Integrand size = 20, antiderivative size = 78 \[ \int \frac {a-c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {\left (a-\frac {c d^2}{e^2}\right ) x}{d \sqrt {d+e x^2}}-\frac {c x \sqrt {d+e x^2}}{2 e^2}+\frac {3 c d \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 e^{5/2}} \] Output:
(a-c*d^2/e^2)*x/d/(e*x^2+d)^(1/2)-1/2*c*x*(e*x^2+d)^(1/2)/e^2+3/2*c*d*arct anh(e^(1/2)*x/(e*x^2+d)^(1/2))/e^(5/2)
Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.99 \[ \int \frac {a-c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {-3 c d^2 x+2 a e^2 x-c d e x^3}{2 d e^2 \sqrt {d+e x^2}}-\frac {3 c d \log \left (-\sqrt {e} x+\sqrt {d+e x^2}\right )}{2 e^{5/2}} \] Input:
Integrate[(a - c*x^4)/(d + e*x^2)^(3/2),x]
Output:
(-3*c*d^2*x + 2*a*e^2*x - c*d*e*x^3)/(2*d*e^2*Sqrt[d + e*x^2]) - (3*c*d*Lo g[-(Sqrt[e]*x) + Sqrt[d + e*x^2]])/(2*e^(5/2))
Time = 0.34 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1472, 25, 27, 299, 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.
\(\displaystyle \int \frac {a-c x^4}{\left (d+e x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1472 |
\(\displaystyle \frac {x \left (a-\frac {c d^2}{e^2}\right )}{d \sqrt {d+e x^2}}-\frac {\int -\frac {c d \left (d-e x^2\right )}{e^2 \sqrt {e x^2+d}}dx}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {c d \left (d-e x^2\right )}{e^2 \sqrt {e x^2+d}}dx}{d}+\frac {x \left (a-\frac {c d^2}{e^2}\right )}{d \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {c \int \frac {d-e x^2}{\sqrt {e x^2+d}}dx}{e^2}+\frac {x \left (a-\frac {c d^2}{e^2}\right )}{d \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {c \left (\frac {3}{2} d \int \frac {1}{\sqrt {e x^2+d}}dx-\frac {1}{2} x \sqrt {d+e x^2}\right )}{e^2}+\frac {x \left (a-\frac {c d^2}{e^2}\right )}{d \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {c \left (\frac {3}{2} d \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}-\frac {1}{2} x \sqrt {d+e x^2}\right )}{e^2}+\frac {x \left (a-\frac {c d^2}{e^2}\right )}{d \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {x \left (a-\frac {c d^2}{e^2}\right )}{d \sqrt {d+e x^2}}+\frac {c \left (\frac {3 d \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {e}}-\frac {1}{2} x \sqrt {d+e x^2}\right )}{e^2}\) |
Input:
Int[(a - c*x^4)/(d + e*x^2)^(3/2),x]
Output:
((a - (c*d^2)/e^2)*x)/(d*Sqrt[d + e*x^2]) + (c*(-1/2*(x*Sqrt[d + e*x^2]) + (3*d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*Sqrt[e])))/e^2
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]))* 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] :> Wi th[{Qx = PolynomialQuotient[(a + c*x^4)^p, d + e*x^2, x], R = Coeff[Polynom ialRemainder[(a + 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, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-\frac {c x \sqrt {e \,x^{2}+d}}{2 e^{2}}-\frac {c d x}{e^{2} \sqrt {e \,x^{2}+d}}+\frac {a x}{d \sqrt {e \,x^{2}+d}}+\frac {3 c d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 e^{\frac {5}{2}}}\) | \(73\) |
pseudoelliptic | \(\frac {\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}}{x \sqrt {e}}\right ) \sqrt {e \,x^{2}+d}\, c \,d^{2} e^{2}}{2}+x \,e^{\frac {5}{2}} \left (-\frac {1}{2} c d \,x^{2} e +a \,e^{2}-\frac {3}{2} c \,d^{2}\right )}{\sqrt {e \,x^{2}+d}\, e^{\frac {9}{2}} d}\) | \(78\) |
default | \(\frac {a x}{d \sqrt {e \,x^{2}+d}}-c \left (\frac {x^{3}}{2 e \sqrt {e \,x^{2}+d}}-\frac {3 d \left (-\frac {x}{e \sqrt {e \,x^{2}+d}}+\frac {\ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{e^{\frac {3}{2}}}\right )}{2 e}\right )\) | \(80\) |
Input:
int((-c*x^4+a)/(e*x^2+d)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2*c*x*(e*x^2+d)^(1/2)/e^2-1/e^2*c*d*x/(e*x^2+d)^(1/2)+a*x/d/(e*x^2+d)^( 1/2)+3/2/e^(5/2)*c*d*ln(x*e^(1/2)+(e*x^2+d)^(1/2))
Time = 0.08 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.58 \[ \int \frac {a-c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (c d^{2} e x^{2} + c d^{3}\right )} \sqrt {e} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - 2 \, {\left (c d e^{2} x^{3} + {\left (3 \, c d^{2} e - 2 \, a e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{4 \, {\left (d e^{4} x^{2} + d^{2} e^{3}\right )}}, -\frac {3 \, {\left (c d^{2} e x^{2} + c d^{3}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (c d e^{2} x^{3} + {\left (3 \, c d^{2} e - 2 \, a e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{2 \, {\left (d e^{4} x^{2} + d^{2} e^{3}\right )}}\right ] \] Input:
integrate((-c*x^4+a)/(e*x^2+d)^(3/2),x, algorithm="fricas")
Output:
[1/4*(3*(c*d^2*e*x^2 + c*d^3)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqr t(e)*x - d) - 2*(c*d*e^2*x^3 + (3*c*d^2*e - 2*a*e^3)*x)*sqrt(e*x^2 + d))/( d*e^4*x^2 + d^2*e^3), -1/2*(3*(c*d^2*e*x^2 + c*d^3)*sqrt(-e)*arctan(sqrt(- e)*x/sqrt(e*x^2 + d)) + (c*d*e^2*x^3 + (3*c*d^2*e - 2*a*e^3)*x)*sqrt(e*x^2 + d))/(d*e^4*x^2 + d^2*e^3)]
Time = 2.88 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21 \[ \int \frac {a-c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {a x}{d^{\frac {3}{2}} \sqrt {1 + \frac {e x^{2}}{d}}} - c \left (\frac {3 \sqrt {d} x}{2 e^{2} \sqrt {1 + \frac {e x^{2}}{d}}} - \frac {3 d \operatorname {asinh}{\left (\frac {\sqrt {e} x}{\sqrt {d}} \right )}}{2 e^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {d} e \sqrt {1 + \frac {e x^{2}}{d}}}\right ) \] Input:
integrate((-c*x**4+a)/(e*x**2+d)**(3/2),x)
Output:
a*x/(d**(3/2)*sqrt(1 + e*x**2/d)) - c*(3*sqrt(d)*x/(2*e**2*sqrt(1 + e*x**2 /d)) - 3*d*asinh(sqrt(e)*x/sqrt(d))/(2*e**(5/2)) + x**3/(2*sqrt(d)*e*sqrt( 1 + e*x**2/d)))
Exception generated. \[ \int \frac {a-c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((-c*x^4+a)/(e*x^2+d)^(3/2),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 = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87 \[ \int \frac {a-c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=-\frac {{\left (\frac {c x^{2}}{e} + \frac {3 \, c d^{2} e - 2 \, a e^{3}}{d e^{3}}\right )} x}{2 \, \sqrt {e x^{2} + d}} - \frac {3 \, c d \log \left ({\left | -\sqrt {e} x + \sqrt {e x^{2} + d} \right |}\right )}{2 \, e^{\frac {5}{2}}} \] Input:
integrate((-c*x^4+a)/(e*x^2+d)^(3/2),x, algorithm="giac")
Output:
-1/2*(c*x^2/e + (3*c*d^2*e - 2*a*e^3)/(d*e^3))*x/sqrt(e*x^2 + d) - 3/2*c*d *log(abs(-sqrt(e)*x + sqrt(e*x^2 + d)))/e^(5/2)
Timed out. \[ \int \frac {a-c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {a-c\,x^4}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \] 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 = 0.19 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.12 \[ \int \frac {a-c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {8 \sqrt {e \,x^{2}+d}\, a \,e^{3} x -12 \sqrt {e \,x^{2}+d}\, c \,d^{2} e x -4 \sqrt {e \,x^{2}+d}\, c d \,e^{2} x^{3}+12 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) c \,d^{3}+12 \sqrt {e}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {e}\, x}{\sqrt {d}}\right ) c \,d^{2} e \,x^{2}+8 \sqrt {e}\, a d \,e^{2}+8 \sqrt {e}\, a \,e^{3} x^{2}-9 \sqrt {e}\, c \,d^{3}-9 \sqrt {e}\, c \,d^{2} e \,x^{2}}{8 d \,e^{3} \left (e \,x^{2}+d \right )} \] Input:
int((-c*x^4+a)/(e*x^2+d)^(3/2),x)
Output:
(8*sqrt(d + e*x**2)*a*e**3*x - 12*sqrt(d + e*x**2)*c*d**2*e*x - 4*sqrt(d + e*x**2)*c*d*e**2*x**3 + 12*sqrt(e)*log((sqrt(d + e*x**2) + sqrt(e)*x)/sqr t(d))*c*d**3 + 12*sqrt(e)*log((sqrt(d + e*x**2) + sqrt(e)*x)/sqrt(d))*c*d* *2*e*x**2 + 8*sqrt(e)*a*d*e**2 + 8*sqrt(e)*a*e**3*x**2 - 9*sqrt(e)*c*d**3 - 9*sqrt(e)*c*d**2*e*x**2)/(8*d*e**3*(d + e*x**2))