3.1.83 \(\int \frac {\sqrt {a+b x+c x^2}}{x^3 (d-f x^2)} \, dx\)

Optimal. Leaf size=353 \[ \frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{3/2} d}-\frac {\sqrt {a} f \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d^2}-\frac {\sqrt {f} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d} \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 d^2}+\frac {\sqrt {f} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d} \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 d^2}-\frac {(2 a+b x) \sqrt {a+b x+c x^2}}{4 a d x^2} \]

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Rubi [A]  time = 0.88, antiderivative size = 353, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6725, 720, 724, 206, 734, 843, 621, 1021, 1078, 1033} \begin {gather*} \frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{3/2} d}-\frac {\sqrt {a} f \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d^2}-\frac {\sqrt {f} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d} \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 d^2}+\frac {\sqrt {f} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d} \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 d^2}-\frac {(2 a+b x) \sqrt {a+b x+c x^2}}{4 a d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x + c*x^2]/(x^3*(d - f*x^2)),x]

[Out]

-((2*a + b*x)*Sqrt[a + b*x + c*x^2])/(4*a*d*x^2) + ((b^2 - 4*a*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x
+ c*x^2])])/(8*a^(3/2)*d) - (Sqrt[a]*f*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/d^2 - (Sqrt[f]*
Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]*ArcTanh[(b*Sqrt[d] - 2*a*Sqrt[f] + (2*c*Sqrt[d] - b*Sqrt[f])*x)/(2*Sqrt[c*
d - b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + b*x + c*x^2])])/(2*d^2) + (Sqrt[f]*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*A
rcTanh[(b*Sqrt[d] + 2*a*Sqrt[f] + (2*c*Sqrt[d] + b*Sqrt[f])*x)/(2*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a +
b*x + c*x^2])])/(2*d^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 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 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 734

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

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 1021

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

Rule 1033

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[-(a*c), 2]}, Dist[h/2 + (c*g)/(2*q), Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/2 - (c*g)
/(2*q), Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[e^2 - 4*d*f
, 0] && PosQ[-(a*c)]

Rule 1078

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Sym
bol] :> Dist[C/c, Int[1/Sqrt[d + e*x + f*x^2], x], x] + Dist[1/c, Int[(A*c - a*C + B*c*x)/((a + c*x^2)*Sqrt[d
+ e*x + f*x^2]), x], x] /; FreeQ[{a, c, d, e, f, A, B, C}, x] && NeQ[e^2 - 4*d*f, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 0.52, size = 316, normalized size = 0.90 \begin {gather*} \frac {x^2 \left (b^2 d-4 a (2 a f+c d)\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {a} \left (-2 a \sqrt {f} x^2 \sqrt {a f+b \sqrt {d} \sqrt {f}+c d} \tanh ^{-1}\left (\frac {2 a \sqrt {f}+b \sqrt {d}+b \sqrt {f} x+2 c \sqrt {d} x}{2 \sqrt {a+x (b+c x)} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )+2 a \sqrt {f} x^2 \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d} \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+b \left (\sqrt {d}-\sqrt {f} x\right )+2 c \sqrt {d} x}{2 \sqrt {a+x (b+c x)} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )+d (2 a+b x) \sqrt {a+x (b+c x)}\right )}{8 a^{3/2} d^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x + c*x^2]/(x^3*(d - f*x^2)),x]

[Out]

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

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IntegrateAlgebraic [C]  time = 0.86, size = 392, normalized size = 1.11 \begin {gather*} -\frac {f \text {RootSum}\left [\text {$\#$1}^4 (-f)+2 \text {$\#$1}^2 a f+4 \text {$\#$1}^2 c d-4 \text {$\#$1} b \sqrt {c} d-a^2 f+b^2 d\&,\frac {\text {$\#$1}^2 c d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+\text {$\#$1}^2 a f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a^2 (-f) \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b^2 d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-a c d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} b \sqrt {c} d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )}{\text {$\#$1}^3 f-\text {$\#$1} a f-2 \text {$\#$1} c d+b \sqrt {c} d}\&\right ]}{2 d^2}+\frac {\left (-8 a^2 f-4 a c d+b^2 d\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x+c x^2}-\sqrt {c} x}{\sqrt {a}}\right )}{4 a^{3/2} d^2}+\frac {(-2 a-b x) \sqrt {a+b x+c x^2}}{4 a d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a - b*x)*Sqrt[a + b*x + c*x^2])/(4*a*d*x^2) + ((b^2*d - 4*a*c*d - 8*a^2*f)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a
+ b*x + c*x^2])/Sqrt[a]])/(4*a^(3/2)*d^2) - (f*RootSum[b^2*d - a^2*f - 4*b*Sqrt[c]*d*#1 + 4*c*d*#1^2 + 2*a*f*
#1^2 - f*#1^4 & , (b^2*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a*c*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*
x + c*x^2] - #1] - a^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*b*Sqrt[c]*d*Log[-(Sqrt[c]*x) + Sqr
t[a + b*x + c*x^2] - #1]*#1 + c*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + a*f*Log[-(Sqrt[c]*x) +
Sqrt[a + b*x + c*x^2] - #1]*#1^2)/(b*Sqrt[c]*d - 2*c*d*#1 - a*f*#1 + f*#1^3) & ])/(2*d^2)

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fricas [B]  time = 133.83, size = 1485, normalized size = 4.21

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/x^3/(-f*x^2+d),x, algorithm="fricas")

[Out]

[1/16*(4*a^2*d^2*x^2*sqrt((d^4*sqrt(b^2*f^3/d^7) + c*d*f + a*f^2)/d^4)*log((2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b
^2*f^3/d^7)*sqrt((d^4*sqrt(b^2*f^3/d^7) + c*d*f + a*f^2)/d^4) + 2*b*c*f^2*x + b^2*f^2 + (b*d^3*f*x + 2*a*d^3*f
)*sqrt(b^2*f^3/d^7))/x) - 4*a^2*d^2*x^2*sqrt((d^4*sqrt(b^2*f^3/d^7) + c*d*f + a*f^2)/d^4)*log(-(2*sqrt(c*x^2 +
b*x + a)*d^5*sqrt(b^2*f^3/d^7)*sqrt((d^4*sqrt(b^2*f^3/d^7) + c*d*f + a*f^2)/d^4) - 2*b*c*f^2*x - b^2*f^2 - (b
*d^3*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) - 4*a^2*d^2*x^2*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4)
*log((2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b^2*f^3/d^7)*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4) + 2*b*c
*f^2*x + b^2*f^2 - (b*d^3*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) + 4*a^2*d^2*x^2*sqrt(-(d^4*sqrt(b^2*f^3/d^7)
- c*d*f - a*f^2)/d^4)*log(-(2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b^2*f^3/d^7)*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f
- a*f^2)/d^4) - 2*b*c*f^2*x - b^2*f^2 + (b*d^3*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) + (8*a^2*f - (b^2 - 4*a
*c)*d)*sqrt(a)*x^2*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^
2) - 4*(a*b*d*x + 2*a^2*d)*sqrt(c*x^2 + b*x + a))/(a^2*d^2*x^2), 1/8*(2*a^2*d^2*x^2*sqrt((d^4*sqrt(b^2*f^3/d^7
) + c*d*f + a*f^2)/d^4)*log((2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b^2*f^3/d^7)*sqrt((d^4*sqrt(b^2*f^3/d^7) + c*d*f
+ a*f^2)/d^4) + 2*b*c*f^2*x + b^2*f^2 + (b*d^3*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) - 2*a^2*d^2*x^2*sqrt((d
^4*sqrt(b^2*f^3/d^7) + c*d*f + a*f^2)/d^4)*log(-(2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b^2*f^3/d^7)*sqrt((d^4*sqrt(
b^2*f^3/d^7) + c*d*f + a*f^2)/d^4) - 2*b*c*f^2*x - b^2*f^2 - (b*d^3*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) - 2
*a^2*d^2*x^2*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4)*log((2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b^2*f^3/
d^7)*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4) + 2*b*c*f^2*x + b^2*f^2 - (b*d^3*f*x + 2*a*d^3*f)*sqrt
(b^2*f^3/d^7))/x) + 2*a^2*d^2*x^2*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4)*log(-(2*sqrt(c*x^2 + b*x
+ a)*d^5*sqrt(b^2*f^3/d^7)*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4) - 2*b*c*f^2*x - b^2*f^2 + (b*d^3
*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) + (8*a^2*f - (b^2 - 4*a*c)*d)*sqrt(-a)*x^2*arctan(1/2*sqrt(c*x^2 + b*x
+ a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) - 2*(a*b*d*x + 2*a^2*d)*sqrt(c*x^2 + b*x + a))/(a^2*d^2*x^
2)]

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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((c*x^2+b*x+a)^(1/2)/x^3/(-f*x^2+d),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 1.36sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.02, size = 1953, normalized size = 5.53

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {\sqrt {c x^{2} + b x + a}}{{\left (f x^{2} - d\right )} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/x^3/(-f*x^2+d),x, algorithm="maxima")

[Out]

-integrate(sqrt(c*x^2 + b*x + a)/((f*x^2 - d)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {c\,x^2+b\,x+a}}{x^3\,\left (d-f\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\sqrt {a + b x + c x^{2}}}{- d x^{3} + f x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(1/2)/x**3/(-f*x**2+d),x)

[Out]

-Integral(sqrt(a + b*x + c*x**2)/(-d*x**3 + f*x**5), x)

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