3.9.70 \(\int \frac {(A+B x) (a+b x+c x^2)^{5/2}}{x^6} \, dx\)
Optimal. Leaf size=346 \[ \frac {\left (a+b x+c x^2\right )^{3/2} \left (4 a \left (-16 a A c-10 a b B+3 A b^2\right )-3 x \left (10 a B \left (4 a c+b^2\right )-A \left (3 b^3-20 a b c\right )\right )\right )}{192 a^2 x^3}+\frac {\sqrt {a+b x+c x^2} \left (-A \left (128 a^2 c^2-28 a b^2 c+3 b^4\right )+2 c x \left (10 a B \left (12 a c+b^2\right )-A \left (3 b^3-28 a b c\right )\right )+10 a b B \left (b^2-20 a c\right )\right )}{128 a^2 x}+\frac {\left (10 a B \left (-48 a^2 c^2-24 a b^2 c+b^4\right )-A \left (240 a^2 b c^2-40 a b^3 c+3 b^5\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{256 a^{5/2}}+\frac {1}{2} c^{3/2} (2 A c+5 b B) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )-\frac {\left (a+b x+c x^2\right )^{5/2} (5 x (2 a B+A b)+8 a A)}{40 a x^5} \]
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Rubi [A] time = 0.49, antiderivative size = 346, normalized size of antiderivative = 1.00,
number of steps used = 8, 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} \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-A \left (128 a^2 c^2-28 a b^2 c+3 b^4\right )+2 c x \left (10 a B \left (12 a c+b^2\right )-A \left (3 b^3-28 a b c\right )\right )+10 a b B \left (b^2-20 a c\right )\right )}{128 a^2 x}+\frac {\left (10 a B \left (-48 a^2 c^2-24 a b^2 c+b^4\right )-A \left (240 a^2 b c^2-40 a b^3 c+3 b^5\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{256 a^{5/2}}+\frac {\left (a+b x+c x^2\right )^{3/2} \left (4 a \left (-16 a A c-10 a b B+3 A b^2\right )-3 x \left (10 a B \left (4 a c+b^2\right )-A \left (3 b^3-20 a b c\right )\right )\right )}{192 a^2 x^3}+\frac {1}{2} c^{3/2} (2 A c+5 b B) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )-\frac {\left (a+b x+c x^2\right )^{5/2} (5 x (2 a B+A b)+8 a A)}{40 a x^5} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^6,x]
[Out]
((10*a*b*B*(b^2 - 20*a*c) - A*(3*b^4 - 28*a*b^2*c + 128*a^2*c^2) + 2*c*(10*a*B*(b^2 + 12*a*c) - A*(3*b^3 - 28*
a*b*c))*x)*Sqrt[a + b*x + c*x^2])/(128*a^2*x) + ((4*a*(3*A*b^2 - 10*a*b*B - 16*a*A*c) - 3*(10*a*B*(b^2 + 4*a*c
) - A*(3*b^3 - 20*a*b*c))*x)*(a + b*x + c*x^2)^(3/2))/(192*a^2*x^3) - ((8*a*A + 5*(A*b + 2*a*B)*x)*(a + b*x +
c*x^2)^(5/2))/(40*a*x^5) + ((10*a*B*(b^4 - 24*a*b^2*c - 48*a^2*c^2) - A*(3*b^5 - 40*a*b^3*c + 240*a^2*b*c^2))*
ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(256*a^(5/2)) + (c^(3/2)*(5*b*B + 2*A*c)*ArcTanh[(b +
2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/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 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]
Rubi steps
\begin {align*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^6} \, dx &=-\frac {(8 a A+5 (A b+2 a B) x) \left (a+b x+c x^2\right )^{5/2}}{40 a x^5}-\frac {\int \frac {\left (\frac {1}{2} \left (3 A b^2-10 a b B-16 a A c\right )-(A b+10 a B) c x\right ) \left (a+b x+c x^2\right )^{3/2}}{x^4} \, dx}{8 a}\\ &=\frac {\left (4 a \left (3 A b^2-10 a b B-16 a A c\right )-3 \left (10 a B \left (b^2+4 a c\right )-A \left (3 b^3-20 a b c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 a^2 x^3}-\frac {(8 a A+5 (A b+2 a B) x) \left (a+b x+c x^2\right )^{5/2}}{40 a x^5}+\frac {\int \frac {\left (\frac {1}{4} \left (-10 a b B \left (b^2-20 a c\right )+4 A \left (\frac {3 b^4}{4}-7 a b^2 c+32 a^2 c^2\right )\right )+\frac {1}{2} c \left (10 a B \left (b^2+12 a c\right )-A \left (3 b^3-28 a b c\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{x^2} \, dx}{32 a^2}\\ &=\frac {\left (10 a b B \left (b^2-20 a c\right )-A \left (3 b^4-28 a b^2 c+128 a^2 c^2\right )+2 c \left (10 a B \left (b^2+12 a c\right )-A \left (3 b^3-28 a b c\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{128 a^2 x}+\frac {\left (4 a \left (3 A b^2-10 a b B-16 a A c\right )-3 \left (10 a B \left (b^2+4 a c\right )-A \left (3 b^3-20 a b c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 a^2 x^3}-\frac {(8 a A+5 (A b+2 a B) x) \left (a+b x+c x^2\right )^{5/2}}{40 a x^5}-\frac {\int \frac {\frac {1}{4} \left (10 a B \left (b^4-24 a b^2 c-48 a^2 c^2\right )-4 A \left (\frac {3 b^5}{4}-10 a b^3 c+60 a^2 b c^2\right )\right )-32 a^2 c^2 (5 b B+2 A c) x}{x \sqrt {a+b x+c x^2}} \, dx}{64 a^2}\\ &=\frac {\left (10 a b B \left (b^2-20 a c\right )-A \left (3 b^4-28 a b^2 c+128 a^2 c^2\right )+2 c \left (10 a B \left (b^2+12 a c\right )-A \left (3 b^3-28 a b c\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{128 a^2 x}+\frac {\left (4 a \left (3 A b^2-10 a b B-16 a A c\right )-3 \left (10 a B \left (b^2+4 a c\right )-A \left (3 b^3-20 a b c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 a^2 x^3}-\frac {(8 a A+5 (A b+2 a B) x) \left (a+b x+c x^2\right )^{5/2}}{40 a x^5}+\frac {1}{2} \left (c^2 (5 b B+2 A c)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx-\frac {\left (10 a B \left (b^4-24 a b^2 c-48 a^2 c^2\right )-A \left (3 b^5-40 a b^3 c+240 a^2 b c^2\right )\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{256 a^2}\\ &=\frac {\left (10 a b B \left (b^2-20 a c\right )-A \left (3 b^4-28 a b^2 c+128 a^2 c^2\right )+2 c \left (10 a B \left (b^2+12 a c\right )-A \left (3 b^3-28 a b c\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{128 a^2 x}+\frac {\left (4 a \left (3 A b^2-10 a b B-16 a A c\right )-3 \left (10 a B \left (b^2+4 a c\right )-A \left (3 b^3-20 a b c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 a^2 x^3}-\frac {(8 a A+5 (A b+2 a B) x) \left (a+b x+c x^2\right )^{5/2}}{40 a x^5}+\left (c^2 (5 b B+2 A c)\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 {\left (10 a B \left (b^4-24 a b^2 c-48 a^2 c^2\right )-A \left (3 b^5-40 a b^3 c+240 a^2 b c^2\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 )}{128 a^2}\\ &=\frac {\left (10 a b B \left (b^2-20 a c\right )-A \left (3 b^4-28 a b^2 c+128 a^2 c^2\right )+2 c \left (10 a B \left (b^2+12 a c\right )-A \left (3 b^3-28 a b c\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{128 a^2 x}+\frac {\left (4 a \left (3 A b^2-10 a b B-16 a A c\right )-3 \left (10 a B \left (b^2+4 a c\right )-A \left (3 b^3-20 a b c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 a^2 x^3}-\frac {(8 a A+5 (A b+2 a B) x) \left (a+b x+c x^2\right )^{5/2}}{40 a x^5}+\frac {\left (10 a B \left (b^4-24 a b^2 c-48 a^2 c^2\right )-A \left (3 b^5-40 a b^3 c+240 a^2 b c^2\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{256 a^{5/2}}+\frac {1}{2} c^{3/2} (5 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*}
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Mathematica [A] time = 0.85, size = 289, normalized size = 0.84 \begin {gather*} -\frac {\left (A \left (240 a^2 b c^2-40 a b^3 c+3 b^5\right )+10 a B \left (48 a^2 c^2+24 a b^2 c-b^4\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{256 a^{5/2}}-\frac {\sqrt {a+x (b+c x)} \left (96 a^4 (4 A+5 B x)+16 a^3 x (A (63 b+88 c x)+5 B x (17 b+27 c x))+4 a^2 x^2 \left (2 A \left (93 b^2+311 b c x+368 c^2 x^2\right )+5 B x \left (59 b^2+278 b c x-96 c^2 x^2\right )\right )+30 a b^2 x^3 (A (b+18 c x)+5 b B x)-45 A b^4 x^4\right )}{1920 a^2 x^5}+\frac {1}{2} c^{3/2} (2 A c+5 b B) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^6,x]
[Out]
-1/1920*(Sqrt[a + x*(b + c*x)]*(-45*A*b^4*x^4 + 96*a^4*(4*A + 5*B*x) + 30*a*b^2*x^3*(5*b*B*x + A*(b + 18*c*x))
+ 16*a^3*x*(5*B*x*(17*b + 27*c*x) + A*(63*b + 88*c*x)) + 4*a^2*x^2*(5*B*x*(59*b^2 + 278*b*c*x - 96*c^2*x^2) +
2*A*(93*b^2 + 311*b*c*x + 368*c^2*x^2))))/(a^2*x^5) - ((10*a*B*(-b^4 + 24*a*b^2*c + 48*a^2*c^2) + A*(3*b^5 -
40*a*b^3*c + 240*a^2*b*c^2))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/(256*a^(5/2)) + (c^(3/2)*
(5*b*B + 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/2
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IntegrateAlgebraic [A] time = 4.11, size = 327, normalized size = 0.95 \begin {gather*} \frac {\left (-480 a^3 B c^2-240 a^2 A b c^2-240 a^2 b^2 B c+40 a A b^3 c+10 a b^4 B-3 A b^5\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x+c x^2}-\sqrt {c} x}{\sqrt {a}}\right )}{128 a^{5/2}}+\frac {\sqrt {a+b x+c x^2} \left (-384 a^4 A-480 a^4 B x-1008 a^3 A b x-1408 a^3 A c x^2-1360 a^3 b B x^2-2160 a^3 B c x^3-744 a^2 A b^2 x^2-2488 a^2 A b c x^3-2944 a^2 A c^2 x^4-1180 a^2 b^2 B x^3-5560 a^2 b B c x^4+1920 a^2 B c^2 x^5-30 a A b^3 x^3-540 a A b^2 c x^4-150 a b^3 B x^4+45 A b^4 x^4\right )}{1920 a^2 x^5}+\frac {1}{2} \left (-2 A c^{5/2}-5 b B c^{3/2}\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^6,x]
[Out]
(Sqrt[a + b*x + c*x^2]*(-384*a^4*A - 1008*a^3*A*b*x - 480*a^4*B*x - 744*a^2*A*b^2*x^2 - 1360*a^3*b*B*x^2 - 140
8*a^3*A*c*x^2 - 30*a*A*b^3*x^3 - 1180*a^2*b^2*B*x^3 - 2488*a^2*A*b*c*x^3 - 2160*a^3*B*c*x^3 + 45*A*b^4*x^4 - 1
50*a*b^3*B*x^4 - 540*a*A*b^2*c*x^4 - 5560*a^2*b*B*c*x^4 - 2944*a^2*A*c^2*x^4 + 1920*a^2*B*c^2*x^5))/(1920*a^2*
x^5) + ((-3*A*b^5 + 10*a*b^4*B + 40*a*A*b^3*c - 240*a^2*b^2*B*c - 240*a^2*A*b*c^2 - 480*a^3*B*c^2)*ArcTanh[(-(
Sqrt[c]*x) + Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(128*a^(5/2)) + ((-5*b*B*c^(3/2) - 2*A*c^(5/2))*Log[b + 2*c*x -
2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/2
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fricas [A] time = 6.85, size = 1445, normalized size = 4.18
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)^(5/2)/x^6,x, algorithm="fricas")
[Out]
[1/7680*(1920*(5*B*a^3*b*c + 2*A*a^3*c^2)*sqrt(c)*x^5*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) - 15*(10*B*a*b^4 - 3*A*b^5 - 240*(2*B*a^3 + A*a^2*b)*c^2 - 40*(6*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*(1920*B*a^3*c^2*x^5 - 384*A*a^5 - (150*B*a^2*b^3 - 45*A*a*b^4 + 2944*A*a^3*c^2 + 20*(278*B*a^3*b + 27
*A*a^2*b^2)*c)*x^4 - 2*(590*B*a^3*b^2 + 15*A*a^2*b^3 + 4*(270*B*a^4 + 311*A*a^3*b)*c)*x^3 - 8*(170*B*a^4*b + 9
3*A*a^3*b^2 + 176*A*a^4*c)*x^2 - 48*(10*B*a^5 + 21*A*a^4*b)*x)*sqrt(c*x^2 + b*x + a))/(a^3*x^5), -1/7680*(3840
*(5*B*a^3*b*c + 2*A*a^3*c^2)*sqrt(-c)*x^5*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)) + 15*(10*B*a*b^4 - 3*A*b^5 - 240*(2*B*a^3 + A*a^2*b)*c^2 - 40*(6*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*(1920*B*a^
3*c^2*x^5 - 384*A*a^5 - (150*B*a^2*b^3 - 45*A*a*b^4 + 2944*A*a^3*c^2 + 20*(278*B*a^3*b + 27*A*a^2*b^2)*c)*x^4
- 2*(590*B*a^3*b^2 + 15*A*a^2*b^3 + 4*(270*B*a^4 + 311*A*a^3*b)*c)*x^3 - 8*(170*B*a^4*b + 93*A*a^3*b^2 + 176*A
*a^4*c)*x^2 - 48*(10*B*a^5 + 21*A*a^4*b)*x)*sqrt(c*x^2 + b*x + a))/(a^3*x^5), -1/3840*(15*(10*B*a*b^4 - 3*A*b^
5 - 240*(2*B*a^3 + A*a^2*b)*c^2 - 40*(6*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)) - 960*(5*B*a^3*b*c + 2*A*a^3*c^2)*sqrt(c)*x^5*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) - 2*(1920*B*a^3*c^2*x^5 - 384*A*a^5 - (150
*B*a^2*b^3 - 45*A*a*b^4 + 2944*A*a^3*c^2 + 20*(278*B*a^3*b + 27*A*a^2*b^2)*c)*x^4 - 2*(590*B*a^3*b^2 + 15*A*a^
2*b^3 + 4*(270*B*a^4 + 311*A*a^3*b)*c)*x^3 - 8*(170*B*a^4*b + 93*A*a^3*b^2 + 176*A*a^4*c)*x^2 - 48*(10*B*a^5 +
21*A*a^4*b)*x)*sqrt(c*x^2 + b*x + a))/(a^3*x^5), -1/3840*(15*(10*B*a*b^4 - 3*A*b^5 - 240*(2*B*a^3 + A*a^2*b)*
c^2 - 40*(6*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)) + 1920*(5*B*a^3*b*c + 2*A*a^3*c^2)*sqrt(-c)*x^5*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*(1920*B*a^3*c^2*x^5 - 384*A*a^5 - (150*B*a^2*b^3 - 45*A*a*b^4 + 2944*A*
a^3*c^2 + 20*(278*B*a^3*b + 27*A*a^2*b^2)*c)*x^4 - 2*(590*B*a^3*b^2 + 15*A*a^2*b^3 + 4*(270*B*a^4 + 311*A*a^3*
b)*c)*x^3 - 8*(170*B*a^4*b + 93*A*a^3*b^2 + 176*A*a^4*c)*x^2 - 48*(10*B*a^5 + 21*A*a^4*b)*x)*sqrt(c*x^2 + b*x
+ a))/(a^3*x^5)]
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giac [B] time = 0.99, size = 1526, normalized size = 4.41
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)^(5/2)/x^6,x, algorithm="giac")
[Out]
sqrt(c*x^2 + b*x + a)*B*c^2 - 1/2*(5*B*b*c^(5/2) + 2*A*c^(7/2))*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))
*c - b*sqrt(c)))/c - 1/128*(10*B*a*b^4 - 3*A*b^5 - 240*B*a^2*b^2*c + 40*A*a*b^3*c - 480*B*a^3*c^2 - 240*A*a^2*
b*c^2)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a^2) + 1/1920*(150*(sqrt(c)*x - sqrt(c*
x^2 + b*x + a))^9*B*a*b^4 - 45*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*b^5 + 7920*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))^9*B*a^2*b^2*c + 600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a*b^3*c + 4320*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))^9*B*a^3*c^2 + 7920*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a^2*b*c^2 + 3840*(sqrt(c)*x - sqrt(c*x^2 +
b*x + a))^8*B*a^2*b^3*sqrt(c) + 23040*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*B*a^3*b*c^(3/2) + 11520*(sqrt(c)*
x - sqrt(c*x^2 + b*x + a))^8*A*a^2*b^2*c^(3/2) + 11520*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*A*a^3*c^(5/2) + 5
80*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^2*b^4 + 210*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a*b^5 - 13920
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^3*b^2*c + 6160*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^2*b^3*c -
4800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^4*c^2 - 2400*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^3*b*c^2
- 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^3*b^3*sqrt(c) + 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*A*
a^2*b^4*sqrt(c) - 57600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^4*b*c^(3/2) - 23040*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^6*A*a^4*c^(5/2) - 1280*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^3*b^4 + 384*(sqrt(c)*x - sqrt(c*x
^2 + b*x + a))^5*A*a^2*b^5 + 19200*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^4*b^2*c + 6400*(sqrt(c)*x - sqrt(
c*x^2 + b*x + a))^5*A*a^3*b^3*c + 19200*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^4*b*c^2 + 70400*(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))^4*B*a^5*b*c^(3/2) + 19200*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^4*b^2*c^(3/2) + 35
840*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^5*c^(5/2) + 700*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^4*b^4
- 210*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^3*b^5 - 16800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^5*b^2*
c + 2800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^4*b^3*c + 4800*(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 - 44800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a
^6*b*c^(3/2) - 17920*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^6*c^(5/2) - 150*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))*B*a^5*b^4 + 45*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^4*b^5 + 3600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B
*a^6*b^2*c - 600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^5*b^3*c - 4320*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*
a^7*c^2 + 3600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^6*b*c^2 + 8960*B*a^7*b*c^(3/2) + 5888*A*a^7*c^(5/2))/((
(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^5*a^2)
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maple [B] time = 0.07, size = 1371, normalized size = 3.96
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^6,x)
[Out]
-21/320*A/a^4*b^3*c*(c*x^2+b*x+a)^(5/2)+15/8*B*c^2*(c*x^2+b*x+a)^(1/2)+A*c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2
+b*x+a)^(1/2))+19/240*A/a^3*b*c/x^2*(c*x^2+b*x+a)^(7/2)-13/96*A/a^3*b^2*c^2*(c*x^2+b*x+a)^(3/2)*x+35/48*B/a^2*
b*c^2*(c*x^2+b*x+a)^(3/2)*x+25/16*B/a*b*c^2*(c*x^2+b*x+a)^(1/2)*x-7/32*A/a^2*b^2*c^2*(c*x^2+b*x+a)^(1/2)*x+11/
160*A/a^4*b^2*c/x*(c*x^2+b*x+a)^(7/2)+3/128*A/a^3*b^4*(c*x^2+b*x+a)^(1/2)*x*c-11/160*A/a^4*b^2*c^2*(c*x^2+b*x+
a)^(5/2)*x+1/128*A/a^4*b^4*c*(c*x^2+b*x+a)^(3/2)*x+3/640*A/a^5*b^4*c*(c*x^2+b*x+a)^(5/2)*x-19/48*B/a^3*b*c/x*(
c*x^2+b*x+a)^(7/2)+19/48*B/a^3*b*c^2*(c*x^2+b*x+a)^(5/2)*x-5/64*B/a^2*b^3*(c*x^2+b*x+a)^(1/2)*x*c-5/192*B/a^3*
b^3*c*(c*x^2+b*x+a)^(3/2)*x-1/64*B/a^4*b^3*c*(c*x^2+b*x+a)^(5/2)*x+109/240*A/a^3*b*c^2*(c*x^2+b*x+a)^(5/2)-5/6
4*B/a^2*b^4*(c*x^2+b*x+a)^(1/2)-5/192*B/a^3*b^4*(c*x^2+b*x+a)^(3/2)-1/64*B/a^4*b^4*(c*x^2+b*x+a)^(5/2)+5/2*B*b
*c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3/8*B*c^2/a^2*(c*x^2+b*x+a)^(5/2)+5/8*B*c^2/a*(c*x^2+b*x+
a)^(3/2)-1/5*A/a/x^5*(c*x^2+b*x+a)^(7/2)-3/256*A/a^(5/2)*b^5*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+3/1
28*A/a^3*b^5*(c*x^2+b*x+a)^(1/2)+1/128*A/a^4*b^5*(c*x^2+b*x+a)^(3/2)+3/640*A/a^5*b^5*(c*x^2+b*x+a)^(5/2)+5/128
*B/a^(3/2)*b^4*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-1/4*B/a/x^4*(c*x^2+b*x+a)^(7/2)-15/8*B*c^2*a^(1/2
)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+23/16*A/a*b*c^2*(c*x^2+b*x+a)^(1/2)+31/48*A/a^2*b*c^2*(c*x^2+b
*x+a)^(3/2)+3/40*A/a^2*b/x^4*(c*x^2+b*x+a)^(7/2)-23/192*A/a^3*b^3*c*(c*x^2+b*x+a)^(3/2)-1/320*A/a^4*b^3/x^2*(c
*x^2+b*x+a)^(7/2)-17/64*A/a^2*b^3*c*(c*x^2+b*x+a)^(1/2)-1/80*A/a^3*b^2/x^3*(c*x^2+b*x+a)^(7/2)-3/640*A/a^5*b^4
/x*(c*x^2+b*x+a)^(7/2)-2/15*A*c/a^2/x^3*(c*x^2+b*x+a)^(7/2)+A*c^3/a*(c*x^2+b*x+a)^(1/2)*x-8/15*A*c^2/a^3/x*(c*
x^2+b*x+a)^(7/2)+2/3*A*c^3/a^2*(c*x^2+b*x+a)^(3/2)*x+8/15*A*c^3/a^3*(c*x^2+b*x+a)^(5/2)*x-15/16*A/a^(1/2)*b*c^
2*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+5/32*A/a^(3/2)*b^3*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2)
)/x)+65/96*B/a^2*b^2*c*(c*x^2+b*x+a)^(3/2)+1/96*B/a^3*b^2/x^2*(c*x^2+b*x+a)^(7/2)+55/32*B/a*b^2*c*(c*x^2+b*x+a
)^(1/2)+1/24*B/a^2*b/x^3*(c*x^2+b*x+a)^(7/2)+1/64*B/a^4*b^3/x*(c*x^2+b*x+a)^(7/2)+37/96*B/a^3*b^2*c*(c*x^2+b*x
+a)^(5/2)-3/8*B*c/a^2/x^2*(c*x^2+b*x+a)^(7/2)-15/16*B/a^(1/2)*b^2*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))
/x)
________________________________________________________________________________________
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)^(5/2)/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?
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{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)^(5/2))/x^6,x)
[Out]
int(((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^6, x)
________________________________________________________________________________________
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 {5}{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)**(5/2)/x**6,x)
[Out]
Integral((A + B*x)*(a + b*x + c*x**2)**(5/2)/x**6, x)
________________________________________________________________________________________