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

Optimal. Leaf size=89 \[ \frac {x \left (a+\frac {d (c d-b e)}{e^2}\right )}{d \sqrt {d+e x^2}}-\frac {(3 c d-2 b e) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 e^{5/2}}+\frac {c x \sqrt {d+e x^2}}{2 e^2} \]

[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

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Rubi [A]  time = 0.07, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1157, 388, 217, 206} \[ \frac {x \left (a+\frac {d (c d-b e)}{e^2}\right )}{d \sqrt {d+e x^2}}-\frac {(3 c d-2 b e) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 e^{5/2}}+\frac {c x \sqrt {d+e x^2}}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + (d*(c*d - b*e))/e^2)*x)/(d*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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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 388

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 1157

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] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rubi steps

\begin {align*} \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 {\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 (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) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{2 e^2}\\ &=\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) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{2 e^2}\\ &=\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) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 e^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 98, normalized size = 1.10 \[ \frac {\sqrt {e} x \left (2 e (a e-b d)+c d \left (3 d+e x^2\right )\right )-d^{3/2} \sqrt {\frac {e x^2}{d}+1} (3 c d-2 b e) \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d e^{5/2} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e]*x*(2*e*(-(b*d) + a*e) + c*d*(3*d + e*x^2)) - d^(3/2)*(3*c*d - 2*b*e)*Sqrt[1 + (e*x^2)/d]*ArcSinh[(Sqr
t[e]*x)/Sqrt[d]])/(2*d*e^(5/2)*Sqrt[d + e*x^2])

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fricas [A]  time = 0.88, size = 249, normalized size = 2.80 \[ \left [-\frac {{\left (3 \, c d^{3} - 2 \, b d^{2} e + {\left (3 \, c d^{2} e - 2 \, b d e^{2}\right )} x^{2}\right )} \sqrt {e} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - 2 \, {\left (c d e^{2} x^{3} + {\left (3 \, c d^{2} e - 2 \, b d e^{2} + 2 \, a e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{4 \, {\left (d e^{4} x^{2} + d^{2} e^{3}\right )}}, \frac {{\left (3 \, c d^{3} - 2 \, b d^{2} e + {\left (3 \, c d^{2} e - 2 \, b d e^{2}\right )} x^{2}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (c d e^{2} x^{3} + {\left (3 \, c d^{2} e - 2 \, b d e^{2} + 2 \, a e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{2 \, {\left (d e^{4} x^{2} + d^{2} e^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.20, size = 80, normalized size = 0.90 \[ \frac {1}{2} \, {\left (3 \, c d - 2 \, b e\right )} e^{\left (-\frac {5}{2}\right )} \log \left ({\left | -x e^{\frac {1}{2}} + \sqrt {x^{2} e + d} \right |}\right ) + \frac {{\left (c x^{2} e^{\left (-1\right )} + \frac {{\left (3 \, c d^{2} e - 2 \, b d e^{2} + 2 \, a e^{3}\right )} e^{\left (-3\right )}}{d}\right )} x}{2 \, \sqrt {x^{2} e + d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 112, normalized size = 1.26 \[ \frac {c \,x^{3}}{2 \sqrt {e \,x^{2}+d}\, e}+\frac {a x}{\sqrt {e \,x^{2}+d}\, d}-\frac {b x}{\sqrt {e \,x^{2}+d}\, e}+\frac {3 c d x}{2 \sqrt {e \,x^{2}+d}\, e^{2}}+\frac {b \ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{e^{\frac {3}{2}}}-\frac {3 c d \ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{2 e^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.13, size = 97, normalized size = 1.09 \[ \frac {c x^{3}}{2 \, \sqrt {e x^{2} + d} e} + \frac {a x}{\sqrt {e x^{2} + d} d} + \frac {3 \, c d x}{2 \, \sqrt {e x^{2} + d} e^{2}} - \frac {b x}{\sqrt {e x^{2} + d} e} - \frac {3 \, c d \operatorname {arsinh}\left (\frac {e x}{\sqrt {d e}}\right )}{2 \, e^{\frac {5}{2}}} + \frac {b \operatorname {arsinh}\left (\frac {e x}{\sqrt {d e}}\right )}{e^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {c\,x^4+b\,x^2+a}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 9.98, size = 134, normalized size = 1.51 \[ \frac {a x}{d^{\frac {3}{2}} \sqrt {1 + \frac {e x^{2}}{d}}} + b \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {e} x}{\sqrt {d}} \right )}}{e^{\frac {3}{2}}} - \frac {x}{\sqrt {d} e \sqrt {1 + \frac {e x^{2}}{d}}}\right ) + c \left (\frac {3 \sqrt {d} x}{2 e^{2} \sqrt {1 + \frac {e x^{2}}{d}}} - \frac {3 d \operatorname {asinh}{\left (\frac {\sqrt {e} x}{\sqrt {d}} \right )}}{2 e^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {d} e \sqrt {1 + \frac {e x^{2}}{d}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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