Integrand size = 19, antiderivative size = 144 \[ \int \left (d+e x^2\right )^{3/2} \left (a+c x^4\right ) \, dx=\frac {3}{128} d \left (16 a+\frac {c d^2}{e^2}\right ) x \sqrt {d+e x^2}+\frac {1}{64} \left (16 a+\frac {c d^2}{e^2}\right ) x \left (d+e x^2\right )^{3/2}-\frac {c d x \left (d+e x^2\right )^{5/2}}{16 e^2}+\frac {c x^3 \left (d+e x^2\right )^{5/2}}{8 e}+\frac {3 d^2 \left (c d^2+16 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{128 e^{5/2}} \] Output:
3/128*d*(16*a+c*d^2/e^2)*x*(e*x^2+d)^(1/2)+1/64*(16*a+c*d^2/e^2)*x*(e*x^2+ d)^(3/2)-1/16*c*d*x*(e*x^2+d)^(5/2)/e^2+1/8*c*x^3*(e*x^2+d)^(5/2)/e+3/128* d^2*(16*a*e^2+c*d^2)*arctanh(e^(1/2)*x/(e*x^2+d)^(1/2))/e^(5/2)
Time = 0.15 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.82 \[ \int \left (d+e x^2\right )^{3/2} \left (a+c x^4\right ) \, dx=\frac {\sqrt {d+e x^2} \left (-3 c d^3 x+80 a d e^2 x+2 c d^2 e x^3+32 a e^3 x^3+24 c d e^2 x^5+16 c e^3 x^7\right )}{128 e^2}-\frac {3 \left (c d^4+16 a d^2 e^2\right ) \log \left (-\sqrt {e} x+\sqrt {d+e x^2}\right )}{128 e^{5/2}} \] Input:
Integrate[(d + e*x^2)^(3/2)*(a + c*x^4),x]
Output:
(Sqrt[d + e*x^2]*(-3*c*d^3*x + 80*a*d*e^2*x + 2*c*d^2*e*x^3 + 32*a*e^3*x^3 + 24*c*d*e^2*x^5 + 16*c*e^3*x^7))/(128*e^2) - (3*(c*d^4 + 16*a*d^2*e^2)*L og[-(Sqrt[e]*x) + Sqrt[d + e*x^2]])/(128*e^(5/2))
Time = 0.40 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1474, 299, 211, 211, 224, 219}
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.
| \(\displaystyle \int \left (a+c x^4\right ) \left (d+e x^2\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 1474 |
| \(\displaystyle \frac {\int \left (8 a e-3 c d x^2\right ) \left (e x^2+d\right )^{3/2}dx}{8 e}+\frac {c x^3 \left (d+e x^2\right )^{5/2}}{8 e}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {\left (16 a e^2+c d^2\right ) \int \left (e x^2+d\right )^{3/2}dx}{2 e}-\frac {c d x \left (d+e x^2\right )^{5/2}}{2 e}}{8 e}+\frac {c x^3 \left (d+e x^2\right )^{5/2}}{8 e}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\left (16 a e^2+c d^2\right ) \left (\frac {3}{4} d \int \sqrt {e x^2+d}dx+\frac {1}{4} x \left (d+e x^2\right )^{3/2}\right )}{2 e}-\frac {c d x \left (d+e x^2\right )^{5/2}}{2 e}}{8 e}+\frac {c x^3 \left (d+e x^2\right )^{5/2}}{8 e}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\left (16 a e^2+c d^2\right ) \left (\frac {3}{4} d \left (\frac {1}{2} d \int \frac {1}{\sqrt {e x^2+d}}dx+\frac {1}{2} x \sqrt {d+e x^2}\right )+\frac {1}{4} x \left (d+e x^2\right )^{3/2}\right )}{2 e}-\frac {c d x \left (d+e x^2\right )^{5/2}}{2 e}}{8 e}+\frac {c x^3 \left (d+e x^2\right )^{5/2}}{8 e}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\left (16 a e^2+c d^2\right ) \left (\frac {3}{4} d \left (\frac {1}{2} d \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}+\frac {1}{2} x \sqrt {d+e x^2}\right )+\frac {1}{4} x \left (d+e x^2\right )^{3/2}\right )}{2 e}-\frac {c d x \left (d+e x^2\right )^{5/2}}{2 e}}{8 e}+\frac {c x^3 \left (d+e x^2\right )^{5/2}}{8 e}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\left (16 a e^2+c d^2\right ) \left (\frac {3}{4} d \left (\frac {d \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {e}}+\frac {1}{2} x \sqrt {d+e x^2}\right )+\frac {1}{4} x \left (d+e x^2\right )^{3/2}\right )}{2 e}-\frac {c d x \left (d+e x^2\right )^{5/2}}{2 e}}{8 e}+\frac {c x^3 \left (d+e x^2\right )^{5/2}}{8 e}\) |
Input:
Int[(d + e*x^2)^(3/2)*(a + c*x^4),x]
Output:
(c*x^3*(d + e*x^2)^(5/2))/(8*e) + (-1/2*(c*d*x*(d + e*x^2)^(5/2))/e + ((c* d^2 + 16*a*e^2)*((x*(d + e*x^2)^(3/2))/4 + (3*d*((x*Sqrt[d + e*x^2])/2 + ( d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*Sqrt[e])))/4))/(2*e))/(8*e)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[((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[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Si mp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))), x] + Simp[1/( e*(4*p + 2*q + 1)) Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 2*q + 1)*(a + c *x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, c, d, e, q}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && !LtQ[q, -1]
Time = 0.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.71
| method | result | size |
| pseudoelliptic | \(\frac {\frac {5 \left (\frac {3}{5} a \,d^{2} e^{2}+\frac {3}{80} d^{4} c \right ) \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}}{x \sqrt {e}}\right )}{8}+\frac {5 x \left (d \left (\frac {3 c \,x^{4}}{10}+a \right ) e^{\frac {5}{2}}+\frac {2 x^{2} \left (\frac {c \,x^{4}}{2}+a \right ) e^{\frac {7}{2}}}{5}-\frac {3 d^{2} c \left (-\frac {2 e^{\frac {3}{2}} x^{2}}{3}+\sqrt {e}\, d \right )}{80}\right ) \sqrt {e \,x^{2}+d}}{8}}{e^{\frac {5}{2}}}\) | \(102\) |
| risch | \(\frac {x \left (16 e^{3} c \,x^{6}+24 c d \,e^{2} x^{4}+32 a \,e^{3} x^{2}+2 c \,d^{2} e \,x^{2}+80 d \,e^{2} a -3 d^{3} c \right ) \sqrt {e \,x^{2}+d}}{128 e^{2}}+\frac {3 d^{2} \left (16 a \,e^{2}+c \,d^{2}\right ) \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{128 e^{\frac {5}{2}}}\) | \(105\) |
| default | \(a \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4}+\frac {3 d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4}\right )+c \left (\frac {x^{3} \left (e \,x^{2}+d \right )^{\frac {5}{2}}}{8 e}-\frac {3 d \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {5}{2}}}{6 e}-\frac {d \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4}+\frac {3 d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4}\right )}{6 e}\right )}{8 e}\right )\) | \(154\) |
Input:
int((e*x^2+d)^(3/2)*(c*x^4+a),x,method=_RETURNVERBOSE)
Output:
5/8/e^(5/2)*((3/5*a*d^2*e^2+3/80*d^4*c)*arctanh((e*x^2+d)^(1/2)/x/e^(1/2)) +x*(d*(3/10*c*x^4+a)*e^(5/2)+2/5*x^2*(1/2*c*x^4+a)*e^(7/2)-3/80*d^2*c*(-2/ 3*e^(3/2)*x^2+e^(1/2)*d))*(e*x^2+d)^(1/2))
Time = 0.10 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.64 \[ \int \left (d+e x^2\right )^{3/2} \left (a+c x^4\right ) \, dx=\left [\frac {3 \, {\left (c d^{4} + 16 \, a d^{2} e^{2}\right )} \sqrt {e} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) + 2 \, {\left (16 \, c e^{4} x^{7} + 24 \, c d e^{3} x^{5} + 2 \, {\left (c d^{2} e^{2} + 16 \, a e^{4}\right )} x^{3} - {\left (3 \, c d^{3} e - 80 \, a d e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{256 \, e^{3}}, -\frac {3 \, {\left (c d^{4} + 16 \, a d^{2} e^{2}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - {\left (16 \, c e^{4} x^{7} + 24 \, c d e^{3} x^{5} + 2 \, {\left (c d^{2} e^{2} + 16 \, a e^{4}\right )} x^{3} - {\left (3 \, c d^{3} e - 80 \, a d e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{128 \, e^{3}}\right ] \] Input:
integrate((e*x^2+d)^(3/2)*(c*x^4+a),x, algorithm="fricas")
Output:
[1/256*(3*(c*d^4 + 16*a*d^2*e^2)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)* sqrt(e)*x - d) + 2*(16*c*e^4*x^7 + 24*c*d*e^3*x^5 + 2*(c*d^2*e^2 + 16*a*e^ 4)*x^3 - (3*c*d^3*e - 80*a*d*e^3)*x)*sqrt(e*x^2 + d))/e^3, -1/128*(3*(c*d^ 4 + 16*a*d^2*e^2)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (16*c*e^4* x^7 + 24*c*d*e^3*x^5 + 2*(c*d^2*e^2 + 16*a*e^4)*x^3 - (3*c*d^3*e - 80*a*d* e^3)*x)*sqrt(e*x^2 + d))/e^3]
Time = 0.35 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.18 \[ \int \left (d+e x^2\right )^{3/2} \left (a+c x^4\right ) \, dx=\begin {cases} \sqrt {d + e x^{2}} \cdot \left (\frac {3 c d x^{5}}{16} + \frac {c e x^{7}}{8} + \frac {x^{3} \left (a e^{2} + \frac {c d^{2}}{16}\right )}{4 e} + \frac {x \left (2 a d e - \frac {3 d \left (a e^{2} + \frac {c d^{2}}{16}\right )}{4 e}\right )}{2 e}\right ) + \left (a d^{2} - \frac {d \left (2 a d e - \frac {3 d \left (a e^{2} + \frac {c d^{2}}{16}\right )}{4 e}\right )}{2 e}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {e} \sqrt {d + e x^{2}} + 2 e x \right )}}{\sqrt {e}} & \text {for}\: d \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {e x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: e \neq 0 \\d^{\frac {3}{2}} \left (a x + \frac {c x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((e*x**2+d)**(3/2)*(c*x**4+a),x)
Output:
Piecewise((sqrt(d + e*x**2)*(3*c*d*x**5/16 + c*e*x**7/8 + x**3*(a*e**2 + c *d**2/16)/(4*e) + x*(2*a*d*e - 3*d*(a*e**2 + c*d**2/16)/(4*e))/(2*e)) + (a *d**2 - d*(2*a*d*e - 3*d*(a*e**2 + c*d**2/16)/(4*e))/(2*e))*Piecewise((log (2*sqrt(e)*sqrt(d + e*x**2) + 2*e*x)/sqrt(e), Ne(d, 0)), (x*log(x)/sqrt(e* x**2), True)), Ne(e, 0)), (d**(3/2)*(a*x + c*x**5/5), True))
Exception generated. \[ \int \left (d+e x^2\right )^{3/2} \left (a+c x^4\right ) \, dx=\text {Exception raised: ValueError} \] Input:
integrate((e*x^2+d)^(3/2)*(c*x^4+a),x, algorithm="maxima")
Output:
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
Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.81 \[ \int \left (d+e x^2\right )^{3/2} \left (a+c x^4\right ) \, dx=\frac {1}{128} \, {\left (2 \, {\left (4 \, {\left (2 \, c e x^{2} + 3 \, c d\right )} x^{2} + \frac {c d^{2} e^{5} + 16 \, a e^{7}}{e^{6}}\right )} x^{2} - \frac {3 \, c d^{3} e^{4} - 80 \, a d e^{6}}{e^{6}}\right )} \sqrt {e x^{2} + d} x - \frac {3 \, {\left (c d^{4} + 16 \, a d^{2} e^{2}\right )} \log \left ({\left | -\sqrt {e} x + \sqrt {e x^{2} + d} \right |}\right )}{128 \, e^{\frac {5}{2}}} \] Input:
integrate((e*x^2+d)^(3/2)*(c*x^4+a),x, algorithm="giac")
Output:
1/128*(2*(4*(2*c*e*x^2 + 3*c*d)*x^2 + (c*d^2*e^5 + 16*a*e^7)/e^6)*x^2 - (3 *c*d^3*e^4 - 80*a*d*e^6)/e^6)*sqrt(e*x^2 + d)*x - 3/128*(c*d^4 + 16*a*d^2* e^2)*log(abs(-sqrt(e)*x + sqrt(e*x^2 + d)))/e^(5/2)
Timed out. \[ \int \left (d+e x^2\right )^{3/2} \left (a+c x^4\right ) \, dx=\int \left (c\,x^4+a\right )\,{\left (e\,x^2+d\right )}^{3/2} \,d x \] Input:
int((a + c*x^4)*(d + e*x^2)^(3/2),x)
Output:
int((a + c*x^4)*(d + e*x^2)^(3/2), x)
Time = 0.25 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.16 \[ \int \left (d+e x^2\right )^{3/2} \left (a+c x^4\right ) \, dx=\frac {80 \sqrt {e \,x^{2}+d}\, a d \,e^{3} x +32 \sqrt {e \,x^{2}+d}\, a \,e^{4} x^{3}-3 \sqrt {e \,x^{2}+d}\, c \,d^{3} e x +2 \sqrt {e \,x^{2}+d}\, c \,d^{2} e^{2} x^{3}+24 \sqrt {e \,x^{2}+d}\, c d \,e^{3} x^{5}+16 \sqrt {e \,x^{2}+d}\, c \,e^{4} x^{7}+48 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a \,d^{2} e^{2}+3 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) c \,d^{4}}{128 e^{3}} \] Input:
int((e*x^2+d)^(3/2)*(c*x^4+a),x)
Output:
(80*sqrt(d + e*x**2)*a*d*e**3*x + 32*sqrt(d + e*x**2)*a*e**4*x**3 - 3*sqrt (d + e*x**2)*c*d**3*e*x + 2*sqrt(d + e*x**2)*c*d**2*e**2*x**3 + 24*sqrt(d + e*x**2)*c*d*e**3*x**5 + 16*sqrt(d + e*x**2)*c*e**4*x**7 + 48*sqrt(e)*log ((sqrt(d + e*x**2) + sqrt(e)*x)/sqrt(d))*a*d**2*e**2 + 3*sqrt(e)*log((sqrt (d + e*x**2) + sqrt(e)*x)/sqrt(d))*c*d**4)/(128*e**3)