3.9.57 \(\int \frac {(A+B x) (a+b x+c x^2)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=170 \[ \frac {3 \left (b^2-4 a c\right )^2 (A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{256 a^{7/2}}-\frac {3 \left (b^2-4 a c\right ) (2 a+b x) (A b-2 a B) \sqrt {a+b x+c x^2}}{128 a^3 x^2}+\frac {(2 a+b x) (A b-2 a B) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac {A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5} \]

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Rubi [A]  time = 0.10, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {806, 720, 724, 206} \begin {gather*} -\frac {3 \left (b^2-4 a c\right ) (2 a+b x) (A b-2 a B) \sqrt {a+b x+c x^2}}{128 a^3 x^2}+\frac {3 \left (b^2-4 a c\right )^2 (A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{256 a^{7/2}}+\frac {(2 a+b x) (A b-2 a B) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac {A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^(3/2))/x^6,x]

[Out]

(-3*(A*b - 2*a*B)*(b^2 - 4*a*c)*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(128*a^3*x^2) + ((A*b - 2*a*B)*(2*a + b*x)*
(a + b*x + c*x^2)^(3/2))/(16*a^2*x^4) - (A*(a + b*x + c*x^2)^(5/2))/(5*a*x^5) + (3*(A*b - 2*a*B)*(b^2 - 4*a*c)
^2*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(256*a^(7/2))

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 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 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 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x^6} \, dx &=-\frac {A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5}-\frac {(A b-2 a B) \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^5} \, dx}{2 a}\\ &=\frac {(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac {A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5}+\frac {\left (3 (A b-2 a B) \left (b^2-4 a c\right )\right ) \int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx}{32 a^2}\\ &=-\frac {3 (A b-2 a B) \left (b^2-4 a c\right ) (2 a+b x) \sqrt {a+b x+c x^2}}{128 a^3 x^2}+\frac {(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac {A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5}-\frac {\left (3 (A b-2 a B) \left (b^2-4 a c\right )^2\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{256 a^3}\\ &=-\frac {3 (A b-2 a B) \left (b^2-4 a c\right ) (2 a+b x) \sqrt {a+b x+c x^2}}{128 a^3 x^2}+\frac {(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac {A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5}+\frac {\left (3 (A b-2 a B) \left (b^2-4 a c\right )^2\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 )}{128 a^3}\\ &=-\frac {3 (A b-2 a B) \left (b^2-4 a c\right ) (2 a+b x) \sqrt {a+b x+c x^2}}{128 a^3 x^2}+\frac {(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac {A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5}+\frac {3 (A b-2 a B) \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{256 a^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 157, normalized size = 0.92 \begin {gather*} \frac {(A b-2 a B) \left (16 a^{3/2} (2 a+b x) (a+x (b+c x))^{3/2}-3 x^2 \left (b^2-4 a c\right ) \left (2 \sqrt {a} (2 a+b x) \sqrt {a+x (b+c x)}-x^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )\right )\right )}{256 a^{7/2} x^4}-\frac {A (a+x (b+c x))^{5/2}}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^(3/2))/x^6,x]

[Out]

-1/5*(A*(a + x*(b + c*x))^(5/2))/(a*x^5) + ((A*b - 2*a*B)*(16*a^(3/2)*(2*a + b*x)*(a + x*(b + c*x))^(3/2) - 3*
(b^2 - 4*a*c)*x^2*(2*Sqrt[a]*(2*a + b*x)*Sqrt[a + x*(b + c*x)] - (b^2 - 4*a*c)*x^2*ArcTanh[(2*a + b*x)/(2*Sqrt
[a]*Sqrt[a + x*(b + c*x)])])))/(256*a^(7/2)*x^4)

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IntegrateAlgebraic [A]  time = 2.51, size = 300, normalized size = 1.76 \begin {gather*} \frac {3 \left (8 a A b c^2+8 a b^2 B c-4 A b^3 c+b^4 (-B)\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x+c x^2}-\sqrt {c} x}{\sqrt {a}}\right )}{64 a^{5/2}}+\frac {3 \left (32 a^3 B c^2-A b^5\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )}{128 a^{7/2}}+\frac {\sqrt {a+b x+c x^2} \left (-128 a^4 A-160 a^4 B x-176 a^3 A b x-256 a^3 A c x^2-240 a^3 b B x^2-400 a^3 B c x^3-8 a^2 A b^2 x^2-56 a^2 A b c x^3-128 a^2 A c^2 x^4-20 a^2 b^2 B x^3-200 a^2 b B c x^4+10 a A b^3 x^3+100 a A b^2 c x^4+30 a b^3 B x^4-15 A b^4 x^4\right )}{640 a^3 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a + b*x + c*x^2)^(3/2))/x^6,x]

[Out]

(Sqrt[a + b*x + c*x^2]*(-128*a^4*A - 176*a^3*A*b*x - 160*a^4*B*x - 8*a^2*A*b^2*x^2 - 240*a^3*b*B*x^2 - 256*a^3
*A*c*x^2 + 10*a*A*b^3*x^3 - 20*a^2*b^2*B*x^3 - 56*a^2*A*b*c*x^3 - 400*a^3*B*c*x^3 - 15*A*b^4*x^4 + 30*a*b^3*B*
x^4 + 100*a*A*b^2*c*x^4 - 200*a^2*b*B*c*x^4 - 128*a^2*A*c^2*x^4))/(640*a^3*x^5) + (3*(-(A*b^5) + 32*a^3*B*c^2)
*ArcTanh[(Sqrt[c]*x - Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(128*a^(7/2)) + (3*(-(b^4*B) - 4*A*b^3*c + 8*a*b^2*B*c
+ 8*a*A*b*c^2)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(64*a^(5/2))

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fricas [A]  time = 1.93, size = 555, normalized size = 3.26 \begin {gather*} \left [\frac {15 \, {\left (2 \, B a b^{4} - A b^{5} + 16 \, {\left (2 \, B a^{3} - A a^{2} b\right )} c^{2} - 8 \, {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} c\right )} \sqrt {a} x^{5} \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 (128 \, A a^{5} - {\left (30 \, B a^{2} b^{3} - 15 \, A a b^{4} - 128 \, A a^{3} c^{2} - 100 \, {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} c\right )} x^{4} + 2 \, {\left (10 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3} + 4 \, {\left (50 \, B a^{4} + 7 \, A a^{3} b\right )} c\right )} x^{3} + 8 \, {\left (30 \, B a^{4} b + A a^{3} b^{2} + 32 \, A a^{4} c\right )} x^{2} + 16 \, {\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2560 \, a^{4} x^{5}}, \frac {15 \, {\left (2 \, B a b^{4} - A b^{5} + 16 \, {\left (2 \, B a^{3} - A a^{2} b\right )} c^{2} - 8 \, {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} c\right )} \sqrt {-a} x^{5} \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 ) - 2 \, {\left (128 \, A a^{5} - {\left (30 \, B a^{2} b^{3} - 15 \, A a b^{4} - 128 \, A a^{3} c^{2} - 100 \, {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} c\right )} x^{4} + 2 \, {\left (10 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3} + 4 \, {\left (50 \, B a^{4} + 7 \, A a^{3} b\right )} c\right )} x^{3} + 8 \, {\left (30 \, B a^{4} b + A a^{3} b^{2} + 32 \, A a^{4} c\right )} x^{2} + 16 \, {\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1280 \, a^{4} x^{5}}\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^6,x, algorithm="fricas")

[Out]

[1/2560*(15*(2*B*a*b^4 - A*b^5 + 16*(2*B*a^3 - A*a^2*b)*c^2 - 8*(2*B*a^2*b^2 - A*a*b^3)*c)*sqrt(a)*x^5*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*(128*A*a^5 - (30*B*
a^2*b^3 - 15*A*a*b^4 - 128*A*a^3*c^2 - 100*(2*B*a^3*b - A*a^2*b^2)*c)*x^4 + 2*(10*B*a^3*b^2 - 5*A*a^2*b^3 + 4*
(50*B*a^4 + 7*A*a^3*b)*c)*x^3 + 8*(30*B*a^4*b + A*a^3*b^2 + 32*A*a^4*c)*x^2 + 16*(10*B*a^5 + 11*A*a^4*b)*x)*sq
rt(c*x^2 + b*x + a))/(a^4*x^5), 1/1280*(15*(2*B*a*b^4 - A*b^5 + 16*(2*B*a^3 - A*a^2*b)*c^2 - 8*(2*B*a^2*b^2 -
A*a*b^3)*c)*sqrt(-a)*x^5*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)) - 2*(1
28*A*a^5 - (30*B*a^2*b^3 - 15*A*a*b^4 - 128*A*a^3*c^2 - 100*(2*B*a^3*b - A*a^2*b^2)*c)*x^4 + 2*(10*B*a^3*b^2 -
5*A*a^2*b^3 + 4*(50*B*a^4 + 7*A*a^3*b)*c)*x^3 + 8*(30*B*a^4*b + A*a^3*b^2 + 32*A*a^4*c)*x^2 + 16*(10*B*a^5 +
11*A*a^4*b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^5)]

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giac [B]  time = 0.29, size = 1357, normalized size = 7.98

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(3/2)/x^6,x, algorithm="giac")

[Out]

3/128*(2*B*a*b^4 - A*b^5 - 16*B*a^2*b^2*c + 8*A*a*b^3*c + 32*B*a^3*c^2 - 16*A*a^2*b*c^2)*arctan(-(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a^3) - 1/640*(30*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a*b^4 - 15
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*b^5 - 240*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a^2*b^2*c + 120*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a*b^3*c - 800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a^3*c^2 - 240*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a^2*b*c^2 - 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*B*a^3*b*c^(3/2) - 12
80*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*A*a^3*c^(5/2) - 140*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^2*b^4 +
70*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a*b^5 - 1440*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^3*b^2*c - 5
60*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^2*b^3*c + 320*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^4*c^2 - 2
720*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^3*b*c^2 - 1280*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^3*b^3*s
qrt(c) + 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^4*b*c^(3/2) - 5120*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))
^6*A*a^3*b^2*c^(3/2) - 128*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^2*b^5 - 2560*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))^5*A*a^3*b^3*c - 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^4*b*c^2 + 1280*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^4*B*a^4*b^3*sqrt(c) - 1280*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^3*b^4*sqrt(c) - 2560*(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))^4*B*a^5*b*c^(3/2) - 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^4*b^2*c^(3/2) -
2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^5*c^(5/2) + 140*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^4*b^
4 - 70*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^3*b^5 + 1440*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^5*b^2*
c - 2000*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^4*b^3*c - 320*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^6*c
^2 - 2400*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^5*b*c^2 + 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^6
*b*c^(3/2) - 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^5*b^2*c^(3/2) - 30*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))*B*a^5*b^4 + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^4*b^5 + 240*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*
a^6*b^2*c - 120*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^5*b^3*c + 800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^
7*c^2 - 1040*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^6*b*c^2 - 256*A*a^7*c^(5/2))/(((sqrt(c)*x - sqrt(c*x^2 +
b*x + a))^2 - a)^5*a^3)

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maple [B]  time = 0.09, size = 978, normalized size = 5.75 \begin {gather*} \frac {3 A b \,c^{2} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {3}{2}}}-\frac {3 A \,b^{3} c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{32 a^{\frac {5}{2}}}+\frac {3 A \,b^{5} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{256 a^{\frac {7}{2}}}-\frac {3 B \,c^{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 B \,b^{2} c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {3}{2}}}-\frac {3 B \,b^{4} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{128 a^{\frac {5}{2}}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2} c^{2} x}{32 a^{3}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{4} c x}{128 a^{4}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, B b \,c^{2} x}{16 a^{2}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B \,b^{3} c x}{64 a^{3}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A b \,c^{2}}{16 a^{2}}+\frac {9 \sqrt {c \,x^{2}+b x +a}\, A \,b^{3} c}{64 a^{3}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{5}}{128 a^{4}}+\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{2} c^{2} x}{32 a^{4}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{4} c x}{128 a^{5}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B \,c^{2}}{8 a}-\frac {9 \sqrt {c \,x^{2}+b x +a}\, B \,b^{2} c}{32 a^{2}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B \,b^{4}}{64 a^{3}}-\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b \,c^{2} x}{16 a^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{3} c x}{64 a^{4}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b \,c^{2}}{16 a^{3}}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{3} c}{64 a^{4}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{5}}{128 a^{5}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,c^{2}}{8 a^{2}}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{2} c}{32 a^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{4}}{64 a^{4}}-\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A \,b^{2} c}{32 a^{4} x}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A \,b^{4}}{128 a^{5} x}+\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B b c}{16 a^{3} x}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B \,b^{3}}{64 a^{4} x}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A b c}{16 a^{3} x^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A \,b^{3}}{64 a^{4} x^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B c}{8 a^{2} x^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B \,b^{2}}{32 a^{3} x^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A \,b^{2}}{16 a^{3} x^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B b}{8 a^{2} x^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A b}{8 a^{2} x^{4}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B}{4 a \,x^{4}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A}{5 a \,x^{5}} \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^6,x)

[Out]

-3/16*B/a^3*b*c^2*(c*x^2+b*x+a)^(3/2)*x-3/16*B/a^2*b*c^2*(c*x^2+b*x+a)^(1/2)*x+3/32*A/a^3*b^2*c^2*(c*x^2+b*x+a
)^(1/2)*x+1/16*A/a^3*b*c/x^2*(c*x^2+b*x+a)^(5/2)+3/32*A/a^4*b^2*c^2*(c*x^2+b*x+a)^(3/2)*x-3/32*A/a^4*b^2*c/x*(
c*x^2+b*x+a)^(5/2)-1/128*A/a^5*b^4*c*(c*x^2+b*x+a)^(3/2)*x-3/128*A/a^4*b^4*c*(c*x^2+b*x+a)^(1/2)*x+1/64*B/a^4*
b^3*c*(c*x^2+b*x+a)^(3/2)*x+3/64*B/a^3*b^3*c*(c*x^2+b*x+a)^(1/2)*x+3/16*B/a^3*b*c/x*(c*x^2+b*x+a)^(5/2)+1/64*A
/a^4*b^3/x^2*(c*x^2+b*x+a)^(5/2)-1/128*A/a^5*b^5*(c*x^2+b*x+a)^(3/2)-3/128*A/a^4*b^5*(c*x^2+b*x+a)^(1/2)+3/256
*A/a^(7/2)*b^5*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-3/8*B*c^2/a^(1/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/
2)*a^(1/2))/x)+1/64*B/a^4*b^4*(c*x^2+b*x+a)^(3/2)+3/64*B/a^3*b^4*(c*x^2+b*x+a)^(1/2)+1/8*B*c^2/a^2*(c*x^2+b*x+
a)^(3/2)+3/8*B*c^2/a*(c*x^2+b*x+a)^(1/2)-1/4*B/a/x^4*(c*x^2+b*x+a)^(5/2)-3/128*B/a^(5/2)*b^4*ln((b*x+2*a+2*(c*
x^2+b*x+a)^(1/2)*a^(1/2))/x)+5/64*A/a^4*b^3*c*(c*x^2+b*x+a)^(3/2)+9/64*A/a^3*b^3*c*(c*x^2+b*x+a)^(1/2)+3/16*A/
a^(3/2)*b*c^2*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-1/16*A/a^3*b*c^2*(c*x^2+b*x+a)^(3/2)-3/16*A/a^2*b*
c^2*(c*x^2+b*x+a)^(1/2)+1/8*A/a^2*b/x^4*(c*x^2+b*x+a)^(5/2)-3/32*A/a^(5/2)*b^3*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(
1/2)*a^(1/2))/x)-1/16*A/a^3*b^2/x^3*(c*x^2+b*x+a)^(5/2)+1/8*B/a^2*b/x^3*(c*x^2+b*x+a)^(5/2)-1/32*B/a^3*b^2/x^2
*(c*x^2+b*x+a)^(5/2)-1/64*B/a^4*b^3/x*(c*x^2+b*x+a)^(5/2)-5/32*B/a^3*b^2*c*(c*x^2+b*x+a)^(3/2)-9/32*B/a^2*b^2*
c*(c*x^2+b*x+a)^(1/2)-1/8*B*c/a^2/x^2*(c*x^2+b*x+a)^(5/2)+3/16*B/a^(3/2)*b^2*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/
2)*a^(1/2))/x)+1/128*A/a^5*b^4/x*(c*x^2+b*x+a)^(5/2)-1/5*A*(c*x^2+b*x+a)^(5/2)/a/x^5

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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^6,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^6} \,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^6,x)

[Out]

int(((A + B*x)*(a + b*x + c*x^2)^(3/2))/x^6, 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^{6}}\, 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**6,x)

[Out]

Integral((A + B*x)*(a + b*x + c*x**2)**(3/2)/x**6, x)

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