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]

((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))

________________________________________________________________________________________

Rubi [A]  time = 0.0726533, antiderivative size = 89, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.107454, size = 98, normalized size = 1.1 \[ \frac{\sqrt{e} x \left (2 e (a e-b d)+c d \left (3 d+e x^2\right )\right )-d^{3/2} \sqrt{\frac{e x^2}{d}+1} (3 c d-2 b e) \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d e^{5/2} \sqrt{d+e x^2}} \]

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])

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 112, normalized size = 1.3 \begin{align*}{\frac{c{x}^{3}}{2\,e}{\frac{1}{\sqrt{e{x}^{2}+d}}}}+{\frac{3\,cdx}{2\,{e}^{2}}{\frac{1}{\sqrt{e{x}^{2}+d}}}}-{\frac{3\,cd}{2}\ln \left ( x\sqrt{e}+\sqrt{e{x}^{2}+d} \right ){e}^{-{\frac{5}{2}}}}-{\frac{bx}{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}}+{b\ln \left ( x\sqrt{e}+\sqrt{e{x}^{2}+d} \right ){e}^{-{\frac{3}{2}}}}+{\frac{ax}{d}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \end{align*}

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 4.29752, size = 552, normalized size = 6.2 \begin{align*} \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 ] \end{align*}

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)]

________________________________________________________________________________________

Sympy [A]  time = 7.82472, size = 134, normalized size = 1.51 \begin{align*} \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 ) \end{align*}

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)))

________________________________________________________________________________________

Giac [A]  time = 1.21141, size = 108, normalized size = 1.21 \begin{align*} \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}} \end{align*}

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)