Integrand size = 23, antiderivative size = 279 \[ \int x^3 \left (d+e x^2\right )^{3/2} (a+b \arctan (c x)) \, dx=\frac {b \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) x \sqrt {d+e x^2}}{560 c^5 e}-\frac {b \left (13 c^2 d-30 e\right ) x \left (d+e x^2\right )^{3/2}}{840 c^3 e}-\frac {b x \left (d+e x^2\right )^{5/2}}{42 c e}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}+\frac {b \left (c^2 d-e\right )^{5/2} \left (2 c^2 d+5 e\right ) \arctan \left (\frac {\sqrt {c^2 d-e} x}{\sqrt {d+e x^2}}\right )}{35 c^7 e^2}+\frac {b \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{560 c^7 e^{3/2}} \]
-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*arctan(c*x))/e^2+1/7*(e*x^2+d)^(7/2)*(a+b*a rctan(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
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.50 \[ \int x^3 \left (d+e x^2\right )^{3/2} (a+b \arctan (c x)) \, dx=-\frac {c^2 \sqrt {d+e x^2} \left (48 a c^5 \left (2 d-5 e x^2\right ) \left (d+e x^2\right )^2+b e x \left (120 e^2-6 c^2 e \left (37 d+10 e x^2\right )+c^4 \left (57 d^2+106 d e x^2+40 e^2 x^4\right )\right )\right )+48 b c^7 \left (2 d-5 e x^2\right ) \left (d+e x^2\right )^{5/2} \arctan (c x)+24 i b \left (c^2 d-e\right )^{5/2} \left (2 c^2 d+5 e\right ) \log \left (-\frac {140 i c^8 e^2 \left (c d-i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \left (c^2 d-e\right )^{7/2} \left (2 c^2 d+5 e\right ) (i+c x)}\right )-24 i b \left (c^2 d-e\right )^{5/2} \left (2 c^2 d+5 e\right ) \log \left (\frac {140 i c^8 e^2 \left (c d+i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \left (c^2 d-e\right )^{7/2} \left (2 c^2 d+5 e\right ) (-i+c x)}\right )-3 b \sqrt {e} \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{1680 c^7 e^2} \]
-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)
Time = 0.61 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {5511, 27, 403, 403, 27, 403, 398, 224, 219, 291, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x^3 \left (d+e x^2\right )^{3/2} (a+b \arctan (c x)) \, dx\) |
| \(\Big \downarrow \) 5511 |
| \(\displaystyle -b c \int -\frac {\left (2 d-5 e x^2\right ) \left (e x^2+d\right )^{5/2}}{35 e^2 \left (c^2 x^2+1\right )}dx+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \int \frac {\left (2 d-5 e x^2\right ) \left (e x^2+d\right )^{5/2}}{c^2 x^2+1}dx}{35 e^2}+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}\) |
| \(\Big \downarrow \) 403 |
| \(\displaystyle \frac {b c \left (\frac {\int \frac {\left (e x^2+d\right )^{3/2} \left (d \left (12 d c^2+5 e\right )-\left (13 c^2 d-30 e\right ) e x^2\right )}{c^2 x^2+1}dx}{6 c^2}-\frac {5 e x \left (d+e x^2\right )^{5/2}}{6 c^2}\right )}{35 e^2}+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}\) |
| \(\Big \downarrow \) 403 |
| \(\displaystyle \frac {b c \left (\frac {\frac {\int \frac {3 \sqrt {e x^2+d} \left (e \left (3 d^2 c^4+54 d e c^2-40 e^2\right ) x^2+d \left (16 d^2 c^4+11 d e c^2-10 e^2\right )\right )}{c^2 x^2+1}dx}{4 c^2}-\frac {e x \left (13 c^2 d-30 e\right ) \left (d+e x^2\right )^{3/2}}{4 c^2}}{6 c^2}-\frac {5 e x \left (d+e x^2\right )^{5/2}}{6 c^2}\right )}{35 e^2}+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \left (\frac {\frac {3 \int \frac {\sqrt {e x^2+d} \left (e \left (3 d^2 c^4+54 d e c^2-40 e^2\right ) x^2+d \left (16 d^2 c^4+11 d e c^2-10 e^2\right )\right )}{c^2 x^2+1}dx}{4 c^2}-\frac {e x \left (13 c^2 d-30 e\right ) \left (d+e x^2\right )^{3/2}}{4 c^2}}{6 c^2}-\frac {5 e x \left (d+e x^2\right )^{5/2}}{6 c^2}\right )}{35 e^2}+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}\) |
| \(\Big \downarrow \) 403 |
| \(\displaystyle \frac {b c \left (\frac {\frac {3 \left (\frac {\int \frac {e \left (35 d^3 c^6+70 d^2 e c^4-168 d e^2 c^2+80 e^3\right ) x^2+d \left (32 d^3 c^6+19 d^2 e c^4-74 d e^2 c^2+40 e^3\right )}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx}{2 c^2}+\frac {e x \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) \sqrt {d+e x^2}}{2 c^2}\right )}{4 c^2}-\frac {e x \left (13 c^2 d-30 e\right ) \left (d+e x^2\right )^{3/2}}{4 c^2}}{6 c^2}-\frac {5 e x \left (d+e x^2\right )^{5/2}}{6 c^2}\right )}{35 e^2}+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {b c \left (\frac {\frac {3 \left (\frac {\frac {16 \left (2 c^2 d+5 e\right ) \left (c^2 d-e\right )^3 \int \frac {1}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx}{c^2}+\frac {e \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) \int \frac {1}{\sqrt {e x^2+d}}dx}{c^2}}{2 c^2}+\frac {e x \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) \sqrt {d+e x^2}}{2 c^2}\right )}{4 c^2}-\frac {e x \left (13 c^2 d-30 e\right ) \left (d+e x^2\right )^{3/2}}{4 c^2}}{6 c^2}-\frac {5 e x \left (d+e x^2\right )^{5/2}}{6 c^2}\right )}{35 e^2}+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {b c \left (\frac {\frac {3 \left (\frac {\frac {16 \left (2 c^2 d+5 e\right ) \left (c^2 d-e\right )^3 \int \frac {1}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx}{c^2}+\frac {e \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}}{c^2}}{2 c^2}+\frac {e x \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) \sqrt {d+e x^2}}{2 c^2}\right )}{4 c^2}-\frac {e x \left (13 c^2 d-30 e\right ) \left (d+e x^2\right )^{3/2}}{4 c^2}}{6 c^2}-\frac {5 e x \left (d+e x^2\right )^{5/2}}{6 c^2}\right )}{35 e^2}+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b c \left (\frac {\frac {3 \left (\frac {\frac {16 \left (2 c^2 d+5 e\right ) \left (c^2 d-e\right )^3 \int \frac {1}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx}{c^2}+\frac {\sqrt {e} \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c^2}}{2 c^2}+\frac {e x \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) \sqrt {d+e x^2}}{2 c^2}\right )}{4 c^2}-\frac {e x \left (13 c^2 d-30 e\right ) \left (d+e x^2\right )^{3/2}}{4 c^2}}{6 c^2}-\frac {5 e x \left (d+e x^2\right )^{5/2}}{6 c^2}\right )}{35 e^2}+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {b c \left (\frac {\frac {3 \left (\frac {\frac {16 \left (2 c^2 d+5 e\right ) \left (c^2 d-e\right )^3 \int \frac {1}{1-\frac {\left (e-c^2 d\right ) x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}}{c^2}+\frac {\sqrt {e} \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c^2}}{2 c^2}+\frac {e x \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) \sqrt {d+e x^2}}{2 c^2}\right )}{4 c^2}-\frac {e x \left (13 c^2 d-30 e\right ) \left (d+e x^2\right )^{3/2}}{4 c^2}}{6 c^2}-\frac {5 e x \left (d+e x^2\right )^{5/2}}{6 c^2}\right )}{35 e^2}+\frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{7/2} (a+b \arctan (c x))}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 e^2}+\frac {b c \left (\frac {\frac {3 \left (\frac {\frac {16 \left (2 c^2 d+5 e\right ) \left (c^2 d-e\right )^{5/2} \arctan \left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{c^2}+\frac {\sqrt {e} \left (35 c^6 d^3+70 c^4 d^2 e-168 c^2 d e^2+80 e^3\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c^2}}{2 c^2}+\frac {e x \left (3 c^4 d^2+54 c^2 d e-40 e^2\right ) \sqrt {d+e x^2}}{2 c^2}\right )}{4 c^2}-\frac {e x \left (13 c^2 d-30 e\right ) \left (d+e x^2\right )^{3/2}}{4 c^2}}{6 c^2}-\frac {5 e x \left (d+e x^2\right )^{5/2}}{6 c^2}\right )}{35 e^2}\) |
-1/5*(d*(d + e*x^2)^(5/2)*(a + b*ArcTan[c*x]))/e^2 + ((d + e*x^2)^(7/2)*(a + b*ArcTan[c*x]))/(7*e^2) + (b*c*((-5*e*x*(d + e*x^2)^(5/2))/(6*c^2) + (- 1/4*((13*c^2*d - 30*e)*e*x*(d + e*x^2)^(3/2))/c^2 + (3*((e*(3*c^4*d^2 + 54 *c^2*d*e - 40*e^2)*x*Sqrt[d + e*x^2])/(2*c^2) + ((16*(c^2*d - e)^(5/2)*(2* c^2*d + 5*e)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/c^2 + (Sqrt[e]*( 35*c^6*d^3 + 70*c^4*d^2*e - 168*c^2*d*e^2 + 80*e^3)*ArcTanh[(Sqrt[e]*x)/Sq rt[d + e*x^2]])/c^2)/(2*c^2)))/(4*c^2))/(6*c^2)))/(35*e^2)
3.12.83.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
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]}, Sim p[(a + b*ArcTan[c*x]) u, x] - Simp[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] && !ILt Q[(m - 1)/2, 0]))
\[\int x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arctan \left (c x \right )\right )d x\]
Time = 8.76 (sec) , antiderivative size = 1566, normalized size of antiderivative = 5.61 \[ \int x^3 \left (d+e x^2\right )^{3/2} (a+b \arctan (c x)) \, dx=\text {Too large to display} \]
[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)*sq rt(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)*sqr t(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*sq rt(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)*...
\[ \int x^3 \left (d+e x^2\right )^{3/2} (a+b \arctan (c x)) \, dx=\int x^{3} \left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{\frac {3}{2}}\, dx \]
Exception generated. \[ \int x^3 \left (d+e x^2\right )^{3/2} (a+b \arctan (c x)) \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int x^3 \left (d+e x^2\right )^{3/2} (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arctan \left (c x\right ) + a\right )} x^{3} \,d x } \]
Timed out. \[ \int x^3 \left (d+e x^2\right )^{3/2} (a+b \arctan (c x)) \, dx=\int x^3\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^{3/2} \,d x \]