3.278 \(\int \sqrt{d+e x^2} (a+b x^2+c x^4) \, dx\)

Optimal. Leaf size=132 \[ \frac{x \sqrt{d+e x^2} \left (8 a e^2-2 b d e+c d^2\right )}{16 e^2}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \left (8 a e^2-2 b d e+c d^2\right )}{16 e^{5/2}}-\frac{x \left (d+e x^2\right )^{3/2} (c d-2 b e)}{8 e^2}+\frac{c x^3 \left (d+e x^2\right )^{3/2}}{6 e} \]

[Out]

((c*d^2 - 2*b*d*e + 8*a*e^2)*x*Sqrt[d + e*x^2])/(16*e^2) - ((c*d - 2*b*e)*x*(d + e*x^2)^(3/2))/(8*e^2) + (c*x^
3*(d + e*x^2)^(3/2))/(6*e) + (d*(c*d^2 - 2*b*d*e + 8*a*e^2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(16*e^(5/2))

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Rubi [A]  time = 0.109218, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {1159, 388, 195, 217, 206} \[ \frac{x \sqrt{d+e x^2} \left (8 a e^2-2 b d e+c d^2\right )}{16 e^2}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \left (8 a e^2-2 b d e+c d^2\right )}{16 e^{5/2}}-\frac{x \left (d+e x^2\right )^{3/2} (c d-2 b e)}{8 e^2}+\frac{c x^3 \left (d+e x^2\right )^{3/2}}{6 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c*d^2 - 2*b*d*e + 8*a*e^2)*x*Sqrt[d + e*x^2])/(16*e^2) - ((c*d - 2*b*e)*x*(d + e*x^2)^(3/2))/(8*e^2) + (c*x^
3*(d + e*x^2)^(3/2))/(6*e) + (d*(c*d^2 - 2*b*d*e + 8*a*e^2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(16*e^(5/2))

Rule 1159

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(c^p*x^(4*p - 1)*
(d + e*x^2)^(q + 1))/(e*(4*p + 2*q + 1)), x] + Dist[1/(e*(4*p + 2*q + 1)), Int[(d + e*x^2)^q*ExpandToSum[e*(4*
p + 2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 2*q + 1)*x^(4*p), x], x], x] /
; FreeQ[{a, b, c, d, e, q}, 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]

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 \sqrt{d+e x^2} \left (a+b x^2+c x^4\right ) \, dx &=\frac{c x^3 \left (d+e x^2\right )^{3/2}}{6 e}+\frac{\int \sqrt{d+e x^2} \left (6 a e-3 (c d-2 b e) x^2\right ) \, dx}{6 e}\\ &=-\frac{(c d-2 b e) x \left (d+e x^2\right )^{3/2}}{8 e^2}+\frac{c x^3 \left (d+e x^2\right )^{3/2}}{6 e}+\frac{1}{8} \left (8 a+\frac{d (c d-2 b e)}{e^2}\right ) \int \sqrt{d+e x^2} \, dx\\ &=\frac{1}{16} \left (8 a+\frac{d (c d-2 b e)}{e^2}\right ) x \sqrt{d+e x^2}-\frac{(c d-2 b e) x \left (d+e x^2\right )^{3/2}}{8 e^2}+\frac{c x^3 \left (d+e x^2\right )^{3/2}}{6 e}+\frac{1}{16} \left (d \left (8 a+\frac{d (c d-2 b e)}{e^2}\right )\right ) \int \frac{1}{\sqrt{d+e x^2}} \, dx\\ &=\frac{1}{16} \left (8 a+\frac{d (c d-2 b e)}{e^2}\right ) x \sqrt{d+e x^2}-\frac{(c d-2 b e) x \left (d+e x^2\right )^{3/2}}{8 e^2}+\frac{c x^3 \left (d+e x^2\right )^{3/2}}{6 e}+\frac{1}{16} \left (d \left (8 a+\frac{d (c d-2 b e)}{e^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )\\ &=\frac{1}{16} \left (8 a+\frac{d (c d-2 b e)}{e^2}\right ) x \sqrt{d+e x^2}-\frac{(c d-2 b e) x \left (d+e x^2\right )^{3/2}}{8 e^2}+\frac{c x^3 \left (d+e x^2\right )^{3/2}}{6 e}+\frac{d \left (c d^2-2 b d e+8 a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{16 e^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.237715, size = 121, normalized size = 0.92 \[ \frac{\sqrt{d+e x^2} \left (\sqrt{e} x \left (6 e \left (4 a e+b \left (d+2 e x^2\right )\right )+c \left (-3 d^2+2 d e x^2+8 e^2 x^4\right )\right )+\frac{3 \sqrt{d} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (8 a e^2-2 b d e+c d^2\right )}{\sqrt{\frac{e x^2}{d}+1}}\right )}{48 e^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x^2]*(Sqrt[e]*x*(c*(-3*d^2 + 2*d*e*x^2 + 8*e^2*x^4) + 6*e*(4*a*e + b*(d + 2*e*x^2))) + (3*Sqrt[d]*
(c*d^2 - 2*b*d*e + 8*a*e^2)*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[1 + (e*x^2)/d]))/(48*e^(5/2))

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Maple [A]  time = 0.01, size = 175, normalized size = 1.3 \begin{align*}{\frac{c{x}^{3}}{6\,e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{cdx}{8\,{e}^{2}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{c{d}^{2}x}{16\,{e}^{2}}\sqrt{e{x}^{2}+d}}+{\frac{c{d}^{3}}{16}\ln \left ( x\sqrt{e}+\sqrt{e{x}^{2}+d} \right ){e}^{-{\frac{5}{2}}}}+{\frac{bx}{4\,e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{bdx}{8\,e}\sqrt{e{x}^{2}+d}}-{\frac{b{d}^{2}}{8}\ln \left ( x\sqrt{e}+\sqrt{e{x}^{2}+d} \right ){e}^{-{\frac{3}{2}}}}+{\frac{ax}{2}\sqrt{e{x}^{2}+d}}+{\frac{ad}{2}\ln \left ( x\sqrt{e}+\sqrt{e{x}^{2}+d} \right ){\frac{1}{\sqrt{e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^(1/2)*(c*x^4+b*x^2+a),x)

[Out]

1/6*c*x^3*(e*x^2+d)^(3/2)/e-1/8*c*d/e^2*x*(e*x^2+d)^(3/2)+1/16*c*d^2/e^2*x*(e*x^2+d)^(1/2)+1/16*c*d^3/e^(5/2)*
ln(x*e^(1/2)+(e*x^2+d)^(1/2))+1/4*b*x*(e*x^2+d)^(3/2)/e-1/8*b*d/e*x*(e*x^2+d)^(1/2)-1/8*b*d^2/e^(3/2)*ln(x*e^(
1/2)+(e*x^2+d)^(1/2))+1/2*a*x*(e*x^2+d)^(1/2)+1/2*a*d/e^(1/2)*ln(x*e^(1/2)+(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((e*x^2+d)^(1/2)*(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.96313, size = 533, normalized size = 4.04 \begin{align*} \left [\frac{3 \,{\left (c d^{3} - 2 \, b d^{2} e + 8 \, a d e^{2}\right )} \sqrt{e} \log \left (-2 \, e x^{2} - 2 \, \sqrt{e x^{2} + d} \sqrt{e} x - d\right ) + 2 \,{\left (8 \, c e^{3} x^{5} + 2 \,{\left (c d e^{2} + 6 \, b e^{3}\right )} x^{3} - 3 \,{\left (c d^{2} e - 2 \, b d e^{2} - 8 \, a e^{3}\right )} x\right )} \sqrt{e x^{2} + d}}{96 \, e^{3}}, -\frac{3 \,{\left (c d^{3} - 2 \, b d^{2} e + 8 \, a d e^{2}\right )} \sqrt{-e} \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right ) -{\left (8 \, c e^{3} x^{5} + 2 \,{\left (c d e^{2} + 6 \, b e^{3}\right )} x^{3} - 3 \,{\left (c d^{2} e - 2 \, b d e^{2} - 8 \, a e^{3}\right )} x\right )} \sqrt{e x^{2} + d}}{48 \, e^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(1/2)*(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

[1/96*(3*(c*d^3 - 2*b*d^2*e + 8*a*d*e^2)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + 2*(8*c*e^3*
x^5 + 2*(c*d*e^2 + 6*b*e^3)*x^3 - 3*(c*d^2*e - 2*b*d*e^2 - 8*a*e^3)*x)*sqrt(e*x^2 + d))/e^3, -1/48*(3*(c*d^3 -
2*b*d^2*e + 8*a*d*e^2)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (8*c*e^3*x^5 + 2*(c*d*e^2 + 6*b*e^3)*x^3
- 3*(c*d^2*e - 2*b*d*e^2 - 8*a*e^3)*x)*sqrt(e*x^2 + d))/e^3]

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Sympy [B]  time = 11.0742, size = 272, normalized size = 2.06 \begin{align*} \frac{a \sqrt{d} x \sqrt{1 + \frac{e x^{2}}{d}}}{2} + \frac{a d \operatorname{asinh}{\left (\frac{\sqrt{e} x}{\sqrt{d}} \right )}}{2 \sqrt{e}} + \frac{b d^{\frac{3}{2}} x}{8 e \sqrt{1 + \frac{e x^{2}}{d}}} + \frac{3 b \sqrt{d} x^{3}}{8 \sqrt{1 + \frac{e x^{2}}{d}}} - \frac{b d^{2} \operatorname{asinh}{\left (\frac{\sqrt{e} x}{\sqrt{d}} \right )}}{8 e^{\frac{3}{2}}} + \frac{b e x^{5}}{4 \sqrt{d} \sqrt{1 + \frac{e x^{2}}{d}}} - \frac{c d^{\frac{5}{2}} x}{16 e^{2} \sqrt{1 + \frac{e x^{2}}{d}}} - \frac{c d^{\frac{3}{2}} x^{3}}{48 e \sqrt{1 + \frac{e x^{2}}{d}}} + \frac{5 c \sqrt{d} x^{5}}{24 \sqrt{1 + \frac{e x^{2}}{d}}} + \frac{c d^{3} \operatorname{asinh}{\left (\frac{\sqrt{e} x}{\sqrt{d}} \right )}}{16 e^{\frac{5}{2}}} + \frac{c e x^{7}}{6 \sqrt{d} \sqrt{1 + \frac{e x^{2}}{d}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**(1/2)*(c*x**4+b*x**2+a),x)

[Out]

a*sqrt(d)*x*sqrt(1 + e*x**2/d)/2 + a*d*asinh(sqrt(e)*x/sqrt(d))/(2*sqrt(e)) + b*d**(3/2)*x/(8*e*sqrt(1 + e*x**
2/d)) + 3*b*sqrt(d)*x**3/(8*sqrt(1 + e*x**2/d)) - b*d**2*asinh(sqrt(e)*x/sqrt(d))/(8*e**(3/2)) + b*e*x**5/(4*s
qrt(d)*sqrt(1 + e*x**2/d)) - c*d**(5/2)*x/(16*e**2*sqrt(1 + e*x**2/d)) - c*d**(3/2)*x**3/(48*e*sqrt(1 + e*x**2
/d)) + 5*c*sqrt(d)*x**5/(24*sqrt(1 + e*x**2/d)) + c*d**3*asinh(sqrt(e)*x/sqrt(d))/(16*e**(5/2)) + c*e*x**7/(6*
sqrt(d)*sqrt(1 + e*x**2/d))

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Giac [A]  time = 1.13256, size = 143, normalized size = 1.08 \begin{align*} -\frac{1}{16} \,{\left (c d^{3} - 2 \, b d^{2} e + 8 \, a d e^{2}\right )} e^{\left (-\frac{5}{2}\right )} \log \left ({\left | -x e^{\frac{1}{2}} + \sqrt{x^{2} e + d} \right |}\right ) + \frac{1}{48} \,{\left (2 \,{\left (4 \, c x^{2} +{\left (c d e^{3} + 6 \, b e^{4}\right )} e^{\left (-4\right )}\right )} x^{2} - 3 \,{\left (c d^{2} e^{2} - 2 \, b d e^{3} - 8 \, a e^{4}\right )} e^{\left (-4\right )}\right )} \sqrt{x^{2} e + d} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(1/2)*(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

-1/16*(c*d^3 - 2*b*d^2*e + 8*a*d*e^2)*e^(-5/2)*log(abs(-x*e^(1/2) + sqrt(x^2*e + d))) + 1/48*(2*(4*c*x^2 + (c*
d*e^3 + 6*b*e^4)*e^(-4))*x^2 - 3*(c*d^2*e^2 - 2*b*d*e^3 - 8*a*e^4)*e^(-4))*sqrt(x^2*e + d)*x