\(\int \sqrt {d+e x^2} (a+b x^2+c x^4) \, dx\) [278]

Optimal result

Integrand size = 24, antiderivative size = 132 \[ \int \sqrt {d+e x^2} \left (a+b x^2+c x^4\right ) \, dx=\frac {\left (c d^2-2 b d e+8 a e^2\right ) x \sqrt {d+e x^2}}{16 e^2}-\frac {(c d-2 b e) x \left (d+e x^2\right )^{3/2}}{8 e^2}+\frac {c x^3 \left (d+e x^2\right )^{3/2}}{6 e}+\frac {d \left (c d^2-2 b d e+8 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{16 e^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1173, 396, 201, 223, 212} \[ \int \sqrt {d+e x^2} \left (a+b x^2+c x^4\right ) \, dx=\frac {d \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \left (8 a e^2-2 b d e+c d^2\right )}{16 e^{5/2}}+\frac {x \sqrt {d+e x^2} \left (8 a e^2-2 b d e+c d^2\right )}{16 e^2}-\frac {x \left (d+e x^2\right )^{3/2} (c d-2 b e)}{8 e^2}+\frac {c x^3 \left (d+e x^2\right )^{3/2}}{6 e} \]

[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 201

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 212

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

Rule 223

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 396

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 1173

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*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.85 \[ \int \sqrt {d+e x^2} \left (a+b x^2+c x^4\right ) \, dx=\frac {x \sqrt {d+e x^2} \left (-3 c d^2+6 b d e+24 a e^2+2 c d e x^2+12 b e^2 x^2+8 c e^2 x^4\right )}{48 e^2}-\frac {d \left (c d^2-2 b d e+8 a e^2\right ) \log \left (-\sqrt {e} x+\sqrt {d+e x^2}\right )}{16 e^{5/2}} \]

[In]

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

[Out]

(x*Sqrt[d + e*x^2]*(-3*c*d^2 + 6*b*d*e + 24*a*e^2 + 2*c*d*e*x^2 + 12*b*e^2*x^2 + 8*c*e^2*x^4))/(48*e^2) - (d*(
c*d^2 - 2*b*d*e + 8*a*e^2)*Log[-(Sqrt[e]*x) + Sqrt[d + e*x^2]])/(16*e^(5/2))

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(\frac {d \left (a \,e^{2}-\frac {1}{4} b d e +\frac {1}{8} c \,d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}}{x \sqrt {e}}\right )+\sqrt {e \,x^{2}+d}\, \left (\left (\frac {1}{3} c \,x^{4}+\frac {1}{2} b \,x^{2}+a \right ) e^{\frac {5}{2}}+\frac {d \left (\left (\frac {c \,x^{2}}{3}+b \right ) e^{\frac {3}{2}}-\frac {c d \sqrt {e}}{2}\right )}{4}\right ) x}{2 e^{\frac {5}{2}}}\)	\(96\)
risch	\(\frac {x \left (8 c \,e^{2} x^{4}+12 b \,e^{2} x^{2}+2 d e \,x^{2} c +24 a \,e^{2}+6 b d e -3 c \,d^{2}\right ) \sqrt {e \,x^{2}+d}}{48 e^{2}}+\frac {d \left (8 a \,e^{2}-2 b d e +c \,d^{2}\right ) \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{16 e^{\frac {5}{2}}}\)	\(100\)
default	\(a \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )+c \left (\frac {x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{6 e}-\frac {d \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 e}-\frac {d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4 e}\right )}{2 e}\right )+b \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 e}-\frac {d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4 e}\right )\)	\(181\)

[In]

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

[Out]

1/2*(d*(a*e^2-1/4*b*d*e+1/8*c*d^2)*arctanh((e*x^2+d)^(1/2)/x/e^(1/2))+(e*x^2+d)^(1/2)*((1/3*c*x^4+1/2*b*x^2+a)
*e^(5/2)+1/4*d*((1/3*c*x^2+b)*e^(3/2)-1/2*c*d*e^(1/2)))*x)/e^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.76 \[ \int \sqrt {d+e x^2} \left (a+b x^2+c x^4\right ) \, dx=\left [\frac {3 \, {\left (c d^{3} - 2 \, b d^{2} e + 8 \, a d e^{2}\right )} \sqrt {e} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) + 2 \, {\left (8 \, c e^{3} x^{5} + 2 \, {\left (c d e^{2} + 6 \, b e^{3}\right )} x^{3} - 3 \, {\left (c d^{2} e - 2 \, b d e^{2} - 8 \, a e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{96 \, e^{3}}, -\frac {3 \, {\left (c d^{3} - 2 \, b d^{2} e + 8 \, a d e^{2}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - {\left (8 \, c e^{3} x^{5} + 2 \, {\left (c d e^{2} + 6 \, b e^{3}\right )} x^{3} - 3 \, {\left (c d^{2} e - 2 \, b d e^{2} - 8 \, a e^{3}\right )} x\right )} \sqrt {e x^{2} + d}}{48 \, e^{3}}\right ] \]

[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]

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.16 \[ \int \sqrt {d+e x^2} \left (a+b x^2+c x^4\right ) \, dx=\begin {cases} \sqrt {d + e x^{2}} \left (\frac {c x^{5}}{6} + \frac {x^{3} \left (b e + \frac {c d}{6}\right )}{4 e} + \frac {x \left (a e + b d - \frac {3 d \left (b e + \frac {c d}{6}\right )}{4 e}\right )}{2 e}\right ) + \left (a d - \frac {d \left (a e + b d - \frac {3 d \left (b e + \frac {c d}{6}\right )}{4 e}\right )}{2 e}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {e} \sqrt {d + e x^{2}} + 2 e x \right )}}{\sqrt {e}} & \text {for}\: d \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {e x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: e \neq 0 \\\sqrt {d} \left (a x + \frac {b x^{3}}{3} + \frac {c x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(d + e*x**2)*(c*x**5/6 + x**3*(b*e + c*d/6)/(4*e) + x*(a*e + b*d - 3*d*(b*e + c*d/6)/(4*e))/(2*
e)) + (a*d - d*(a*e + b*d - 3*d*(b*e + c*d/6)/(4*e))/(2*e))*Piecewise((log(2*sqrt(e)*sqrt(d + e*x**2) + 2*e*x)
/sqrt(e), Ne(d, 0)), (x*log(x)/sqrt(e*x**2), True)), Ne(e, 0)), (sqrt(d)*(a*x + b*x**3/3 + c*x**5/5), True))

Maxima [F(-2)]

Exception generated. \[ \int \sqrt {d+e x^2} \left (a+b x^2+c x^4\right ) \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86 \[ \int \sqrt {d+e x^2} \left (a+b x^2+c x^4\right ) \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, c x^{2} + \frac {c d e^{3} + 6 \, b e^{4}}{e^{4}}\right )} x^{2} - \frac {3 \, {\left (c d^{2} e^{2} - 2 \, b d e^{3} - 8 \, a e^{4}\right )}}{e^{4}}\right )} \sqrt {e x^{2} + d} x - \frac {{\left (c d^{3} - 2 \, b d^{2} e + 8 \, a d e^{2}\right )} \log \left ({\left | -\sqrt {e} x + \sqrt {e x^{2} + d} \right |}\right )}{16 \, e^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {d+e x^2} \left (a+b x^2+c x^4\right ) \, dx=\int \sqrt {e\,x^2+d}\,\left (c\,x^4+b\,x^2+a\right ) \,d x \]

[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)