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

Optimal. Leaf size=179 \[ -\frac {3 \left (A \left (4 a c+b^2\right )+4 a b B\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {a}}+\frac {3 \left (4 a B c+4 A b c+b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}}-\frac {(A-B x) \left (a+b x+c x^2\right )^{3/2}}{2 x^2}-\frac {3 \sqrt {a+b x+c x^2} (2 a B-x (2 A c+b B)+A b)}{4 x} \]

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Rubi [A]  time = 0.17, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {812, 843, 621, 206, 724} \begin {gather*} -\frac {3 \left (A \left (4 a c+b^2\right )+4 a b B\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {a}}+\frac {3 \left (4 a B c+4 A b c+b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c}}-\frac {(A-B x) \left (a+b x+c x^2\right )^{3/2}}{2 x^2}-\frac {3 \sqrt {a+b x+c x^2} (2 a B-x (2 A c+b B)+A b)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.44, size = 162, normalized size = 0.91 \begin {gather*} \frac {1}{8} \left (-\frac {3 \left (A \left (4 a c+b^2\right )+4 a b B\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{\sqrt {a}}+\frac {3 \left (4 a B c+4 A b c+b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}+\frac {2 \sqrt {a+x (b+c x)} (x (A (4 c x-5 b)+B x (5 b+2 c x))-2 a (A+2 B x))}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[a + x*(b + c*x)]*(-2*a*(A + 2*B*x) + x*(B*x*(5*b + 2*c*x) + A*(-5*b + 4*c*x))))/x^2 - (3*(4*a*b*B + A
*(b^2 + 4*a*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/Sqrt[a] + (3*(b^2*B + 4*A*b*c + 4*a*B*
c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/Sqrt[c])/8

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IntegrateAlgebraic [A]  time = 1.28, size = 166, normalized size = 0.93 \begin {gather*} -\frac {3 \left (4 a B c+4 A b c+b^2 B\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{8 \sqrt {c}}-\frac {3 \left (4 a A c+4 a b B+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x+c x^2}-\sqrt {c} x}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {\sqrt {a+b x+c x^2} \left (-2 a A-4 a B x-5 A b x+4 A c x^2+5 b B x^2+2 B c x^3\right )}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-2*a*A - 5*A*b*x - 4*a*B*x + 5*b*B*x^2 + 4*A*c*x^2 + 2*B*c*x^3))/(4*x^2) - (3*(A*b^2 +
4*a*b*B + 4*a*A*c)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(4*Sqrt[a]) - (3*(b^2*B + 4*A*b*c
+ 4*a*B*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(8*Sqrt[c])

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fricas [A]  time = 1.90, size = 921, normalized size = 5.15 \begin {gather*} \left [\frac {3 \, {\left (B a b^{2} + 4 \, {\left (B a^{2} + A a b\right )} c\right )} \sqrt {c} x^{2} \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 ) + 3 \, {\left (4 \, A a c^{2} + {\left (4 \, B a b + A b^{2}\right )} c\right )} \sqrt {a} x^{2} \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 ) + 4 \, {\left (2 \, B a c^{2} x^{3} - 2 \, A a^{2} c - {\left (4 \, B a^{2} + 5 \, A a b\right )} c x + {\left (5 \, B a b c + 4 \, A a c^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + b x + a}}{16 \, a c x^{2}}, -\frac {6 \, {\left (B a b^{2} + 4 \, {\left (B a^{2} + A a b\right )} c\right )} \sqrt {-c} x^{2} \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 ) - 3 \, {\left (4 \, A a c^{2} + {\left (4 \, B a b + A b^{2}\right )} c\right )} \sqrt {a} x^{2} \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 ) - 4 \, {\left (2 \, B a c^{2} x^{3} - 2 \, A a^{2} c - {\left (4 \, B a^{2} + 5 \, A a b\right )} c x + {\left (5 \, B a b c + 4 \, A a c^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + b x + a}}{16 \, a c x^{2}}, \frac {6 \, {\left (4 \, A a c^{2} + {\left (4 \, B a b + A b^{2}\right )} c\right )} \sqrt {-a} x^{2} \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 a b^{2} + 4 \, {\left (B a^{2} + A a b\right )} c\right )} \sqrt {c} x^{2} \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 (2 \, B a c^{2} x^{3} - 2 \, A a^{2} c - {\left (4 \, B a^{2} + 5 \, A a b\right )} c x + {\left (5 \, B a b c + 4 \, A a c^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + b x + a}}{16 \, a c x^{2}}, \frac {3 \, {\left (4 \, A a c^{2} + {\left (4 \, B a b + A b^{2}\right )} c\right )} \sqrt {-a} x^{2} \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 a b^{2} + 4 \, {\left (B a^{2} + A a b\right )} c\right )} \sqrt {-c} x^{2} \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 (2 \, B a c^{2} x^{3} - 2 \, A a^{2} c - {\left (4 \, B a^{2} + 5 \, A a b\right )} c x + {\left (5 \, B a b c + 4 \, A a c^{2}\right )} x^{2}\right )} \sqrt {c x^{2} + b x + a}}{8 \, a c x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(B*a*b^2 + 4*(B*a^2 + A*a*b)*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) + 3*(4*A*a*c^2 + (4*B*a*b + A*b^2)*c)*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*(2*B*a*c^2*x^3 - 2*A*a^2*c - (4*B*a^2 + 5*A
*a*b)*c*x + (5*B*a*b*c + 4*A*a*c^2)*x^2)*sqrt(c*x^2 + b*x + a))/(a*c*x^2), -1/16*(6*(B*a*b^2 + 4*(B*a^2 + A*a*
b)*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)) - 3*(4*A*a*c
^2 + (4*B*a*b + A*b^2)*c)*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*(2*B*a*c^2*x^3 - 2*A*a^2*c - (4*B*a^2 + 5*A*a*b)*c*x + (5*B*a*b*c + 4*A*a*c^2)*x^2)*
sqrt(c*x^2 + b*x + a))/(a*c*x^2), 1/16*(6*(4*A*a*c^2 + (4*B*a*b + A*b^2)*c)*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)) + 3*(B*a*b^2 + 4*(B*a^2 + A*a*b)*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*(2*B*a*c^2*x^3 - 2*A*a^2
*c - (4*B*a^2 + 5*A*a*b)*c*x + (5*B*a*b*c + 4*A*a*c^2)*x^2)*sqrt(c*x^2 + b*x + a))/(a*c*x^2), 1/8*(3*(4*A*a*c^
2 + (4*B*a*b + A*b^2)*c)*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)) - 3*(B*a*b^2 + 4*(B*a^2 + A*a*b)*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*(2*B*a*c^2*x^3 - 2*A*a^2*c - (4*B*a^2 + 5*A*a*b)*c*x + (5*B*a*b*c + 4*A*a*c^2)*x^
2)*sqrt(c*x^2 + b*x + a))/(a*c*x^2)]

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giac [B]  time = 0.38, size = 412, normalized size = 2.30 \begin {gather*} \frac {1}{4} \, {\left (2 \, B c x + \frac {5 \, B b c + 4 \, A c^{2}}{c}\right )} \sqrt {c x^{2} + b x + a} + \frac {3 \, {\left (4 \, B a b + A b^{2} + 4 \, A a c\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a}} - \frac {3 \, {\left (B b^{2} + 4 \, B a c + 4 \, A b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{8 \, \sqrt {c}} + \frac {4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} B a b + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A b^{2} + 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a c + 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} B a^{2} \sqrt {c} + 16 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} A a b \sqrt {c} - 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{2} b - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a b^{2} + 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a^{2} c - 8 \, B a^{3} \sqrt {c} - 8 \, A a^{2} b \sqrt {c}}{4 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*B*c*x + (5*B*b*c + 4*A*c^2)/c)*sqrt(c*x^2 + b*x + a) + 3/4*(4*B*a*b + A*b^2 + 4*A*a*c)*arctan(-(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/sqrt(-a) - 3/8*(B*b^2 + 4*B*a*c + 4*A*b*c)*log(abs(2*(sqrt(c)*x - sqrt(c
*x^2 + b*x + a))*sqrt(c) + b))/sqrt(c) + 1/4*(4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a*b + 5*(sqrt(c)*x - s
qrt(c*x^2 + b*x + a))^3*A*b^2 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a*c + 8*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))^2*B*a^2*sqrt(c) + 16*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a*b*sqrt(c) - 4*(sqrt(c)*x - sqrt(c*x^2 +
b*x + a))*B*a^2*b - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a*b^2 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a
^2*c - 8*B*a^3*sqrt(c) - 8*A*a^2*b*sqrt(c))/((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^2

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maple [B]  time = 0.06, size = 463, normalized size = 2.59 \begin {gather*} -\frac {3 A \sqrt {a}\, c \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 {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{8 \sqrt {a}}+\frac {3 A b \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2}+\frac {3 B a \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2}-\frac {3 B \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 B \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A b c x}{4 a}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B c x}{2}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2}}{4 a}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b c x}{4 a^{2}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A c}{2}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B c x}{a}+\frac {9 \sqrt {c \,x^{2}+b x +a}\, B b}{4}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A c}{2 a}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{2}}{4 a^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b}{a}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A b}{4 a^{2} x}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B}{a x}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A}{2 a \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(3/2)/x^3,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 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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