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

Optimal. Leaf size=193 \[ -\frac{\left (-24 a A c^2-12 a b B c-6 A b^2 c+b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2}}+\frac{\sqrt{a+b x+c x^2} \left (8 a B c+2 c x (6 A c+b B)+18 A b c+b^2 B\right )}{8 c}-\frac{(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}-\frac{1}{2} \sqrt{a} (2 a B+3 A b) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right ) \]

[Out]

((b^2*B + 18*A*b*c + 8*a*B*c + 2*c*(b*B + 6*A*c)*x)*Sqrt[a + b*x + c*x^2])/(8*c) - ((3*A - B*x)*(a + b*x + c*x
^2)^(3/2))/(3*x) - (Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/2 - ((b^3*
B - 6*A*b^2*c - 12*a*b*B*c - 24*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(3/2))

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Rubi [A]  time = 0.176743, antiderivative size = 193, 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 = {812, 814, 843, 621, 206, 724} \[ -\frac{\left (-24 a A c^2-12 a b B c-6 A b^2 c+b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2}}+\frac{\sqrt{a+b x+c x^2} \left (8 a B c+2 c x (6 A c+b B)+18 A b c+b^2 B\right )}{8 c}-\frac{(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}-\frac{1}{2} \sqrt{a} (2 a B+3 A b) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*B + 18*A*b*c + 8*a*B*c + 2*c*(b*B + 6*A*c)*x)*Sqrt[a + b*x + c*x^2])/(8*c) - ((3*A - B*x)*(a + b*x + c*x
^2)^(3/2))/(3*x) - (Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/2 - ((b^3*
B - 6*A*b^2*c - 12*a*b*B*c - 24*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*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 )^{3/2}}{x^2} \, dx &=-\frac{(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}-\frac{1}{2} \int \frac{(-3 A b-2 a B-(b B+6 A c) x) \sqrt{a+b x+c x^2}}{x} \, dx\\ &=\frac{\left (b^2 B+18 A b c+8 a B c+2 c (b B+6 A c) x\right ) \sqrt{a+b x+c x^2}}{8 c}-\frac{(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}+\frac{\int \frac{4 a (3 A b+2 a B) c-\frac{1}{2} \left (b^3 B-6 A b^2 c-12 a b B c-24 a A c^2\right ) x}{x \sqrt{a+b x+c x^2}} \, dx}{8 c}\\ &=\frac{\left (b^2 B+18 A b c+8 a B c+2 c (b B+6 A c) x\right ) \sqrt{a+b x+c x^2}}{8 c}-\frac{(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}+\frac{1}{2} (a (3 A b+2 a B)) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx+\frac{\left (-b^3 B+6 A b^2 c+12 a b B c+24 a A c^2\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{16 c}\\ &=\frac{\left (b^2 B+18 A b c+8 a B c+2 c (b B+6 A c) x\right ) \sqrt{a+b x+c x^2}}{8 c}-\frac{(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}-(a (3 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 )+\frac{\left (-b^3 B+6 A b^2 c+12 a b B c+24 a A c^2\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 )}{8 c}\\ &=\frac{\left (b^2 B+18 A b c+8 a B c+2 c (b B+6 A c) x\right ) \sqrt{a+b x+c x^2}}{8 c}-\frac{(3 A-B x) \left (a+b x+c x^2\right )^{3/2}}{3 x}-\frac{1}{2} \sqrt{a} (3 A b+2 a B) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )-\frac{\left (b^3 B-6 A b^2 c-12 a b B c-24 a A c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.340255, size = 183, normalized size = 0.95 \[ \frac{1}{48} \left (\frac{2 \sqrt{a+x (b+c x)} \left (x \left (2 b c (15 A+7 B x)+4 c^2 x (3 A+2 B x)+3 b^2 B\right )-8 a c (3 A-4 B x)\right )}{c x}+\frac{3 \left (24 a A c^2+12 a b B c+6 A b^2 c+b^3 (-B)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{c^{3/2}}-24 \sqrt{a} (2 a B+3 A b) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[a + x*(b + c*x)]*(-8*a*c*(3*A - 4*B*x) + x*(3*b^2*B + 4*c^2*x*(3*A + 2*B*x) + 2*b*c*(15*A + 7*B*x))))
/(c*x) - 24*Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])] + (3*(-(b^3*B) + 6*
A*b^2*c + 12*a*b*B*c + 24*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/c^(3/2))/48

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Maple [B]  time = 0.01, size = 365, normalized size = 1.9 \begin{align*}{\frac{B}{3} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{bBx}{4}\sqrt{c{x}^{2}+bx+a}}+{\frac{{b}^{2}B}{8\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,abB}{4}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{b}^{3}B}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+Ba\sqrt{c{x}^{2}+bx+a}-B{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ) -{\frac{A}{ax} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{2}}}}+{\frac{Ab}{a} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,A{b}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}+{\frac{9\,Ab}{4}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,Ab}{2}\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ) }+{\frac{Acx}{a} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,Acx}{2}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,aA}{2}\sqrt{c}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ) } \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^2,x)

[Out]

1/3*B*(c*x^2+b*x+a)^(3/2)+1/4*B*b*(c*x^2+b*x+a)^(1/2)*x+1/8*B/c*(c*x^2+b*x+a)^(1/2)*b^2+3/4*B*b/c^(1/2)*ln((1/
2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/16*B/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^3+B*a*(c*
x^2+b*x+a)^(1/2)-B*a^(3/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-A/a/x*(c*x^2+b*x+a)^(5/2)+A/a*b*(c*x^
2+b*x+a)^(3/2)+3/8*A*b^2/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+9/4*A*b*(c*x^2+b*x+a)^(1/2)-3/2*A
*a^(1/2)*b*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+A/a*c*(c*x^2+b*x+a)^(3/2)*x+3/2*A*c*(c*x^2+b*x+a)^(1/
2)*x+3/2*A*a*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))

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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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 14.9874, size = 2187, normalized size = 11.33 \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^2,x, algorithm="fricas")

[Out]

[1/96*(24*(2*B*a + 3*A*b)*sqrt(a)*c^2*x*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) - 3*(B*b^3 - 24*A*a*c^2 - 6*(2*B*a*b + A*b^2)*c)*sqrt(c)*x*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*(8*B*c^3*x^3 - 24*A*a*c^2 + 2*(7*B*b*c^2 + 6*A*
c^3)*x^2 + (3*B*b^2*c + 2*(16*B*a + 15*A*b)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^2*x), 1/48*(12*(2*B*a + 3*A*b)*s
qrt(a)*c^2*x*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) + 3
*(B*b^3 - 24*A*a*c^2 - 6*(2*B*a*b + A*b^2)*c)*sqrt(-c)*x*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*(8*B*c^3*x^3 - 24*A*a*c^2 + 2*(7*B*b*c^2 + 6*A*c^3)*x^2 + (3*B*b^2*c + 2*(16*B*a
+ 15*A*b)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^2*x), 1/96*(48*(2*B*a + 3*A*b)*sqrt(-a)*c^2*x*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)) - 3*(B*b^3 - 24*A*a*c^2 - 6*(2*B*a*b + A*b^2)*c)*s
qrt(c)*x*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*(8*B*c^3*x^
3 - 24*A*a*c^2 + 2*(7*B*b*c^2 + 6*A*c^3)*x^2 + (3*B*b^2*c + 2*(16*B*a + 15*A*b)*c^2)*x)*sqrt(c*x^2 + b*x + a))
/(c^2*x), 1/48*(24*(2*B*a + 3*A*b)*sqrt(-a)*c^2*x*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)) + 3*(B*b^3 - 24*A*a*c^2 - 6*(2*B*a*b + A*b^2)*c)*sqrt(-c)*x*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*(8*B*c^3*x^3 - 24*A*a*c^2 + 2*(7*B*b*c^2 + 6*A*c^3)*x^2 +
(3*B*b^2*c + 2*(16*B*a + 15*A*b)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^2*x)]

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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^{2}}\, 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**2,x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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^2,x, algorithm="giac")

[Out]

Exception raised: TypeError