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

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

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Rubi [A]  time = 0.11, antiderivative size = 132, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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]

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*}

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Mathematica [A]  time = 0.23, size = 121, normalized size = 0.92 \begin {gather*} \frac {\sqrt {d+e x^2} \left (\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}}+\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 )\right )}{48 e^{5/2}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.19, size = 117, normalized size = 0.89 \begin {gather*} \frac {\sqrt {d+e x^2} \left (24 a e^2 x+6 b d e x+12 b e^2 x^3-3 c d^2 x+2 c d e x^3+8 c e^2 x^5\right )}{48 e^2}+\frac {\log \left (\sqrt {d+e x^2}-\sqrt {e} x\right ) \left (-8 a d e^2+2 b d^2 e-c d^3\right )}{16 e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x^2]*(-3*c*d^2*x + 6*b*d*e*x + 24*a*e^2*x + 2*c*d*e*x^3 + 12*b*e^2*x^3 + 8*c*e^2*x^5))/(48*e^2) +
((-(c*d^3) + 2*b*d^2*e - 8*a*d*e^2)*Log[-(Sqrt[e]*x) + Sqrt[d + e*x^2]])/(16*e^(5/2))

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fricas [A]  time = 1.00, size = 232, normalized size = 1.76 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.22, size = 106, normalized size = 0.80 \begin {gather*} -\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 {gather*}

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

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maple [A]  time = 0.01, size = 175, normalized size = 1.33 \begin {gather*} \frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}} c \,x^{3}}{6 e}+\frac {a d \ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}-\frac {b \,d^{2} \ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{8 e^{\frac {3}{2}}}+\frac {c \,d^{3} \ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{16 e^{\frac {5}{2}}}+\frac {\sqrt {e \,x^{2}+d}\, a x}{2}-\frac {\sqrt {e \,x^{2}+d}\, b d x}{8 e}+\frac {\sqrt {e \,x^{2}+d}\, c \,d^{2} x}{16 e^{2}}+\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}} b x}{4 e}-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}} c d x}{8 e^{2}} \end {gather*}

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

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maxima [A]  time = 0.98, size = 153, normalized size = 1.16 \begin {gather*} \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} c x^{3}}{6 \, e} + \frac {1}{2} \, \sqrt {e x^{2} + d} a x - \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} c d x}{8 \, e^{2}} + \frac {\sqrt {e x^{2} + d} c d^{2} x}{16 \, e^{2}} + \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} b x}{4 \, e} - \frac {\sqrt {e x^{2} + d} b d x}{8 \, e} + \frac {c d^{3} \operatorname {arsinh}\left (\frac {e x}{\sqrt {d e}}\right )}{16 \, e^{\frac {5}{2}}} - \frac {b d^{2} \operatorname {arsinh}\left (\frac {e x}{\sqrt {d e}}\right )}{8 \, e^{\frac {3}{2}}} + \frac {a d \operatorname {arsinh}\left (\frac {e x}{\sqrt {d e}}\right )}{2 \, \sqrt {e}} \end {gather*}

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {e\,x^2+d}\,\left (c\,x^4+b\,x^2+a\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 12.27, size = 272, normalized size = 2.06 \begin {gather*} \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 {gather*}

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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