3.1183 \(\int x^3 (d+e x^2)^{3/2} (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=279 \[ \frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e^2}+\frac {b \left (c^2 d-e\right )^{5/2} \left (2 c^2 d+5 e\right ) \tan ^{-1}\left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{35 c^7 e^2}-\frac {b x \left (13 c^2 d-30 e\right ) \left (d+e x^2\right )^{3/2}}{840 c^3 e}+\frac {b x \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) \sqrt {d+e x^2}}{560 c^5 e}+\frac {b \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{560 c^7 e^{3/2}}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c e} \]

[Out]

-1/840*b*(13*c^2*d-30*e)*x*(e*x^2+d)^(3/2)/c^3/e-1/42*b*x*(e*x^2+d)^(5/2)/c/e-1/5*d*(e*x^2+d)^(5/2)*(a+b*arcta
n(c*x))/e^2+1/7*(e*x^2+d)^(7/2)*(a+b*arctan(c*x))/e^2+1/35*b*(c^2*d-e)^(5/2)*(2*c^2*d+5*e)*arctan(x*(c^2*d-e)^
(1/2)/(e*x^2+d)^(1/2))/c^7/e^2+1/560*b*(35*c^6*d^3+70*c^4*d^2*e-168*c^2*d*e^2+80*e^3)*arctanh(x*e^(1/2)/(e*x^2
+d)^(1/2))/c^7/e^(3/2)+1/560*b*(3*c^4*d^2+54*c^2*d*e-40*e^2)*x*(e*x^2+d)^(1/2)/c^5/e

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Rubi [A]  time = 0.46, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {266, 43, 4976, 12, 528, 523, 217, 206, 377, 203} \[ \frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e^2}+\frac {b x \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) \sqrt {d+e x^2}}{560 c^5 e}+\frac {b \left (70 c^4 d^2 e+35 c^6 d^3-168 c^2 d e^2+80 e^3\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{560 c^7 e^{3/2}}+\frac {b \left (c^2 d-e\right )^{5/2} \left (2 c^2 d+5 e\right ) \tan ^{-1}\left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{35 c^7 e^2}-\frac {b x \left (13 c^2 d-30 e\right ) \left (d+e x^2\right )^{3/2}}{840 c^3 e}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(3*c^4*d^2 + 54*c^2*d*e - 40*e^2)*x*Sqrt[d + e*x^2])/(560*c^5*e) - (b*(13*c^2*d - 30*e)*x*(d + e*x^2)^(3/2)
)/(840*c^3*e) - (b*x*(d + e*x^2)^(5/2))/(42*c*e) - (d*(d + e*x^2)^(5/2)*(a + b*ArcTan[c*x]))/(5*e^2) + ((d + e
*x^2)^(7/2)*(a + b*ArcTan[c*x]))/(7*e^2) + (b*(c^2*d - e)^(5/2)*(2*c^2*d + 5*e)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqr
t[d + e*x^2]])/(35*c^7*e^2) + (b*(35*c^6*d^3 + 70*c^4*d^2*e - 168*c^2*d*e^2 + 80*e^3)*ArcTanh[(Sqrt[e]*x)/Sqrt
[d + e*x^2]])/(560*c^7*e^(3/2))

Rule 12

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 266

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

Rule 377

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 528

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]

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 x^3 \left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx &=-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e^2}-(b c) \int \frac {\left (d+e x^2\right )^{5/2} \left (-2 d+5 e x^2\right )}{35 e^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e^2}-\frac {(b c) \int \frac {\left (d+e x^2\right )^{5/2} \left (-2 d+5 e x^2\right )}{1+c^2 x^2} \, dx}{35 e^2}\\ &=-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e^2}-\frac {b \int \frac {\left (d+e x^2\right )^{3/2} \left (-d \left (12 c^2 d+5 e\right )+\left (13 c^2 d-30 e\right ) e x^2\right )}{1+c^2 x^2} \, dx}{210 c e^2}\\ &=-\frac {b \left (13 c^2 d-30 e\right ) x \left (d+e x^2\right )^{3/2}}{840 c^3 e}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e^2}-\frac {b \int \frac {\sqrt {d+e x^2} \left (-3 d \left (16 c^4 d^2+11 c^2 d e-10 e^2\right )-3 e \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) x^2\right )}{1+c^2 x^2} \, dx}{840 c^3 e^2}\\ &=\frac {b \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) x \sqrt {d+e x^2}}{560 c^5 e}-\frac {b \left (13 c^2 d-30 e\right ) x \left (d+e x^2\right )^{3/2}}{840 c^3 e}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e^2}-\frac {b \int \frac {-3 d \left (32 c^6 d^3+19 c^4 d^2 e-74 c^2 d e^2+40 e^3\right )-3 e \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) x^2}{\left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx}{1680 c^5 e^2}\\ &=\frac {b \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) x \sqrt {d+e x^2}}{560 c^5 e}-\frac {b \left (13 c^2 d-30 e\right ) x \left (d+e x^2\right )^{3/2}}{840 c^3 e}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e^2}+\frac {\left (b \left (c^2 d-e\right )^3 \left (2 c^2 d+5 e\right )\right ) \int \frac {1}{\left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx}{35 c^7 e^2}+\frac {\left (b \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right )\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{560 c^7 e}\\ &=\frac {b \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) x \sqrt {d+e x^2}}{560 c^5 e}-\frac {b \left (13 c^2 d-30 e\right ) x \left (d+e x^2\right )^{3/2}}{840 c^3 e}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e^2}+\frac {\left (b \left (c^2 d-e\right )^3 \left (2 c^2 d+5 e\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\left (-c^2 d+e\right ) x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{35 c^7 e^2}+\frac {\left (b \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{560 c^7 e}\\ &=\frac {b \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) x \sqrt {d+e x^2}}{560 c^5 e}-\frac {b \left (13 c^2 d-30 e\right ) x \left (d+e x^2\right )^{3/2}}{840 c^3 e}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \tan ^{-1}(c x)\right )}{7 e^2}+\frac {b \left (c^2 d-e\right )^{5/2} \left (2 c^2 d+5 e\right ) \tan ^{-1}\left (\frac {\sqrt {c^2 d-e} x}{\sqrt {d+e x^2}}\right )}{35 c^7 e^2}+\frac {b \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{560 c^7 e^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.71, size = 418, normalized size = 1.50 \[ -\frac {c^2 \sqrt {d+e x^2} \left (48 a c^5 \left (2 d-5 e x^2\right ) \left (d+e x^2\right )^2+b e x \left (c^4 \left (57 d^2+106 d e x^2+40 e^2 x^4\right )-6 c^2 e \left (37 d+10 e x^2\right )+120 e^2\right )\right )+48 b c^7 \tan ^{-1}(c x) \left (2 d-5 e x^2\right ) \left (d+e x^2\right )^{5/2}+24 i b \left (c^2 d-e\right )^{5/2} \left (2 c^2 d+5 e\right ) \log \left (-\frac {140 i c^8 e^2 \left (\sqrt {c^2 d-e} \sqrt {d+e x^2}+c d-i e x\right )}{b (c x+i) \left (c^2 d-e\right )^{7/2} \left (2 c^2 d+5 e\right )}\right )-24 i b \left (c^2 d-e\right )^{5/2} \left (2 c^2 d+5 e\right ) \log \left (\frac {140 i c^8 e^2 \left (\sqrt {c^2 d-e} \sqrt {d+e x^2}+c d+i e x\right )}{b (c x-i) \left (c^2 d-e\right )^{7/2} \left (2 c^2 d+5 e\right )}\right )-3 b \sqrt {e} \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) \log \left (\sqrt {e} \sqrt {d+e x^2}+e x\right )}{1680 c^7 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1680*(c^2*Sqrt[d + e*x^2]*(48*a*c^5*(2*d - 5*e*x^2)*(d + e*x^2)^2 + b*e*x*(120*e^2 - 6*c^2*e*(37*d + 10*e*x
^2) + c^4*(57*d^2 + 106*d*e*x^2 + 40*e^2*x^4))) + 48*b*c^7*(2*d - 5*e*x^2)*(d + e*x^2)^(5/2)*ArcTan[c*x] + (24
*I)*b*(c^2*d - e)^(5/2)*(2*c^2*d + 5*e)*Log[((-140*I)*c^8*e^2*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))
/(b*(c^2*d - e)^(7/2)*(2*c^2*d + 5*e)*(I + c*x))] - (24*I)*b*(c^2*d - e)^(5/2)*(2*c^2*d + 5*e)*Log[((140*I)*c^
8*e^2*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(7/2)*(2*c^2*d + 5*e)*(-I + c*x))] - 3*b
*Sqrt[e]*(35*c^6*d^3 + 70*c^4*d^2*e - 168*c^2*d*e^2 + 80*e^3)*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(c^7*e^2)

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fricas [A]  time = 14.79, size = 1566, normalized size = 5.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3360*(3*(35*b*c^6*d^3 + 70*b*c^4*d^2*e - 168*b*c^2*d*e^2 + 80*b*e^3)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 +
d)*sqrt(e)*x - d) + 24*(2*b*c^6*d^3 + b*c^4*d^2*e - 8*b*c^2*d*e^2 + 5*b*e^3)*sqrt(-c^2*d + e)*log(((c^4*d^2 -
8*c^2*d*e + 8*e^2)*x^4 - 2*(3*c^2*d^2 - 4*d*e)*x^2 + 4*((c^2*d - 2*e)*x^3 - d*x)*sqrt(-c^2*d + e)*sqrt(e*x^2 +
d) + d^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) + 2*(240*a*c^7*e^3*x^6 + 384*a*c^7*d*e^2*x^4 - 40*b*c^6*e^3*x^5 + 48*a*c
^7*d^2*e*x^2 - 96*a*c^7*d^3 - 2*(53*b*c^6*d*e^2 - 30*b*c^4*e^3)*x^3 - 3*(19*b*c^6*d^2*e - 74*b*c^4*d*e^2 + 40*
b*c^2*e^3)*x + 48*(5*b*c^7*e^3*x^6 + 8*b*c^7*d*e^2*x^4 + b*c^7*d^2*e*x^2 - 2*b*c^7*d^3)*arctan(c*x))*sqrt(e*x^
2 + d))/(c^7*e^2), 1/3360*(48*(2*b*c^6*d^3 + b*c^4*d^2*e - 8*b*c^2*d*e^2 + 5*b*e^3)*sqrt(c^2*d - e)*arctan(1/2
*sqrt(c^2*d - e)*((c^2*d - 2*e)*x^2 - d)*sqrt(e*x^2 + d)/((c^2*d*e - e^2)*x^3 + (c^2*d^2 - d*e)*x)) + 3*(35*b*
c^6*d^3 + 70*b*c^4*d^2*e - 168*b*c^2*d*e^2 + 80*b*e^3)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d)
+ 2*(240*a*c^7*e^3*x^6 + 384*a*c^7*d*e^2*x^4 - 40*b*c^6*e^3*x^5 + 48*a*c^7*d^2*e*x^2 - 96*a*c^7*d^3 - 2*(53*b
*c^6*d*e^2 - 30*b*c^4*e^3)*x^3 - 3*(19*b*c^6*d^2*e - 74*b*c^4*d*e^2 + 40*b*c^2*e^3)*x + 48*(5*b*c^7*e^3*x^6 +
8*b*c^7*d*e^2*x^4 + b*c^7*d^2*e*x^2 - 2*b*c^7*d^3)*arctan(c*x))*sqrt(e*x^2 + d))/(c^7*e^2), -1/1680*(3*(35*b*c
^6*d^3 + 70*b*c^4*d^2*e - 168*b*c^2*d*e^2 + 80*b*e^3)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - 12*(2*b*c^
6*d^3 + b*c^4*d^2*e - 8*b*c^2*d*e^2 + 5*b*e^3)*sqrt(-c^2*d + e)*log(((c^4*d^2 - 8*c^2*d*e + 8*e^2)*x^4 - 2*(3*
c^2*d^2 - 4*d*e)*x^2 + 4*((c^2*d - 2*e)*x^3 - d*x)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d) + d^2)/(c^4*x^4 + 2*c^2*x^
2 + 1)) - (240*a*c^7*e^3*x^6 + 384*a*c^7*d*e^2*x^4 - 40*b*c^6*e^3*x^5 + 48*a*c^7*d^2*e*x^2 - 96*a*c^7*d^3 - 2*
(53*b*c^6*d*e^2 - 30*b*c^4*e^3)*x^3 - 3*(19*b*c^6*d^2*e - 74*b*c^4*d*e^2 + 40*b*c^2*e^3)*x + 48*(5*b*c^7*e^3*x
^6 + 8*b*c^7*d*e^2*x^4 + b*c^7*d^2*e*x^2 - 2*b*c^7*d^3)*arctan(c*x))*sqrt(e*x^2 + d))/(c^7*e^2), 1/1680*(24*(2
*b*c^6*d^3 + b*c^4*d^2*e - 8*b*c^2*d*e^2 + 5*b*e^3)*sqrt(c^2*d - e)*arctan(1/2*sqrt(c^2*d - e)*((c^2*d - 2*e)*
x^2 - d)*sqrt(e*x^2 + d)/((c^2*d*e - e^2)*x^3 + (c^2*d^2 - d*e)*x)) - 3*(35*b*c^6*d^3 + 70*b*c^4*d^2*e - 168*b
*c^2*d*e^2 + 80*b*e^3)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (240*a*c^7*e^3*x^6 + 384*a*c^7*d*e^2*x^4
- 40*b*c^6*e^3*x^5 + 48*a*c^7*d^2*e*x^2 - 96*a*c^7*d^3 - 2*(53*b*c^6*d*e^2 - 30*b*c^4*e^3)*x^3 - 3*(19*b*c^6*d
^2*e - 74*b*c^4*d*e^2 + 40*b*c^2*e^3)*x + 48*(5*b*c^7*e^3*x^6 + 8*b*c^7*d*e^2*x^4 + b*c^7*d^2*e*x^2 - 2*b*c^7*
d^3)*arctan(c*x))*sqrt(e*x^2 + d))/(c^7*e^2)]

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

[Out]

sage0*x

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maple [F]  time = 1.11, size = 0, normalized size = 0.00 \[ \int x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arctan \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{35} \, {\left (\frac {5 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{e} - \frac {2 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d}{e^{2}}\right )} a + \frac {1}{2} \, b \int 2 \, {\left (e x^{5} + d x^{3}\right )} \sqrt {e x^{2} + d} \arctan \left (c x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(5*(e*x^2 + d)^(5/2)*x^2/e - 2*(e*x^2 + d)^(5/2)*d/e^2)*a + 1/2*b*integrate(2*(e*x^5 + d*x^3)*sqrt(e*x^2
+ d)*arctan(c*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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