3.1.8 \(\int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx\)

Optimal. Leaf size=288 \[ -\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (4 a c d-a e^2+8 b c^2\right ) \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{8 c^{3/2} \left (a+b x^2\right )}-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{2 c x^2 \left (a+b x^2\right )}+\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2+e x} (2 x (a d+2 b c)+a e)}{4 c x \left (a+b x^2\right )}+\frac {b e \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{2 \sqrt {d} \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.78, antiderivative size = 288, 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 = {6744, 1650, 812, 843, 621, 206, 724} \begin {gather*} -\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (4 a c d-a e^2+8 b c^2\right ) \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{8 c^{3/2} \left (a+b x^2\right )}-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{2 c x^2 \left (a+b x^2\right )}+\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2+e x} (2 x (a d+2 b c)+a e)}{4 c x \left (a+b x^2\right )}+\frac {b e \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{2 \sqrt {d} \left (a+b x^2\right )} \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^3,x]

[Out]

((a*e + 2*(2*b*c + a*d)*x)*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*c*x*(a + b*x^2)) - (a*(c
+ e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*c*x^2*(a + b*x^2)) + (b*e*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])])/(2*Sqrt[d]*(a + b*x^2)) - ((8*b*c^2 + 4*a*c*d - a
*e^2)*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])])/(8*c^(3/2)*(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 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 724

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 812

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

Rule 843

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 1650

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomia
lQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*
x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^
(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Q + c*d*R*(m + 1) - b*e*R*(m + p + 2)
- c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]

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

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Mathematica [A]  time = 0.34, size = 177, normalized size = 0.61 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (4 b c^{3/2} e x^2 \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+x (d x+e)}}\right )-\sqrt {d} \left (x^2 \left (4 a c d-a e^2+8 b c^2\right ) \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+x (d x+e)}}\right )+2 \sqrt {c} \sqrt {c+x (d x+e)} \left (2 a c+a e x-4 b c x^2\right )\right )\right )}{8 c^{3/2} \sqrt {d} x^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^3,x]

[Out]

(Sqrt[(a + b*x^2)^2]*(4*b*c^(3/2)*e*x^2*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + x*(e + d*x)])] - Sqrt[d]*(2*Sq
rt[c]*(2*a*c + a*e*x - 4*b*c*x^2)*Sqrt[c + x*(e + d*x)] + (8*b*c^2 + 4*a*c*d - a*e^2)*x^2*ArcTanh[(2*c + e*x)/
(2*Sqrt[c]*Sqrt[c + x*(e + d*x)])])))/(8*c^(3/2)*Sqrt[d]*x^2*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 0.84, size = 160, normalized size = 0.56 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (\frac {\left (-4 a c d+a e^2-8 b c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2+e x}-\sqrt {d} x}{\sqrt {c}}\right )}{4 c^{3/2}}+\frac {\sqrt {c+d x^2+e x} \left (-2 a c-a e x+4 b c x^2\right )}{4 c x^2}-\frac {b e \log \left (-2 \sqrt {d} \sqrt {c+d x^2+e x}+2 d x+e\right )}{2 \sqrt {d}}\right )}{a+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x^3,x]

[Out]

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

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fricas [A]  time = 3.42, size = 749, normalized size = 2.60 \begin {gather*} \left [\frac {4 \, b c^{2} \sqrt {d} e x^{2} \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 ) - {\left (8 \, b c^{2} d + 4 \, a c d^{2} - a d e^{2}\right )} \sqrt {c} x^{2} \log \left (\frac {8 \, c e x + {\left (4 \, c d + e^{2}\right )} x^{2} + 4 \, \sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) + 4 \, {\left (4 \, b c^{2} d x^{2} - a c d e x - 2 \, a c^{2} d\right )} \sqrt {d x^{2} + e x + c}}{16 \, c^{2} d x^{2}}, -\frac {8 \, b c^{2} \sqrt {-d} e x^{2} \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 ) + {\left (8 \, b c^{2} d + 4 \, a c d^{2} - a d e^{2}\right )} \sqrt {c} x^{2} \log \left (\frac {8 \, c e x + {\left (4 \, c d + e^{2}\right )} x^{2} + 4 \, \sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) - 4 \, {\left (4 \, b c^{2} d x^{2} - a c d e x - 2 \, a c^{2} d\right )} \sqrt {d x^{2} + e x + c}}{16 \, c^{2} d x^{2}}, \frac {2 \, b c^{2} \sqrt {d} e x^{2} \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 ) + {\left (8 \, b c^{2} d + 4 \, a c d^{2} - a d e^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c e x + c^{2}\right )}}\right ) + 2 \, {\left (4 \, b c^{2} d x^{2} - a c d e x - 2 \, a c^{2} d\right )} \sqrt {d x^{2} + e x + c}}{8 \, c^{2} d x^{2}}, -\frac {4 \, b c^{2} \sqrt {-d} e x^{2} \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 ) - {\left (8 \, b c^{2} d + 4 \, a c d^{2} - a d e^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c e x + c^{2}\right )}}\right ) - 2 \, {\left (4 \, b c^{2} d x^{2} - a c d e x - 2 \, a c^{2} d\right )} \sqrt {d x^{2} + e x + c}}{8 \, c^{2} d x^{2}}\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^3,x, algorithm="fricas")

[Out]

[1/16*(4*b*c^2*sqrt(d)*e*x^2*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) - (8*b*c^2*d + 4*a*c*d^2 - a*d*e^2)*sqrt(c)*x^2*log((8*c*e*x + (4*c*d + e^2)*x^2 + 4*sqrt(d*x^2 + e*x + c)
*(e*x + 2*c)*sqrt(c) + 8*c^2)/x^2) + 4*(4*b*c^2*d*x^2 - a*c*d*e*x - 2*a*c^2*d)*sqrt(d*x^2 + e*x + c))/(c^2*d*x
^2), -1/16*(8*b*c^2*sqrt(-d)*e*x^2*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)) + (8*b*c^2*d + 4*a*c*d^2 - a*d*e^2)*sqrt(c)*x^2*log((8*c*e*x + (4*c*d + e^2)*x^2 + 4*sqrt(d*x^2 + e*x + c)
*(e*x + 2*c)*sqrt(c) + 8*c^2)/x^2) - 4*(4*b*c^2*d*x^2 - a*c*d*e*x - 2*a*c^2*d)*sqrt(d*x^2 + e*x + c))/(c^2*d*x
^2), 1/8*(2*b*c^2*sqrt(d)*e*x^2*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) + (8*b*c^2*d + 4*a*c*d^2 - a*d*e^2)*sqrt(-c)*x^2*arctan(1/2*sqrt(d*x^2 + e*x + c)*(e*x + 2*c)*sqrt(-c)/
(c*d*x^2 + c*e*x + c^2)) + 2*(4*b*c^2*d*x^2 - a*c*d*e*x - 2*a*c^2*d)*sqrt(d*x^2 + e*x + c))/(c^2*d*x^2), -1/8*
(4*b*c^2*sqrt(-d)*e*x^2*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)) - (8*b*
c^2*d + 4*a*c*d^2 - a*d*e^2)*sqrt(-c)*x^2*arctan(1/2*sqrt(d*x^2 + e*x + c)*(e*x + 2*c)*sqrt(-c)/(c*d*x^2 + c*e
*x + c^2)) - 2*(4*b*c^2*d*x^2 - a*c*d*e*x - 2*a*c^2*d)*sqrt(d*x^2 + e*x + c))/(c^2*d*x^2)]

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giac [A]  time = 0.89, size = 370, normalized size = 1.28 \begin {gather*} -\frac {b e \log \left ({\left | -2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} d - \sqrt {d} e \right |}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, \sqrt {d}} + \sqrt {d x^{2} + x e + c} b \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {{\left (8 \, b c^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, a c d \mathrm {sgn}\left (b x^{2} + a\right ) - a e^{2} \mathrm {sgn}\left (b x^{2} + a\right )\right )} \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + x e + c}}{\sqrt {-c}}\right )}{4 \, \sqrt {-c} c} + \frac {4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )}^{3} a c d \mathrm {sgn}\left (b x^{2} + a\right ) + 8 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )}^{2} a c \sqrt {d} e \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} a c^{2} d \mathrm {sgn}\left (b x^{2} + a\right ) + {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )}^{3} a e^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} a c e^{2} \mathrm {sgn}\left (b x^{2} + a\right )}{4 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )}^{2} - c\right )}^{2} c} \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^3,x, algorithm="giac")

[Out]

-1/2*b*e*log(abs(-2*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))*d - sqrt(d)*e))*sgn(b*x^2 + a)/sqrt(d) + sqrt(d*x^2 +
x*e + c)*b*sgn(b*x^2 + a) + 1/4*(8*b*c^2*sgn(b*x^2 + a) + 4*a*c*d*sgn(b*x^2 + a) - a*e^2*sgn(b*x^2 + a))*arcta
n(-(sqrt(d)*x - sqrt(d*x^2 + x*e + c))/sqrt(-c))/(sqrt(-c)*c) + 1/4*(4*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))^3*a
*c*d*sgn(b*x^2 + a) + 8*(sqrt(d)*x - sqrt(d*x^2 + x*e + c))^2*a*c*sqrt(d)*e*sgn(b*x^2 + a) + 4*(sqrt(d)*x - sq
rt(d*x^2 + x*e + c))*a*c^2*d*sgn(b*x^2 + a) + (sqrt(d)*x - sqrt(d*x^2 + x*e + c))^3*a*e^2*sgn(b*x^2 + a) + (sq
rt(d)*x - sqrt(d*x^2 + x*e + c))*a*c*e^2*sgn(b*x^2 + a))/(((sqrt(d)*x - sqrt(d*x^2 + x*e + c))^2 - c)^2*c)

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maple [A]  time = 0.01, size = 329, normalized size = 1.14 \begin {gather*} \frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-4 a \,c^{\frac {3}{2}} d^{\frac {5}{2}} x^{2} \ln \left (\frac {e x +2 c +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {c}}{x}\right )+a \sqrt {c}\, d^{\frac {3}{2}} e^{2} x^{2} \ln \left (\frac {e x +2 c +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {c}}{x}\right )-8 b \,c^{\frac {5}{2}} d^{\frac {3}{2}} x^{2} \ln \left (\frac {e x +2 c +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {c}}{x}\right )+4 b \,c^{2} d e \,x^{2} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-2 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {5}{2}} e \,x^{3}+4 \sqrt {d \,x^{2}+e x +c}\, a c \,d^{\frac {5}{2}} x^{2}-2 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {3}{2}} e^{2} x^{2}+8 \sqrt {d \,x^{2}+e x +c}\, b \,c^{2} d^{\frac {3}{2}} x^{2}+2 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,d^{\frac {3}{2}} e x -4 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a c \,d^{\frac {3}{2}}\right )}{8 \left (b \,x^{2}+a \right ) c^{2} d^{\frac {3}{2}} x^{2}} \end {gather*}

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^3,x)

[Out]

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

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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^3,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 \frac {\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {d\,x^2+e\,x+c}}{x^3} \,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^3,x)

[Out]

int((((a + b*x^2)^2)^(1/2)*(c + e*x + d*x^2)^(1/2))/x^3, 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((d*x**2+e*x+c)**(1/2)*((b*x**2+a)**2)**(1/2)/x**3,x)

[Out]

Timed out

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