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

Optimal. Leaf size=294 \[ \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (8 a d^2+4 b c d-b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{8 d^{3/2} \left (a+b x^2\right )}-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{c x \left (a+b x^2\right )}+\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2+e x} (e (4 a d+b c)+2 d x (2 a d+b c))}{4 c d \left (a+b x^2\right )}-\frac {a e \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{2 \sqrt {c} \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.77, antiderivative size = 294, 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, 814, 843, 621, 206, 724} \[ \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (8 a d^2+4 b c d-b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{8 d^{3/2} \left (a+b x^2\right )}-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{c x \left (a+b x^2\right )}+\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2+e x} (e (4 a d+b c)+2 d x (2 a d+b c))}{4 c d \left (a+b x^2\right )}-\frac {a e \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{2 \sqrt {c} \left (a+b x^2\right )} \]

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^2,x]

[Out]

(((b*c + 4*a*d)*e + 2*d*(b*c + 2*a*d)*x)*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*c*d*(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])/(c*x*(a + b*x^2)) + ((4*b*c*d + 8*a*d^2 -
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])])/(8*d^(3/2)*(a +
b*x^2)) - (a*e*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])])/(2*Sqr
t[c]*(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 814

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

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Mathematica [A]  time = 0.25, size = 171, normalized size = 0.58 \[ \frac {\sqrt {\left (a+b x^2\right )^2} \left (\sqrt {c} x \left (8 a d^2+4 b c d-b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+x (d x+e)}}\right )+2 \sqrt {d} \left (\sqrt {c} \sqrt {c+x (d x+e)} (b x (2 d x+e)-4 a d)-2 a d e x \tanh ^{-1}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+x (d x+e)}}\right )\right )\right )}{8 \sqrt {c} d^{3/2} x \left (a+b x^2\right )} \]

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^2,x]

[Out]

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

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fricas [A]  time = 2.44, size = 731, normalized size = 2.49 \[ \left [\frac {4 \, a \sqrt {c} d^{2} e x \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 ) - {\left (4 \, b c^{2} d + 8 \, a c d^{2} - b c e^{2}\right )} \sqrt {d} x \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 (2 \, b c d^{2} x^{2} + b c d e x - 4 \, a c d^{2}\right )} \sqrt {d x^{2} + e x + c}}{16 \, c d^{2} x}, \frac {2 \, a \sqrt {c} d^{2} e x \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 ) - {\left (4 \, b c^{2} d + 8 \, a c d^{2} - b c e^{2}\right )} \sqrt {-d} x \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 (2 \, b c d^{2} x^{2} + b c d e x - 4 \, a c d^{2}\right )} \sqrt {d x^{2} + e x + c}}{8 \, c d^{2} x}, \frac {8 \, a \sqrt {-c} d^{2} e x \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 ) - {\left (4 \, b c^{2} d + 8 \, a c d^{2} - b c e^{2}\right )} \sqrt {d} x \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 (2 \, b c d^{2} x^{2} + b c d e x - 4 \, a c d^{2}\right )} \sqrt {d x^{2} + e x + c}}{16 \, c d^{2} x}, \frac {4 \, a \sqrt {-c} d^{2} e x \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 ) - {\left (4 \, b c^{2} d + 8 \, a c d^{2} - b c e^{2}\right )} \sqrt {-d} x \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 (2 \, b c d^{2} x^{2} + b c d e x - 4 \, a c d^{2}\right )} \sqrt {d x^{2} + e x + c}}{8 \, c d^{2} x}\right ] \]

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^2,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(b*x^2+a)]Unable to divide, perhaps due to rounding error%%%{%%{[1,0]:[1,0,%%%{-1,[1]%%%}]%%},[4,0,0,0,0,0]%%
%}+%%%{%%{[-2,0]:[1,0,%%%{-1,[1]%%%}]%%},[2,0,0,0,1,0]%%%}+%%%{%%{[1,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,0,0,0,2,0]%
%%} / %%%{%%%{1,[1]%%%},[4,0,0,0,0,0]%%%}+%%%{%%%{-2,[1]%%%},[2,0,0,0,1,0]%%%}+%%%{%%%{1,[1]%%%},[0,0,0,0,2,0]
%%%} Error: Bad Argument Value

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maple [A]  time = 0.01, size = 288, normalized size = 0.98 \[ -\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-8 a c \,d^{3} x \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+4 a \sqrt {c}\, d^{\frac {5}{2}} e x \ln \left (\frac {e x +2 c +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {c}}{x}\right )-4 b \,c^{2} d^{2} x \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+b c d \,e^{2} x \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-8 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {7}{2}} x^{2}-4 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {5}{2}} x^{2}-8 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {5}{2}} e x -2 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {3}{2}} e x +8 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,d^{\frac {5}{2}}\right )}{8 \left (b \,x^{2}+a \right ) c \,d^{\frac {5}{2}} x} \]

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^2,x)

[Out]

-1/8*((b*x^2+a)^2)^(1/2)*(-8*(d*x^2+e*x+c)^(1/2)*d^(7/2)*x^2*a-4*(d*x^2+e*x+c)^(1/2)*d^(5/2)*x^2*b*c+4*c^(1/2)
*ln((e*x+2*c+2*(d*x^2+e*x+c)^(1/2)*c^(1/2))/x)*d^(5/2)*x*a*e+8*(d*x^2+e*x+c)^(3/2)*d^(5/2)*a-8*(d*x^2+e*x+c)^(
1/2)*d^(5/2)*x*a*e-2*(d*x^2+e*x+c)^(1/2)*d^(3/2)*x*b*c*e-8*ln(1/2*(2*d*x+e+2*(d*x^2+e*x+c)^(1/2)*d^(1/2))/d^(1
/2))*x*a*c*d^3-4*ln(1/2*(2*d*x+e+2*(d*x^2+e*x+c)^(1/2)*d^(1/2))/d^(1/2))*x*b*c^2*d^2+ln(1/2*(2*d*x+e+2*(d*x^2+
e*x+c)^(1/2)*d^(1/2))/d^(1/2))*x*b*c*d*e^2)/(b*x^2+a)/c/x/d^(5/2)

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

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

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^2,x)

[Out]

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

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

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**2,x)

[Out]

Timed out

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