3.942 \(\int \frac{(A+B x) (a+b x+c x^2)^{5/2}}{x^3} \, dx\)

Optimal. Leaf size=273 \[ -\frac{5 \left (-48 a^2 B c^2-96 a A b c^2-24 a b^2 B c-8 A b^3 c+b^4 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{3/2}}+\frac{5 \sqrt{a+b x+c x^2} \left (2 c x \left (12 a B c+16 A b c+b^2 B\right )+32 a A c^2+44 a b B c+40 A b^2 c+b^3 B\right )}{64 c}-\frac{5}{8} \sqrt{a} \left (4 a A c+4 a b B+3 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )-\frac{(2 A-B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^2}-\frac{5 \left (a+b x+c x^2\right )^{3/2} (6 (a B+A b)-x (4 A c+b B))}{24 x} \]

[Out]

(5*(b^3*B + 40*A*b^2*c + 44*a*b*B*c + 32*a*A*c^2 + 2*c*(b^2*B + 16*A*b*c + 12*a*B*c)*x)*Sqrt[a + b*x + c*x^2])
/(64*c) - (5*(6*(A*b + a*B) - (b*B + 4*A*c)*x)*(a + b*x + c*x^2)^(3/2))/(24*x) - ((2*A - B*x)*(a + b*x + c*x^2
)^(5/2))/(4*x^2) - (5*Sqrt[a]*(3*A*b^2 + 4*a*b*B + 4*a*A*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^
2])])/8 - (5*(b^4*B - 8*A*b^3*c - 24*a*b^2*B*c - 96*a*A*b*c^2 - 48*a^2*B*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*S
qrt[a + b*x + c*x^2])])/(128*c^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.258943, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {812, 814, 843, 621, 206, 724} \[ -\frac{5 \left (-48 a^2 B c^2-96 a A b c^2-24 a b^2 B c-8 A b^3 c+b^4 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{3/2}}+\frac{5 \sqrt{a+b x+c x^2} \left (2 c x \left (12 a B c+16 A b c+b^2 B\right )+32 a A c^2+44 a b B c+40 A b^2 c+b^3 B\right )}{64 c}-\frac{5}{8} \sqrt{a} \left (4 a A c+4 a b B+3 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )-\frac{(2 A-B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^2}-\frac{5 \left (a+b x+c x^2\right )^{3/2} (6 (a B+A b)-x (4 A c+b B))}{24 x} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^3,x]

[Out]

(5*(b^3*B + 40*A*b^2*c + 44*a*b*B*c + 32*a*A*c^2 + 2*c*(b^2*B + 16*A*b*c + 12*a*B*c)*x)*Sqrt[a + b*x + c*x^2])
/(64*c) - (5*(6*(A*b + a*B) - (b*B + 4*A*c)*x)*(a + b*x + c*x^2)^(3/2))/(24*x) - ((2*A - B*x)*(a + b*x + c*x^2
)^(5/2))/(4*x^2) - (5*Sqrt[a]*(3*A*b^2 + 4*a*b*B + 4*a*A*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^
2])])/8 - (5*(b^4*B - 8*A*b^3*c - 24*a*b^2*B*c - 96*a*A*b*c^2 - 48*a^2*B*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*S
qrt[a + b*x + c*x^2])])/(128*c^(3/2))

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*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] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 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 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]

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^3} \, dx &=-\frac{(2 A-B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^2}-\frac{5}{16} \int \frac{(-4 (A b+a B)-2 (b B+4 A c) x) \left (a+b x+c x^2\right )^{3/2}}{x^2} \, dx\\ &=-\frac{5 (6 (A b+a B)-(b B+4 A c) x) \left (a+b x+c x^2\right )^{3/2}}{24 x}-\frac{(2 A-B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^2}+\frac{5}{32} \int \frac{\left (4 \left (3 A b^2+4 a b B+4 a A c\right )+2 \left (b^2 B+16 A b c+12 a B c\right ) x\right ) \sqrt{a+b x+c x^2}}{x} \, dx\\ &=\frac{5 \left (b^3 B+40 A b^2 c+44 a b B c+32 a A c^2+2 c \left (b^2 B+16 A b c+12 a B c\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c}-\frac{5 (6 (A b+a B)-(b B+4 A c) x) \left (a+b x+c x^2\right )^{3/2}}{24 x}-\frac{(2 A-B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^2}-\frac{5 \int \frac{-16 a c \left (3 A b^2+4 a b B+4 a A c\right )+\left (b^4 B-8 A b^3 c-24 a b^2 B c-96 a A b c^2-48 a^2 B c^2\right ) x}{x \sqrt{a+b x+c x^2}} \, dx}{128 c}\\ &=\frac{5 \left (b^3 B+40 A b^2 c+44 a b B c+32 a A c^2+2 c \left (b^2 B+16 A b c+12 a B c\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c}-\frac{5 (6 (A b+a B)-(b B+4 A c) x) \left (a+b x+c x^2\right )^{3/2}}{24 x}-\frac{(2 A-B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^2}+\frac{1}{8} \left (5 a \left (3 A b^2+4 a b B+4 a A c\right )\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx-\frac{\left (5 \left (b^4 B-8 A b^3 c-24 a b^2 B c-96 a A b c^2-48 a^2 B c^2\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{128 c}\\ &=\frac{5 \left (b^3 B+40 A b^2 c+44 a b B c+32 a A c^2+2 c \left (b^2 B+16 A b c+12 a B c\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c}-\frac{5 (6 (A b+a B)-(b B+4 A c) x) \left (a+b x+c x^2\right )^{3/2}}{24 x}-\frac{(2 A-B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^2}-\frac{1}{4} \left (5 a \left (3 A b^2+4 a b B+4 a A c\right )\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 )-\frac{\left (5 \left (b^4 B-8 A b^3 c-24 a b^2 B c-96 a A b c^2-48 a^2 B c^2\right )\right ) \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 )}{64 c}\\ &=\frac{5 \left (b^3 B+40 A b^2 c+44 a b B c+32 a A c^2+2 c \left (b^2 B+16 A b c+12 a B c\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c}-\frac{5 (6 (A b+a B)-(b B+4 A c) x) \left (a+b x+c x^2\right )^{3/2}}{24 x}-\frac{(2 A-B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^2}-\frac{5}{8} \sqrt{a} \left (3 A b^2+4 a b B+4 a A c\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )-\frac{5 \left (b^4 B-8 A b^3 c-24 a b^2 B c-96 a A b c^2-48 a^2 B c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.495488, size = 254, normalized size = 0.93 \[ \frac{\sqrt{a+x (b+c x)} \left (-96 a^2 c (A+2 B x)+4 a c x (B x (139 b+54 c x)-4 A (27 b-28 c x))+x^2 \left (2 b^2 c (132 A+59 B x)+8 b c^2 x (26 A+17 B x)+16 c^3 x^2 (4 A+3 B x)+15 b^3 B\right )\right )}{192 c x^2}+\frac{5 \left (48 a^2 B c^2+96 a A b c^2+24 a b^2 B c+8 A b^3 c+b^4 (-B)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{128 c^{3/2}}-\frac{5}{8} \sqrt{a} \left (4 a A c+4 a b B+3 A b^2\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^3,x]

[Out]

(Sqrt[a + x*(b + c*x)]*(-96*a^2*c*(A + 2*B*x) + x^2*(15*b^3*B + 16*c^3*x^2*(4*A + 3*B*x) + 8*b*c^2*x*(26*A + 1
7*B*x) + 2*b^2*c*(132*A + 59*B*x)) + 4*a*c*x*(-4*A*(27*b - 28*c*x) + B*x*(139*b + 54*c*x))))/(192*c*x^2) - (5*
Sqrt[a]*(3*A*b^2 + 4*a*b*B + 4*a*A*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/8 + (5*(-(b^4*B)
+ 8*A*b^3*c + 24*a*b^2*B*c + 96*a*A*b*c^2 + 48*a^2*B*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)
])])/(128*c^(3/2))

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Maple [B]  time = 0.01, size = 663, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^3,x)

[Out]

3/4*A/a^2*b*c*(c*x^2+b*x+a)^(5/2)*x+5/4*A/a*b*c*(c*x^2+b*x+a)^(3/2)*x+5/4*A/a*b^2*(c*x^2+b*x+a)^(3/2)+3/4*A/a^
2*b^2*(c*x^2+b*x+a)^(5/2)-B/a/x*(c*x^2+b*x+a)^(7/2)+55/16*B*(c*x^2+b*x+a)^(1/2)*b*a+5/32*B*(c*x^2+b*x+a)^(1/2)
*x*b^2+5/64*B/c*(c*x^2+b*x+a)^(1/2)*b^3-5/128*B/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^4+B/a*b*
(c*x^2+b*x+a)^(5/2)+15/8*B*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2-5/2*B*a^(3/2)*b*ln((2*a+b*x
+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+5/4*B*c*(c*x^2+b*x+a)^(3/2)*x+5/16*A*b^3/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*
x^2+b*x+a)^(1/2))-1/2*A/a/x^2*(c*x^2+b*x+a)^(7/2)-5/2*A*a^(3/2)*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x
)-15/8*A*a^(1/2)*b^2*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+1/2*A/a*c*(c*x^2+b*x+a)^(5/2)+5/2*A*a*c*(c*
x^2+b*x+a)^(1/2)+35/24*B*b*(c*x^2+b*x+a)^(3/2)+25/8*A*b^2*(c*x^2+b*x+a)^(1/2)+5/6*A*c*(c*x^2+b*x+a)^(3/2)+5/2*
A*b*(c*x^2+b*x+a)^(1/2)*x*c+15/4*A*a*b*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3/4*A/a^2*b/x*(c*x^
2+b*x+a)^(7/2)+15/16*B/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a*b^2+15/8*B*(c*x^2+b*x+a)^(1/2)*x*
a*c+B/a*c*(c*x^2+b*x+a)^(5/2)*x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 27.8384, size = 2993, normalized size = 10.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^3,x, algorithm="fricas")

[Out]

[-1/768*(15*(B*b^4 - 48*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*sqrt(c)*x^2*log(-8*c^2*x^2 - 8*b*c*x
- b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 240*(4*A*a*c^3 + (4*B*a*b + 3*A*b^2)*c^2)*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*(48*B
*c^4*x^5 - 96*A*a^2*c^2 + 8*(17*B*b*c^3 + 8*A*c^4)*x^4 - 48*(4*B*a^2 + 9*A*a*b)*c^2*x + 2*(59*B*b^2*c^2 + 4*(2
7*B*a + 26*A*b)*c^3)*x^3 + (15*B*b^3*c + 448*A*a*c^3 + 4*(139*B*a*b + 66*A*b^2)*c^2)*x^2)*sqrt(c*x^2 + b*x + a
))/(c^2*x^2), 1/384*(15*(B*b^4 - 48*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*sqrt(-c)*x^2*arctan(1/2*s
qrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 120*(4*A*a*c^3 + (4*B*a*b + 3*A*b^2)*c^2)
*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) + 2
*(48*B*c^4*x^5 - 96*A*a^2*c^2 + 8*(17*B*b*c^3 + 8*A*c^4)*x^4 - 48*(4*B*a^2 + 9*A*a*b)*c^2*x + 2*(59*B*b^2*c^2
+ 4*(27*B*a + 26*A*b)*c^3)*x^3 + (15*B*b^3*c + 448*A*a*c^3 + 4*(139*B*a*b + 66*A*b^2)*c^2)*x^2)*sqrt(c*x^2 + b
*x + a))/(c^2*x^2), 1/768*(480*(4*A*a*c^3 + (4*B*a*b + 3*A*b^2)*c^2)*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)) - 15*(B*b^4 - 48*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b
^3)*c)*sqrt(c)*x^2*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(
48*B*c^4*x^5 - 96*A*a^2*c^2 + 8*(17*B*b*c^3 + 8*A*c^4)*x^4 - 48*(4*B*a^2 + 9*A*a*b)*c^2*x + 2*(59*B*b^2*c^2 +
4*(27*B*a + 26*A*b)*c^3)*x^3 + (15*B*b^3*c + 448*A*a*c^3 + 4*(139*B*a*b + 66*A*b^2)*c^2)*x^2)*sqrt(c*x^2 + b*x
+ a))/(c^2*x^2), 1/384*(240*(4*A*a*c^3 + (4*B*a*b + 3*A*b^2)*c^2)*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)) + 15*(B*b^4 - 48*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3
)*c)*sqrt(-c)*x^2*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(48*B*c^4
*x^5 - 96*A*a^2*c^2 + 8*(17*B*b*c^3 + 8*A*c^4)*x^4 - 48*(4*B*a^2 + 9*A*a*b)*c^2*x + 2*(59*B*b^2*c^2 + 4*(27*B*
a + 26*A*b)*c^3)*x^3 + (15*B*b^3*c + 448*A*a*c^3 + 4*(139*B*a*b + 66*A*b^2)*c^2)*x^2)*sqrt(c*x^2 + b*x + a))/(
c^2*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac{5}{2}}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**(5/2)/x**3,x)

[Out]

Integral((A + B*x)*(a + b*x + c*x**2)**(5/2)/x**3, x)

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Giac [B]  time = 1.37671, size = 711, normalized size = 2.6 \begin{align*} \frac{1}{192} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (6 \, B c^{2} x + \frac{17 \, B b c^{4} + 8 \, A c^{5}}{c^{3}}\right )} x + \frac{59 \, B b^{2} c^{3} + 108 \, B a c^{4} + 104 \, A b c^{4}}{c^{3}}\right )} x + \frac{15 \, B b^{3} c^{2} + 556 \, B a b c^{3} + 264 \, A b^{2} c^{3} + 448 \, A a c^{4}}{c^{3}}\right )} + \frac{5 \,{\left (4 \, B a^{2} b + 3 \, A a b^{2} + 4 \, A a^{2} c\right )} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + b x + a}}{\sqrt{-a}}\right )}{4 \, \sqrt{-a}} + \frac{5 \,{\left (B b^{4} - 24 \, B a b^{2} c - 8 \, A b^{3} c - 48 \, B a^{2} c^{2} - 96 \, A a b c^{2}\right )} \log \left ({\left | 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} + b \right |}\right )}{128 \, c^{\frac{3}{2}}} + \frac{4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} B a^{2} b + 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} A a b^{2} + 4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} A a^{2} c + 8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} B a^{3} \sqrt{c} + 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} A a^{2} b \sqrt{c} - 4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} B a^{3} b - 7 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} A a^{2} b^{2} + 4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} A a^{3} c - 8 \, B a^{4} \sqrt{c} - 16 \, A a^{3} b \sqrt{c}}{4 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} - a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^3,x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*B*c^2*x + (17*B*b*c^4 + 8*A*c^5)/c^3)*x + (59*B*b^2*c^3 + 108*B*a*c^4 + 1
04*A*b*c^4)/c^3)*x + (15*B*b^3*c^2 + 556*B*a*b*c^3 + 264*A*b^2*c^3 + 448*A*a*c^4)/c^3) + 5/4*(4*B*a^2*b + 3*A*
a*b^2 + 4*A*a^2*c)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/sqrt(-a) + 5/128*(B*b^4 - 24*B*a*b^2*
c - 8*A*b^3*c - 48*B*a^2*c^2 - 96*A*a*b*c^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(3/
2) + 1/4*(4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^2*b + 9*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a*b^2 +
4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^2*c + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^3*sqrt(c) + 24*(
sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^2*b*sqrt(c) - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^3*b - 7*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))*A*a^2*b^2 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^3*c - 8*B*a^4*sqrt(c) - 1
6*A*a^3*b*sqrt(c))/((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^2