3.9.71 \(\int \frac {(A+B x) (a+b x+c x^2)^{5/2}}{x^7} \, dx\)

Optimal. Leaf size=332 \[ -\frac {\left (a+b x+c x^2\right )^{3/2} \left (2 a \left (12 a b B-5 A \left (b^2-4 a c\right )\right )+x \left (4 a B \left (16 a c+3 b^2\right )-5 A \left (b^3-4 a b c\right )\right )\right )}{192 a^2 x^4}+\frac {\left (5 A \left (b^2-4 a c\right )^3-4 a b B \left (240 a^2 c^2-40 a b^2 c+3 b^4\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{1024 a^{7/2}}+\frac {\sqrt {a+b x+c x^2} \left (2 a \left (4 a b B \left (3 b^2-28 a c\right )-5 A \left (b^2-4 a c\right )^2\right )-x \left (5 A b \left (b^2-4 a c\right )^2-4 a B \left (-128 a^2 c^2-28 a b^2 c+3 b^4\right )\right )\right )}{512 a^3 x^2}-\frac {\left (a+b x+c x^2\right )^{5/2} (x (12 a B+5 A b)+10 a A)}{60 a x^6}+B c^{5/2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \]

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Rubi [A]  time = 0.47, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {810, 843, 621, 206, 724} \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (2 a \left (4 a b B \left (3 b^2-28 a c\right )-5 A \left (b^2-4 a c\right )^2\right )-x \left (5 A b \left (b^2-4 a c\right )^2-4 a B \left (-128 a^2 c^2-28 a b^2 c+3 b^4\right )\right )\right )}{512 a^3 x^2}+\frac {\left (5 A \left (b^2-4 a c\right )^3-4 a b B \left (240 a^2 c^2-40 a b^2 c+3 b^4\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{1024 a^{7/2}}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (x \left (4 a B \left (16 a c+3 b^2\right )-5 A \left (b^3-4 a b c\right )\right )+2 a \left (12 a b B-5 A \left (b^2-4 a c\right )\right )\right )}{192 a^2 x^4}-\frac {\left (a+b x+c x^2\right )^{5/2} (x (12 a B+5 A b)+10 a A)}{60 a x^6}+B c^{5/2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^7,x]

[Out]

((2*a*(4*a*b*B*(3*b^2 - 28*a*c) - 5*A*(b^2 - 4*a*c)^2) - (5*A*b*(b^2 - 4*a*c)^2 - 4*a*B*(3*b^4 - 28*a*b^2*c -
128*a^2*c^2))*x)*Sqrt[a + b*x + c*x^2])/(512*a^3*x^2) - ((2*a*(12*a*b*B - 5*A*(b^2 - 4*a*c)) + (4*a*B*(3*b^2 +
16*a*c) - 5*A*(b^3 - 4*a*b*c))*x)*(a + b*x + c*x^2)^(3/2))/(192*a^2*x^4) - ((10*a*A + (5*A*b + 12*a*B)*x)*(a
+ b*x + c*x^2)^(5/2))/(60*a*x^6) + ((5*A*(b^2 - 4*a*c)^3 - 4*a*b*B*(3*b^4 - 40*a*b^2*c + 240*a^2*c^2))*ArcTanh
[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(1024*a^(7/2)) + B*c^(5/2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr
t[a + b*x + c*x^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 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^7} \, dx &=-\frac {(10 a A+(5 A b+12 a B) x) \left (a+b x+c x^2\right )^{5/2}}{60 a x^6}-\frac {\int \frac {\left (\frac {1}{2} \left (-12 a b B+5 A \left (b^2-4 a c\right )\right )-12 a B c x\right ) \left (a+b x+c x^2\right )^{3/2}}{x^5} \, dx}{12 a}\\ &=-\frac {\left (2 a \left (12 a b B-5 A \left (b^2-4 a c\right )\right )+\left (4 a B \left (3 b^2+16 a c\right )-5 A \left (b^3-4 a b c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 a^2 x^4}-\frac {(10 a A+(5 A b+12 a B) x) \left (a+b x+c x^2\right )^{5/2}}{60 a x^6}+\frac {\int \frac {\left (-\frac {3}{4} \left (4 a b B \left (3 b^2-28 a c\right )-5 A \left (b^2-4 a c\right )^2\right )+96 a^2 B c^2 x\right ) \sqrt {a+b x+c x^2}}{x^3} \, dx}{96 a^2}\\ &=\frac {\left (2 a \left (4 a b B \left (3 b^2-28 a c\right )-5 A \left (b^2-4 a c\right )^2\right )-\left (5 A b \left (b^2-4 a c\right )^2-4 a B \left (3 b^4-28 a b^2 c-128 a^2 c^2\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{512 a^3 x^2}-\frac {\left (2 a \left (12 a b B-5 A \left (b^2-4 a c\right )\right )+\left (4 a B \left (3 b^2+16 a c\right )-5 A \left (b^3-4 a b c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 a^2 x^4}-\frac {(10 a A+(5 A b+12 a B) x) \left (a+b x+c x^2\right )^{5/2}}{60 a x^6}-\frac {\int \frac {\frac {3}{8} \left (5 A \left (b^2-4 a c\right )^3-4 a b B \left (3 b^4-40 a b^2 c+240 a^2 c^2\right )\right )-384 a^3 B c^3 x}{x \sqrt {a+b x+c x^2}} \, dx}{384 a^3}\\ &=\frac {\left (2 a \left (4 a b B \left (3 b^2-28 a c\right )-5 A \left (b^2-4 a c\right )^2\right )-\left (5 A b \left (b^2-4 a c\right )^2-4 a B \left (3 b^4-28 a b^2 c-128 a^2 c^2\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{512 a^3 x^2}-\frac {\left (2 a \left (12 a b B-5 A \left (b^2-4 a c\right )\right )+\left (4 a B \left (3 b^2+16 a c\right )-5 A \left (b^3-4 a b c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 a^2 x^4}-\frac {(10 a A+(5 A b+12 a B) x) \left (a+b x+c x^2\right )^{5/2}}{60 a x^6}+\left (B c^3\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx-\frac {\left (5 A \left (b^2-4 a c\right )^3-4 a b B \left (3 b^4-40 a b^2 c+240 a^2 c^2\right )\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{1024 a^3}\\ &=\frac {\left (2 a \left (4 a b B \left (3 b^2-28 a c\right )-5 A \left (b^2-4 a c\right )^2\right )-\left (5 A b \left (b^2-4 a c\right )^2-4 a B \left (3 b^4-28 a b^2 c-128 a^2 c^2\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{512 a^3 x^2}-\frac {\left (2 a \left (12 a b B-5 A \left (b^2-4 a c\right )\right )+\left (4 a B \left (3 b^2+16 a c\right )-5 A \left (b^3-4 a b c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 a^2 x^4}-\frac {(10 a A+(5 A b+12 a B) x) \left (a+b x+c x^2\right )^{5/2}}{60 a x^6}+\left (2 B c^3\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 (5 A \left (b^2-4 a c\right )^3-4 a b B \left (3 b^4-40 a b^2 c+240 a^2 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 )}{512 a^3}\\ &=\frac {\left (2 a \left (4 a b B \left (3 b^2-28 a c\right )-5 A \left (b^2-4 a c\right )^2\right )-\left (5 A b \left (b^2-4 a c\right )^2-4 a B \left (3 b^4-28 a b^2 c-128 a^2 c^2\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{512 a^3 x^2}-\frac {\left (2 a \left (12 a b B-5 A \left (b^2-4 a c\right )\right )+\left (4 a B \left (3 b^2+16 a c\right )-5 A \left (b^3-4 a b c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 a^2 x^4}-\frac {(10 a A+(5 A b+12 a B) x) \left (a+b x+c x^2\right )^{5/2}}{60 a x^6}+\frac {\left (5 A \left (b^2-4 a c\right )^3-4 a b B \left (3 b^4-40 a b^2 c+240 a^2 c^2\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{1024 a^{7/2}}+B c^{5/2} \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 = 0.98, size = 308, normalized size = 0.93 \begin {gather*} \frac {\left (5 A \left (b^2-4 a c\right )^3-4 a b B \left (240 a^2 c^2-40 a b^2 c+3 b^4\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{1024 a^{7/2}}-\frac {\sqrt {a+x (b+c x)} \left (256 a^5 (5 A+6 B x)+64 a^4 x (A (50 b+65 c x)+B x (63 b+88 c x))+16 a^3 x^2 \left (15 A \left (9 b^2+26 b c x+22 c^2 x^2\right )+2 B x \left (93 b^2+311 b c x+368 c^2 x^2\right )\right )+40 a^2 b x^3 \left (A \left (b^2+12 b c x+66 c^2 x^2\right )+3 b B x (b+18 c x)\right )-10 a b^3 x^4 (5 A (b+16 c x)+18 b B x)+75 A b^5 x^5\right )}{7680 a^3 x^6}+B c^{5/2} \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^7,x]

[Out]

-1/7680*(Sqrt[a + x*(b + c*x)]*(75*A*b^5*x^5 + 256*a^5*(5*A + 6*B*x) - 10*a*b^3*x^4*(18*b*B*x + 5*A*(b + 16*c*
x)) + 64*a^4*x*(A*(50*b + 65*c*x) + B*x*(63*b + 88*c*x)) + 40*a^2*b*x^3*(3*b*B*x*(b + 18*c*x) + A*(b^2 + 12*b*
c*x + 66*c^2*x^2)) + 16*a^3*x^2*(15*A*(9*b^2 + 26*b*c*x + 22*c^2*x^2) + 2*B*x*(93*b^2 + 311*b*c*x + 368*c^2*x^
2))))/(a^3*x^6) + ((5*A*(b^2 - 4*a*c)^3 - 4*a*b*B*(3*b^4 - 40*a*b^2*c + 240*a^2*c^2))*ArcTanh[(2*a + b*x)/(2*S
qrt[a]*Sqrt[a + x*(b + c*x)])])/(1024*a^(7/2)) + B*c^(5/2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)
])]

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IntegrateAlgebraic [A]  time = 3.62, size = 388, normalized size = 1.17 \begin {gather*} \frac {\left (-320 a^3 A c^3-960 a^3 b B c^2+240 a^2 A b^2 c^2+160 a^2 b^3 B c-60 a A b^4 c-12 a b^5 B+5 A b^6\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x+c x^2}-\sqrt {c} x}{\sqrt {a}}\right )}{512 a^{7/2}}+\frac {\sqrt {a+b x+c x^2} \left (-1280 a^5 A-1536 a^5 B x-3200 a^4 A b x-4160 a^4 A c x^2-4032 a^4 b B x^2-5632 a^4 B c x^3-2160 a^3 A b^2 x^2-6240 a^3 A b c x^3-5280 a^3 A c^2 x^4-2976 a^3 b^2 B x^3-9952 a^3 b B c x^4-11776 a^3 B c^2 x^5-40 a^2 A b^3 x^3-480 a^2 A b^2 c x^4-2640 a^2 A b c^2 x^5-120 a^2 b^3 B x^4-2160 a^2 b^2 B c x^5+50 a A b^4 x^4+800 a A b^3 c x^5+180 a b^4 B x^5-75 A b^5 x^5\right )}{7680 a^3 x^6}-B c^{5/2} \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^7,x]

[Out]

(Sqrt[a + b*x + c*x^2]*(-1280*a^5*A - 3200*a^4*A*b*x - 1536*a^5*B*x - 2160*a^3*A*b^2*x^2 - 4032*a^4*b*B*x^2 -
4160*a^4*A*c*x^2 - 40*a^2*A*b^3*x^3 - 2976*a^3*b^2*B*x^3 - 6240*a^3*A*b*c*x^3 - 5632*a^4*B*c*x^3 + 50*a*A*b^4*
x^4 - 120*a^2*b^3*B*x^4 - 480*a^2*A*b^2*c*x^4 - 9952*a^3*b*B*c*x^4 - 5280*a^3*A*c^2*x^4 - 75*A*b^5*x^5 + 180*a
*b^4*B*x^5 + 800*a*A*b^3*c*x^5 - 2160*a^2*b^2*B*c*x^5 - 2640*a^2*A*b*c^2*x^5 - 11776*a^3*B*c^2*x^5))/(7680*a^3
*x^6) + ((5*A*b^6 - 12*a*b^5*B - 60*a*A*b^4*c + 160*a^2*b^3*B*c + 240*a^2*A*b^2*c^2 - 960*a^3*b*B*c^2 - 320*a^
3*A*c^3)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(512*a^(7/2)) - B*c^(5/2)*Log[b + 2*c*x - 2*
Sqrt[c]*Sqrt[a + b*x + c*x^2]]

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fricas [A]  time = 8.72, size = 1659, normalized size = 5.00

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^7,x, algorithm="fricas")

[Out]

[1/30720*(15360*B*a^4*c^(5/2)*x^6*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*(12*B*a*b^5 - 5*A*b^6 + 320*A*a^3*c^3 + 240*(4*B*a^3*b - A*a^2*b^2)*c^2 - 20*(8*B*a^2*b^3 - 3*A
*a*b^4)*c)*sqrt(a)*x^6*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*(1280*A*a^6 - (180*B*a^2*b^4 - 75*A*a*b^5 - 16*(736*B*a^4 + 165*A*a^3*b)*c^2 - 80*(27*B*a^3*b^2 - 1
0*A*a^2*b^3)*c)*x^5 + 2*(60*B*a^3*b^3 - 25*A*a^2*b^4 + 2640*A*a^4*c^2 + 16*(311*B*a^4*b + 15*A*a^3*b^2)*c)*x^4
+ 8*(372*B*a^4*b^2 + 5*A*a^3*b^3 + 4*(176*B*a^5 + 195*A*a^4*b)*c)*x^3 + 16*(252*B*a^5*b + 135*A*a^4*b^2 + 260
*A*a^5*c)*x^2 + 128*(12*B*a^6 + 25*A*a^5*b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^6), -1/30720*(30720*B*a^4*sqrt(-c
)*c^2*x^6*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*(12*B*a*b^5 - 5*
A*b^6 + 320*A*a^3*c^3 + 240*(4*B*a^3*b - A*a^2*b^2)*c^2 - 20*(8*B*a^2*b^3 - 3*A*a*b^4)*c)*sqrt(a)*x^6*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*(1280*A*a^6 - (180*B
*a^2*b^4 - 75*A*a*b^5 - 16*(736*B*a^4 + 165*A*a^3*b)*c^2 - 80*(27*B*a^3*b^2 - 10*A*a^2*b^3)*c)*x^5 + 2*(60*B*a
^3*b^3 - 25*A*a^2*b^4 + 2640*A*a^4*c^2 + 16*(311*B*a^4*b + 15*A*a^3*b^2)*c)*x^4 + 8*(372*B*a^4*b^2 + 5*A*a^3*b
^3 + 4*(176*B*a^5 + 195*A*a^4*b)*c)*x^3 + 16*(252*B*a^5*b + 135*A*a^4*b^2 + 260*A*a^5*c)*x^2 + 128*(12*B*a^6 +
25*A*a^5*b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^6), 1/15360*(7680*B*a^4*c^(5/2)*x^6*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*(12*B*a*b^5 - 5*A*b^6 + 320*A*a^3*c^3 + 240*(4*
B*a^3*b - A*a^2*b^2)*c^2 - 20*(8*B*a^2*b^3 - 3*A*a*b^4)*c)*sqrt(-a)*x^6*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*(1280*A*a^6 - (180*B*a^2*b^4 - 75*A*a*b^5 - 16*(736*B*a^4 + 165*A
*a^3*b)*c^2 - 80*(27*B*a^3*b^2 - 10*A*a^2*b^3)*c)*x^5 + 2*(60*B*a^3*b^3 - 25*A*a^2*b^4 + 2640*A*a^4*c^2 + 16*(
311*B*a^4*b + 15*A*a^3*b^2)*c)*x^4 + 8*(372*B*a^4*b^2 + 5*A*a^3*b^3 + 4*(176*B*a^5 + 195*A*a^4*b)*c)*x^3 + 16*
(252*B*a^5*b + 135*A*a^4*b^2 + 260*A*a^5*c)*x^2 + 128*(12*B*a^6 + 25*A*a^5*b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x
^6), -1/15360*(15360*B*a^4*sqrt(-c)*c^2*x^6*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*(12*B*a*b^5 - 5*A*b^6 + 320*A*a^3*c^3 + 240*(4*B*a^3*b - A*a^2*b^2)*c^2 - 20*(8*B*a^2*b^3 -
3*A*a*b^4)*c)*sqrt(-a)*x^6*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*
(1280*A*a^6 - (180*B*a^2*b^4 - 75*A*a*b^5 - 16*(736*B*a^4 + 165*A*a^3*b)*c^2 - 80*(27*B*a^3*b^2 - 10*A*a^2*b^3
)*c)*x^5 + 2*(60*B*a^3*b^3 - 25*A*a^2*b^4 + 2640*A*a^4*c^2 + 16*(311*B*a^4*b + 15*A*a^3*b^2)*c)*x^4 + 8*(372*B
*a^4*b^2 + 5*A*a^3*b^3 + 4*(176*B*a^5 + 195*A*a^4*b)*c)*x^3 + 16*(252*B*a^5*b + 135*A*a^4*b^2 + 260*A*a^5*c)*x
^2 + 128*(12*B*a^6 + 25*A*a^5*b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^6)]

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giac [B]  time = 1.45, size = 2089, normalized size = 6.29

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^7,x, algorithm="giac")

[Out]

-B*c^(5/2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c - b*sqrt(c))) + 1/512*(12*B*a*b^5 - 5*A*b^6 - 160*
B*a^2*b^3*c + 60*A*a*b^4*c + 960*B*a^3*b*c^2 - 240*A*a^2*b^2*c^2 + 320*A*a^3*c^3)*arctan(-(sqrt(c)*x - sqrt(c*
x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a^3) - 1/7680*(180*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^11*B*a*b^5 - 75*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))^11*A*b^6 - 2400*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^11*B*a^2*b^3*c + 900*(sqr
t(c)*x - sqrt(c*x^2 + b*x + a))^11*A*a*b^4*c - 31680*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^11*B*a^3*b*c^2 - 3600
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^11*A*a^2*b^2*c^2 - 10560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^11*A*a^3*c^3
- 46080*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^10*B*a^3*b^2*c^(3/2) - 46080*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^
10*B*a^4*c^(5/2) - 46080*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^10*A*a^3*b*c^(5/2) - 1020*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^9*B*a^2*b^5 + 425*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a*b^6 - 22240*(sqrt(c)*x - sqrt(c*x^2 +
b*x + a))^9*B*a^3*b^3*c - 5100*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a^2*b^4*c + 41280*(sqrt(c)*x - sqrt(c*
x^2 + b*x + a))^9*B*a^4*b*c^2 - 56400*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a^3*b^2*c^2 - 1600*(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))^9*A*a^4*c^3 - 15360*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*B*a^3*b^4*sqrt(c) + 46080*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))^8*B*a^4*b^2*c^(3/2) - 76800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*A*a^3*b^3*c
^(3/2) + 138240*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*B*a^5*c^(5/2) - 696*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^
7*B*a^3*b^5 - 990*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^2*b^6 - 960*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*
B*a^4*b^3*c - 34200*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^3*b^4*c - 86400*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^7*B*a^5*b*c^2 - 93600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^4*b^2*c^2 - 28800*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^7*A*a^5*c^3 + 15360*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^4*b^4*sqrt(c) - 15360*(sqrt(c)*x - s
qrt(c*x^2 + b*x + a))^6*A*a^3*b^5*sqrt(c) - 76800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^5*b^2*c^(3/2) - 51
200*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*A*a^4*b^3*c^(3/2) - 235520*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a
^6*c^(5/2) - 153600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*A*a^5*b*c^(5/2) + 2376*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))^5*B*a^4*b^5 - 990*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^3*b^6 + 14400*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))^5*B*a^5*b^3*c - 34200*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^4*b^4*c + 67200*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^5*B*a^6*b*c^2 - 93600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^5*b^2*c^2 - 28800*(sqrt(c)*x - sq
rt(c*x^2 + b*x + a))^5*A*a^6*c^3 + 76800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^6*b^2*c^(3/2) - 76800*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^5*b^3*c^(3/2) + 215040*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^7*c^(5/2
) - 1020*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^5*b^5 + 425*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^4*b^6
+ 13600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^6*b^3*c - 5100*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^5*
b^4*c - 4800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^7*b*c^2 - 56400*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A
*a^6*b^2*c^2 - 1600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^7*c^3 - 95232*(sqrt(c)*x - sqrt(c*x^2 + b*x + a)
)^2*B*a^8*c^(5/2) - 46080*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^7*b*c^(5/2) + 180*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))*B*a^6*b^5 - 75*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^5*b^6 - 2400*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))*B*a^7*b^3*c + 900*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^6*b^4*c + 14400*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))*B*a^8*b*c^2 - 3600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^7*b^2*c^2 - 10560*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))*A*a^8*c^3 + 23552*B*a^9*c^(5/2))/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^6*a^3)

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maple [B]  time = 0.09, size = 1677, normalized size = 5.05

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^7,x)

[Out]

-1/32*A/a^4*b^2*c/x^2*(c*x^2+b*x+a)^(7/2)+5/96*A/a^4*b^3*c^2*(c*x^2+b*x+a)^(3/2)*x-5/1536*A/a^5*b^5*c*(c*x^2+b
*x+a)^(3/2)*x-1/512*A/a^6*b^5*c*(c*x^2+b*x+a)^(5/2)*x+5/64*A/a^3*b^3*c^2*(c*x^2+b*x+a)^(1/2)*x+5/192*A/a^5*b^3
*c^2*(c*x^2+b*x+a)^(5/2)*x-5/512*A/a^4*b^5*(c*x^2+b*x+a)^(1/2)*x*c-5/192*A/a^5*b^3*c/x*(c*x^2+b*x+a)^(7/2)+5/3
2*A/a^4*b*c^2/x*(c*x^2+b*x+a)^(7/2)+19/240*B/a^3*b*c/x^2*(c*x^2+b*x+a)^(7/2)-13/96*B/a^3*b^2*c^2*(c*x^2+b*x+a)
^(3/2)*x+11/160*B/a^4*b^2*c/x*(c*x^2+b*x+a)^(7/2)+3/640*B/a^5*b^4*c*(c*x^2+b*x+a)^(5/2)*x+1/128*B/a^4*b^4*c*(c
*x^2+b*x+a)^(3/2)*x-11/160*B/a^4*b^2*c^2*(c*x^2+b*x+a)^(5/2)*x-7/32*B/a^2*b^2*c^2*(c*x^2+b*x+a)^(1/2)*x+3/128*
B/a^3*b^4*(c*x^2+b*x+a)^(1/2)*x*c-5/32*A/a^3*b*c^3*(c*x^2+b*x+a)^(3/2)*x-5/32*A/a^2*b*c^3*(c*x^2+b*x+a)^(1/2)*
x-5/32*A/a^4*b*c^3*(c*x^2+b*x+a)^(5/2)*x+1/16*A/a^3*b*c/x^3*(c*x^2+b*x+a)^(7/2)+B*c^(5/2)*ln((c*x+1/2*b)/c^(1/
2)+(c*x^2+b*x+a)^(1/2))+1/128*B/a^4*b^5*(c*x^2+b*x+a)^(3/2)+3/640*B/a^5*b^5*(c*x^2+b*x+a)^(5/2)+3/128*B/a^3*b^
5*(c*x^2+b*x+a)^(1/2)-1/5*B/a/x^5*(c*x^2+b*x+a)^(7/2)-3/256*B/a^(5/2)*b^5*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^
(1/2))/x)+5/1024*A/a^(7/2)*b^6*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-5/16*A*c^3/a^(1/2)*ln((b*x+2*a+2*
(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-5/1536*A/a^5*b^6*(c*x^2+b*x+a)^(3/2)-1/512*A/a^6*b^6*(c*x^2+b*x+a)^(5/2)-5/512
*A/a^4*b^6*(c*x^2+b*x+a)^(1/2)+5/48*A*c^3/a^2*(c*x^2+b*x+a)^(3/2)+1/16*A*c^3/a^3*(c*x^2+b*x+a)^(5/2)+5/16*A*c^
3/a*(c*x^2+b*x+a)^(1/2)-1/6*A/a/x^6*(c*x^2+b*x+a)^(7/2)+31/48*B/a^2*b*c^2*(c*x^2+b*x+a)^(3/2)+109/240*B/a^3*b*
c^2*(c*x^2+b*x+a)^(5/2)+1/512*A/a^6*b^5/x*(c*x^2+b*x+a)^(7/2)+1/12*A/a^2*b/x^5*(c*x^2+b*x+a)^(7/2)-1/16*A*c^2/
a^3/x^2*(c*x^2+b*x+a)^(7/2)-15/256*A/a^(5/2)*b^4*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+15/64*A/a^(3/
2)*b^2*c^2*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+5/32*B/a^(3/2)*b^3*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2
)*a^(1/2))/x)-15/16*B/a^(1/2)*b*c^2*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-8/15*B*c^2/a^3/x*(c*x^2+b*x+
a)^(7/2)+2/3*B*c^3/a^2*(c*x^2+b*x+a)^(3/2)*x+B*c^3/a*(c*x^2+b*x+a)^(1/2)*x+8/15*B*c^3/a^3*(c*x^2+b*x+a)^(5/2)*
x-2/15*B*c/a^2/x^3*(c*x^2+b*x+a)^(7/2)-23/192*B/a^3*b^3*c*(c*x^2+b*x+a)^(3/2)-21/320*B/a^4*b^3*c*(c*x^2+b*x+a)
^(5/2)-1/320*B/a^4*b^3/x^2*(c*x^2+b*x+a)^(7/2)-3/640*B/a^5*b^4/x*(c*x^2+b*x+a)^(7/2)+23/16*B/a*b*c^2*(c*x^2+b*
x+a)^(1/2)+3/40*B/a^2*b/x^4*(c*x^2+b*x+a)^(7/2)-1/80*B/a^3*b^2/x^3*(c*x^2+b*x+a)^(7/2)-17/64*B/a^2*b^3*c*(c*x^
2+b*x+a)^(1/2)-1/24*A*c/a^2/x^4*(c*x^2+b*x+a)^(7/2)-5/32*A/a^3*b^2*c^2*(c*x^2+b*x+a)^(3/2)-1/8*A/a^4*b^2*c^2*(
c*x^2+b*x+a)^(5/2)-5/16*A/a^2*b^2*c^2*(c*x^2+b*x+a)^(1/2)-1/32*A/a^3*b^2/x^4*(c*x^2+b*x+a)^(7/2)+1/192*A/a^4*b
^3/x^3*(c*x^2+b*x+a)^(7/2)+25/256*A/a^3*b^4*c*(c*x^2+b*x+a)^(1/2)+35/768*A/a^4*b^4*c*(c*x^2+b*x+a)^(3/2)+19/76
8*A/a^5*b^4*c*(c*x^2+b*x+a)^(5/2)+1/768*A/a^5*b^4/x^2*(c*x^2+b*x+a)^(7/2)

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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)^(5/2)/x^7,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 = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{x^7} \,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^7,x)

[Out]

int(((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^7, 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 {5}{2}}}{x^{7}}\, 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**7,x)

[Out]

Integral((A + B*x)*(a + b*x + c*x**2)**(5/2)/x**7, x)

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