3.1.6 \(\int \frac {(A+B x) \sqrt {a+b x+c x^2}}{d-f x^2} \, dx\)
Optimal. Leaf size=331 \[ -\frac {\left (B \sqrt {d}-A \sqrt {f}\right ) \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d} \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 \sqrt {d} f^{3/2}}+\frac {\left (A \sqrt {f}+B \sqrt {d}\right ) \sqrt {a f+b \sqrt {d} \sqrt {f}+c d} \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 \sqrt {d} f^{3/2}}-\frac {(2 A c+b B) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f}-\frac {B \sqrt {a+b x+c x^2}}{f} \]
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Rubi [A] time = 0.60, antiderivative size = 331, normalized size of antiderivative = 1.00,
number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{1021, 1078, 621, 206, 1033, 724} \begin {gather*} -\frac {\left (B \sqrt {d}-A \sqrt {f}\right ) \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 \sqrt {d} f^{3/2}}+\frac {\left (A \sqrt {f}+B \sqrt {d}\right ) \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 \sqrt {d} f^{3/2}}-\frac {(2 A c+b B) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f}-\frac {B \sqrt {a+b x+c x^2}}{f} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((A + B*x)*Sqrt[a + b*x + c*x^2])/(d - f*x^2),x]
[Out]
-((B*Sqrt[a + b*x + c*x^2])/f) - ((b*B + 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*Sqr
t[c]*f) - ((B*Sqrt[d] - A*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*Sqrt[d]*f^(3/2)) +
((B*Sqrt[d] + A*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*Sqrt[d]*f^(3/2))
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 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]
Rubi steps
\begin {align*} \int \frac {(A+B x) \sqrt {a+b x+c x^2}}{d-f x^2} \, dx &=-\frac {B \sqrt {a+b x+c x^2}}{f}+\frac {\int \frac {\frac {1}{2} (b B d+2 a A f)+(B c d+A b f+a B f) x+\frac {1}{2} (b B+2 A c) f x^2}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{f}\\ &=-\frac {B \sqrt {a+b x+c x^2}}{f}-\frac {\int \frac {-\frac {1}{2} (b B+2 A c) d f-\frac {1}{2} f (b B d+2 a A f)-f (B c d+A b f+a B f) x}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{f^2}-\frac {(b B+2 A c) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 f}\\ &=-\frac {B \sqrt {a+b x+c x^2}}{f}-\frac {(b B+2 A c) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{f}+\frac {\left (\left (B \sqrt {d}-A \sqrt {f}\right ) \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 \sqrt {d} f}+\frac {\left (\left (B \sqrt {d}+A \sqrt {f}\right ) \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 \sqrt {d} f}\\ &=-\frac {B \sqrt {a+b x+c x^2}}{f}-\frac {(b B+2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f}-\frac {\left (\left (B \sqrt {d}-A \sqrt {f}\right ) \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 )}{\sqrt {d} f}-\frac {\left (\left (B \sqrt {d}+A \sqrt {f}\right ) \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 )}{\sqrt {d} f}\\ &=-\frac {B \sqrt {a+b x+c x^2}}{f}-\frac {(b B+2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f}-\frac {\left (B \sqrt {d}-A \sqrt {f}\right ) \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 \sqrt {d} f^{3/2}}+\frac {\left (B \sqrt {d}+A \sqrt {f}\right ) \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 \sqrt {d} f^{3/2}}\\ \end {align*}
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Mathematica [A] time = 0.80, size = 322, normalized size = 0.97 \begin {gather*} \frac {\left (A \sqrt {f}+B \sqrt {d}\right ) \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 )-\left (B \sqrt {d}-A \sqrt {f}\right ) \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 )-2 B \sqrt {d} \sqrt {f} \sqrt {a+x (b+c x)}}{2 \sqrt {d} f^{3/2}}-\frac {(2 A c+b B) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{2 \sqrt {c} f} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((A + B*x)*Sqrt[a + b*x + c*x^2])/(d - f*x^2),x]
[Out]
-1/2*((b*B + 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(Sqrt[c]*f) + (-2*B*Sqrt[d]*Sqrt[f
]*Sqrt[a + x*(b + c*x)] + (B*Sqrt[d] + A*Sqrt[f])*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*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)])] - (B*Sq
rt[d] - A*Sqrt[f])*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]*ArcTanh[(-2*a*Sqrt[f] + 2*c*Sqrt[d]*x + b*(Sqrt[d] - Sq
rt[f]*x))/(2*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + x*(b + c*x)])])/(2*Sqrt[d]*f^(3/2))
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IntegrateAlgebraic [C] time = 0.78, size = 523, normalized size = 1.58 \begin {gather*} -\frac {\text {RootSum}\left [\text {$\#$1}^4 (-f)+2 \text {$\#$1}^2 a f+4 \text {$\#$1}^2 c d-4 \text {$\#$1} b \sqrt {c} d-a^2 f+b^2 d\&,\frac {\text {$\#$1}^2 A b f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+\text {$\#$1}^2 B c d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+\text {$\#$1}^2 a B f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a^2 (-B) f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} A c^{3/2} d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+A b c d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} a A \sqrt {c} f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b^2 B d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-a B c d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} b B \sqrt {c} d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )}{\text {$\#$1}^3 f-\text {$\#$1} a f-2 \text {$\#$1} c d+b \sqrt {c} d}\&\right ]}{2 f}+\frac {(2 A c+b B) \log \left (-2 \sqrt {c} f \sqrt {a+b x+c x^2}+b f+2 c f x\right )}{2 \sqrt {c} f}-\frac {B \sqrt {a+b x+c x^2}}{f} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[((A + B*x)*Sqrt[a + b*x + c*x^2])/(d - f*x^2),x]
[Out]
-((B*Sqrt[a + b*x + c*x^2])/f) + ((b*B + 2*A*c)*Log[b*f + 2*c*f*x - 2*Sqrt[c]*f*Sqrt[a + b*x + c*x^2]])/(2*Sqr
t[c]*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*B*d*Log[-(Sqrt[
c]*x) + Sqrt[a + b*x + c*x^2] - #1] + A*b*c*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a*B*c*d*Log[-(S
qrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a^2*B*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*b*B*Sqrt[
c]*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 2*A*c^(3/2)*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2
] - #1]*#1 - 2*a*A*Sqrt[c]*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + B*c*d*Log[-(Sqrt[c]*x) + Sqrt
[a + b*x + c*x^2] - #1]*#1^2 + A*b*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + a*B*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*f)
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fricas [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((B*x+A)*(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x, algorithm="fricas")
[Out]
Timed out
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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((B*x+A)*(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type
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maple [B] time = 0.08, size = 3358, normalized size = 10.15 \begin {gather*} \text {output too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*x+A)*(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x)
[Out]
1/2/(d*f)^(1/2)*((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+a*f+c*
d))^(1/2)*A-1/2/f*((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+a*f+
c*d))^(1/2)*B-1/2/f*ln((1/2/f*(-2*c*(d*f)^(1/2)+b*f)+c*(x+(d*f)^(1/2)/f))/c^(1/2)+((x+(d*f)^(1/2)/f)^2*c+1/f*(
-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2))*c^(1/2)*A+1/2*(d*f)^(1/2)/f^2*ln(
(1/2/f*(-2*c*(d*f)^(1/2)+b*f)+c*(x+(d*f)^(1/2)/f))/c^(1/2)+((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(
x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2))*c^(1/2)*B+1/4/(d*f)^(1/2)*ln((1/2/f*(-2*c*(d*f)^(1/2)+b*
f)+c*(x+(d*f)^(1/2)/f))/c^(1/2)+((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d
*f)^(1/2)+a*f+c*d))^(1/2))/c^(1/2)*b*A-1/4/f*ln((1/2/f*(-2*c*(d*f)^(1/2)+b*f)+c*(x+(d*f)^(1/2)/f))/c^(1/2)+((x
+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2))/c^(1/2)*
b*B+1/2/f/(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*ln((2/f*(-b*(d*f)^(1/2)+a*f+c*d)+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+
(d*f)^(1/2)/f)+2*(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*
f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2))/(x+(d*f)^(1/2)/f))*b*A-1/2*(d*f)^(1/2)/f^2/(1/f*(-b*(d*f)^(1/
2)+a*f+c*d))^(1/2)*ln((2/f*(-b*(d*f)^(1/2)+a*f+c*d)+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+2*(1/f*(-b*(d
*f)^(1/2)+a*f+c*d))^(1/2)*((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1
/2)+a*f+c*d))^(1/2))/(x+(d*f)^(1/2)/f))*b*B-1/2/(d*f)^(1/2)/(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*ln((2/f*(-b*(
d*f)^(1/2)+a*f+c*d)+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+2*(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*((x+(d
*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2))/(x+(d*f)^(1
/2)/f))*a*A+1/2/f/(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*ln((2/f*(-b*(d*f)^(1/2)+a*f+c*d)+1/f*(-2*c*(d*f)^(1/2)+
b*f)*(x+(d*f)^(1/2)/f)+2*(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f
)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2))/(x+(d*f)^(1/2)/f))*a*B-1/2/(d*f)^(1/2)/f/(1/f*(-b*(d*
f)^(1/2)+a*f+c*d))^(1/2)*ln((2/f*(-b*(d*f)^(1/2)+a*f+c*d)+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+2*(1/f*
(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d
*f)^(1/2)+a*f+c*d))^(1/2))/(x+(d*f)^(1/2)/f))*c*d*A+1/2/f^2/(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*ln((2/f*(-b*(
d*f)^(1/2)+a*f+c*d)+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+2*(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*((x+(d
*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2))/(x+(d*f)^(1
/2)/f))*c*d*B-1/2/(d*f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+
a*f+c*d)/f)^(1/2)*A-1/2/f*((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+a*f+
c*d)/f)^(1/2)*B-1/2/f*ln((1/2*(2*c*(d*f)^(1/2)+b*f)/f+c*(x-(d*f)^(1/2)/f))/c^(1/2)+((x-(d*f)^(1/2)/f)^2*c+(2*c
*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2))*c^(1/2)*A-1/2*(d*f)^(1/2)/f^2*ln((1/2*
(2*c*(d*f)^(1/2)+b*f)/f+c*(x-(d*f)^(1/2)/f))/c^(1/2)+((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(
1/2)/f)+(b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2))*c^(1/2)*B-1/4/(d*f)^(1/2)*ln((1/2*(2*c*(d*f)^(1/2)+b*f)/f+c*(x-(d*f)
^(1/2)/f))/c^(1/2)+((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+a*f+c*d)/f)
^(1/2))/c^(1/2)*b*A-1/4/f*ln((1/2*(2*c*(d*f)^(1/2)+b*f)/f+c*(x-(d*f)^(1/2)/f))/c^(1/2)+((x-(d*f)^(1/2)/f)^2*c+
(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2))/c^(1/2)*b*B+1/2/f/((b*(d*f)^(1/2)+
a*f+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+a*f+c*d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+2*((b*(d*f)^(1/2)+
a*f+c*d)/f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+a*f+c*d)/f)^
(1/2))/(x-(d*f)^(1/2)/f))*b*A+1/2*(d*f)^(1/2)/f^2/((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+a*f+c
*d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+2*((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c+(2*
c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))*b*B+1/2/(d*f)^(1/2
)/((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+a*f+c*d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+
2*((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^
(1/2)+a*f+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))*a*A+1/2/f/((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+a
*f+c*d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+2*((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c
+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))*a*B+1/2/(d*f)^
(1/2)/f/((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+a*f+c*d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/
2)/f)+2*((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*
(d*f)^(1/2)+a*f+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))*c*d*A+1/2/f^2/((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*ln((2*(b*(d*
f)^(1/2)+a*f+c*d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+2*((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*((x-(d*f)^(1
/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))*c*d*
B
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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((B*x+A)*(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),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(((c*sqrt(4*d*f))/(2*f^2)>0)',
see `assume?` for more details)Is ((c*sqrt(4*d*f))/(2*f^2) +b/(2*f)) ^2 -(c*((b*sqrt(4*d*f))
/(2*f) +(c*d)/f+a)) /f^2 positive, negative or zero?
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{d-f\,x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*x)*(a + b*x + c*x^2)^(1/2))/(d - f*x^2),x)
[Out]
int(((A + B*x)*(a + b*x + c*x^2)^(1/2))/(d - f*x^2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {A \sqrt {a + b x + c x^{2}}}{- d + f x^{2}}\, dx - \int \frac {B x \sqrt {a + b x + c x^{2}}}{- d + f x^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(c*x**2+b*x+a)**(1/2)/(-f*x**2+d),x)
[Out]
-Integral(A*sqrt(a + b*x + c*x**2)/(-d + f*x**2), x) - Integral(B*x*sqrt(a + b*x + c*x**2)/(-d + f*x**2), x)
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