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

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

[Out]

-1/3*(-B*x+3*A)*(c*x^2+b*x+a)^(3/2)/x-1/16*(-24*A*a*c^2-6*A*b^2*c-12*B*a*b*c+B*b^3)*arctanh(1/2*(2*c*x+b)/c^(1
/2)/(c*x^2+b*x+a)^(1/2))/c^(3/2)-1/2*(3*A*b+2*B*a)*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))*a^(1/2)+
1/8*(b^2*B+18*A*b*c+8*a*B*c+2*c*(6*A*c+B*b)*x)*(c*x^2+b*x+a)^(1/2)/c

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Rubi [A]  time = 0.18, antiderivative size = 193, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.32, 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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fricas [A]  time = 5.66, size = 917, normalized size = 4.75 \[ \left [\frac {24 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {a} c^{2} x \log \left (-\frac {8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{2}}\right ) - 3 \, {\left (B b^{3} - 24 \, A a c^{2} - 6 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {c} x \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, B c^{3} x^{3} - 24 \, A a c^{2} + 2 \, {\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2} + {\left (3 \, B b^{2} c + 2 \, {\left (16 \, B a + 15 \, A b\right )} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{96 \, c^{2} x}, \frac {12 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {a} c^{2} x \log \left (-\frac {8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{2}}\right ) + 3 \, {\left (B b^{3} - 24 \, A a c^{2} - 6 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (8 \, B c^{3} x^{3} - 24 \, A a c^{2} + 2 \, {\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2} + {\left (3 \, B b^{2} c + 2 \, {\left (16 \, B a + 15 \, A b\right )} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{48 \, c^{2} x}, \frac {48 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {-a} c^{2} x \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{2} + a b x + a^{2}\right )}}\right ) - 3 \, {\left (B b^{3} - 24 \, A a c^{2} - 6 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {c} x \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, B c^{3} x^{3} - 24 \, A a c^{2} + 2 \, {\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2} + {\left (3 \, B b^{2} c + 2 \, {\left (16 \, B a + 15 \, A b\right )} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{96 \, c^{2} x}, \frac {24 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {-a} c^{2} x \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{2} + a b x + a^{2}\right )}}\right ) + 3 \, {\left (B b^{3} - 24 \, A a c^{2} - 6 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (8 \, B c^{3} x^{3} - 24 \, A a c^{2} + 2 \, {\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2} + {\left (3 \, B b^{2} c + 2 \, {\left (16 \, B a + 15 \, A b\right )} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{48 \, c^{2} x}\right ] \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{%%{[1,0]:[1,0,%%%{-1,[1]%%%}]%%},[4,0,0,0,0]%%%}+%%%{%%{[-2,0]:
[1,0,%%%{-1,[1]%%%}]%%},[2,0,1,0,0]%%%}+%%%{%%{[1,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,0,2,0,0]%%%} / %%%{%%%{1,[1]%%
%},[4,0,0,0,0]%%%}+%%%{%%%{-2,[1]%%%},[2,0,1,0,0]%%%}+%%%{%%%{1,[1]%%%},[0,0,2,0,0]%%%} Error: Bad Argument Va
lue

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maple [B]  time = 0.06, size = 365, normalized size = 1.89 \[ \frac {3 A a \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2}-\frac {3 A \sqrt {a}\, b \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2}+\frac {3 A \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}-B \,a^{\frac {3}{2}} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )+\frac {3 B a b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 \sqrt {c}}-\frac {B \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {3}{2}}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A c x}{2}+\frac {\sqrt {c \,x^{2}+b x +a}\, B b x}{4}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A c x}{a}+\frac {9 \sqrt {c \,x^{2}+b x +a}\, A b}{4}+\sqrt {c \,x^{2}+b x +a}\, B a +\frac {\sqrt {c \,x^{2}+b x +a}\, B \,b^{2}}{8 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b}{a}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B}{3}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A}{a x} \]

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]

-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((c*x+1/2*b)/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((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+A*c/a*(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*c^(1/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+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((c*x+1/2*b
)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/16*B/c^(3/2)*ln((c*x+1/2*b)/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((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)*(c*x^2+b*x+a)^(3/2)/x^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 = -0.01 \[ \int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{x^{2}}\, dx \]

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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