3.976 \(\int \frac {A+B x}{x^2 (a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=288 \[ \frac {(5 A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{7/2}}-\frac {2 \left (-A \left (32 a^2 c^2-32 a b^2 c+5 b^4\right )-c x \left (24 a^2 B c-28 a A b c-2 a b^2 B+5 A b^3\right )+2 a b B \left (b^2-8 a c\right )\right )}{3 a^2 x \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {\sqrt {a+b x+c x^2} \left (2 a b B \left (3 b^2-20 a c\right )-A \left (128 a^2 c^2-100 a b^2 c+15 b^4\right )\right )}{3 a^3 x \left (b^2-4 a c\right )^2}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

2/3*(A*b^2-a*b*B-2*a*A*c+(A*b-2*B*a)*c*x)/a/(-4*a*c+b^2)/x/(c*x^2+b*x+a)^(3/2)+1/2*(5*A*b-2*B*a)*arctanh(1/2*(
b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(7/2)-2/3*(2*a*b*B*(-8*a*c+b^2)-A*(32*a^2*c^2-32*a*b^2*c+5*b^4)-c*(-28
*A*a*b*c+5*A*b^3+24*B*a^2*c-2*B*a*b^2)*x)/a^2/(-4*a*c+b^2)^2/x/(c*x^2+b*x+a)^(1/2)+1/3*(2*a*b*B*(-20*a*c+3*b^2
)-A*(128*a^2*c^2-100*a*b^2*c+15*b^4))*(c*x^2+b*x+a)^(1/2)/a^3/(-4*a*c+b^2)^2/x

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Rubi [A]  time = 0.31, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {822, 806, 724, 206} \[ \frac {\sqrt {a+b x+c x^2} \left (2 a b B \left (3 b^2-20 a c\right )-A \left (128 a^2 c^2-100 a b^2 c+15 b^4\right )\right )}{3 a^3 x \left (b^2-4 a c\right )^2}-\frac {2 \left (-c x \left (24 a^2 B c-28 a A b c-2 a b^2 B+5 A b^3\right )-A \left (32 a^2 c^2-32 a b^2 c+5 b^4\right )+2 a b B \left (b^2-8 a c\right )\right )}{3 a^2 x \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {(5 A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{7/2}}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^2*(a + b*x + c*x^2)^(5/2)),x]

[Out]

(2*(A*b^2 - a*b*B - 2*a*A*c + (A*b - 2*a*B)*c*x))/(3*a*(b^2 - 4*a*c)*x*(a + b*x + c*x^2)^(3/2)) - (2*(2*a*b*B*
(b^2 - 8*a*c) - A*(5*b^4 - 32*a*b^2*c + 32*a^2*c^2) - c*(5*A*b^3 - 2*a*b^2*B - 28*a*A*b*c + 24*a^2*B*c)*x))/(3
*a^2*(b^2 - 4*a*c)^2*x*Sqrt[a + b*x + c*x^2]) + ((2*a*b*B*(3*b^2 - 20*a*c) - A*(15*b^4 - 100*a*b^2*c + 128*a^2
*c^2))*Sqrt[a + b*x + c*x^2])/(3*a^3*(b^2 - 4*a*c)^2*x) + ((5*A*b - 2*a*B)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt
[a + b*x + c*x^2])])/(2*a^(7/2))

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 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 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((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[Sim
plify[m + 2*p + 3], 0]

Rule 822

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}{x^2 \left (a+b x+c x^2\right )^{5/2}} \, dx &=\frac {2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{3 a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (-5 A b^2+2 a b B+16 a A c\right )-3 (A b-2 a B) c x}{x^2 \left (a+b x+c x^2\right )^{3/2}} \, dx}{3 a \left (b^2-4 a c\right )}\\ &=\frac {2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{3 a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (2 a b B \left (b^2-8 a c\right )-A \left (5 b^4-32 a b^2 c+32 a^2 c^2\right )-c \left (5 A b^3-2 a b^2 B-28 a A b c+24 a^2 B c\right ) x\right )}{3 a^2 \left (b^2-4 a c\right )^2 x \sqrt {a+b x+c x^2}}+\frac {4 \int \frac {\frac {1}{4} \left (-2 a b B \left (3 b^2-20 a c\right )+4 A \left (\frac {15 b^4}{4}-25 a b^2 c+32 a^2 c^2\right )\right )-\frac {1}{2} c \left (2 a B \left (b^2-12 a c\right )-A \left (5 b^3-28 a b c\right )\right ) x}{x^2 \sqrt {a+b x+c x^2}} \, dx}{3 a^2 \left (b^2-4 a c\right )^2}\\ &=\frac {2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{3 a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (2 a b B \left (b^2-8 a c\right )-A \left (5 b^4-32 a b^2 c+32 a^2 c^2\right )-c \left (5 A b^3-2 a b^2 B-28 a A b c+24 a^2 B c\right ) x\right )}{3 a^2 \left (b^2-4 a c\right )^2 x \sqrt {a+b x+c x^2}}+\frac {\left (2 a b B \left (3 b^2-20 a c\right )-A \left (15 b^4-100 a b^2 c+128 a^2 c^2\right )\right ) \sqrt {a+b x+c x^2}}{3 a^3 \left (b^2-4 a c\right )^2 x}-\frac {(5 A b-2 a B) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{2 a^3}\\ &=\frac {2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{3 a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (2 a b B \left (b^2-8 a c\right )-A \left (5 b^4-32 a b^2 c+32 a^2 c^2\right )-c \left (5 A b^3-2 a b^2 B-28 a A b c+24 a^2 B c\right ) x\right )}{3 a^2 \left (b^2-4 a c\right )^2 x \sqrt {a+b x+c x^2}}+\frac {\left (2 a b B \left (3 b^2-20 a c\right )-A \left (15 b^4-100 a b^2 c+128 a^2 c^2\right )\right ) \sqrt {a+b x+c x^2}}{3 a^3 \left (b^2-4 a c\right )^2 x}+\frac {(5 A b-2 a B) \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 )}{a^3}\\ &=\frac {2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{3 a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (2 a b B \left (b^2-8 a c\right )-A \left (5 b^4-32 a b^2 c+32 a^2 c^2\right )-c \left (5 A b^3-2 a b^2 B-28 a A b c+24 a^2 B c\right ) x\right )}{3 a^2 \left (b^2-4 a c\right )^2 x \sqrt {a+b x+c x^2}}+\frac {\left (2 a b B \left (3 b^2-20 a c\right )-A \left (15 b^4-100 a b^2 c+128 a^2 c^2\right )\right ) \sqrt {a+b x+c x^2}}{3 a^3 \left (b^2-4 a c\right )^2 x}+\frac {(5 A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 285, normalized size = 0.99 \[ \frac {2 \left (\frac {3 \left (b^2-4 a c\right ) (5 A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{4 a^{5/2}}-\frac {\sqrt {a+x (b+c x)} \left (A \left (128 a^2 c^2-100 a b^2 c+15 b^4\right )+2 a b B \left (20 a c-3 b^2\right )\right )}{2 a^2 x \left (b^2-4 a c\right )}+\frac {A \left (-32 a^2 c^2+32 a b^2 c+28 a b c^2 x-5 b^4-5 b^3 c x\right )+2 a B \left (-8 a b c-12 a c^2 x+b^3+b^2 c x\right )}{a x \left (4 a c-b^2\right ) \sqrt {a+x (b+c x)}}+\frac {A \left (-2 a c+b^2+b c x\right )-a B (b+2 c x)}{x (a+x (b+c x))^{3/2}}\right )}{3 a \left (b^2-4 a c\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^2*(a + b*x + c*x^2)^(5/2)),x]

[Out]

(2*(-1/2*((2*a*b*B*(-3*b^2 + 20*a*c) + A*(15*b^4 - 100*a*b^2*c + 128*a^2*c^2))*Sqrt[a + x*(b + c*x)])/(a^2*(b^
2 - 4*a*c)*x) + (-(a*B*(b + 2*c*x)) + A*(b^2 - 2*a*c + b*c*x))/(x*(a + x*(b + c*x))^(3/2)) + (2*a*B*(b^3 - 8*a
*b*c + b^2*c*x - 12*a*c^2*x) + A*(-5*b^4 + 32*a*b^2*c - 32*a^2*c^2 - 5*b^3*c*x + 28*a*b*c^2*x))/(a*(-b^2 + 4*a
*c)*x*Sqrt[a + x*(b + c*x)]) + (3*(5*A*b - 2*a*B)*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b +
c*x)])])/(4*a^(5/2))))/(3*a*(b^2 - 4*a*c))

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fricas [B]  time = 8.42, size = 1655, normalized size = 5.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^2/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*((16*(2*B*a^3 - 5*A*a^2*b)*c^4 - 8*(2*B*a^2*b^2 - 5*A*a*b^3)*c^3 + (2*B*a*b^4 - 5*A*b^5)*c^2)*x^5 +
2*(16*(2*B*a^3*b - 5*A*a^2*b^2)*c^3 - 8*(2*B*a^2*b^3 - 5*A*a*b^4)*c^2 + (2*B*a*b^5 - 5*A*b^6)*c)*x^4 + (2*B*a*
b^6 - 5*A*b^7 + 32*(2*B*a^4 - 5*A*a^3*b)*c^3 - 6*(2*B*a^2*b^4 - 5*A*a*b^5)*c)*x^3 + 2*(2*B*a^2*b^5 - 5*A*a*b^6
+ 16*(2*B*a^4*b - 5*A*a^3*b^2)*c^2 - 8*(2*B*a^3*b^3 - 5*A*a^2*b^4)*c)*x^2 + (2*B*a^3*b^4 - 5*A*a^2*b^5 + 16*(
2*B*a^5 - 5*A*a^4*b)*c^2 - 8*(2*B*a^4*b^2 - 5*A*a^3*b^3)*c)*x)*sqrt(a)*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 + 4*s
qrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) + 4*(3*A*a^3*b^4 - 24*A*a^4*b^2*c + 48*A*a^5*c^2 + (128
*A*a^3*c^4 + 20*(2*B*a^3*b - 5*A*a^2*b^2)*c^3 - 3*(2*B*a^2*b^3 - 5*A*a*b^4)*c^2)*x^4 - 6*(4*(2*B*a^4 - 13*A*a^
3*b)*c^3 - 7*(2*B*a^3*b^2 - 5*A*a^2*b^3)*c^2 + (2*B*a^2*b^4 - 5*A*a*b^5)*c)*x^3 - 3*(2*B*a^2*b^5 - 5*A*a*b^6 -
16*A*a^3*b^2*c^2 - 64*A*a^4*c^3 - 6*(2*B*a^3*b^3 - 5*A*a^2*b^4)*c)*x^2 - 4*(2*B*a^3*b^4 - 5*A*a^2*b^5 + 16*(B
*a^5 - 4*A*a^4*b)*c^2 - (14*B*a^4*b^2 - 37*A*a^3*b^3)*c)*x)*sqrt(c*x^2 + b*x + a))/((a^4*b^4*c^2 - 8*a^5*b^2*c
^3 + 16*a^6*c^4)*x^5 + 2*(a^4*b^5*c - 8*a^5*b^3*c^2 + 16*a^6*b*c^3)*x^4 + (a^4*b^6 - 6*a^5*b^4*c + 32*a^7*c^3)
*x^3 + 2*(a^5*b^5 - 8*a^6*b^3*c + 16*a^7*b*c^2)*x^2 + (a^6*b^4 - 8*a^7*b^2*c + 16*a^8*c^2)*x), 1/6*(3*((16*(2*
B*a^3 - 5*A*a^2*b)*c^4 - 8*(2*B*a^2*b^2 - 5*A*a*b^3)*c^3 + (2*B*a*b^4 - 5*A*b^5)*c^2)*x^5 + 2*(16*(2*B*a^3*b -
5*A*a^2*b^2)*c^3 - 8*(2*B*a^2*b^3 - 5*A*a*b^4)*c^2 + (2*B*a*b^5 - 5*A*b^6)*c)*x^4 + (2*B*a*b^6 - 5*A*b^7 + 32
*(2*B*a^4 - 5*A*a^3*b)*c^3 - 6*(2*B*a^2*b^4 - 5*A*a*b^5)*c)*x^3 + 2*(2*B*a^2*b^5 - 5*A*a*b^6 + 16*(2*B*a^4*b -
5*A*a^3*b^2)*c^2 - 8*(2*B*a^3*b^3 - 5*A*a^2*b^4)*c)*x^2 + (2*B*a^3*b^4 - 5*A*a^2*b^5 + 16*(2*B*a^5 - 5*A*a^4*
b)*c^2 - 8*(2*B*a^4*b^2 - 5*A*a^3*b^3)*c)*x)*sqrt(-a)*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*(3*A*a^3*b^4 - 24*A*a^4*b^2*c + 48*A*a^5*c^2 + (128*A*a^3*c^4 + 20*(2*B*a^3*b - 5*A
*a^2*b^2)*c^3 - 3*(2*B*a^2*b^3 - 5*A*a*b^4)*c^2)*x^4 - 6*(4*(2*B*a^4 - 13*A*a^3*b)*c^3 - 7*(2*B*a^3*b^2 - 5*A*
a^2*b^3)*c^2 + (2*B*a^2*b^4 - 5*A*a*b^5)*c)*x^3 - 3*(2*B*a^2*b^5 - 5*A*a*b^6 - 16*A*a^3*b^2*c^2 - 64*A*a^4*c^3
- 6*(2*B*a^3*b^3 - 5*A*a^2*b^4)*c)*x^2 - 4*(2*B*a^3*b^4 - 5*A*a^2*b^5 + 16*(B*a^5 - 4*A*a^4*b)*c^2 - (14*B*a^
4*b^2 - 37*A*a^3*b^3)*c)*x)*sqrt(c*x^2 + b*x + a))/((a^4*b^4*c^2 - 8*a^5*b^2*c^3 + 16*a^6*c^4)*x^5 + 2*(a^4*b^
5*c - 8*a^5*b^3*c^2 + 16*a^6*b*c^3)*x^4 + (a^4*b^6 - 6*a^5*b^4*c + 32*a^7*c^3)*x^3 + 2*(a^5*b^5 - 8*a^6*b^3*c
+ 16*a^7*b*c^2)*x^2 + (a^6*b^4 - 8*a^7*b^2*c + 16*a^8*c^2)*x)]

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giac [A]  time = 0.27, size = 483, normalized size = 1.68 \[ \frac {2 \, {\left ({\left ({\left (\frac {{\left (3 \, B a^{9} b^{3} c^{2} - 6 \, A a^{8} b^{4} c^{2} - 20 \, B a^{10} b c^{3} + 38 \, A a^{9} b^{2} c^{3} - 40 \, A a^{10} c^{4}\right )} x}{a^{11} b^{4} - 8 \, a^{12} b^{2} c + 16 \, a^{13} c^{2}} + \frac {3 \, {\left (2 \, B a^{9} b^{4} c - 4 \, A a^{8} b^{5} c - 14 \, B a^{10} b^{2} c^{2} + 27 \, A a^{9} b^{3} c^{2} + 8 \, B a^{11} c^{3} - 36 \, A a^{10} b c^{3}\right )}}{a^{11} b^{4} - 8 \, a^{12} b^{2} c + 16 \, a^{13} c^{2}}\right )} x + \frac {3 \, {\left (B a^{9} b^{5} - 2 \, A a^{8} b^{6} - 6 \, B a^{10} b^{3} c + 12 \, A a^{9} b^{4} c - 8 \, A a^{10} b^{2} c^{2} - 16 \, A a^{11} c^{3}\right )}}{a^{11} b^{4} - 8 \, a^{12} b^{2} c + 16 \, a^{13} c^{2}}\right )} x + \frac {4 \, B a^{10} b^{4} - 7 \, A a^{9} b^{5} - 28 \, B a^{11} b^{2} c + 50 \, A a^{10} b^{3} c + 32 \, B a^{12} c^{2} - 80 \, A a^{11} b c^{2}}{a^{11} b^{4} - 8 \, a^{12} b^{2} c + 16 \, a^{13} c^{2}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} + \frac {{\left (2 \, B a - 5 \, A b\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A b + 2 \, A a \sqrt {c}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^2/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

2/3*((((3*B*a^9*b^3*c^2 - 6*A*a^8*b^4*c^2 - 20*B*a^10*b*c^3 + 38*A*a^9*b^2*c^3 - 40*A*a^10*c^4)*x/(a^11*b^4 -
8*a^12*b^2*c + 16*a^13*c^2) + 3*(2*B*a^9*b^4*c - 4*A*a^8*b^5*c - 14*B*a^10*b^2*c^2 + 27*A*a^9*b^3*c^2 + 8*B*a^
11*c^3 - 36*A*a^10*b*c^3)/(a^11*b^4 - 8*a^12*b^2*c + 16*a^13*c^2))*x + 3*(B*a^9*b^5 - 2*A*a^8*b^6 - 6*B*a^10*b
^3*c + 12*A*a^9*b^4*c - 8*A*a^10*b^2*c^2 - 16*A*a^11*c^3)/(a^11*b^4 - 8*a^12*b^2*c + 16*a^13*c^2))*x + (4*B*a^
10*b^4 - 7*A*a^9*b^5 - 28*B*a^11*b^2*c + 50*A*a^10*b^3*c + 32*B*a^12*c^2 - 80*A*a^11*b*c^2)/(a^11*b^4 - 8*a^12
*b^2*c + 16*a^13*c^2))/(c*x^2 + b*x + a)^(3/2) + (2*B*a - 5*A*b)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/s
qrt(-a))/(sqrt(-a)*a^3) + ((sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*b + 2*A*a*sqrt(c))/(((sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^2 - a)*a^3)

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maple [B]  time = 0.07, size = 709, normalized size = 2.46 \[ -\frac {128 A \,c^{3} x}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, a}+\frac {40 A \,b^{2} c^{2} x}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, a^{2}}-\frac {16 B b \,c^{2} x}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, a}-\frac {64 A b \,c^{2}}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, a}-\frac {16 A \,c^{2} x}{3 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a}+\frac {20 A \,b^{3} c}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, a^{2}}+\frac {5 A \,b^{2} c x}{3 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2}}-\frac {8 B \,b^{2} c}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, a}-\frac {2 B b c x}{3 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a}-\frac {8 A b c}{3 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a}+\frac {5 A \,b^{3}}{6 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2}}+\frac {5 A \,b^{2} c x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{3}}-\frac {B \,b^{2}}{3 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a}-\frac {2 B b c x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2}}+\frac {5 A \,b^{3}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{3}}-\frac {B \,b^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2}}-\frac {5 A b}{6 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2}}+\frac {B}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a}+\frac {5 A b \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {7}{2}}}-\frac {B \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{a^{\frac {5}{2}}}-\frac {A}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a x}-\frac {5 A b}{2 \sqrt {c \,x^{2}+b x +a}\, a^{3}}+\frac {B}{\sqrt {c \,x^{2}+b x +a}\, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^2/(c*x^2+b*x+a)^(5/2),x)

[Out]

-A/a/x/(c*x^2+b*x+a)^(3/2)-5/6*A/a^2*b/(c*x^2+b*x+a)^(3/2)+5/3*A/a^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*c*x+5
/6*A/a^2*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+40/3*A/a^2*b^2*c^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x+20/3*A/a^2
*b^3*c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)-5/2*A/a^3*b/(c*x^2+b*x+a)^(1/2)+5*A/a^3*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)
^(1/2)*c*x+5/2*A/a^3*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+5/2*A/a^(7/2)*b*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(
1/2))/x)-16/3*A*c^2/a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x-8/3*A*c/a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*b-128/3*A*c^
3/a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-64/3*A*c^2/a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*b+1/3*B/a/(c*x^2+b*x+a)
^(3/2)-2/3*B/a*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*c*x-1/3*B/a*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)-16/3*B/a*b*c^
2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-8/3*B/a*b^2*c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+B/a^2/(c*x^2+b*x+a)^(1/2
)-2*B/a^2*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*c*x-B/a^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-B/a^(5/2)*ln((b*x+2*
a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)

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maxima [F(-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((B*x+A)/x^2/(c*x^2+b*x+a)^(5/2),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(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x}{x^2\,{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^2*(a + b*x + c*x^2)^(5/2)),x)

[Out]

int((A + B*x)/(x^2*(a + b*x + c*x^2)^(5/2)), 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((B*x+A)/x**2/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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