3.947 \(\int \frac {(A+B x) (a+b x+c x^2)^{5/2}}{x^8} \, dx\)
Optimal. Leaf size=219 \[ -\frac {5 \left (b^2-4 a c\right )^3 (A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2048 a^{9/2}}+\frac {5 \left (b^2-4 a c\right )^2 (2 a+b x) (A b-2 a B) \sqrt {a+b x+c x^2}}{1024 a^4 x^2}-\frac {5 \left (b^2-4 a c\right ) (2 a+b x) (A b-2 a B) \left (a+b x+c x^2\right )^{3/2}}{384 a^3 x^4}+\frac {(2 a+b x) (A b-2 a B) \left (a+b x+c x^2\right )^{5/2}}{24 a^2 x^6}-\frac {A \left (a+b x+c x^2\right )^{7/2}}{7 a x^7} \]
[Out]
-5/384*(A*b-2*B*a)*(-4*a*c+b^2)*(b*x+2*a)*(c*x^2+b*x+a)^(3/2)/a^3/x^4+1/24*(A*b-2*B*a)*(b*x+2*a)*(c*x^2+b*x+a)
^(5/2)/a^2/x^6-1/7*A*(c*x^2+b*x+a)^(7/2)/a/x^7-5/2048*(A*b-2*B*a)*(-4*a*c+b^2)^3*arctanh(1/2*(b*x+2*a)/a^(1/2)
/(c*x^2+b*x+a)^(1/2))/a^(9/2)+5/1024*(A*b-2*B*a)*(-4*a*c+b^2)^2*(b*x+2*a)*(c*x^2+b*x+a)^(1/2)/a^4/x^2
________________________________________________________________________________________
Rubi [A] time = 0.13, antiderivative size = 219, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used =
{806, 720, 724, 206} \[ -\frac {5 \left (b^2-4 a c\right ) (2 a+b x) (A b-2 a B) \left (a+b x+c x^2\right )^{3/2}}{384 a^3 x^4}+\frac {5 \left (b^2-4 a c\right )^2 (2 a+b x) (A b-2 a B) \sqrt {a+b x+c x^2}}{1024 a^4 x^2}-\frac {5 \left (b^2-4 a c\right )^3 (A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2048 a^{9/2}}+\frac {(2 a+b x) (A b-2 a B) \left (a+b x+c x^2\right )^{5/2}}{24 a^2 x^6}-\frac {A \left (a+b x+c x^2\right )^{7/2}}{7 a x^7} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^8,x]
[Out]
(5*(A*b - 2*a*B)*(b^2 - 4*a*c)^2*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(1024*a^4*x^2) - (5*(A*b - 2*a*B)*(b^2 - 4
*a*c)*(2*a + b*x)*(a + b*x + c*x^2)^(3/2))/(384*a^3*x^4) + ((A*b - 2*a*B)*(2*a + b*x)*(a + b*x + c*x^2)^(5/2))
/(24*a^2*x^6) - (A*(a + b*x + c*x^2)^(7/2))/(7*a*x^7) - (5*(A*b - 2*a*B)*(b^2 - 4*a*c)^3*ArcTanh[(2*a + b*x)/(
2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(2048*a^(9/2))
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 )^{5/2}}{x^8} \, dx &=-\frac {A \left (a+b x+c x^2\right )^{7/2}}{7 a x^7}-\frac {(A b-2 a B) \int \frac {\left (a+b x+c x^2\right )^{5/2}}{x^7} \, dx}{2 a}\\ &=\frac {(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{5/2}}{24 a^2 x^6}-\frac {A \left (a+b x+c x^2\right )^{7/2}}{7 a x^7}+\frac {\left (5 (A b-2 a B) \left (b^2-4 a c\right )\right ) \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^5} \, dx}{48 a^2}\\ &=-\frac {5 (A b-2 a B) \left (b^2-4 a c\right ) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{384 a^3 x^4}+\frac {(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{5/2}}{24 a^2 x^6}-\frac {A \left (a+b x+c x^2\right )^{7/2}}{7 a x^7}-\frac {\left (5 (A b-2 a B) \left (b^2-4 a c\right )^2\right ) \int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx}{256 a^3}\\ &=\frac {5 (A b-2 a B) \left (b^2-4 a c\right )^2 (2 a+b x) \sqrt {a+b x+c x^2}}{1024 a^4 x^2}-\frac {5 (A b-2 a B) \left (b^2-4 a c\right ) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{384 a^3 x^4}+\frac {(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{5/2}}{24 a^2 x^6}-\frac {A \left (a+b x+c x^2\right )^{7/2}}{7 a x^7}+\frac {\left (5 (A b-2 a B) \left (b^2-4 a c\right )^3\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{2048 a^4}\\ &=\frac {5 (A b-2 a B) \left (b^2-4 a c\right )^2 (2 a+b x) \sqrt {a+b x+c x^2}}{1024 a^4 x^2}-\frac {5 (A b-2 a B) \left (b^2-4 a c\right ) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{384 a^3 x^4}+\frac {(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{5/2}}{24 a^2 x^6}-\frac {A \left (a+b x+c x^2\right )^{7/2}}{7 a x^7}-\frac {\left (5 (A b-2 a B) \left (b^2-4 a c\right )^3\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 )}{1024 a^4}\\ &=\frac {5 (A b-2 a B) \left (b^2-4 a c\right )^2 (2 a+b x) \sqrt {a+b x+c x^2}}{1024 a^4 x^2}-\frac {5 (A b-2 a B) \left (b^2-4 a c\right ) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{384 a^3 x^4}+\frac {(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{5/2}}{24 a^2 x^6}-\frac {A \left (a+b x+c x^2\right )^{7/2}}{7 a x^7}-\frac {5 (A b-2 a B) \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2048 a^{9/2}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.47, size = 198, normalized size = 0.90 \[ \frac {(A b-2 a B) \left (256 a^{5/2} (2 a+b x) (a+x (b+c x))^{5/2}-5 x^2 \left (b^2-4 a c\right ) \left (16 a^{3/2} (2 a+b x) (a+x (b+c x))^{3/2}-3 x^2 \left (b^2-4 a c\right ) \left (2 \sqrt {a} (2 a+b x) \sqrt {a+x (b+c x)}-x^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )\right )\right )\right )}{6144 a^{9/2} x^6}-\frac {A (a+x (b+c x))^{7/2}}{7 a x^7} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^8,x]
[Out]
-1/7*(A*(a + x*(b + c*x))^(7/2))/(a*x^7) + ((A*b - 2*a*B)*(256*a^(5/2)*(2*a + b*x)*(a + x*(b + c*x))^(5/2) - 5
*(b^2 - 4*a*c)*x^2*(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)])]))))/(6144*a
^(9/2)*x^6)
________________________________________________________________________________________
fricas [B] time = 10.91, size = 887, normalized size = 4.05 \[ \left [-\frac {105 \, {\left (2 \, B a b^{6} - A b^{7} - 64 \, {\left (2 \, B a^{4} - A a^{3} b\right )} c^{3} + 48 \, {\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} c^{2} - 12 \, {\left (2 \, B a^{2} b^{4} - A a b^{5}\right )} c\right )} \sqrt {a} x^{7} \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 (3072 \, A a^{7} + {\left (210 \, B a^{2} b^{5} - 105 \, A a b^{6} + 3072 \, A a^{4} c^{3} + 3696 \, {\left (2 \, B a^{4} b - A a^{3} b^{2}\right )} c^{2} - 1120 \, {\left (2 \, B a^{3} b^{3} - A a^{2} b^{4}\right )} c\right )} x^{6} - 2 \, {\left (70 \, B a^{3} b^{4} - 35 \, A a^{2} b^{5} - 48 \, {\left (154 \, B a^{5} + 19 \, A a^{4} b\right )} c^{2} - 336 \, {\left (2 \, B a^{4} b^{2} - A a^{3} b^{3}\right )} c\right )} x^{5} + 8 \, {\left (14 \, B a^{4} b^{3} - 7 \, A a^{3} b^{4} + 1152 \, A a^{5} c^{2} + 12 \, {\left (182 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} c\right )} x^{4} + 16 \, {\left (378 \, B a^{5} b^{2} + 3 \, A a^{4} b^{3} + 4 \, {\left (182 \, B a^{6} + 197 \, A a^{5} b\right )} c\right )} x^{3} + 128 \, {\left (70 \, B a^{6} b + 37 \, A a^{5} b^{2} + 72 \, A a^{6} c\right )} x^{2} + 256 \, {\left (14 \, B a^{7} + 29 \, A a^{6} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{86016 \, a^{5} x^{7}}, -\frac {105 \, {\left (2 \, B a b^{6} - A b^{7} - 64 \, {\left (2 \, B a^{4} - A a^{3} b\right )} c^{3} + 48 \, {\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} c^{2} - 12 \, {\left (2 \, B a^{2} b^{4} - A a b^{5}\right )} c\right )} \sqrt {-a} x^{7} \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 (3072 \, A a^{7} + {\left (210 \, B a^{2} b^{5} - 105 \, A a b^{6} + 3072 \, A a^{4} c^{3} + 3696 \, {\left (2 \, B a^{4} b - A a^{3} b^{2}\right )} c^{2} - 1120 \, {\left (2 \, B a^{3} b^{3} - A a^{2} b^{4}\right )} c\right )} x^{6} - 2 \, {\left (70 \, B a^{3} b^{4} - 35 \, A a^{2} b^{5} - 48 \, {\left (154 \, B a^{5} + 19 \, A a^{4} b\right )} c^{2} - 336 \, {\left (2 \, B a^{4} b^{2} - A a^{3} b^{3}\right )} c\right )} x^{5} + 8 \, {\left (14 \, B a^{4} b^{3} - 7 \, A a^{3} b^{4} + 1152 \, A a^{5} c^{2} + 12 \, {\left (182 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} c\right )} x^{4} + 16 \, {\left (378 \, B a^{5} b^{2} + 3 \, A a^{4} b^{3} + 4 \, {\left (182 \, B a^{6} + 197 \, A a^{5} b\right )} c\right )} x^{3} + 128 \, {\left (70 \, B a^{6} b + 37 \, A a^{5} b^{2} + 72 \, A a^{6} c\right )} x^{2} + 256 \, {\left (14 \, B a^{7} + 29 \, A a^{6} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{43008 \, a^{5} x^{7}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^8,x, algorithm="fricas")
[Out]
[-1/86016*(105*(2*B*a*b^6 - A*b^7 - 64*(2*B*a^4 - A*a^3*b)*c^3 + 48*(2*B*a^3*b^2 - A*a^2*b^3)*c^2 - 12*(2*B*a^
2*b^4 - A*a*b^5)*c)*sqrt(a)*x^7*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*(3072*A*a^7 + (210*B*a^2*b^5 - 105*A*a*b^6 + 3072*A*a^4*c^3 + 3696*(2*B*a^4*b - A*a^3*b^2)
*c^2 - 1120*(2*B*a^3*b^3 - A*a^2*b^4)*c)*x^6 - 2*(70*B*a^3*b^4 - 35*A*a^2*b^5 - 48*(154*B*a^5 + 19*A*a^4*b)*c^
2 - 336*(2*B*a^4*b^2 - A*a^3*b^3)*c)*x^5 + 8*(14*B*a^4*b^3 - 7*A*a^3*b^4 + 1152*A*a^5*c^2 + 12*(182*B*a^5*b +
5*A*a^4*b^2)*c)*x^4 + 16*(378*B*a^5*b^2 + 3*A*a^4*b^3 + 4*(182*B*a^6 + 197*A*a^5*b)*c)*x^3 + 128*(70*B*a^6*b +
37*A*a^5*b^2 + 72*A*a^6*c)*x^2 + 256*(14*B*a^7 + 29*A*a^6*b)*x)*sqrt(c*x^2 + b*x + a))/(a^5*x^7), -1/43008*(1
05*(2*B*a*b^6 - A*b^7 - 64*(2*B*a^4 - A*a^3*b)*c^3 + 48*(2*B*a^3*b^2 - A*a^2*b^3)*c^2 - 12*(2*B*a^2*b^4 - A*a*
b^5)*c)*sqrt(-a)*x^7*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*(3072*
A*a^7 + (210*B*a^2*b^5 - 105*A*a*b^6 + 3072*A*a^4*c^3 + 3696*(2*B*a^4*b - A*a^3*b^2)*c^2 - 1120*(2*B*a^3*b^3 -
A*a^2*b^4)*c)*x^6 - 2*(70*B*a^3*b^4 - 35*A*a^2*b^5 - 48*(154*B*a^5 + 19*A*a^4*b)*c^2 - 336*(2*B*a^4*b^2 - A*a
^3*b^3)*c)*x^5 + 8*(14*B*a^4*b^3 - 7*A*a^3*b^4 + 1152*A*a^5*c^2 + 12*(182*B*a^5*b + 5*A*a^4*b^2)*c)*x^4 + 16*(
378*B*a^5*b^2 + 3*A*a^4*b^3 + 4*(182*B*a^6 + 197*A*a^5*b)*c)*x^3 + 128*(70*B*a^6*b + 37*A*a^5*b^2 + 72*A*a^6*c
)*x^2 + 256*(14*B*a^7 + 29*A*a^6*b)*x)*sqrt(c*x^2 + b*x + a))/(a^5*x^7)]
________________________________________________________________________________________
giac [B] time = 0.43, size = 2598, normalized size = 11.86 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^8,x, algorithm="giac")
[Out]
-5/1024*(2*B*a*b^6 - A*b^7 - 24*B*a^2*b^4*c + 12*A*a*b^5*c + 96*B*a^3*b^2*c^2 - 48*A*a^2*b^3*c^2 - 128*B*a^4*c
^3 + 64*A*a^3*b*c^3)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a^4) + 1/21504*(210*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))^13*B*a*b^6 - 105*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^13*A*b^7 - 2520*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^13*B*a^2*b^4*c + 1260*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^13*A*a*b^5*c + 10080*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))^13*B*a^3*b^2*c^2 - 5040*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^13*A*a^2*b^3*c^2 +
29568*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^13*B*a^4*c^3 + 6720*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^13*A*a^3*b*c
^3 + 129024*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^12*B*a^4*b*c^(5/2) + 43008*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))
^12*A*a^4*c^(7/2) - 1400*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^11*B*a^2*b^6 + 700*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))^11*A*a*b^7 + 16800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^11*B*a^3*b^4*c - 8400*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))^11*A*a^2*b^5*c + 147840*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^11*B*a^4*b^2*c^2 + 33600*(sqrt(c)*x - sqr
t(c*x^2 + b*x + a))^11*A*a^3*b^3*c^2 - 25088*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^11*B*a^5*c^3 + 141568*(sqrt(c
)*x - sqrt(c*x^2 + b*x + a))^11*A*a^4*b*c^3 + 215040*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^10*B*a^4*b^3*c^(3/2)
- 129024*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^10*B*a^5*b*c^(5/2) + 387072*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^1
0*A*a^4*b^2*c^(5/2) + 3962*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a^3*b^6 - 1981*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))^9*A*a^2*b^7 + 81480*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a^4*b^4*c + 23772*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^9*A*a^3*b^5*c + 104160*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a^5*b^2*c^2 + 378000*(sqrt(c)*x - s
qrt(c*x^2 + b*x + a))^9*A*a^4*b^3*c^2 + 76160*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a^6*c^3 + 284480*(sqrt(c
)*x - sqrt(c*x^2 + b*x + a))^9*A*a^5*b*c^3 + 43008*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*B*a^4*b^5*sqrt(c) - 7
1680*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*B*a^5*b^3*c^(3/2) + 358400*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*A*
a^4*b^4*c^(3/2) + 430080*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*B*a^6*b*c^(5/2) + 430080*(sqrt(c)*x - sqrt(c*x^
2 + b*x + a))^8*A*a^5*b^2*c^(5/2) + 215040*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*A*a^6*c^(7/2) + 3072*(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))^7*A*a^3*b^7 + 129024*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^4*b^5*c + 645120*(s
qrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^5*b^3*c^2 + 430080*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^6*b*c^3 -
43008*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^5*b^5*sqrt(c) + 43008*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*A
*a^4*b^6*sqrt(c) + 71680*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^6*b^3*c^(3/2) + 286720*(sqrt(c)*x - sqrt(c*
x^2 + b*x + a))^6*A*a^5*b^4*c^(3/2) - 430080*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^7*b*c^(5/2) + 860160*(s
qrt(c)*x - sqrt(c*x^2 + b*x + a))^6*A*a^6*b^2*c^(5/2) - 3962*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^5*b^6 +
1981*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^4*b^7 - 81480*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^6*b^4*
c + 105252*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^5*b^5*c - 104160*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*
a^7*b^2*c^2 + 482160*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^6*b^3*c^2 - 76160*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))^5*B*a^8*c^3 + 360640*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^7*b*c^3 - 215040*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^4*B*a^7*b^3*c^(3/2) + 215040*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^6*b^4*c^(3/2) + 129024*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^8*b*c^(5/2) + 258048*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^7*b^2*c^
(5/2) + 129024*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^8*c^(7/2) + 1400*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^
3*B*a^6*b^6 - 700*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^5*b^7 - 16800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^
3*B*a^7*b^4*c + 8400*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^6*b^5*c - 147840*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))^3*B*a^8*b^2*c^2 + 181440*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^7*b^3*c^2 + 25088*(sqrt(c)*x - sqrt(c
*x^2 + b*x + a))^3*B*a^9*c^3 + 116480*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^8*b*c^3 - 129024*(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))^2*B*a^9*b*c^(5/2) + 129024*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^8*b^2*c^(5/2) - 21
0*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^7*b^6 + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^6*b^7 + 2520*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^8*b^4*c - 1260*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^7*b^5*c - 10080*(s
qrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^9*b^2*c^2 + 5040*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^8*b^3*c^2 - 295
68*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^10*c^3 + 36288*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^9*b*c^3 + 61
44*A*a^10*c^(7/2))/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^7*a^4)
________________________________________________________________________________________
maple [B] time = 0.11, size = 1874, normalized size = 8.56 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^8,x)
[Out]
-1/512*B/a^6*b^5*c*(c*x^2+b*x+a)^(5/2)*x-5/1536*B/a^5*b^5*c*(c*x^2+b*x+a)^(3/2)*x+5/64*B/a^3*b^3*c^2*(c*x^2+b*
x+a)^(1/2)*x+1/16*B/a^3*b*c/x^3*(c*x^2+b*x+a)^(7/2)-5/192*B/a^5*b^3*c/x*(c*x^2+b*x+a)^(7/2)-5/32*B/a^2*b*c^3*(
c*x^2+b*x+a)^(1/2)*x+5/32*B/a^4*b*c^2/x*(c*x^2+b*x+a)^(7/2)-5/32*B/a^4*b*c^3*(c*x^2+b*x+a)^(5/2)*x-5/32*B/a^3*
b*c^3*(c*x^2+b*x+a)^(3/2)*x-1/32*B/a^4*b^2*c/x^2*(c*x^2+b*x+a)^(7/2)-5/512*B/a^4*b^5*(c*x^2+b*x+a)^(1/2)*x*c-5
/192*A/a^5*b^4*c^2*(c*x^2+b*x+a)^(3/2)*x-5/64*A/a^5*b^2*c^2/x*(c*x^2+b*x+a)^(7/2)+5/64*A/a^3*b^2*c^3*(c*x^2+b*
x+a)^(1/2)*x+1/1024*A/a^7*b^6*c*(c*x^2+b*x+a)^(5/2)*x-1/32*A/a^4*b^2*c/x^3*(c*x^2+b*x+a)^(7/2)+1/32*A/a^4*b*c^
2/x^2*(c*x^2+b*x+a)^(7/2)+1/48*A/a^3*b*c/x^4*(c*x^2+b*x+a)^(7/2)+5/64*A/a^5*b^2*c^3*(c*x^2+b*x+a)^(5/2)*x+5/19
2*B/a^5*b^3*c^2*(c*x^2+b*x+a)^(5/2)*x+5/96*B/a^4*b^3*c^2*(c*x^2+b*x+a)^(3/2)*x-25/512*A/a^4*b^5*c*(c*x^2+b*x+a
)^(1/2)-1/1024*A/a^7*b^6/x*(c*x^2+b*x+a)^(7/2)-19/1536*A/a^6*b^5*c*(c*x^2+b*x+a)^(5/2)-1/1536*A/a^6*b^5/x^2*(c
*x^2+b*x+a)^(7/2)-35/1536*A/a^5*b^5*c*(c*x^2+b*x+a)^(3/2)+1/16*A/a^5*b^3*c^2*(c*x^2+b*x+a)^(5/2)+5/32*A/a^3*b^
3*c^2*(c*x^2+b*x+a)^(1/2)+5/64*A/a^4*b^3*c^2*(c*x^2+b*x+a)^(3/2)+1/64*A/a^4*b^3/x^4*(c*x^2+b*x+a)^(7/2)-1/32*A
/a^4*b*c^3*(c*x^2+b*x+a)^(5/2)-5/32*A/a^2*b*c^3*(c*x^2+b*x+a)^(1/2)-5/96*A/a^3*b*c^3*(c*x^2+b*x+a)^(3/2)+1/12*
A/a^2*b/x^6*(c*x^2+b*x+a)^(7/2)+5/32*A/a^(3/2)*b*c^3*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-15/128*A/a^
(5/2)*b^3*c^2*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+15/512*A/a^(7/2)*b^5*c*ln((b*x+2*a+2*(c*x^2+b*x+a)
^(1/2)*a^(1/2))/x)+5/64*A/a^4*b^2*c^3*(c*x^2+b*x+a)^(3/2)*x+1/64*A/a^5*b^3*c/x^2*(c*x^2+b*x+a)^(7/2)+5/1024*A/
a^5*b^6*(c*x^2+b*x+a)^(1/2)*x*c-5/128*A/a^4*b^4*c^2*(c*x^2+b*x+a)^(1/2)*x+5/3072*A/a^6*b^6*c*(c*x^2+b*x+a)^(3/
2)*x+5/384*A/a^6*b^4*c/x*(c*x^2+b*x+a)^(7/2)-5/384*A/a^6*b^4*c^2*(c*x^2+b*x+a)^(5/2)*x-1/32*B/a^3*b^2/x^4*(c*x
^2+b*x+a)^(7/2)+15/64*B/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)-15/256*B/a^(5/2)*b^4*c*l
n((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-1/24*B*c/a^2/x^4*(c*x^2+b*x+a)^(7/2)-1/16*B*c^2/a^3/x^2*(c*x^2+b*
x+a)^(7/2)+1/12*B/a^2*b/x^5*(c*x^2+b*x+a)^(7/2)+1/192*B/a^4*b^3/x^3*(c*x^2+b*x+a)^(7/2)+25/256*B/a^3*b^4*c*(c*
x^2+b*x+a)^(1/2)+1/512*B/a^6*b^5/x*(c*x^2+b*x+a)^(7/2)+19/768*B/a^5*b^4*c*(c*x^2+b*x+a)^(5/2)+1/768*B/a^5*b^4/
x^2*(c*x^2+b*x+a)^(7/2)+35/768*B/a^4*b^4*c*(c*x^2+b*x+a)^(3/2)-1/8*B/a^4*b^2*c^2*(c*x^2+b*x+a)^(5/2)-1/24*A/a^
3*b^2/x^5*(c*x^2+b*x+a)^(7/2)-1/384*A/a^5*b^4/x^3*(c*x^2+b*x+a)^(7/2)-5/16*B/a^2*b^2*c^2*(c*x^2+b*x+a)^(1/2)-5
/32*B/a^3*b^2*c^2*(c*x^2+b*x+a)^(3/2)+5/1024*A/a^5*b^7*(c*x^2+b*x+a)^(1/2)+5/3072*A/a^6*b^7*(c*x^2+b*x+a)^(3/2
)-5/2048*A/a^(9/2)*b^7*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+1/1024*A/a^7*b^7*(c*x^2+b*x+a)^(5/2)+1/16
*B*c^3/a^3*(c*x^2+b*x+a)^(5/2)+5/16*B*c^3/a*(c*x^2+b*x+a)^(1/2)+5/48*B*c^3/a^2*(c*x^2+b*x+a)^(3/2)+5/1024*B/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*B*c^3/a^(1/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a
^(1/2))/x)-1/6*B/a/x^6*(c*x^2+b*x+a)^(7/2)-1/512*B/a^6*b^6*(c*x^2+b*x+a)^(5/2)-5/512*B/a^4*b^6*(c*x^2+b*x+a)^(
1/2)-5/1536*B/a^5*b^6*(c*x^2+b*x+a)^(3/2)-1/7*A*(c*x^2+b*x+a)^(7/2)/a/x^7
________________________________________________________________________________________
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)^(5/2)/x^8,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 \[ \int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{x^8} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^8,x)
[Out]
int(((A + B*x)*(a + b*x + c*x^2)^(5/2))/x^8, x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{x^{8}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(c*x**2+b*x+a)**(5/2)/x**8,x)
[Out]
Integral((A + B*x)*(a + b*x + c*x**2)**(5/2)/x**8, x)
________________________________________________________________________________________