3.1.39 \(\int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x} \, dx\) [39]

Optimal. Leaf size=286 \[ \frac {\left (8 a d^2-b e^2-2 b d e x\right ) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 d^2 \left (a+b x^2\right )}+\frac {b \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac {e \left (8 a d^2-b \left (4 c d-e^2\right )\right ) \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{16 d^{5/2} \left (a+b x^2\right )}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+e x+d x^2}}\right )}{a+b x^2} \]

[Out]

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

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Rubi [A]
time = 0.50, antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {6876, 1667, 828, 857, 635, 212, 738} \begin {gather*} \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2+e x} \left (8 a d^2-2 b d e x-b e^2\right )}{8 d^2 \left (a+b x^2\right )}+\frac {e \sqrt {a^2+2 a b x^2+b^2 x^4} \left (8 a d^2-b \left (4 c d-e^2\right )\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{16 d^{5/2} \left (a+b x^2\right )}+\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{3 d \left (a+b x^2\right )}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{a+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*d^2 - b*e^2 - 2*b*d*e*x)*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*d^2*(a + b*x^2)) + (b
*(c + e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d*(a + b*x^2)) + (e*(8*a*d^2 - b*(4*c*d - e^2))*S
qrt[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])])/(16*d^(5/2)*(a + b*x^2)
) - (a*Sqrt[c]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(2*c + e*x)/(2*Sqrt[c]*Sqrt[c + e*x + d*x^2])])/(a + b*
x^2)

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 635

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 738

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

Rule 828

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^
2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 857

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

Rule 1667

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 6876

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 \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (2 a b+2 b^2 x^2\right ) \sqrt {c+e x+d x^2}}{x} \, dx}{2 a b+2 b^2 x^2}\\ &=\frac {b \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (6 a b d-3 b^2 e x\right ) \sqrt {c+e x+d x^2}}{x} \, dx}{3 d \left (2 a b+2 b^2 x^2\right )}\\ &=\frac {\left (8 a d^2-b e^2-2 b d e x\right ) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 d^2 \left (a+b x^2\right )}+\frac {b \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}-\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {-24 a b c d^2-\frac {3}{2} b e \left (8 a d^2-b \left (4 c d-e^2\right )\right ) x}{x \sqrt {c+e x+d x^2}} \, dx}{12 d^2 \left (2 a b+2 b^2 x^2\right )}\\ &=\frac {\left (8 a d^2-b e^2-2 b d e x\right ) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 d^2 \left (a+b x^2\right )}+\frac {b \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac {\left (2 a b c \sqrt {a^2+2 a b x^2+b^2 x^4}\right ) \int \frac {1}{x \sqrt {c+e x+d x^2}} \, dx}{2 a b+2 b^2 x^2}+\frac {\left (b e \left (8 a d^2-b \left (4 c d-e^2\right )\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}{8 d^2 \left (2 a b+2 b^2 x^2\right )}\\ &=\frac {\left (8 a d^2-b e^2-2 b d e x\right ) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 d^2 \left (a+b x^2\right )}+\frac {b \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}-\frac {\left (4 a b c \sqrt {a^2+2 a b x^2+b^2 x^4}\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {2 c+e x}{\sqrt {c+e x+d x^2}}\right )}{2 a b+2 b^2 x^2}+\frac {\left (b e \left (8 a d^2-b \left (4 c d-e^2\right )\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}\right ) \text {Subst}\left (\int \frac {1}{4 d-x^2} \, dx,x,\frac {e+2 d x}{\sqrt {c+e x+d x^2}}\right )}{4 d^2 \left (2 a b+2 b^2 x^2\right )}\\ &=\frac {\left (8 a d^2-b e^2-2 b d e x\right ) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 d^2 \left (a+b x^2\right )}+\frac {b \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac {e \left (8 a d^2-b \left (4 c d-e^2\right )\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 )}{16 d^{5/2} \left (a+b x^2\right )}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+e x+d x^2}}\right )}{a+b x^2}\\ \end {align*}

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Mathematica [A]
time = 0.49, size = 174, normalized size = 0.61 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (2 \sqrt {d} \sqrt {c+x (e+d x)} \left (24 a d^2+b \left (8 c d-3 e^2+2 d e x+8 d^2 x^2\right )\right )+96 a \sqrt {c} d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} x-\sqrt {c+x (e+d x)}}{\sqrt {c}}\right )-3 e \left (8 a d^2+b \left (-4 c d+e^2\right )\right ) \log \left (e+2 d x-2 \sqrt {d} \sqrt {c+x (e+d x)}\right )\right )}{48 d^{5/2} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(2*Sqrt[d]*Sqrt[c + x*(e + d*x)]*(24*a*d^2 + b*(8*c*d - 3*e^2 + 2*d*e*x + 8*d^2*x^2)) + 9
6*a*Sqrt[c]*d^(5/2)*ArcTanh[(Sqrt[d]*x - Sqrt[c + x*(e + d*x)])/Sqrt[c]] - 3*e*(8*a*d^2 + b*(-4*c*d + e^2))*Lo
g[e + 2*d*x - 2*Sqrt[d]*Sqrt[c + x*(e + d*x)]]))/(48*d^(5/2)*(a + b*x^2))

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Maple [A]
time = 0.20, size = 251, normalized size = 0.88

method	result	size
default	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (16 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {5}{2}} b -12 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b e x -48 d^{\frac {7}{2}} \sqrt {c}\, \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a +48 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {7}{2}} a -6 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} b \,e^{2}+24 d^{3} \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a e -12 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b c \,d^{2} e +3 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b d \,e^{3}\right )}{48 \left (b \,x^{2}+a \right ) d^{\frac {7}{2}}}\)	\(251\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*((b*x^2+a)^2)^(1/2)*(16*(d*x^2+e*x+c)^(3/2)*d^(5/2)*b-12*(d*x^2+e*x+c)^(1/2)*d^(5/2)*b*e*x-48*d^(7/2)*c^(
1/2)*ln((2*c+e*x+2*c^(1/2)*(d*x^2+e*x+c)^(1/2))/x)*a+48*(d*x^2+e*x+c)^(1/2)*d^(7/2)*a-6*(d*x^2+e*x+c)^(1/2)*d^
(3/2)*b*e^2+24*d^3*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*a*e-12*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*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^3)/(b*x^
2+a)/d^(7/2)

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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((d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x,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(4*c*d-%e^2>0)', see `assume?`
for more det

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Fricas [A]
time = 0.84, size = 771, normalized size = 2.70 \begin {gather*} \left [\frac {48 \, a \sqrt {c} d^{3} \log \left (\frac {4 \, c d x^{2} + x^{2} e^{2} + 8 \, c x e - 4 \, \sqrt {d x^{2} + x e + c} {\left (x e + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) + 3 \, {\left (b e^{3} - 4 \, {\left (b c d - 2 \, a d^{2}\right )} e\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d x e + 4 \, \sqrt {d x^{2} + x e + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (8 \, b d^{3} x^{2} + 2 \, b d^{2} x e + 8 \, b c d^{2} + 24 \, a d^{3} - 3 \, b d e^{2}\right )} \sqrt {d x^{2} + x e + c}}{96 \, d^{3}}, \frac {24 \, a \sqrt {c} d^{3} \log \left (\frac {4 \, c d x^{2} + x^{2} e^{2} + 8 \, c x e - 4 \, \sqrt {d x^{2} + x e + c} {\left (x e + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) - 3 \, {\left (b e^{3} - 4 \, {\left (b c d - 2 \, a d^{2}\right )} e\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + x e + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d x e + c d\right )}}\right ) + 2 \, {\left (8 \, b d^{3} x^{2} + 2 \, b d^{2} x e + 8 \, b c d^{2} + 24 \, a d^{3} - 3 \, b d e^{2}\right )} \sqrt {d x^{2} + x e + c}}{48 \, d^{3}}, \frac {96 \, a \sqrt {-c} d^{3} \arctan \left (\frac {\sqrt {d x^{2} + x e + c} {\left (x e + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c x e + c^{2}\right )}}\right ) + 3 \, {\left (b e^{3} - 4 \, {\left (b c d - 2 \, a d^{2}\right )} e\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d x e + 4 \, \sqrt {d x^{2} + x e + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (8 \, b d^{3} x^{2} + 2 \, b d^{2} x e + 8 \, b c d^{2} + 24 \, a d^{3} - 3 \, b d e^{2}\right )} \sqrt {d x^{2} + x e + c}}{96 \, d^{3}}, \frac {48 \, a \sqrt {-c} d^{3} \arctan \left (\frac {\sqrt {d x^{2} + x e + c} {\left (x e + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c x e + c^{2}\right )}}\right ) - 3 \, {\left (b e^{3} - 4 \, {\left (b c d - 2 \, a d^{2}\right )} e\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + x e + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d x e + c d\right )}}\right ) + 2 \, {\left (8 \, b d^{3} x^{2} + 2 \, b d^{2} x e + 8 \, b c d^{2} + 24 \, a d^{3} - 3 \, b d e^{2}\right )} \sqrt {d x^{2} + x e + c}}{48 \, d^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(48*a*sqrt(c)*d^3*log((4*c*d*x^2 + x^2*e^2 + 8*c*x*e - 4*sqrt(d*x^2 + x*e + c)*(x*e + 2*c)*sqrt(c) + 8*c
^2)/x^2) + 3*(b*e^3 - 4*(b*c*d - 2*a*d^2)*e)*sqrt(d)*log(8*d^2*x^2 + 8*d*x*e + 4*sqrt(d*x^2 + x*e + c)*(2*d*x
+ e)*sqrt(d) + 4*c*d + e^2) + 4*(8*b*d^3*x^2 + 2*b*d^2*x*e + 8*b*c*d^2 + 24*a*d^3 - 3*b*d*e^2)*sqrt(d*x^2 + x*
e + c))/d^3, 1/48*(24*a*sqrt(c)*d^3*log((4*c*d*x^2 + x^2*e^2 + 8*c*x*e - 4*sqrt(d*x^2 + x*e + c)*(x*e + 2*c)*s
qrt(c) + 8*c^2)/x^2) - 3*(b*e^3 - 4*(b*c*d - 2*a*d^2)*e)*sqrt(-d)*arctan(1/2*sqrt(d*x^2 + x*e + c)*(2*d*x + e)
*sqrt(-d)/(d^2*x^2 + d*x*e + c*d)) + 2*(8*b*d^3*x^2 + 2*b*d^2*x*e + 8*b*c*d^2 + 24*a*d^3 - 3*b*d*e^2)*sqrt(d*x
^2 + x*e + c))/d^3, 1/96*(96*a*sqrt(-c)*d^3*arctan(1/2*sqrt(d*x^2 + x*e + c)*(x*e + 2*c)*sqrt(-c)/(c*d*x^2 + c
*x*e + c^2)) + 3*(b*e^3 - 4*(b*c*d - 2*a*d^2)*e)*sqrt(d)*log(8*d^2*x^2 + 8*d*x*e + 4*sqrt(d*x^2 + x*e + c)*(2*
d*x + e)*sqrt(d) + 4*c*d + e^2) + 4*(8*b*d^3*x^2 + 2*b*d^2*x*e + 8*b*c*d^2 + 24*a*d^3 - 3*b*d*e^2)*sqrt(d*x^2
+ x*e + c))/d^3, 1/48*(48*a*sqrt(-c)*d^3*arctan(1/2*sqrt(d*x^2 + x*e + c)*(x*e + 2*c)*sqrt(-c)/(c*d*x^2 + c*x*
e + c^2)) - 3*(b*e^3 - 4*(b*c*d - 2*a*d^2)*e)*sqrt(-d)*arctan(1/2*sqrt(d*x^2 + x*e + c)*(2*d*x + e)*sqrt(-d)/(
d^2*x^2 + d*x*e + c*d)) + 2*(8*b*d^3*x^2 + 2*b*d^2*x*e + 8*b*c*d^2 + 24*a*d^3 - 3*b*d*e^2)*sqrt(d*x^2 + x*e +
c))/d^3]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c + d x^{2} + e x} \sqrt {\left (a + b x^{2}\right )^{2}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c + d*x**2 + e*x)*sqrt((a + b*x**2)**2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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