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

Optimal. Leaf size=309 \[ \frac {e \sqrt {a^2+2 a b x^2+b^2 x^4} \left (4 c d-e^2\right ) \left (-16 a d^2+12 b c d-7 b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{256 d^{9/2} \left (a+b x^2\right )}+\frac {e \sqrt {a^2+2 a b x^2+b^2 x^4} (2 d x+e) \sqrt {c+d x^2+e x} \left (-16 a d^2+12 b c d-7 b e^2\right )}{128 d^4 \left (a+b x^2\right )}-\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2} \left (-80 a d^2+32 b c d+42 b d e x-35 b e^2\right )}{240 d^3 \left (a+b x^2\right )}+\frac {b x^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{5 d \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.63, antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {6744, 1653, 779, 612, 621, 206} \begin {gather*} -\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2} \left (-80 a d^2+32 b c d+42 b d e x-35 b e^2\right )}{240 d^3 \left (a+b x^2\right )}+\frac {e \sqrt {a^2+2 a b x^2+b^2 x^4} (2 d x+e) \sqrt {c+d x^2+e x} \left (-16 a d^2+12 b c d-7 b e^2\right )}{128 d^4 \left (a+b x^2\right )}+\frac {e \sqrt {a^2+2 a b x^2+b^2 x^4} \left (4 c d-e^2\right ) \left (-16 a d^2+12 b c d-7 b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{256 d^{9/2} \left (a+b x^2\right )}+\frac {b x^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{5 d \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(12*b*c*d - 16*a*d^2 - 7*b*e^2)*(e + 2*d*x)*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(128*d^4
*(a + b*x^2)) + (b*x^2*(c + e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d*(a + b*x^2)) - ((32*b*c*d
- 80*a*d^2 - 35*b*e^2 + 42*b*d*e*x)*(c + e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(240*d^3*(a + b*
x^2)) + (e*(4*c*d - e^2)*(12*b*c*d - 16*a*d^2 - 7*b*e^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(e + 2*d*x)/(
2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/(256*d^(9/2)*(a + b*x^2))

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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 6744

Int[(u_)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[Sqrt[a + b*x^n + c*x^(2*n)]/((4
*c)^(p - 1/2)*(b + 2*c*x^n)), Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] &
& EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int x \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x \left (2 a b+2 b^2 x^2\right ) \sqrt {c+e x+d x^2} \, dx}{2 a b+2 b^2 x^2}\\ &=\frac {b x^2 \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}+\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x \left (-2 b (2 b c-5 a d)-7 b^2 e x\right ) \sqrt {c+e x+d x^2} \, dx}{5 d \left (2 a b+2 b^2 x^2\right )}\\ &=\frac {b x^2 \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}-\frac {\left (32 b c d-80 a d^2-35 b e^2+42 b d e x\right ) \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{240 d^3 \left (a+b x^2\right )}+\frac {\left (b e \left (12 b c d-16 a d^2-7 b e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}\right ) \int \sqrt {c+e x+d x^2} \, dx}{16 d^3 \left (2 a b+2 b^2 x^2\right )}\\ &=\frac {e \left (12 b c d-16 a d^2-7 b e^2\right ) (e+2 d x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{128 d^4 \left (a+b x^2\right )}+\frac {b x^2 \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}-\frac {\left (32 b c d-80 a d^2-35 b e^2+42 b d e x\right ) \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{240 d^3 \left (a+b x^2\right )}+\frac {\left (b e \left (4 c d-e^2\right ) \left (12 b c d-16 a d^2-7 b e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}\right ) \int \frac {1}{\sqrt {c+e x+d x^2}} \, dx}{128 d^4 \left (2 a b+2 b^2 x^2\right )}\\ &=\frac {e \left (12 b c d-16 a d^2-7 b e^2\right ) (e+2 d x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{128 d^4 \left (a+b x^2\right )}+\frac {b x^2 \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}-\frac {\left (32 b c d-80 a d^2-35 b e^2+42 b d e x\right ) \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{240 d^3 \left (a+b x^2\right )}+\frac {\left (b e \left (4 c d-e^2\right ) \left (12 b c d-16 a d^2-7 b e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{4 d-x^2} \, dx,x,\frac {e+2 d x}{\sqrt {c+e x+d x^2}}\right )}{64 d^4 \left (2 a b+2 b^2 x^2\right )}\\ &=\frac {e \left (12 b c d-16 a d^2-7 b e^2\right ) (e+2 d x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{128 d^4 \left (a+b x^2\right )}+\frac {b x^2 \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}-\frac {\left (32 b c d-80 a d^2-35 b e^2+42 b d e x\right ) \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{240 d^3 \left (a+b x^2\right )}+\frac {e \left (4 c d-e^2\right ) \left (12 b c d-16 a d^2-7 b e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{256 d^{9/2} \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 214, normalized size = 0.69 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (2 \sqrt {d} \sqrt {c+x (d x+e)} \left (80 a d^2 \left (8 c d+8 d^2 x^2+2 d e x-3 e^2\right )+b \left (-256 c^2 d^2+4 c d \left (32 d^2 x^2-58 d e x+115 e^2\right )+384 d^4 x^4+48 d^3 e x^3-56 d^2 e^2 x^2+70 d e^3 x-105 e^4\right )\right )+15 e \left (e^2-4 c d\right ) \left (16 a d^2-12 b c d+7 b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+x (d x+e)}}\right )\right )}{3840 d^{9/2} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(2*Sqrt[d]*Sqrt[c + x*(e + d*x)]*(80*a*d^2*(8*c*d - 3*e^2 + 2*d*e*x + 8*d^2*x^2) + b*(-25
6*c^2*d^2 - 105*e^4 + 70*d*e^3*x - 56*d^2*e^2*x^2 + 48*d^3*e*x^3 + 384*d^4*x^4 + 4*c*d*(115*e^2 - 58*d*e*x + 3
2*d^2*x^2))) + 15*e*(-4*c*d + e^2)*(-12*b*c*d + 16*a*d^2 + 7*b*e^2)*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + x*
(e + d*x)])]))/(3840*d^(9/2)*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 0.82, size = 239, normalized size = 0.77 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (\frac {\left (64 a c d^3 e-16 a d^2 e^3-48 b c^2 d^2 e+40 b c d e^3-7 b e^5\right ) \log \left (-2 \sqrt {d} \sqrt {c+d x^2+e x}+2 d x+e\right )}{256 d^{9/2}}+\frac {\sqrt {c+d x^2+e x} \left (640 a c d^3+640 a d^4 x^2+160 a d^3 e x-240 a d^2 e^2-256 b c^2 d^2+128 b c d^3 x^2-232 b c d^2 e x+460 b c d e^2+384 b d^4 x^4+48 b d^3 e x^3-56 b d^2 e^2 x^2+70 b d e^3 x-105 b e^4\right )}{1920 d^4}\right )}{a+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*((Sqrt[c + e*x + d*x^2]*(-256*b*c^2*d^2 + 640*a*c*d^3 + 460*b*c*d*e^2 - 240*a*d^2*e^2 - 1
05*b*e^4 - 232*b*c*d^2*e*x + 160*a*d^3*e*x + 70*b*d*e^3*x + 128*b*c*d^3*x^2 + 640*a*d^4*x^2 - 56*b*d^2*e^2*x^2
+ 48*b*d^3*e*x^3 + 384*b*d^4*x^4))/(1920*d^4) + ((-48*b*c^2*d^2*e + 64*a*c*d^3*e + 40*b*c*d*e^3 - 16*a*d^2*e^
3 - 7*b*e^5)*Log[e + 2*d*x - 2*Sqrt[d]*Sqrt[c + e*x + d*x^2]])/(256*d^(9/2))))/(a + b*x^2)

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fricas [A]  time = 1.96, size = 469, normalized size = 1.52 \begin {gather*} \left [\frac {15 \, {\left (7 \, b e^{5} - 8 \, {\left (5 \, b c d - 2 \, a d^{2}\right )} e^{3} + 16 \, {\left (3 \, b c^{2} d^{2} - 4 \, a c d^{3}\right )} e\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x + 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (384 \, b d^{5} x^{4} + 48 \, b d^{4} e x^{3} - 256 \, b c^{2} d^{3} + 640 \, a c d^{4} - 105 \, b d e^{4} + 20 \, {\left (23 \, b c d^{2} - 12 \, a d^{3}\right )} e^{2} + 8 \, {\left (16 \, b c d^{4} + 80 \, a d^{5} - 7 \, b d^{3} e^{2}\right )} x^{2} + 2 \, {\left (35 \, b d^{2} e^{3} - 4 \, {\left (29 \, b c d^{3} - 20 \, a d^{4}\right )} e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{7680 \, d^{5}}, -\frac {15 \, {\left (7 \, b e^{5} - 8 \, {\left (5 \, b c d - 2 \, a d^{2}\right )} e^{3} + 16 \, {\left (3 \, b c^{2} d^{2} - 4 \, a c d^{3}\right )} e\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) - 2 \, {\left (384 \, b d^{5} x^{4} + 48 \, b d^{4} e x^{3} - 256 \, b c^{2} d^{3} + 640 \, a c d^{4} - 105 \, b d e^{4} + 20 \, {\left (23 \, b c d^{2} - 12 \, a d^{3}\right )} e^{2} + 8 \, {\left (16 \, b c d^{4} + 80 \, a d^{5} - 7 \, b d^{3} e^{2}\right )} x^{2} + 2 \, {\left (35 \, b d^{2} e^{3} - 4 \, {\left (29 \, b c d^{3} - 20 \, a d^{4}\right )} e\right )} x\right )} \sqrt {d x^{2} + e x + c}}{3840 \, d^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/7680*(15*(7*b*e^5 - 8*(5*b*c*d - 2*a*d^2)*e^3 + 16*(3*b*c^2*d^2 - 4*a*c*d^3)*e)*sqrt(d)*log(8*d^2*x^2 + 8*d
*e*x + 4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(d) + 4*c*d + e^2) + 4*(384*b*d^5*x^4 + 48*b*d^4*e*x^3 - 256*b*
c^2*d^3 + 640*a*c*d^4 - 105*b*d*e^4 + 20*(23*b*c*d^2 - 12*a*d^3)*e^2 + 8*(16*b*c*d^4 + 80*a*d^5 - 7*b*d^3*e^2)
*x^2 + 2*(35*b*d^2*e^3 - 4*(29*b*c*d^3 - 20*a*d^4)*e)*x)*sqrt(d*x^2 + e*x + c))/d^5, -1/3840*(15*(7*b*e^5 - 8*
(5*b*c*d - 2*a*d^2)*e^3 + 16*(3*b*c^2*d^2 - 4*a*c*d^3)*e)*sqrt(-d)*arctan(1/2*sqrt(d*x^2 + e*x + c)*(2*d*x + e
)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) - 2*(384*b*d^5*x^4 + 48*b*d^4*e*x^3 - 256*b*c^2*d^3 + 640*a*c*d^4 - 105*b*
d*e^4 + 20*(23*b*c*d^2 - 12*a*d^3)*e^2 + 8*(16*b*c*d^4 + 80*a*d^5 - 7*b*d^3*e^2)*x^2 + 2*(35*b*d^2*e^3 - 4*(29
*b*c*d^3 - 20*a*d^4)*e)*x)*sqrt(d*x^2 + e*x + c))/d^5]

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giac [A]  time = 0.60, size = 360, normalized size = 1.17 \begin {gather*} \frac {1}{1920} \, \sqrt {d x^{2} + x e + c} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, b x \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {b e \mathrm {sgn}\left (b x^{2} + a\right )}{d}\right )} x + \frac {16 \, b c d^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 80 \, a d^{4} \mathrm {sgn}\left (b x^{2} + a\right ) - 7 \, b d^{2} e^{2} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{4}}\right )} x - \frac {116 \, b c d^{2} e \mathrm {sgn}\left (b x^{2} + a\right ) - 80 \, a d^{3} e \mathrm {sgn}\left (b x^{2} + a\right ) - 35 \, b d e^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{4}}\right )} x - \frac {256 \, b c^{2} d^{2} \mathrm {sgn}\left (b x^{2} + a\right ) - 640 \, a c d^{3} \mathrm {sgn}\left (b x^{2} + a\right ) - 460 \, b c d e^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 240 \, a d^{2} e^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 105 \, b e^{4} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{4}}\right )} - \frac {{\left (48 \, b c^{2} d^{2} e \mathrm {sgn}\left (b x^{2} + a\right ) - 64 \, a c d^{3} e \mathrm {sgn}\left (b x^{2} + a\right ) - 40 \, b c d e^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 16 \, a d^{2} e^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 7 \, b e^{5} \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | -2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} \sqrt {d} - e \right |}\right )}{256 \, d^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(d*x^2 + x*e + c)*(2*(4*(6*(8*b*x*sgn(b*x^2 + a) + b*e*sgn(b*x^2 + a)/d)*x + (16*b*c*d^3*sgn(b*x^2
+ a) + 80*a*d^4*sgn(b*x^2 + a) - 7*b*d^2*e^2*sgn(b*x^2 + a))/d^4)*x - (116*b*c*d^2*e*sgn(b*x^2 + a) - 80*a*d^3
*e*sgn(b*x^2 + a) - 35*b*d*e^3*sgn(b*x^2 + a))/d^4)*x - (256*b*c^2*d^2*sgn(b*x^2 + a) - 640*a*c*d^3*sgn(b*x^2
+ a) - 460*b*c*d*e^2*sgn(b*x^2 + a) + 240*a*d^2*e^2*sgn(b*x^2 + a) + 105*b*e^4*sgn(b*x^2 + a))/d^4) - 1/256*(4
8*b*c^2*d^2*e*sgn(b*x^2 + a) - 64*a*c*d^3*e*sgn(b*x^2 + a) - 40*b*c*d*e^3*sgn(b*x^2 + a) + 16*a*d^2*e^3*sgn(b*
x^2 + a) + 7*b*e^5*sgn(b*x^2 + a))*log(abs(-2*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))*sqrt(d) - e))/d^(9/2)

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maple [A]  time = 0.01, size = 442, normalized size = 1.43 \begin {gather*} \frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-960 a c \,d^{4} e \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+240 a \,d^{3} e^{3} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+720 b \,c^{2} d^{3} e \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-600 b c \,d^{2} e^{3} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+105 b d \,e^{5} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-960 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {9}{2}} e x +720 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {7}{2}} e x +768 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,d^{\frac {9}{2}} x^{2}-420 \sqrt {d \,x^{2}+e x +c}\, b \,d^{\frac {5}{2}} e^{3} x -480 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {7}{2}} e^{2}+360 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {5}{2}} e^{2}-672 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,d^{\frac {7}{2}} e x -210 \sqrt {d \,x^{2}+e x +c}\, b \,d^{\frac {3}{2}} e^{4}+1280 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,d^{\frac {9}{2}}-512 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b c \,d^{\frac {7}{2}}+560 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,d^{\frac {5}{2}} e^{2}\right )}{3840 \left (b \,x^{2}+a \right ) d^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3840*((b*x^2+a)^2)^(1/2)*(768*(d*x^2+e*x+c)^(3/2)*d^(9/2)*x^2*b-672*(d*x^2+e*x+c)^(3/2)*d^(7/2)*x*b*e+1280*(
d*x^2+e*x+c)^(3/2)*d^(9/2)*a-512*(d*x^2+e*x+c)^(3/2)*d^(7/2)*b*c+560*(d*x^2+e*x+c)^(3/2)*d^(5/2)*b*e^2-960*(d*
x^2+e*x+c)^(1/2)*d^(9/2)*x*a*e+720*(d*x^2+e*x+c)^(1/2)*d^(7/2)*x*b*c*e-420*(d*x^2+e*x+c)^(1/2)*d^(5/2)*x*b*e^3
-480*(d*x^2+e*x+c)^(1/2)*d^(7/2)*a*e^2+360*(d*x^2+e*x+c)^(1/2)*d^(5/2)*b*c*e^2-210*(d*x^2+e*x+c)^(1/2)*d^(3/2)
*b*e^4-960*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*a*c*d^4*e+240*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)
*d^(1/2)+2*d*x+e)/d^(1/2))*a*d^3*e^3+720*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*b*c^2*d^3*e-6
00*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*b*c*d^2*e^3+105*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/
2)+2*d*x+e)/d^(1/2))*b*d*e^5)/(b*x^2+a)/d^(11/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1/2),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^2-4*c*d>0)', see `assume?` f
or more details)Is e^2-4*c*d positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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