3.9.86 \(\int \frac {x^4 (A+B x)}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=280 \[ \frac {\sqrt {a+b x+c x^2} \left (256 a^2 B c^2-2 c x \left (72 a A c^2-116 a b B c-30 A b^2 c+35 b^3 B\right )+312 a A b c^2-460 a b^2 B c-90 A b^3 c+105 b^4 B\right )}{24 c^4 \left (b^2-4 a c\right )}+\frac {x^2 \sqrt {a+b x+c x^2} \left (-16 a B c-6 A b c+7 b^2 B\right )}{3 c^2 \left (b^2-4 a c\right )}-\frac {2 x^3 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {\left (24 a A c^2-60 a b B c-30 A b^2 c+35 b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{9/2}} \]

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Rubi [A]  time = 0.27, antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {818, 832, 779, 621, 206} \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (256 a^2 B c^2-2 c x \left (72 a A c^2-116 a b B c-30 A b^2 c+35 b^3 B\right )+312 a A b c^2-460 a b^2 B c-90 A b^3 c+105 b^4 B\right )}{24 c^4 \left (b^2-4 a c\right )}+\frac {x^2 \sqrt {a+b x+c x^2} \left (-16 a B c-6 A b c+7 b^2 B\right )}{3 c^2 \left (b^2-4 a c\right )}-\frac {\left (24 a A c^2-60 a b B c-30 A b^2 c+35 b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{9/2}}-\frac {2 x^3 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/(a + b*x + c*x^2)^(3/2),x]

[Out]

(-2*x^3*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x))/(c*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + ((7*b^2*B -
6*A*b*c - 16*a*B*c)*x^2*Sqrt[a + b*x + c*x^2])/(3*c^2*(b^2 - 4*a*c)) + ((105*b^4*B - 90*A*b^3*c - 460*a*b^2*B
*c + 312*a*A*b*c^2 + 256*a^2*B*c^2 - 2*c*(35*b^3*B - 30*A*b^2*c - 116*a*b*B*c + 72*a*A*c^2)*x)*Sqrt[a + b*x +
c*x^2])/(24*c^4*(b^2 - 4*a*c)) - ((35*b^3*B - 30*A*b^2*c - 60*a*b*B*c + 24*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqr
t[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(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 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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 818

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 + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 832

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

Rubi steps

\begin {align*} \int \frac {x^4 (A+B x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 x^3 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 \int \frac {x^2 \left (3 a (b B-2 A c)+\frac {1}{2} \left (7 b^2 B-6 A b c-16 a B c\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{c \left (b^2-4 a c\right )}\\ &=-\frac {2 x^3 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {\left (7 b^2 B-6 A b c-16 a B c\right ) x^2 \sqrt {a+b x+c x^2}}{3 c^2 \left (b^2-4 a c\right )}+\frac {2 \int \frac {x \left (-a \left (7 b^2 B-6 A b c-16 a B c\right )-\frac {1}{4} \left (35 b^3 B-30 A b^2 c-116 a b B c+72 a A c^2\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{3 c^2 \left (b^2-4 a c\right )}\\ &=-\frac {2 x^3 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {\left (7 b^2 B-6 A b c-16 a B c\right ) x^2 \sqrt {a+b x+c x^2}}{3 c^2 \left (b^2-4 a c\right )}+\frac {\left (105 b^4 B-90 A b^3 c-460 a b^2 B c+312 a A b c^2+256 a^2 B c^2-2 c \left (35 b^3 B-30 A b^2 c-116 a b B c+72 a A c^2\right ) x\right ) \sqrt {a+b x+c x^2}}{24 c^4 \left (b^2-4 a c\right )}-\frac {\left (35 b^3 B-30 A b^2 c-60 a b B c+24 a A c^2\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c^4}\\ &=-\frac {2 x^3 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {\left (7 b^2 B-6 A b c-16 a B c\right ) x^2 \sqrt {a+b x+c x^2}}{3 c^2 \left (b^2-4 a c\right )}+\frac {\left (105 b^4 B-90 A b^3 c-460 a b^2 B c+312 a A b c^2+256 a^2 B c^2-2 c \left (35 b^3 B-30 A b^2 c-116 a b B c+72 a A c^2\right ) x\right ) \sqrt {a+b x+c x^2}}{24 c^4 \left (b^2-4 a c\right )}-\frac {\left (35 b^3 B-30 A b^2 c-60 a b B c+24 a A c^2\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 )}{8 c^4}\\ &=-\frac {2 x^3 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {\left (7 b^2 B-6 A b c-16 a B c\right ) x^2 \sqrt {a+b x+c x^2}}{3 c^2 \left (b^2-4 a c\right )}+\frac {\left (105 b^4 B-90 A b^3 c-460 a b^2 B c+312 a A b c^2+256 a^2 B c^2-2 c \left (35 b^3 B-30 A b^2 c-116 a b B c+72 a A c^2\right ) x\right ) \sqrt {a+b x+c x^2}}{24 c^4 \left (b^2-4 a c\right )}-\frac {\left (35 b^3 B-30 A b^2 c-60 a b B c+24 a A c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 300, normalized size = 1.07 \begin {gather*} \frac {2 \sqrt {c} \left (-256 a^3 B c^2+4 a^2 c \left (-2 b c (39 A+61 B x)+4 c^2 x (9 A-8 B x)+115 b^2 B\right )+a \left (10 b^3 c (9 A+53 B x)+4 b^2 c^2 x (43 B x-93 A)-8 b c^3 x^2 (15 A+7 B x)+16 c^4 x^3 (3 A+2 B x)-105 b^4 B\right )+b^2 x \left (5 b^2 c (18 A-7 B x)+2 b c^2 x (15 A+7 B x)-4 c^3 x^2 (3 A+2 B x)-105 b^3 B\right )\right )+3 \left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \left (24 a A c^2-60 a b B c-30 A b^2 c+35 b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{48 c^{9/2} \left (4 a c-b^2\right ) \sqrt {a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/(a + b*x + c*x^2)^(3/2),x]

[Out]

(2*Sqrt[c]*(-256*a^3*B*c^2 + b^2*x*(-105*b^3*B + 5*b^2*c*(18*A - 7*B*x) - 4*c^3*x^2*(3*A + 2*B*x) + 2*b*c^2*x*
(15*A + 7*B*x)) + a*(-105*b^4*B + 16*c^4*x^3*(3*A + 2*B*x) - 8*b*c^3*x^2*(15*A + 7*B*x) + 4*b^2*c^2*x*(-93*A +
43*B*x) + 10*b^3*c*(9*A + 53*B*x)) + 4*a^2*c*(115*b^2*B + 4*c^2*x*(9*A - 8*B*x) - 2*b*c*(39*A + 61*B*x))) + 3
*(b^2 - 4*a*c)*(35*b^3*B - 30*A*b^2*c - 60*a*b*B*c + 24*a*A*c^2)*Sqrt[a + x*(b + c*x)]*ArcTanh[(b + 2*c*x)/(2*
Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(48*c^(9/2)*(-b^2 + 4*a*c)*Sqrt[a + x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 1.35, size = 324, normalized size = 1.16 \begin {gather*} \frac {-256 a^3 B c^2-312 a^2 A b c^2+144 a^2 A c^3 x+460 a^2 b^2 B c-488 a^2 b B c^2 x-128 a^2 B c^3 x^2+90 a A b^3 c-372 a A b^2 c^2 x-120 a A b c^3 x^2+48 a A c^4 x^3-105 a b^4 B+530 a b^3 B c x+172 a b^2 B c^2 x^2-56 a b B c^3 x^3+32 a B c^4 x^4+90 A b^4 c x+30 A b^3 c^2 x^2-12 A b^2 c^3 x^3-105 b^5 B x-35 b^4 B c x^2+14 b^3 B c^2 x^3-8 b^2 B c^3 x^4}{24 c^4 \left (4 a c-b^2\right ) \sqrt {a+b x+c x^2}}+\frac {\left (24 a A c^2-60 a b B c-30 A b^2 c+35 b^3 B\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{16 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(A + B*x))/(a + b*x + c*x^2)^(3/2),x]

[Out]

(-105*a*b^4*B + 90*a*A*b^3*c + 460*a^2*b^2*B*c - 312*a^2*A*b*c^2 - 256*a^3*B*c^2 - 105*b^5*B*x + 90*A*b^4*c*x
+ 530*a*b^3*B*c*x - 372*a*A*b^2*c^2*x - 488*a^2*b*B*c^2*x + 144*a^2*A*c^3*x - 35*b^4*B*c*x^2 + 30*A*b^3*c^2*x^
2 + 172*a*b^2*B*c^2*x^2 - 120*a*A*b*c^3*x^2 - 128*a^2*B*c^3*x^2 + 14*b^3*B*c^2*x^3 - 12*A*b^2*c^3*x^3 - 56*a*b
*B*c^3*x^3 + 48*a*A*c^4*x^3 - 8*b^2*B*c^3*x^4 + 32*a*B*c^4*x^4)/(24*c^4*(-b^2 + 4*a*c)*Sqrt[a + b*x + c*x^2])
+ ((35*b^3*B - 30*A*b^2*c - 60*a*b*B*c + 24*a*A*c^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(16*c^(
9/2))

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fricas [A]  time = 0.81, size = 1035, normalized size = 3.70

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[1/96*(3*(35*B*a*b^5 - 96*A*a^3*c^3 + 48*(5*B*a^3*b + 3*A*a^2*b^2)*c^2 + (35*B*b^5*c - 96*A*a^2*c^4 + 48*(5*B*
a^2*b + 3*A*a*b^2)*c^3 - 10*(20*B*a*b^3 + 3*A*b^4)*c^2)*x^2 - 10*(20*B*a^2*b^3 + 3*A*a*b^4)*c + (35*B*b^6 - 96
*A*a^2*b*c^3 + 48*(5*B*a^2*b^2 + 3*A*a*b^3)*c^2 - 10*(20*B*a*b^4 + 3*A*b^5)*c)*x)*sqrt(c)*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*(105*B*a*b^4*c + 8*(B*b^2*c^4 - 4*B*a*c^
5)*x^4 + 8*(32*B*a^3 + 39*A*a^2*b)*c^3 - 2*(7*B*b^3*c^3 + 24*A*a*c^5 - 2*(14*B*a*b + 3*A*b^2)*c^4)*x^3 - 10*(4
6*B*a^2*b^2 + 9*A*a*b^3)*c^2 + (35*B*b^4*c^2 + 8*(16*B*a^2 + 15*A*a*b)*c^4 - 2*(86*B*a*b^2 + 15*A*b^3)*c^3)*x^
2 + (105*B*b^5*c - 144*A*a^2*c^4 + 4*(122*B*a^2*b + 93*A*a*b^2)*c^3 - 10*(53*B*a*b^3 + 9*A*b^4)*c^2)*x)*sqrt(c
*x^2 + b*x + a))/(a*b^2*c^5 - 4*a^2*c^6 + (b^2*c^6 - 4*a*c^7)*x^2 + (b^3*c^5 - 4*a*b*c^6)*x), 1/48*(3*(35*B*a*
b^5 - 96*A*a^3*c^3 + 48*(5*B*a^3*b + 3*A*a^2*b^2)*c^2 + (35*B*b^5*c - 96*A*a^2*c^4 + 48*(5*B*a^2*b + 3*A*a*b^2
)*c^3 - 10*(20*B*a*b^3 + 3*A*b^4)*c^2)*x^2 - 10*(20*B*a^2*b^3 + 3*A*a*b^4)*c + (35*B*b^6 - 96*A*a^2*b*c^3 + 48
*(5*B*a^2*b^2 + 3*A*a*b^3)*c^2 - 10*(20*B*a*b^4 + 3*A*b^5)*c)*x)*sqrt(-c)*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*(105*B*a*b^4*c + 8*(B*b^2*c^4 - 4*B*a*c^5)*x^4 + 8*(32*B*a^3 +
39*A*a^2*b)*c^3 - 2*(7*B*b^3*c^3 + 24*A*a*c^5 - 2*(14*B*a*b + 3*A*b^2)*c^4)*x^3 - 10*(46*B*a^2*b^2 + 9*A*a*b^3
)*c^2 + (35*B*b^4*c^2 + 8*(16*B*a^2 + 15*A*a*b)*c^4 - 2*(86*B*a*b^2 + 15*A*b^3)*c^3)*x^2 + (105*B*b^5*c - 144*
A*a^2*c^4 + 4*(122*B*a^2*b + 93*A*a*b^2)*c^3 - 10*(53*B*a*b^3 + 9*A*b^4)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2
*c^5 - 4*a^2*c^6 + (b^2*c^6 - 4*a*c^7)*x^2 + (b^3*c^5 - 4*a*b*c^6)*x)]

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giac [A]  time = 0.31, size = 367, normalized size = 1.31 \begin {gather*} \frac {{\left ({\left (2 \, {\left (\frac {4 \, {\left (B b^{2} c^{3} - 4 \, B a c^{4}\right )} x}{b^{2} c^{4} - 4 \, a c^{5}} - \frac {7 \, B b^{3} c^{2} - 28 \, B a b c^{3} - 6 \, A b^{2} c^{3} + 24 \, A a c^{4}}{b^{2} c^{4} - 4 \, a c^{5}}\right )} x + \frac {35 \, B b^{4} c - 172 \, B a b^{2} c^{2} - 30 \, A b^{3} c^{2} + 128 \, B a^{2} c^{3} + 120 \, A a b c^{3}}{b^{2} c^{4} - 4 \, a c^{5}}\right )} x + \frac {105 \, B b^{5} - 530 \, B a b^{3} c - 90 \, A b^{4} c + 488 \, B a^{2} b c^{2} + 372 \, A a b^{2} c^{2} - 144 \, A a^{2} c^{3}}{b^{2} c^{4} - 4 \, a c^{5}}\right )} x + \frac {105 \, B a b^{4} - 460 \, B a^{2} b^{2} c - 90 \, A a b^{3} c + 256 \, B a^{3} c^{2} + 312 \, A a^{2} b c^{2}}{b^{2} c^{4} - 4 \, a c^{5}}}{24 \, \sqrt {c x^{2} + b x + a}} + \frac {{\left (35 \, B b^{3} - 60 \, B a b c - 30 \, A b^{2} c + 24 \, A a c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

1/24*(((2*(4*(B*b^2*c^3 - 4*B*a*c^4)*x/(b^2*c^4 - 4*a*c^5) - (7*B*b^3*c^2 - 28*B*a*b*c^3 - 6*A*b^2*c^3 + 24*A*
a*c^4)/(b^2*c^4 - 4*a*c^5))*x + (35*B*b^4*c - 172*B*a*b^2*c^2 - 30*A*b^3*c^2 + 128*B*a^2*c^3 + 120*A*a*b*c^3)/
(b^2*c^4 - 4*a*c^5))*x + (105*B*b^5 - 530*B*a*b^3*c - 90*A*b^4*c + 488*B*a^2*b*c^2 + 372*A*a*b^2*c^2 - 144*A*a
^2*c^3)/(b^2*c^4 - 4*a*c^5))*x + (105*B*a*b^4 - 460*B*a^2*b^2*c - 90*A*a*b^3*c + 256*B*a^3*c^2 + 312*A*a^2*b*c
^2)/(b^2*c^4 - 4*a*c^5))/sqrt(c*x^2 + b*x + a) + 1/16*(35*B*b^3 - 60*B*a*b*c - 30*A*b^2*c + 24*A*a*c^2)*log(ab
s(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(9/2)

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maple [B]  time = 0.07, size = 800, normalized size = 2.86 \begin {gather*} \frac {B \,x^{4}}{3 \sqrt {c \,x^{2}+b x +a}\, c}-\frac {13 A a \,b^{2} x}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {15 A \,b^{4} x}{8 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {A \,x^{3}}{2 \sqrt {c \,x^{2}+b x +a}\, c}-\frac {16 B \,a^{2} b x}{3 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {115 B a \,b^{3} x}{12 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}-\frac {35 B \,b^{5} x}{16 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{4}}-\frac {7 B b \,x^{3}}{12 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {13 A a \,b^{3}}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {15 A \,b^{5}}{16 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{4}}-\frac {5 A b \,x^{2}}{4 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {8 B \,a^{2} b^{2}}{3 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {115 B a \,b^{4}}{24 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{4}}-\frac {4 B a \,x^{2}}{3 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {35 B \,b^{6}}{32 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{5}}+\frac {35 B \,b^{2} x^{2}}{24 \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {3 A a x}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {15 A \,b^{2} x}{8 \sqrt {c \,x^{2}+b x +a}\, c^{3}}-\frac {15 B a b x}{4 \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {35 B \,b^{3} x}{16 \sqrt {c \,x^{2}+b x +a}\, c^{4}}-\frac {3 A a \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {5}{2}}}+\frac {15 A \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {7}{2}}}+\frac {15 B a b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {7}{2}}}-\frac {35 B \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {9}{2}}}-\frac {13 A a b}{4 \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {15 A \,b^{3}}{16 \sqrt {c \,x^{2}+b x +a}\, c^{4}}-\frac {8 B \,a^{2}}{3 \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {115 B a \,b^{2}}{24 \sqrt {c \,x^{2}+b x +a}\, c^{4}}-\frac {35 B \,b^{4}}{32 \sqrt {c \,x^{2}+b x +a}\, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)/(c*x^2+b*x+a)^(3/2),x)

[Out]

15/16*A*b^5/c^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-13/4*A*b/c^3*a/(c*x^2+b*x+a)^(1/2)+3/2*A*a/c^2*x/(c*x^2+b*x+a)
^(1/2)-7/12*B*b/c^2*x^3/(c*x^2+b*x+a)^(1/2)-4/3*B*a/c^2*x^2/(c*x^2+b*x+a)^(1/2)+35/24*B*b^2/c^3*x^2/(c*x^2+b*x
+a)^(1/2)+35/16*B*b^3/c^4*x/(c*x^2+b*x+a)^(1/2)-35/32*B*b^6/c^5/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+15/8*A*b^4/c^3
/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-16/3*B*a^2/c^2*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+115/12*B*b^3/c^3*a/(4*a*
c-b^2)/(c*x^2+b*x+a)^(1/2)*x-13/2*A*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+1/2*A*x^3/c/(c*x^2+b*x+a)^(1/2
)+15/16*A*b^3/c^4/(c*x^2+b*x+a)^(1/2)+15/8*A*b^2/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/3*B*x^4
/c/(c*x^2+b*x+a)^(1/2)-35/32*B*b^4/c^5/(c*x^2+b*x+a)^(1/2)-35/16*B*b^3/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b
*x+a)^(1/2))-8/3*B*a^2/c^3/(c*x^2+b*x+a)^(1/2)-3/2*A*a/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-5/4
*A*b/c^2*x^2/(c*x^2+b*x+a)^(1/2)-15/8*A*b^2/c^3*x/(c*x^2+b*x+a)^(1/2)+115/24*B*b^2/c^4*a/(c*x^2+b*x+a)^(1/2)+1
5/4*B*b/c^(7/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-13/4*A*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)
-35/16*B*b^5/c^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+115/24*B*b^4/c^4*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-15/4*B*b
/c^3*a*x/(c*x^2+b*x+a)^(1/2)-8/3*B*a^2/c^3*b^2/(4*a*c-b^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(x^4*(B*x+A)/(c*x^2+b*x+a)^(3/2),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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x))/(a + b*x + c*x^2)^(3/2),x)

[Out]

int((x^4*(A + B*x))/(a + b*x + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \left (A + B x\right )}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(B*x+A)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral(x**4*(A + B*x)/(a + b*x + c*x**2)**(3/2), x)

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