3.281 \(\int \frac{a+b x^2+c x^4}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=101 \[ -\frac{x \left (4 c d^2-e (2 a e+b d)\right )}{3 d^2 e^2 \sqrt{d+e x^2}}+\frac{x \left (a+\frac{d (c d-b e)}{e^2}\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{c \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{e^{5/2}} \]

[Out]

((a + (d*(c*d - b*e))/e^2)*x)/(3*d*(d + e*x^2)^(3/2)) - ((4*c*d^2 - e*(b*d + 2*a*e))*x)/(3*d^2*e^2*Sqrt[d + e*
x^2]) + (c*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/e^(5/2)

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Rubi [A]  time = 0.0713267, antiderivative size = 101, 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, 385, 217, 206} \[ -\frac{x \left (4 c d^2-e (2 a e+b d)\right )}{3 d^2 e^2 \sqrt{d+e x^2}}+\frac{x \left (a+\frac{d (c d-b e)}{e^2}\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{c \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{e^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)/(d + e*x^2)^(5/2),x]

[Out]

((a + (d*(c*d - b*e))/e^2)*x)/(3*d*(d + e*x^2)^(3/2)) - ((4*c*d^2 - e*(b*d + 2*a*e))*x)/(3*d^2*e^2*Sqrt[d + e*
x^2]) + (c*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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 )^{5/2}} \, dx &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{3 d \left (d+e x^2\right )^{3/2}}-\frac{\int \frac{-2 a+\frac{d (c d-b e)}{e^2}-\frac{3 c d x^2}{e}}{\left (d+e x^2\right )^{3/2}} \, dx}{3 d}\\ &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{3 d \left (d+e x^2\right )^{3/2}}-\frac{\left (4 c d^2-e (b d+2 a e)\right ) x}{3 d^2 e^2 \sqrt{d+e x^2}}+\frac{c \int \frac{1}{\sqrt{d+e x^2}} \, dx}{e^2}\\ &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{3 d \left (d+e x^2\right )^{3/2}}-\frac{\left (4 c d^2-e (b d+2 a e)\right ) x}{3 d^2 e^2 \sqrt{d+e x^2}}+\frac{c \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{e^2}\\ &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{3 d \left (d+e x^2\right )^{3/2}}-\frac{\left (4 c d^2-e (b d+2 a e)\right ) x}{3 d^2 e^2 \sqrt{d+e x^2}}+\frac{c \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{e^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.193527, size = 112, normalized size = 1.11 \[ \frac{\sqrt{e} x \left (e^2 \left (3 a d+2 a e x^2+b d x^2\right )-c d^2 \left (3 d+4 e x^2\right )\right )+3 c d^{5/2} \left (d+e x^2\right ) \sqrt{\frac{e x^2}{d}+1} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{3 d^2 e^{5/2} \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)/(d + e*x^2)^(5/2),x]

[Out]

(Sqrt[e]*x*(-(c*d^2*(3*d + 4*e*x^2)) + e^2*(3*a*d + b*d*x^2 + 2*a*e*x^2)) + 3*c*d^(5/2)*(d + e*x^2)*Sqrt[1 + (
e*x^2)/d]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])/(3*d^2*e^(5/2)*(d + e*x^2)^(3/2))

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Maple [A]  time = 0.009, size = 124, normalized size = 1.2 \begin{align*} -{\frac{c{x}^{3}}{3\,e} \left ( e{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}-{\frac{cx}{{e}^{2}}{\frac{1}{\sqrt{e{x}^{2}+d}}}}+{c\ln \left ( x\sqrt{e}+\sqrt{e{x}^{2}+d} \right ){e}^{-{\frac{5}{2}}}}-{\frac{bx}{3\,e} \left ( e{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}+{\frac{bx}{3\,de}{\frac{1}{\sqrt{e{x}^{2}+d}}}}+{\frac{ax}{3\,d} \left ( e{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,ax}{3\,{d}^{2}}{\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)^(5/2),x)

[Out]

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

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.88764, size = 612, normalized size = 6.06 \begin{align*} \left [\frac{3 \,{\left (c d^{2} e^{2} x^{4} + 2 \, c d^{3} e x^{2} + c d^{4}\right )} \sqrt{e} \log \left (-2 \, e x^{2} - 2 \, \sqrt{e x^{2} + d} \sqrt{e} x - d\right ) - 2 \,{\left ({\left (4 \, c d^{2} e^{2} - b d e^{3} - 2 \, a e^{4}\right )} x^{3} + 3 \,{\left (c d^{3} e - a d e^{3}\right )} x\right )} \sqrt{e x^{2} + d}}{6 \,{\left (d^{2} e^{5} x^{4} + 2 \, d^{3} e^{4} x^{2} + d^{4} e^{3}\right )}}, -\frac{3 \,{\left (c d^{2} e^{2} x^{4} + 2 \, c d^{3} e x^{2} + c d^{4}\right )} \sqrt{-e} \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right ) +{\left ({\left (4 \, c d^{2} e^{2} - b d e^{3} - 2 \, a e^{4}\right )} x^{3} + 3 \,{\left (c d^{3} e - a d e^{3}\right )} x\right )} \sqrt{e x^{2} + d}}{3 \,{\left (d^{2} e^{5} x^{4} + 2 \, d^{3} e^{4} x^{2} + d^{4} 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)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(c*d^2*e^2*x^4 + 2*c*d^3*e*x^2 + c*d^4)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - 2*((
4*c*d^2*e^2 - b*d*e^3 - 2*a*e^4)*x^3 + 3*(c*d^3*e - a*d*e^3)*x)*sqrt(e*x^2 + d))/(d^2*e^5*x^4 + 2*d^3*e^4*x^2
+ d^4*e^3), -1/3*(3*(c*d^2*e^2*x^4 + 2*c*d^3*e*x^2 + c*d^4)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + ((4*
c*d^2*e^2 - b*d*e^3 - 2*a*e^4)*x^3 + 3*(c*d^3*e - a*d*e^3)*x)*sqrt(e*x^2 + d))/(d^2*e^5*x^4 + 2*d^3*e^4*x^2 +
d^4*e^3)]

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Sympy [B]  time = 18.4519, size = 450, normalized size = 4.46 \begin{align*} a \left (\frac{3 d x}{3 d^{\frac{7}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 3 d^{\frac{5}{2}} e x^{2} \sqrt{1 + \frac{e x^{2}}{d}}} + \frac{2 e x^{3}}{3 d^{\frac{7}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 3 d^{\frac{5}{2}} e x^{2} \sqrt{1 + \frac{e x^{2}}{d}}}\right ) + \frac{b x^{3}}{3 d^{\frac{5}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 3 d^{\frac{3}{2}} e x^{2} \sqrt{1 + \frac{e x^{2}}{d}}} + c \left (\frac{3 d^{\frac{39}{2}} e^{11} \sqrt{1 + \frac{e x^{2}}{d}} \operatorname{asinh}{\left (\frac{\sqrt{e} x}{\sqrt{d}} \right )}}{3 d^{\frac{39}{2}} e^{\frac{27}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 3 d^{\frac{37}{2}} e^{\frac{29}{2}} x^{2} \sqrt{1 + \frac{e x^{2}}{d}}} + \frac{3 d^{\frac{37}{2}} e^{12} x^{2} \sqrt{1 + \frac{e x^{2}}{d}} \operatorname{asinh}{\left (\frac{\sqrt{e} x}{\sqrt{d}} \right )}}{3 d^{\frac{39}{2}} e^{\frac{27}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 3 d^{\frac{37}{2}} e^{\frac{29}{2}} x^{2} \sqrt{1 + \frac{e x^{2}}{d}}} - \frac{3 d^{19} e^{\frac{23}{2}} x}{3 d^{\frac{39}{2}} e^{\frac{27}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 3 d^{\frac{37}{2}} e^{\frac{29}{2}} x^{2} \sqrt{1 + \frac{e x^{2}}{d}}} - \frac{4 d^{18} e^{\frac{25}{2}} x^{3}}{3 d^{\frac{39}{2}} e^{\frac{27}{2}} \sqrt{1 + \frac{e x^{2}}{d}} + 3 d^{\frac{37}{2}} e^{\frac{29}{2}} x^{2} \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)**(5/2),x)

[Out]

a*(3*d*x/(3*d**(7/2)*sqrt(1 + e*x**2/d) + 3*d**(5/2)*e*x**2*sqrt(1 + e*x**2/d)) + 2*e*x**3/(3*d**(7/2)*sqrt(1
+ e*x**2/d) + 3*d**(5/2)*e*x**2*sqrt(1 + e*x**2/d))) + b*x**3/(3*d**(5/2)*sqrt(1 + e*x**2/d) + 3*d**(3/2)*e*x*
*2*sqrt(1 + e*x**2/d)) + c*(3*d**(39/2)*e**11*sqrt(1 + e*x**2/d)*asinh(sqrt(e)*x/sqrt(d))/(3*d**(39/2)*e**(27/
2)*sqrt(1 + e*x**2/d) + 3*d**(37/2)*e**(29/2)*x**2*sqrt(1 + e*x**2/d)) + 3*d**(37/2)*e**12*x**2*sqrt(1 + e*x**
2/d)*asinh(sqrt(e)*x/sqrt(d))/(3*d**(39/2)*e**(27/2)*sqrt(1 + e*x**2/d) + 3*d**(37/2)*e**(29/2)*x**2*sqrt(1 +
e*x**2/d)) - 3*d**19*e**(23/2)*x/(3*d**(39/2)*e**(27/2)*sqrt(1 + e*x**2/d) + 3*d**(37/2)*e**(29/2)*x**2*sqrt(1
+ e*x**2/d)) - 4*d**18*e**(25/2)*x**3/(3*d**(39/2)*e**(27/2)*sqrt(1 + e*x**2/d) + 3*d**(37/2)*e**(29/2)*x**2*
sqrt(1 + e*x**2/d)))

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Giac [A]  time = 1.12376, size = 119, normalized size = 1.18 \begin{align*} -c e^{\left (-\frac{5}{2}\right )} \log \left ({\left | -x e^{\frac{1}{2}} + \sqrt{x^{2} e + d} \right |}\right ) - \frac{{\left (\frac{{\left (4 \, c d^{2} e^{2} - b d e^{3} - 2 \, a e^{4}\right )} x^{2} e^{\left (-3\right )}}{d^{2}} + \frac{3 \,{\left (c d^{3} e - a d e^{3}\right )} e^{\left (-3\right )}}{d^{2}}\right )} x}{3 \,{\left (x^{2} e + d\right )}^{\frac{3}{2}}} \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)^(5/2),x, algorithm="giac")

[Out]

-c*e^(-5/2)*log(abs(-x*e^(1/2) + sqrt(x^2*e + d))) - 1/3*((4*c*d^2*e^2 - b*d*e^3 - 2*a*e^4)*x^2*e^(-3)/d^2 + 3
*(c*d^3*e - a*d*e^3)*e^(-3)/d^2)*x/(x^2*e + d)^(3/2)