3.930 \(\int \frac{(A+B x) (a+b x+c x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=206 \[ -\frac{\left (6 a B \left (4 a c+b^2\right )-A \left (b^3-12 a b c\right )\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 a^{3/2}}+\frac{\sqrt{a+b x+c x^2} \left (2 c x (6 a B+A b)-8 a A c-6 a b B+A b^2\right )}{8 a x}-\frac{\left (a+b x+c x^2\right )^{3/2} (3 x (2 a B+A b)+4 a A)}{12 a x^3}+\frac{1}{2} \sqrt{c} (2 A c+3 b B) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \]

[Out]

((A*b^2 - 6*a*b*B - 8*a*A*c + 2*(A*b + 6*a*B)*c*x)*Sqrt[a + b*x + c*x^2])/(8*a*x) - ((4*a*A + 3*(A*b + 2*a*B)*
x)*(a + b*x + c*x^2)^(3/2))/(12*a*x^3) - ((6*a*B*(b^2 + 4*a*c) - A*(b^3 - 12*a*b*c))*ArcTanh[(2*a + b*x)/(2*Sq
rt[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(3/2)) + (Sqrt[c]*(3*b*B + 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a +
b*x + c*x^2])])/2

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Rubi [A]  time = 0.222676, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {810, 812, 843, 621, 206, 724} \[ -\frac{\left (6 a B \left (4 a c+b^2\right )-A \left (b^3-12 a b c\right )\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 a^{3/2}}+\frac{\sqrt{a+b x+c x^2} \left (2 c x (6 a B+A b)-8 a A c-6 a b B+A b^2\right )}{8 a x}-\frac{\left (a+b x+c x^2\right )^{3/2} (3 x (2 a B+A b)+4 a A)}{12 a x^3}+\frac{1}{2} \sqrt{c} (2 A c+3 b B) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^(3/2))/x^4,x]

[Out]

((A*b^2 - 6*a*b*B - 8*a*A*c + 2*(A*b + 6*a*B)*c*x)*Sqrt[a + b*x + c*x^2])/(8*a*x) - ((4*a*A + 3*(A*b + 2*a*B)*
x)*(a + b*x + c*x^2)^(3/2))/(12*a*x^3) - ((6*a*B*(b^2 + 4*a*c) - A*(b^3 - 12*a*b*c))*ArcTanh[(2*a + b*x)/(2*Sq
rt[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(3/2)) + (Sqrt[c]*(3*b*B + 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a +
b*x + c*x^2])])/2

Rule 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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 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 )^{3/2}}{x^4} \, dx &=-\frac{(4 a A+3 (A b+2 a B) x) \left (a+b x+c x^2\right )^{3/2}}{12 a x^3}-\frac{\int \frac{\left (\frac{1}{2} \left (-6 a b B+A \left (b^2-8 a c\right )\right )-(A b+6 a B) c x\right ) \sqrt{a+b x+c x^2}}{x^2} \, dx}{4 a}\\ &=\frac{\left (A b^2-6 a b B-8 a A c+2 (A b+6 a B) c x\right ) \sqrt{a+b x+c x^2}}{8 a x}-\frac{(4 a A+3 (A b+2 a B) x) \left (a+b x+c x^2\right )^{3/2}}{12 a x^3}+\frac{\int \frac{\frac{1}{2} \left (6 a B \left (b^2+4 a c\right )-2 A \left (\frac{b^3}{2}-6 a b c\right )\right )+4 a c (3 b B+2 A c) x}{x \sqrt{a+b x+c x^2}} \, dx}{8 a}\\ &=\frac{\left (A b^2-6 a b B-8 a A c+2 (A b+6 a B) c x\right ) \sqrt{a+b x+c x^2}}{8 a x}-\frac{(4 a A+3 (A b+2 a B) x) \left (a+b x+c x^2\right )^{3/2}}{12 a x^3}+\frac{1}{2} (c (3 b B+2 A c)) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx+\frac{\left (6 a B \left (b^2+4 a c\right )-A \left (b^3-12 a b c\right )\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx}{16 a}\\ &=\frac{\left (A b^2-6 a b B-8 a A c+2 (A b+6 a B) c x\right ) \sqrt{a+b x+c x^2}}{8 a x}-\frac{(4 a A+3 (A b+2 a B) x) \left (a+b x+c x^2\right )^{3/2}}{12 a x^3}+(c (3 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 )-\frac{\left (6 a B \left (b^2+4 a c\right )-A \left (b^3-12 a b 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 )}{8 a}\\ &=\frac{\left (A b^2-6 a b B-8 a A c+2 (A b+6 a B) c x\right ) \sqrt{a+b x+c x^2}}{8 a x}-\frac{(4 a A+3 (A b+2 a B) x) \left (a+b x+c x^2\right )^{3/2}}{12 a x^3}-\frac{\left (6 a B \left (b^2+4 a c\right )-A \left (b^3-12 a b c\right )\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 a^{3/2}}+\frac{1}{2} \sqrt{c} (3 b B+2 A c) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.53673, size = 182, normalized size = 0.88 \[ \frac{1}{48} \left (-\frac{2 \sqrt{a+x (b+c x)} \left (4 a^2 (2 A+3 B x)+2 a x (A (7 b+16 c x)+3 B x (5 b-4 c x))+3 A b^2 x^2\right )}{a x^3}+\frac{3 \left (A \left (b^3-12 a b c\right )-6 a B \left (4 a c+b^2\right )\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{a^{3/2}}+24 \sqrt{c} (2 A c+3 b B) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^(3/2))/x^4,x]

[Out]

((-2*Sqrt[a + x*(b + c*x)]*(3*A*b^2*x^2 + 4*a^2*(2*A + 3*B*x) + 2*a*x*(3*B*x*(5*b - 4*c*x) + A*(7*b + 16*c*x))
))/(a*x^3) + (3*(-6*a*B*(b^2 + 4*a*c) + A*(b^3 - 12*a*b*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x
)])])/a^(3/2) + 24*Sqrt[c]*(3*b*B + 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/48

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Maple [B]  time = 0.01, size = 635, normalized size = 3.1 \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)^(3/2)/x^4,x)

[Out]

5/4*A/a*b*c*(c*x^2+b*x+a)^(1/2)+1/12*A/a^2*b/x^2*(c*x^2+b*x+a)^(5/2)-1/8*A/a^2*b^2*c*(c*x^2+b*x+a)^(1/2)*x-1/2
4*A/a^3*b^2*c*(c*x^2+b*x+a)^(3/2)*x+1/4*B/a^2*b*c*(c*x^2+b*x+a)^(3/2)*x+3/4*B/a*b*c*(c*x^2+b*x+a)^(1/2)*x+1/4*
B/a^2*b^2*(c*x^2+b*x+a)^(3/2)-3/8*B/a^(1/2)*b^2*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+3/2*B*b*c^(1/2)*
ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/2*B/a*c*(c*x^2+b*x+a)^(3/2)-3/2*B*a^(1/2)*c*ln((2*a+b*x+2*a^(1/2
)*(c*x^2+b*x+a)^(1/2))/x)-1/2*B/a/x^2*(c*x^2+b*x+a)^(5/2)+3/4*B/a*b^2*(c*x^2+b*x+a)^(1/2)-1/3*A/a/x^3*(c*x^2+b
*x+a)^(5/2)-1/24*A/a^3*b^3*(c*x^2+b*x+a)^(3/2)-1/8*A/a^2*b^3*(c*x^2+b*x+a)^(1/2)+1/16*A/a^(3/2)*b^3*ln((2*a+b*
x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+3/2*B*c*(c*x^2+b*x+a)^(1/2)+A*c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)
^(1/2))+A/a*c^2*(c*x^2+b*x+a)^(1/2)*x+1/24*A/a^3*b^2/x*(c*x^2+b*x+a)^(5/2)+7/12*A/a^2*b*c*(c*x^2+b*x+a)^(3/2)-
1/4*B/a^2*b/x*(c*x^2+b*x+a)^(5/2)-3/4*A/a^(1/2)*b*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-2/3*A/a^2*c/
x*(c*x^2+b*x+a)^(5/2)+2/3*A/a^2*c^2*(c*x^2+b*x+a)^(3/2)*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)^(3/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 9.99363, size = 2242, normalized size = 10.88 \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)^(3/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(24*(3*B*a^2*b + 2*A*a^2*c)*sqrt(c)*x^3*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) + 3*(6*B*a*b^2 - A*b^3 + 12*(2*B*a^2 + A*a*b)*c)*sqrt(a)*x^3*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*(24*B*a^2*c*x^3 - 8*A*a^3 - (30*B*a^2*b
+ 3*A*a*b^2 + 32*A*a^2*c)*x^2 - 2*(6*B*a^3 + 7*A*a^2*b)*x)*sqrt(c*x^2 + b*x + a))/(a^2*x^3), -1/96*(48*(3*B*a
^2*b + 2*A*a^2*c)*sqrt(-c)*x^3*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))
- 3*(6*B*a*b^2 - A*b^3 + 12*(2*B*a^2 + A*a*b)*c)*sqrt(a)*x^3*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*(24*B*a^2*c*x^3 - 8*A*a^3 - (30*B*a^2*b + 3*A*a*b^2 + 32*A*a^
2*c)*x^2 - 2*(6*B*a^3 + 7*A*a^2*b)*x)*sqrt(c*x^2 + b*x + a))/(a^2*x^3), 1/48*(3*(6*B*a*b^2 - A*b^3 + 12*(2*B*a
^2 + A*a*b)*c)*sqrt(-a)*x^3*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)) + 1
2*(3*B*a^2*b + 2*A*a^2*c)*sqrt(c)*x^3*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqr
t(c) - 4*a*c) + 2*(24*B*a^2*c*x^3 - 8*A*a^3 - (30*B*a^2*b + 3*A*a*b^2 + 32*A*a^2*c)*x^2 - 2*(6*B*a^3 + 7*A*a^2
*b)*x)*sqrt(c*x^2 + b*x + a))/(a^2*x^3), 1/48*(3*(6*B*a*b^2 - A*b^3 + 12*(2*B*a^2 + A*a*b)*c)*sqrt(-a)*x^3*arc
tan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) - 24*(3*B*a^2*b + 2*A*a^2*c)*sqrt(
-c)*x^3*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*(24*B*a^2*c*x^3 - 8
*A*a^3 - (30*B*a^2*b + 3*A*a*b^2 + 32*A*a^2*c)*x^2 - 2*(6*B*a^3 + 7*A*a^2*b)*x)*sqrt(c*x^2 + b*x + a))/(a^2*x^
3)]

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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{3}{2}}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**(3/2)/x**4,x)

[Out]

Integral((A + B*x)*(a + b*x + c*x**2)**(3/2)/x**4, x)

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Giac [B]  time = 1.59287, size = 872, normalized size = 4.23 \begin{align*} \sqrt{c x^{2} + b x + a} B c - \frac{{\left (3 \, B b c + 2 \, A c^{2}\right )} \log \left ({\left | 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} + b \right |}\right )}{2 \, \sqrt{c}} + \frac{{\left (6 \, B a b^{2} - A b^{3} + 24 \, B a^{2} c + 12 \, A a b c\right )} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + b x + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a} + \frac{30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} B a b^{2} \sqrt{c} + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} A b^{3} \sqrt{c} + 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} B a^{2} c^{\frac{3}{2}} + 60 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} A a b c^{\frac{3}{2}} + 96 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{4} B a^{2} b c + 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{4} A a b^{2} c + 96 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{4} A a^{2} c^{2} - 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} B a^{2} b^{2} \sqrt{c} + 8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} A a b^{3} \sqrt{c} - 144 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} B a^{3} b c - 96 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} A a^{3} c^{2} + 18 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} B a^{3} b^{2} \sqrt{c} - 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} A a^{2} b^{3} \sqrt{c} - 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} B a^{4} c^{\frac{3}{2}} + 36 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} A a^{3} b c^{\frac{3}{2}} + 48 \, B a^{4} b c + 64 \, A a^{4} c^{2}}{24 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} - a\right )}^{3} a \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(3/2)/x^4,x, algorithm="giac")

[Out]

sqrt(c*x^2 + b*x + a)*B*c - 1/2*(3*B*b*c + 2*A*c^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b)
)/sqrt(c) + 1/8*(6*B*a*b^2 - A*b^3 + 24*B*a^2*c + 12*A*a*b*c)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt
(-a))/(sqrt(-a)*a) + 1/24*(30*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a*b^2*sqrt(c) + 3*(sqrt(c)*x - sqrt(c*x^
2 + b*x + a))^5*A*b^3*sqrt(c) + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^2*c^(3/2) + 60*(sqrt(c)*x - sqrt(
c*x^2 + b*x + a))^5*A*a*b*c^(3/2) + 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^2*b*c + 48*(sqrt(c)*x - sqrt(
c*x^2 + b*x + a))^4*A*a*b^2*c + 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^2*c^2 - 48*(sqrt(c)*x - sqrt(c*x^
2 + b*x + a))^3*B*a^2*b^2*sqrt(c) + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a*b^3*sqrt(c) - 144*(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))^2*B*a^3*b*c - 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^3*c^2 + 18*(sqrt(c)*x - sqr
t(c*x^2 + b*x + a))*B*a^3*b^2*sqrt(c) - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^2*b^3*sqrt(c) - 24*(sqrt(c)*
x - sqrt(c*x^2 + b*x + a))*B*a^4*c^(3/2) + 36*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^3*b*c^(3/2) + 48*B*a^4*b
*c + 64*A*a^4*c^2)/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^3*a*sqrt(c))