Integrand size = 26, antiderivative size = 199 \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=-\frac {c (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d^3}-\frac {(9 b c-5 a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 d^3}+\frac {b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^3}+\frac {3 (b c-a d) (5 b c-a d) e^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 \sqrt {b} d^{7/2}} \]
3/8*(-a*d+b*c)*(-a*d+5*b*c)*e^(3/2)*arctanh(d^(1/2)*(e*(b*x^2+a)/(d*x^2+c) )^(1/2)/b^(1/2)/e^(1/2))/d^(7/2)/b^(1/2)-c*(-a*d+b*c)*e*(e*(b*x^2+a)/(d*x^ 2+c))^(1/2)/d^3-1/8*(-5*a*d+9*b*c)*e*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/ 2)/d^3+1/4*b*e*(d*x^2+c)^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/d^3
Time = 2.85 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.89 \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {b} \sqrt {d} \sqrt {a+b x^2} \left (a d \left (13 c+5 d x^2\right )+b \left (-15 c^2-5 c d x^2+2 d^2 x^4\right )\right )+3 \left (5 b^2 c^2-6 a b c d+a^2 d^2\right ) \sqrt {c+d x^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )\right )}{8 \sqrt {b} d^{7/2} \sqrt {a+b x^2}} \]
(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[b]*Sqrt[d]*Sqrt[a + b*x^2]*(a*d *(13*c + 5*d*x^2) + b*(-15*c^2 - 5*c*d*x^2 + 2*d^2*x^4)) + 3*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*Sqrt[c + d*x^2]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/(Sq rt[d]*Sqrt[a + b*x^2])]))/(8*Sqrt[b]*d^(7/2)*Sqrt[a + b*x^2])
Time = 0.41 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2053, 2052, 25, 360, 1471, 27, 299, 221}
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.
| \(\displaystyle \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 2053 |
| \(\displaystyle \frac {1}{2} \int x^2 \left (\frac {e \left (b x^2+a\right )}{d x^2+c}\right )^{3/2}dx^2\) |
| \(\Big \downarrow \) 2052 |
| \(\displaystyle e (b c-a d) \int -\frac {x^8 \left (a e-c x^4\right )}{\left (b e-d x^4\right )^3}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\left (e (b c-a d) \int \frac {x^8 \left (a e-c x^4\right )}{\left (b e-d x^4\right )^3}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}\right )\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle e (b c-a d) \left (\frac {b e^2 (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3 \left (b e-d x^4\right )^2}-\frac {\int \frac {4 c d^2 x^8+4 d (b c-a d) e x^4+b (b c-a d) e^2}{\left (b e-d x^4\right )^2}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{4 d^3}\right )\) |
| \(\Big \downarrow \) 1471 |
| \(\displaystyle e (b c-a d) \left (\frac {b e^2 (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3 \left (b e-d x^4\right )^2}-\frac {\frac {e (9 b c-5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 \left (b e-d x^4\right )}-\frac {\int \frac {b e \left (8 c d x^4+(7 b c-3 a d) e\right )}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{2 b e}}{4 d^3}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e (b c-a d) \left (\frac {b e^2 (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3 \left (b e-d x^4\right )^2}-\frac {\frac {e (9 b c-5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 \left (b e-d x^4\right )}-\frac {1}{2} \int \frac {8 c d x^4+(7 b c-3 a d) e}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{4 d^3}\right )\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle e (b c-a d) \left (\frac {b e^2 (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3 \left (b e-d x^4\right )^2}-\frac {\frac {1}{2} \left (8 c \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}-3 e (5 b c-a d) \int \frac {1}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}\right )+\frac {e (9 b c-5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 \left (b e-d x^4\right )}}{4 d^3}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle e (b c-a d) \left (\frac {b e^2 (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3 \left (b e-d x^4\right )^2}-\frac {\frac {1}{2} \left (8 c \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}-\frac {3 \sqrt {e} (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{\sqrt {b} \sqrt {d}}\right )+\frac {e (9 b c-5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 \left (b e-d x^4\right )}}{4 d^3}\right )\) |
(b*c - a*d)*e*((b*(b*c - a*d)*e^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(4*d^ 3*(b*e - d*x^4)^2) - (((9*b*c - 5*a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)] )/(2*(b*e - d*x^4)) + (8*c*Sqrt[(e*(a + b*x^2))/(c + d*x^2)] - (3*(5*b*c - a*d)*Sqrt[e]*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b] *Sqrt[e])])/(Sqrt[b]*Sqrt[d]))/2)/(4*d^3))
3.3.77.3.1 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[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d) Subst[Int[x^(q* (p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.) ))^(p_), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(e*( (a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.18 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.27
| method | result | size |
| risch | \(\frac {\left (2 b d \,x^{2}+5 a d -7 b c \right ) \left (d \,x^{2}+c \right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{8 d^{3}}+\frac {\left (\frac {3 \left (a d -5 b c \right ) \left (a d -b c \right ) \ln \left (\frac {\frac {1}{2} e d a +\frac {1}{2} e b c +b d e \,x^{2}}{\sqrt {b d e}}+\sqrt {b d e \,x^{4}+\left (e d a +e b c \right ) x^{2}+a c e}\right )}{2 \sqrt {b d e}}+\frac {8 c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (b \,x^{2}+a \right )}{\left (a d -b c \right ) \sqrt {b d e \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{8 d^{3} \left (b \,x^{2}+a \right )}\) | \(252\) |
| default | \(\frac {\left (4 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b \,d^{2} x^{4}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{3} x^{2}-18 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c \,d^{2} x^{2}+15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2} d \,x^{2}+10 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a \,d^{2} x^{2}-10 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b c d \,x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} c \,d^{2}-18 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,c^{2} d +15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{3}+16 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a c d -16 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b \,c^{2}+10 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a c d -14 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b \,c^{2}\right ) \left (d \,x^{2}+c \right ) {\left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )}^{\frac {3}{2}}}{16 d^{3} \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (b \,x^{2}+a \right )}\) | \(679\) |
1/8*(2*b*d*x^2+5*a*d-7*b*c)*(d*x^2+c)/d^3*e*(e*(b*x^2+a)/(d*x^2+c))^(1/2)+ 1/8/d^3*(3/2*(a*d-5*b*c)*(a*d-b*c)*ln((1/2*e*d*a+1/2*e*b*c+b*d*e*x^2)/(b*d *e)^(1/2)+(b*d*e*x^4+(a*d*e+b*c*e)*x^2+a*c*e)^(1/2))/(b*d*e)^(1/2)+8*c*(a^ 2*d^2-2*a*b*c*d+b^2*c^2)*(b*x^2+a)/(a*d-b*c)/(b*d*e*x^4+a*d*e*x^2+b*c*e*x^ 2+a*c*e)^(1/2))*e/(b*x^2+a)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)*((d*x^2+c)*e*(b* x^2+a))^(1/2)
Time = 0.68 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.10 \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\left [\frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} e \sqrt {\frac {e}{b d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} + {\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{b d}}\right ) + 4 \, {\left (2 \, b d^{2} e x^{4} - 5 \, {\left (b c d - a d^{2}\right )} e x^{2} - {\left (15 \, b c^{2} - 13 \, a c d\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{32 \, d^{3}}, -\frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} e \sqrt {-\frac {e}{b d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{b d}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) - 2 \, {\left (2 \, b d^{2} e x^{4} - 5 \, {\left (b c d - a d^{2}\right )} e x^{2} - {\left (15 \, b c^{2} - 13 \, a c d\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{16 \, d^{3}}\right ] \]
[1/32*(3*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*e*sqrt(e/(b*d))*log(8*b^2*d^2*e *x^4 + 8*(b^2*c*d + a*b*d^2)*e*x^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*e + 4 *(2*b^2*d^3*x^4 + b^2*c^2*d + a*b*c*d^2 + (3*b^2*c*d^2 + a*b*d^3)*x^2)*sqr t((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(e/(b*d))) + 4*(2*b*d^2*e*x^4 - 5*(b*c* d - a*d^2)*e*x^2 - (15*b*c^2 - 13*a*c*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^3, -1/16*(3*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*e*sqrt(-e/(b*d))*arct an(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(-e/( b*d))/(b*e*x^2 + a*e)) - 2*(2*b*d^2*e*x^4 - 5*(b*c*d - a*d^2)*e*x^2 - (15* b*c^2 - 13*a*c*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^3]
Timed out. \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\text {Timed out} \]
Exception generated. \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]
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
Exception generated. \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%%{2,[0,4,0]%%%},[2,0,0,0]%%%}+%%%{%%{[%%%{-4,[0,3,0]%%%} ,0]:[1,0,
Timed out. \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\int x^3\,{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2} \,d x \]