3.1.8 \(\int \frac {A+B x}{(a+b x+c x^2)^{3/2} (d-f x^2)} \, dx\) [8]

Optimal. Leaf size=381 \[ -\frac {2 \left (a B \left (2 c^2 d-b^2 f+2 a c f\right )+A \left (b^3 f-b c (c d+3 a f)\right )+c \left (A b^2 f+b B (c d-a f)-2 A c (c d+a f)\right ) x\right )}{\left (b^2-4 a c\right ) \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}-\frac {\left (B \sqrt {d}-A \sqrt {f}\right ) \sqrt {f} \tanh ^{-1}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {d} \left (c d-b \sqrt {d} \sqrt {f}+a f\right )^{3/2}}+\frac {\left (B \sqrt {d}+A \sqrt {f}\right ) \sqrt {f} \tanh ^{-1}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {d} \left (c d+b \sqrt {d} \sqrt {f}+a f\right )^{3/2}} \]

[Out]

-1/2*arctanh(1/2*(b*d^(1/2)-2*a*f^(1/2)+x*(2*c*d^(1/2)-b*f^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c*d+a*f-b*d^(1/2)*f^(1
/2))^(1/2))*f^(1/2)*(B*d^(1/2)-A*f^(1/2))/d^(1/2)/(c*d+a*f-b*d^(1/2)*f^(1/2))^(3/2)+1/2*arctanh(1/2*(b*d^(1/2)
+2*a*f^(1/2)+x*(2*c*d^(1/2)+b*f^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c*d+a*f+b*d^(1/2)*f^(1/2))^(1/2))*f^(1/2)*(B*d^(1
/2)+A*f^(1/2))/d^(1/2)/(c*d+a*f+b*d^(1/2)*f^(1/2))^(3/2)-2*(a*B*(2*a*c*f-b^2*f+2*c^2*d)+A*(b^3*f-b*c*(3*a*f+c*
d))+c*(A*b^2*f+b*B*(-a*f+c*d)-2*A*c*(a*f+c*d))*x)/(-4*a*c+b^2)/(b^2*d*f-(a*f+c*d)^2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]
time = 0.49, antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1032, 1047, 738, 212} \begin {gather*} -\frac {2 \left (c x \left (-2 A c (a f+c d)+b B (c d-a f)+A b^2 f\right )-A b c (3 a f+c d)+a B \left (2 a c f+b^2 (-f)+2 c^2 d\right )+A b^3 f\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}-\frac {\sqrt {f} \left (B \sqrt {d}-A \sqrt {f}\right ) \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} \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {\sqrt {f} \left (A \sqrt {f}+B \sqrt {d}\right ) \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} \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A*b^3*f - A*b*c*(c*d + 3*a*f) + a*B*(2*c^2*d - b^2*f + 2*a*c*f) + c*(A*b^2*f + b*B*(c*d - a*f) - 2*A*c*(c
*d + a*f))*x))/((b^2 - 4*a*c)*(b^2*d*f - (c*d + a*f)^2)*Sqrt[a + b*x + c*x^2]) - ((B*Sqrt[d] - A*Sqrt[f])*Sqrt
[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]*Sqr
t[a + b*x + c*x^2])])/(2*Sqrt[d]*(c*d - b*Sqrt[d]*Sqrt[f] + a*f)^(3/2)) + ((B*Sqrt[d] + A*Sqrt[f])*Sqrt[f]*Arc
Tanh[(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]*(c*d + b*Sqrt[d]*Sqrt[f] + a*f)^(3/2))

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 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 1032

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[(a + b*x + c*x^2)^(p + 1)*((d + f*x^2)^(q + 1)/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)*(p + 1)))*((g*c)*((-b
)*(c*d + a*f)) + (g*b - a*h)*(2*c^2*d + b^2*f - c*(2*a*f)) + c*(g*(2*c^2*d + b^2*f - c*(2*a*f)) - h*(b*c*d + a
*b*f))*x), x] + Dist[1/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1)*(d + f
*x^2)^q*Simp[(b*h - 2*g*c)*((c*d - a*f)^2 - (b*d)*((-b)*f))*(p + 1) + (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*
(c*d - a*f)))*(a*f*(p + 1) - c*d*(p + 2)) - (2*f*((g*c)*((-b)*(c*d + a*f)) + (g*b - a*h)*(2*c^2*d + b^2*f - c*
(2*a*f)))*(p + q + 2) - (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(b*f*(p + 1)))*x - c*f*(b^2*(g*f
) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, g, h, q}
, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[b^2*d*f + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1
])

Rule 1047

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]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 1.15, size = 641, normalized size = 1.68 \begin {gather*} \frac {4 A \left (-b^3 f+b c (c d+3 a f)-b^2 c f x+2 c^2 (c d+a f) x\right )+4 B \left (-2 a^2 c f-b c^2 d x+a \left (-2 c^2 d+b^2 f+b c f x\right )\right )-\left (b^2-4 a c\right ) f \sqrt {a+x (b+c x)} \text {RootSum}\left [b^2 d-a^2 f-4 b \sqrt {c} d \text {$\#$1}+4 c d \text {$\#$1}^2+2 a f \text {$\#$1}^2-f \text {$\#$1}^4\&,\frac {b^2 B d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-A b c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+a B c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 a A b f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+a^2 B f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 b B \sqrt {c} d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 A c^{3/2} d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 a A \sqrt {c} f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-B c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+A b f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a B f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{b \sqrt {c} d-2 c d \text {$\#$1}-a f \text {$\#$1}+f \text {$\#$1}^3}\&\right ]}{2 \left (b^2-4 a c\right ) \left (-c^2 d^2-2 a c d f+f \left (b^2 d-a^2 f\right )\right ) \sqrt {a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*A*(-(b^3*f) + b*c*(c*d + 3*a*f) - b^2*c*f*x + 2*c^2*(c*d + a*f)*x) + 4*B*(-2*a^2*c*f - b*c^2*d*x + a*(-2*c^
2*d + b^2*f + b*c*f*x)) - (b^2 - 4*a*c)*f*Sqrt[a + x*(b + c*x)]*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[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*a*A*b*f*Log[
-(Sqrt[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*Sq
rt[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) + S
qrt[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*(b^2 - 4*a*c)*(-
(c^2*d^2) - 2*a*c*d*f + f*(b^2*d - a^2*f))*Sqrt[a + x*(b + c*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(933\) vs. \(2(311)=622\).
time = 0.14, size = 934, normalized size = 2.45

method	result	size
default	\(\frac {\left (A f -B \sqrt {d f}\right ) \left (\frac {f}{\left (-b \sqrt {d f}+f a +c d \right ) \sqrt {\left (x +\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {-b \sqrt {d f}+f a +c d}{f}}}-\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (2 c \left (x +\frac {\sqrt {d f}}{f}\right )+\frac {-2 c \sqrt {d f}+b f}{f}\right )}{\left (-b \sqrt {d f}+f a +c d \right ) \left (\frac {4 c \left (-b \sqrt {d f}+f a +c d \right )}{f}-\frac {\left (-2 c \sqrt {d f}+b f \right )^{2}}{f^{2}}\right ) \sqrt {\left (x +\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {-b \sqrt {d f}+f a +c d}{f}}}-\frac {f \ln \left (\frac {\frac {-2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x +\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {-b \sqrt {d f}+f a +c d}{f}}}{x +\frac {\sqrt {d f}}{f}}\right )}{\left (-b \sqrt {d f}+f a +c d \right ) \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}}\right )}{2 \sqrt {d f}\, f}+\frac {\left (-A f -B \sqrt {d f}\right ) \left (\frac {f}{\left (b \sqrt {d f}+f a +c d \right ) \sqrt {\left (x -\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {b \sqrt {d f}+f a +c d}{f}}}-\frac {\left (2 c \sqrt {d f}+b f \right ) \left (2 c \left (x -\frac {\sqrt {d f}}{f}\right )+\frac {2 c \sqrt {d f}+b f}{f}\right )}{\left (b \sqrt {d f}+f a +c d \right ) \left (\frac {4 c \left (b \sqrt {d f}+f a +c d \right )}{f}-\frac {\left (2 c \sqrt {d f}+b f \right )^{2}}{f^{2}}\right ) \sqrt {\left (x -\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {b \sqrt {d f}+f a +c d}{f}}}-\frac {f \ln \left (\frac {\frac {2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x -\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {b \sqrt {d f}+f a +c d}{f}}}{x -\frac {\sqrt {d f}}{f}}\right )}{\left (b \sqrt {d f}+f a +c d \right ) \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}}\right )}{2 \sqrt {d f}\, f}\)	\(934\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(c*x^2+b*x+a)^(3/2)/(-f*x^2+d),x,method=_RETURNVERBOSE)

[Out]

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

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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)^(3/2)/(-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?

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Fricas [F(-1)] Timed out
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)^(3/2)/(-f*x^2+d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
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)**(3/2)/(-f*x**2+d),x)

[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)^(3/2)/(-f*x^2+d),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueEvaluation time:
3.53Done

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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