3.25.78 \(\int \frac {A+B x}{(d+e x)^2 (a+b x+c x^2)^{3/2}} \, dx\) [2478]

Optimal. Leaf size=334 \[ \frac {2 \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}+\frac {e \left (b^2 e (B d-3 A e)-4 c \left (A c d^2+3 a B d e-2 a A e^2\right )+2 b \left (B c d^2+2 A c d e+a B e^2\right )\right ) \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac {e \left (3 A e (2 c d-b e)-B \left (4 c d^2-e (b d+2 a e)\right )\right ) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{2 \left (c d^2-b d e+a e^2\right )^{5/2}} \]

[Out]

1/2*e*(3*A*e*(-b*e+2*c*d)-B*(4*c*d^2-e*(2*a*e+b*d)))*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2
)^(1/2)/(c*x^2+b*x+a)^(1/2))/(a*e^2-b*d*e+c*d^2)^(5/2)+2*(a*B*(-b*e+2*c*d)-A*(2*a*c*e-b^2*e+b*c*d)+c*(A*b*e-2*
A*c*d-2*B*a*e+B*b*d)*x)/(-4*a*c+b^2)/(a*e^2-b*d*e+c*d^2)/(e*x+d)/(c*x^2+b*x+a)^(1/2)+e*(b^2*e*(-3*A*e+B*d)-4*c
*(-2*A*a*e^2+A*c*d^2+3*B*a*d*e)+2*b*(2*A*c*d*e+B*a*e^2+B*c*d^2))*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)/(a*e^2-b*d*e
+c*d^2)^2/(e*x+d)

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Rubi [A]
time = 0.27, antiderivative size = 332, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {836, 820, 738, 212} \begin {gather*} \frac {e \sqrt {a+b x+c x^2} \left (2 b \left (a B e^2+2 A c d e+B c d^2\right )-4 c \left (-2 a A e^2+3 a B d e+A c d^2\right )+b^2 e (B d-3 A e)\right )}{\left (b^2-4 a c\right ) (d+e x) \left (a e^2-b d e+c d^2\right )^2}+\frac {2 \left (-A \left (2 a c e+b^2 (-e)+b c d\right )+c x (-2 a B e+A b e-2 A c d+b B d)+a B (2 c d-b e)\right )}{\left (b^2-4 a c\right ) (d+e x) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {e \left (-B e (2 a e+b d)-3 A e (2 c d-b e)+4 B c d^2\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*B*(2*c*d - b*e) - A*(b*c*d - b^2*e + 2*a*c*e) + c*(b*B*d - 2*A*c*d + A*b*e - 2*a*B*e)*x))/((b^2 - 4*a*c)
*(c*d^2 - b*d*e + a*e^2)*(d + e*x)*Sqrt[a + b*x + c*x^2]) + (e*(b^2*e*(B*d - 3*A*e) - 4*c*(A*c*d^2 + 3*a*B*d*e
- 2*a*A*e^2) + 2*b*(B*c*d^2 + 2*A*c*d*e + a*B*e^2))*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*
e^2)^2*(d + e*x)) - (e*(4*B*c*d^2 - B*e*(b*d + 2*a*e) - 3*A*e*(2*c*d - b*e))*ArcTanh[(b*d - 2*a*e + (2*c*d - b
*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(2*(c*d^2 - b*d*e + a*e^2)^(5/2))

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 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Dist[
(b*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && EqQ[S
implify[m + 2*p + 3], 0]

Rule 836

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*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] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {A+B x}{(d+e x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx &=\frac {2 \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {\frac {1}{2} e \left (b^2 (B d-3 A e)-8 a c (B d-A e)+2 b (A c d+a B e)\right )-c e (b B d-2 A c d+A b e-2 a B e) x}{(d+e x)^2 \sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=\frac {2 \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}+\frac {e \left (b^2 e (B d-3 A e)-4 c \left (A c d^2+3 a B d e-2 a A e^2\right )+2 b \left (B c d^2+2 A c d e+a B e^2\right )\right ) \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 (d+e x)}-\frac {\left (e \left (4 B c d^2-B e (b d+2 a e)-3 A e (2 c d-b e)\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 \left (c d^2-b d e+a e^2\right )^2}\\ &=\frac {2 \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}+\frac {e \left (b^2 e (B d-3 A e)-4 c \left (A c d^2+3 a B d e-2 a A e^2\right )+2 b \left (B c d^2+2 A c d e+a B e^2\right )\right ) \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac {\left (e \left (4 B c d^2-B e (b d+2 a e)-3 A e (2 c d-b e)\right )\right ) \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{\left (c d^2-b d e+a e^2\right )^2}\\ &=\frac {2 \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}+\frac {e \left (b^2 e (B d-3 A e)-4 c \left (A c d^2+3 a B d e-2 a A e^2\right )+2 b \left (B c d^2+2 A c d e+a B e^2\right )\right ) \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 (d+e x)}-\frac {e \left (4 B c d^2-B e (b d+2 a e)-3 A e (2 c d-b e)\right ) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{2 \left (c d^2-b d e+a e^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 10.62, size = 326, normalized size = 0.98 \begin {gather*} \frac {e \left (b^2 e (B d-3 A e)-4 c \left (A c d^2+3 a B d e-2 a A e^2\right )+2 b \left (B c d^2+2 A c d e+a B e^2\right )\right ) \sqrt {a+x (b+c x)}}{\left (b^2-4 a c\right ) \left (c d^2+e (-b d+a e)\right )^2 (d+e x)}+\frac {2 \left (A b^2 e+b B c d x-2 A c (a e+c d x)+A b c (-d+e x)+a B (-b e+2 c (d-e x))\right )}{\left (b^2-4 a c\right ) \left (c d^2+e (-b d+a e)\right ) (d+e x) \sqrt {a+x (b+c x)}}+\frac {e \left (4 B c d^2-B e (b d+2 a e)+3 A e (-2 c d+b e)\right ) \tanh ^{-1}\left (\frac {-b d+2 a e-2 c d x+b e x}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )}{2 \left (c d^2+e (-b d+a e)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(b^2*e*(B*d - 3*A*e) - 4*c*(A*c*d^2 + 3*a*B*d*e - 2*a*A*e^2) + 2*b*(B*c*d^2 + 2*A*c*d*e + a*B*e^2))*Sqrt[a
+ x*(b + c*x)])/((b^2 - 4*a*c)*(c*d^2 + e*(-(b*d) + a*e))^2*(d + e*x)) + (2*(A*b^2*e + b*B*c*d*x - 2*A*c*(a*e
+ c*d*x) + A*b*c*(-d + e*x) + a*B*(-(b*e) + 2*c*(d - e*x))))/((b^2 - 4*a*c)*(c*d^2 + e*(-(b*d) + a*e))*(d + e*
x)*Sqrt[a + x*(b + c*x)]) + (e*(4*B*c*d^2 - B*e*(b*d + 2*a*e) + 3*A*e*(-2*c*d + b*e))*ArcTanh[(-(b*d) + 2*a*e
- 2*c*d*x + b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])])/(2*(c*d^2 + e*(-(b*d) + a*e))^(5
/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1063\) vs. \(2(320)=640\).
time = 0.08, size = 1064, normalized size = 3.19

method	result	size
default	\(\frac {B \left (\frac {e^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {\left (b e -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{e^{2}}+\frac {\left (A e -B d \right ) \left (-\frac {e^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (x +\frac {d}{e}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {3 \left (b e -2 c d \right ) e \left (\frac {e^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {\left (b e -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )}-\frac {4 c \,e^{2} \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{e^{3}}\)	\(1064\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(e*x+d)^2/(c*x^2+b*x+a)^(3/2),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(c*d^2-%e*b*d+%e^2*a>0)', see `
assume?` for

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2012 vs. \(2 (337) = 674\).
time = 38.94, size = 4067, normalized size = 12.18 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(sqrt(c*d^2 - b*d*e + a*e^2)*(((4*(2*B*a^2 - 3*A*a*b)*c^2 - (2*B*a*b^2 - 3*A*b^3)*c)*x^3 - (2*B*a*b^3 - 3
*A*b^4 - 4*(2*B*a^2*b - 3*A*a*b^2)*c)*x^2 - (2*B*a^2*b^2 - 3*A*a*b^3 - 4*(2*B*a^3 - 3*A*a^2*b)*c)*x)*e^4 - ((B
*b^3*c - 24*A*a*c^3 - 2*(2*B*a*b - 3*A*b^2)*c^2)*d*x^3 + (B*b^4 - 4*(2*B*a^2 + 3*A*a*b)*c^2 - (2*B*a*b^2 - 3*A
*b^3)*c)*d*x^2 + 3*(B*a*b^3 - A*b^4 - 8*A*a^2*c^2 - 2*(2*B*a^2*b - 3*A*a*b^2)*c)*d*x + (2*B*a^2*b^2 - 3*A*a*b^
3 - 4*(2*B*a^3 - 3*A*a^2*b)*c)*d)*e^3 + (4*(B*b^2*c^2 - 4*B*a*c^3)*d^2*x^3 + 3*(B*b^3*c + 8*A*a*c^3 - 2*(2*B*a
*b + A*b^2)*c^2)*d^2*x^2 - (B*b^4 + 8*(2*B*a^2 - 3*A*a*b)*c^2 - 2*(4*B*a*b^2 - 3*A*b^3)*c)*d^2*x - (B*a*b^3 -
24*A*a^2*c^2 - 2*(2*B*a^2*b - 3*A*a*b^2)*c)*d^2)*e^2 + 4*((B*b^2*c^2 - 4*B*a*c^3)*d^3*x^2 + (B*b^3*c - 4*B*a*b
*c^2)*d^3*x + (B*a*b^2*c - 4*B*a^2*c^2)*d^3)*e)*log(-(8*c^2*d^2*x^2 + 8*b*c*d^2*x + (b^2 + 4*a*c)*d^2 - 4*sqrt
(c*d^2 - b*d*e + a*e^2)*(2*c*d*x + b*d - (b*x + 2*a)*e)*sqrt(c*x^2 + b*x + a) + (8*a*b*x + (b^2 + 4*a*c)*x^2 +
8*a^2)*e^2 - 2*(4*b*c*d*x^2 + 4*a*b*d + (3*b^2 + 4*a*c)*d*x)*e)/(x^2*e^2 + 2*d*x*e + d^2)) + 4*(2*(2*B*a - A*
b)*c^3*d^5 + 2*(B*b*c^3 - 2*A*c^4)*d^5*x - (A*a^2*b^2 - 4*A*a^3*c - (8*A*a^2*c^2 + (2*B*a^2*b - 3*A*a*b^2)*c)*
x^2 - (2*B*a^2*b^2 - 3*A*a*b^3 - 2*(2*B*a^3 - 5*A*a^2*b)*c)*x)*e^5 - ((4*(3*B*a^2 + A*a*b)*c^2 + (B*a*b^2 - 3*
A*b^3)*c)*d*x^2 + (B*a*b^3 - 3*A*b^4 + 4*A*a^2*c^2 + 2*(B*a^2*b + 4*A*a*b^2)*c)*d*x - (3*B*a^2*b^2 - A*a*b^3 -
2*(4*B*a^3 - A*a^2*b)*c)*d)*e^4 - ((B*b^3*c - 4*A*a*c^3 - (16*B*a*b - 7*A*b^2)*c^2)*d^2*x^2 + (B*b^4 + 8*(B*a
^2 - 2*A*a*b)*c^2 - (8*B*a*b^2 - 5*A*b^3)*c)*d^2*x + (3*B*a*b^3 - 2*A*b^4 + 4*A*a^2*c^2 - (4*B*a^2*b - 3*A*a*b
^2)*c)*d^2)*e^3 - ((B*b^2*c^2 + 4*(3*B*a - 2*A*b)*c^3)*d^3*x^2 - (B*b^3*c - 8*A*a*c^3)*d^3*x + (4*(B*a^2 - 3*A
*a*b)*c^2 - (7*B*a*b^2 - 6*A*b^3)*c)*d^3)*e^2 + 2*((B*b*c^3 - 2*A*c^4)*d^4*x^2 - (B*b^2*c^2 + (2*B*a - 3*A*b)*
c^3)*d^4*x - (4*A*a*c^3 + (4*B*a*b - 3*A*b^2)*c^2)*d^4)*e)*sqrt(c*x^2 + b*x + a))/((b^2*c^4 - 4*a*c^5)*d^7*x^2
+ (b^3*c^3 - 4*a*b*c^4)*d^7*x + (a*b^2*c^3 - 4*a^2*c^4)*d^7 + ((a^3*b^2*c - 4*a^4*c^2)*x^3 + (a^3*b^3 - 4*a^4
*b*c)*x^2 + (a^4*b^2 - 4*a^5*c)*x)*e^7 - (3*(a^2*b^3*c - 4*a^3*b*c^2)*d*x^3 + (3*a^2*b^4 - 13*a^3*b^2*c + 4*a^
4*c^2)*d*x^2 + 2*(a^3*b^3 - 4*a^4*b*c)*d*x - (a^4*b^2 - 4*a^5*c)*d)*e^6 + 3*((a*b^4*c - 3*a^2*b^2*c^2 - 4*a^3*
c^3)*d^2*x^3 + (a*b^5 - 4*a^2*b^3*c)*d^2*x^2 + (a^3*b^2*c - 4*a^4*c^2)*d^2*x - (a^3*b^3 - 4*a^4*b*c)*d^2)*e^5
- ((b^5*c + 2*a*b^3*c^2 - 24*a^2*b*c^3)*d^3*x^3 + (b^6 - a*b^4*c - 15*a^2*b^2*c^2 + 12*a^3*c^3)*d^3*x^2 - (2*a
*b^5 - 11*a^2*b^3*c + 12*a^3*b*c^2)*d^3*x - 3*(a^2*b^4 - 3*a^3*b^2*c - 4*a^4*c^2)*d^3)*e^4 + (3*(b^4*c^2 - 3*a
*b^2*c^3 - 4*a^2*c^4)*d^4*x^3 + (2*b^5*c - 11*a*b^3*c^2 + 12*a^2*b*c^3)*d^4*x^2 - (b^6 - a*b^4*c - 15*a^2*b^2*
c^2 + 12*a^3*c^3)*d^4*x - (a*b^5 + 2*a^2*b^3*c - 24*a^3*b*c^2)*d^4)*e^3 - 3*((b^3*c^3 - 4*a*b*c^4)*d^5*x^3 - (
a*b^2*c^3 - 4*a^2*c^4)*d^5*x^2 - (b^5*c - 4*a*b^3*c^2)*d^5*x - (a*b^4*c - 3*a^2*b^2*c^2 - 4*a^3*c^3)*d^5)*e^2
+ ((b^2*c^4 - 4*a*c^5)*d^6*x^3 - 2*(b^3*c^3 - 4*a*b*c^4)*d^6*x^2 - (3*b^4*c^2 - 13*a*b^2*c^3 + 4*a^2*c^4)*d^6*
x - 3*(a*b^3*c^2 - 4*a^2*b*c^3)*d^6)*e), -1/2*(sqrt(-c*d^2 + b*d*e - a*e^2)*(((4*(2*B*a^2 - 3*A*a*b)*c^2 - (2*
B*a*b^2 - 3*A*b^3)*c)*x^3 - (2*B*a*b^3 - 3*A*b^4 - 4*(2*B*a^2*b - 3*A*a*b^2)*c)*x^2 - (2*B*a^2*b^2 - 3*A*a*b^3
- 4*(2*B*a^3 - 3*A*a^2*b)*c)*x)*e^4 - ((B*b^3*c - 24*A*a*c^3 - 2*(2*B*a*b - 3*A*b^2)*c^2)*d*x^3 + (B*b^4 - 4*
(2*B*a^2 + 3*A*a*b)*c^2 - (2*B*a*b^2 - 3*A*b^3)*c)*d*x^2 + 3*(B*a*b^3 - A*b^4 - 8*A*a^2*c^2 - 2*(2*B*a^2*b - 3
*A*a*b^2)*c)*d*x + (2*B*a^2*b^2 - 3*A*a*b^3 - 4*(2*B*a^3 - 3*A*a^2*b)*c)*d)*e^3 + (4*(B*b^2*c^2 - 4*B*a*c^3)*d
^2*x^3 + 3*(B*b^3*c + 8*A*a*c^3 - 2*(2*B*a*b + A*b^2)*c^2)*d^2*x^2 - (B*b^4 + 8*(2*B*a^2 - 3*A*a*b)*c^2 - 2*(4
*B*a*b^2 - 3*A*b^3)*c)*d^2*x - (B*a*b^3 - 24*A*a^2*c^2 - 2*(2*B*a^2*b - 3*A*a*b^2)*c)*d^2)*e^2 + 4*((B*b^2*c^2
- 4*B*a*c^3)*d^3*x^2 + (B*b^3*c - 4*B*a*b*c^2)*d^3*x + (B*a*b^2*c - 4*B*a^2*c^2)*d^3)*e)*arctan(-1/2*sqrt(-c*
d^2 + b*d*e - a*e^2)*(2*c*d*x + b*d - (b*x + 2*a)*e)*sqrt(c*x^2 + b*x + a)/(c^2*d^2*x^2 + b*c*d^2*x + a*c*d^2
+ (a*c*x^2 + a*b*x + a^2)*e^2 - (b*c*d*x^2 + b^2*d*x + a*b*d)*e)) - 2*(2*(2*B*a - A*b)*c^3*d^5 + 2*(B*b*c^3 -
2*A*c^4)*d^5*x - (A*a^2*b^2 - 4*A*a^3*c - (8*A*a^2*c^2 + (2*B*a^2*b - 3*A*a*b^2)*c)*x^2 - (2*B*a^2*b^2 - 3*A*a
*b^3 - 2*(2*B*a^3 - 5*A*a^2*b)*c)*x)*e^5 - ((4*(3*B*a^2 + A*a*b)*c^2 + (B*a*b^2 - 3*A*b^3)*c)*d*x^2 + (B*a*b^3
- 3*A*b^4 + 4*A*a^2*c^2 + 2*(B*a^2*b + 4*A*a*b^2)*c)*d*x - (3*B*a^2*b^2 - A*a*b^3 - 2*(4*B*a^3 - A*a^2*b)*c)*
d)*e^4 - ((B*b^3*c - 4*A*a*c^3 - (16*B*a*b - 7*A*b^2)*c^2)*d^2*x^2 + (B*b^4 + 8*(B*a^2 - 2*A*a*b)*c^2 - (8*B*a
*b^2 - 5*A*b^3)*c)*d^2*x + (3*B*a*b^3 - 2*A*b^4 + 4*A*a^2*c^2 - (4*B*a^2*b - 3*A*a*b^2)*c)*d^2)*e^3 - ((B*b^2*
c^2 + 4*(3*B*a - 2*A*b)*c^3)*d^3*x^2 - (B*b^3*c - 8*A*a*c^3)*d^3*x + (4*(B*a^2 - 3*A*a*b)*c^2 - (7*B*a*b^2 - 6
*A*b^3)*c)*d^3)*e^2 + 2*((B*b*c^3 - 2*A*c^4)*d^4*x^2 - (B*b^2*c^2 + (2*B*a - 3*A*b)*c^3)*d^4*x - (4*A*a*c^3 +
(4*B*a*b - 3*A*b^2)*c^2)*d^4)*e)*sqrt(c*x^2 + b...

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

[Out]

Timed out

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Giac [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)/(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

Timed out

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

[Out]

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

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