^2\,e\,x^3\,\mathrm {atan}\left (c\,x\right )+\frac {3\,b\,d\,e^2\,x^5\,\mathrm {atan}\left (c\,x\right )}{5}+\frac {b\,d^2\,e\,\ln \left (c^2\,x^2+1\right )}{2\,c^3}-\frac {3\,b\,d\,e^2\,\ln \left (c^2\,x^2+1\right )}{10\,c^5}-\frac {b\,d^2\,e\,x^2}{2\,c}-\frac {3\,b\,d\,e^2\,x^4}{20\,c}+\frac {3\,b\,d\,e^2\,x^2}{10\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*e^3*x^7)/7 + a*d^3*x - (b*d^3*log(c^2*x^2 + 1))/(2*c) + (b*e^3*log(c^2*x^2 + 1))/(14*c^7) - (b*e^3*x^6)/(42
*c) + (b*e^3*x^4)/(28*c^3) - (b*e^3*x^2)/(14*c^5) + b*d^3*x*atan(c*x) + a*d^2*e*x^3 + (3*a*d*e^2*x^5)/5 + (b*e
^3*x^7*atan(c*x))/7 + b*d^2*e*x^3*atan(c*x) + (3*b*d*e^2*x^5*atan(c*x))/5 + (b*d^2*e*log(c^2*x^2 + 1))/(2*c^3)
- (3*b*d*e^2*log(c^2*x^2 + 1))/(10*c^5) - (b*d^2*e*x^2)/(2*c) - (3*b*d*e^2*x^4)/(20*c) + (3*b*d*e^2*x^2)/(10*
c^3)

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sympy [A]  time = 3.10, size = 306, normalized size = 1.63 \[ \begin {cases} a d^{3} x + a d^{2} e x^{3} + \frac {3 a d e^{2} x^{5}}{5} + \frac {a e^{3} x^{7}}{7} + b d^{3} x \operatorname {atan}{\left (c x \right )} + b d^{2} e x^{3} \operatorname {atan}{\left (c x \right )} + \frac {3 b d e^{2} x^{5} \operatorname {atan}{\left (c x \right )}}{5} + \frac {b e^{3} x^{7} \operatorname {atan}{\left (c x \right )}}{7} - \frac {b d^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} - \frac {b d^{2} e x^{2}}{2 c} - \frac {3 b d e^{2} x^{4}}{20 c} - \frac {b e^{3} x^{6}}{42 c} + \frac {b d^{2} e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c^{3}} + \frac {3 b d e^{2} x^{2}}{10 c^{3}} + \frac {b e^{3} x^{4}}{28 c^{3}} - \frac {3 b d e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10 c^{5}} - \frac {b e^{3} x^{2}}{14 c^{5}} + \frac {b e^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{14 c^{7}} & \text {for}\: c \neq 0 \\a \left (d^{3} x + d^{2} e x^{3} + \frac {3 d e^{2} x^{5}}{5} + \frac {e^{3} x^{7}}{7}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d**3*x + a*d**2*e*x**3 + 3*a*d*e**2*x**5/5 + a*e**3*x**7/7 + b*d**3*x*atan(c*x) + b*d**2*e*x**3*a
tan(c*x) + 3*b*d*e**2*x**5*atan(c*x)/5 + b*e**3*x**7*atan(c*x)/7 - b*d**3*log(x**2 + c**(-2))/(2*c) - b*d**2*e
*x**2/(2*c) - 3*b*d*e**2*x**4/(20*c) - b*e**3*x**6/(42*c) + b*d**2*e*log(x**2 + c**(-2))/(2*c**3) + 3*b*d*e**2
*x**2/(10*c**3) + b*e**3*x**4/(28*c**3) - 3*b*d*e**2*log(x**2 + c**(-2))/(10*c**5) - b*e**3*x**2/(14*c**5) + b
*e**3*log(x**2 + c**(-2))/(14*c**7), Ne(c, 0)), (a*(d**3*x + d**2*e*x**3 + 3*d*e**2*x**5/5 + e**3*x**7/7), Tru
e))

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