\(\int \frac {a+b x^2+c x^4}{(d+e x^2)^{3/2}} \, dx\) [280]

Optimal result

Integrand size = 24, antiderivative size = 89 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) x}{d \sqrt {d+e x^2}}+\frac {c x \sqrt {d+e x^2}}{2 e^2}-\frac {(3 c d-2 b e) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 e^{5/2}} \]

[Out]

-1/2*(-2*b*e+3*c*d)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/e^(5/2)+(a+d*(-b*e+c*d)/e^2)*x/d/(e*x^2+d)^(1/2)+1/2*c*
x*(e*x^2+d)^(1/2)/e^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1171, 396, 223, 212} \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {x \left (a e^2-b d e+c d^2\right )}{d e^2 \sqrt {d+e x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) (3 c d-2 b e)}{2 e^{5/2}}+\frac {c x \sqrt {d+e x^2}}{2 e^2} \]

[In]

Int[(a + b*x^2 + c*x^4)/(d + e*x^2)^(3/2),x]

[Out]

((c*d^2 - b*d*e + a*e^2)*x)/(d*e^2*Sqrt[d + e*x^2]) + (c*x*Sqrt[d + e*x^2])/(2*e^2) - ((3*c*d - 2*b*e)*ArcTanh
[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*e^(5/2))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 396

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

Rule 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c d^2-b d e+a e^2\right ) x}{d e^2 \sqrt {d+e x^2}}-\frac {\int \frac {\frac {d (c d-b e)}{e^2}-\frac {c d x^2}{e}}{\sqrt {d+e x^2}} \, dx}{d} \\ & = \frac {\left (c d^2-b d e+a e^2\right ) x}{d e^2 \sqrt {d+e x^2}}+\frac {c x \sqrt {d+e x^2}}{2 e^2}-\frac {(3 c d-2 b e) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{2 e^2} \\ & = \frac {\left (c d^2-b d e+a e^2\right ) x}{d e^2 \sqrt {d+e x^2}}+\frac {c x \sqrt {d+e x^2}}{2 e^2}-\frac {(3 c d-2 b e) \text {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{2 e^2} \\ & = \frac {\left (c d^2-b d e+a e^2\right ) x}{d e^2 \sqrt {d+e x^2}}+\frac {c x \sqrt {d+e x^2}}{2 e^2}-\frac {(3 c d-2 b e) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 e^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {x \left (3 c d^2-2 b d e+2 a e^2+c d e x^2\right )}{2 d e^2 \sqrt {d+e x^2}}+\frac {(3 c d-2 b e) \log \left (-\sqrt {e} x+\sqrt {d+e x^2}\right )}{2 e^{5/2}} \]

[In]

Integrate[(a + b*x^2 + c*x^4)/(d + e*x^2)^(3/2),x]

[Out]

(x*(3*c*d^2 - 2*b*d*e + 2*a*e^2 + c*d*e*x^2))/(2*d*e^2*Sqrt[d + e*x^2]) + ((3*c*d - 2*b*e)*Log[-(Sqrt[e]*x) +
Sqrt[d + e*x^2]])/(2*e^(5/2))

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96

method	result	size
pseudoelliptic	\(\frac {\sqrt {e \,x^{2}+d}\, d \left (b e -\frac {3 c d}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}}{x \sqrt {e}}\right )+\left (-\left (-\frac {c \,x^{2}}{2}+b \right ) d \,e^{\frac {3}{2}}+\frac {3 c \,d^{2} \sqrt {e}}{2}+a \,e^{\frac {5}{2}}\right ) x}{\sqrt {e \,x^{2}+d}\, e^{\frac {5}{2}} d}\)	\(85\)
risch	\(\frac {c x \sqrt {e \,x^{2}+d}}{2 e^{2}}+\frac {\frac {2 a \,e^{2} x}{d \sqrt {e \,x^{2}+d}}-\frac {c d x}{\sqrt {e \,x^{2}+d}}+\left (2 b \,e^{2}-3 d c e \right ) \left (-\frac {x}{e \sqrt {e \,x^{2}+d}}+\frac {\ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{e^{\frac {3}{2}}}\right )}{2 e^{2}}\)	\(106\)
default	\(\frac {a x}{d \sqrt {e \,x^{2}+d}}+c \left (\frac {x^{3}}{2 e \sqrt {e \,x^{2}+d}}-\frac {3 d \left (-\frac {x}{e \sqrt {e \,x^{2}+d}}+\frac {\ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{e^{\frac {3}{2}}}\right )}{2 e}\right )+b \left (-\frac {x}{e \sqrt {e \,x^{2}+d}}+\frac {\ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{e^{\frac {3}{2}}}\right )\)	\(117\)

[In]

int((c*x^4+b*x^2+a)/(e*x^2+d)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/(e*x^2+d)^(1/2)/e^(5/2)*((e*x^2+d)^(1/2)*d*(b*e-3/2*c*d)*arctanh((e*x^2+d)^(1/2)/x/e^(1/2))+(-(-1/2*c*x^2+b)
*d*e^(3/2)+3/2*c*d^2*e^(1/2)+a*e^(5/2))*x)/d

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.80 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (3 \, c d^{3} - 2 \, b d^{2} e + {\left (3 \, c d^{2} e - 2 \, b d e^{2}\right )} x^{2}\right )} \sqrt {e} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - 2 \, {\left (c d e^{2} x^{3} + {\left (3 \, c d^{2} e - 2 \, b d e^{2} + 2 \, a e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{4 \, {\left (d e^{4} x^{2} + d^{2} e^{3}\right )}}, \frac {{\left (3 \, c d^{3} - 2 \, b d^{2} e + {\left (3 \, c d^{2} e - 2 \, b d e^{2}\right )} x^{2}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (c d e^{2} x^{3} + {\left (3 \, c d^{2} e - 2 \, b d e^{2} + 2 \, a e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{2 \, {\left (d e^{4} x^{2} + d^{2} e^{3}\right )}}\right ] \]

[In]

integrate((c*x^4+b*x^2+a)/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*((3*c*d^3 - 2*b*d^2*e + (3*c*d^2*e - 2*b*d*e^2)*x^2)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x
- d) - 2*(c*d*e^2*x^3 + (3*c*d^2*e - 2*b*d*e^2 + 2*a*e^3)*x)*sqrt(e*x^2 + d))/(d*e^4*x^2 + d^2*e^3), 1/2*((3*c
*d^3 - 2*b*d^2*e + (3*c*d^2*e - 2*b*d*e^2)*x^2)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (c*d*e^2*x^3 + (
3*c*d^2*e - 2*b*d*e^2 + 2*a*e^3)*x)*sqrt(e*x^2 + d))/(d*e^4*x^2 + d^2*e^3)]

Sympy [A] (verification not implemented)

Time = 3.90 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.51 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {a x}{d^{\frac {3}{2}} \sqrt {1 + \frac {e x^{2}}{d}}} + b \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {e} x}{\sqrt {d}} \right )}}{e^{\frac {3}{2}}} - \frac {x}{\sqrt {d} e \sqrt {1 + \frac {e x^{2}}{d}}}\right ) + c \left (\frac {3 \sqrt {d} x}{2 e^{2} \sqrt {1 + \frac {e x^{2}}{d}}} - \frac {3 d \operatorname {asinh}{\left (\frac {\sqrt {e} x}{\sqrt {d}} \right )}}{2 e^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {d} e \sqrt {1 + \frac {e x^{2}}{d}}}\right ) \]

[In]

integrate((c*x**4+b*x**2+a)/(e*x**2+d)**(3/2),x)

[Out]

a*x/(d**(3/2)*sqrt(1 + e*x**2/d)) + b*(asinh(sqrt(e)*x/sqrt(d))/e**(3/2) - x/(sqrt(d)*e*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*sq
rt(1 + e*x**2/d)))

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*x^4+b*x^2+a)/(e*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {{\left (\frac {c x^{2}}{e} + \frac {3 \, c d^{2} e - 2 \, b d e^{2} + 2 \, a e^{3}}{d e^{3}}\right )} x}{2 \, \sqrt {e x^{2} + d}} + \frac {{\left (3 \, c d - 2 \, b e\right )} \log \left ({\left | -\sqrt {e} x + \sqrt {e x^{2} + d} \right |}\right )}{2 \, e^{\frac {5}{2}}} \]

[In]

integrate((c*x^4+b*x^2+a)/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

1/2*(c*x^2/e + (3*c*d^2*e - 2*b*d*e^2 + 2*a*e^3)/(d*e^3))*x/sqrt(e*x^2 + d) + 1/2*(3*c*d - 2*b*e)*log(abs(-sqr
t(e)*x + sqrt(e*x^2 + d)))/e^(5/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {c\,x^4+b\,x^2+a}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

[In]

int((a + b*x^2 + c*x^4)/(d + e*x^2)^(3/2),x)

[Out]

int((a + b*x^2 + c*x^4)/(d + e*x^2)^(3/2), x)