3.9.95 \(\int \frac {x^4 (A+B x)}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=285 \[ -\frac {2 x \left (x \left (32 a^2 B c^2+16 a A b c^2-32 a b^2 B c-2 A b^3 c+5 b^4 B\right )+a \left (24 a A c^2-28 a b B c-2 A b^2 c+5 b^3 B\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {\sqrt {a+b x+c x^2} \left (128 a^2 B c^2+40 a A b c^2-100 a b^2 B c-6 A b^3 c+15 b^4 B\right )}{3 c^3 \left (b^2-4 a c\right )^2}-\frac {2 x^3 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {(5 b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{7/2}} \]

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Rubi [A]  time = 0.27, antiderivative size = 285, 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 = {818, 640, 621, 206} \begin {gather*} -\frac {2 x \left (x \left (32 a^2 B c^2+16 a A b c^2-32 a b^2 B c-2 A b^3 c+5 b^4 B\right )+a \left (24 a A c^2-28 a b B c-2 A b^2 c+5 b^3 B\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {\sqrt {a+b x+c x^2} \left (128 a^2 B c^2+40 a A b c^2-100 a b^2 B c-6 A b^3 c+15 b^4 B\right )}{3 c^3 \left (b^2-4 a c\right )^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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {(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 )}{2 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*x^3*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x))/(3*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) - (2*x*(a
*(5*b^3*B - 2*A*b^2*c - 28*a*b*B*c + 24*a*A*c^2) + (5*b^4*B - 2*A*b^3*c - 32*a*b^2*B*c + 16*a*A*b*c^2 + 32*a^2
*B*c^2)*x))/(3*c^2*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2]) + ((15*b^4*B - 6*A*b^3*c - 100*a*b^2*B*c + 40*a*A*b*
c^2 + 128*a^2*B*c^2)*Sqrt[a + b*x + c*x^2])/(3*c^3*(b^2 - 4*a*c)^2) - ((5*b*B - 2*A*c)*ArcTanh[(b + 2*c*x)/(2*
Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*c^(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 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 640

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

Rubi steps

\begin {align*} \int \frac {x^4 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, 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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \int \frac {x^2 \left (3 a (b B-2 A c)+\frac {1}{2} \left (5 b^2 B-2 A b c-16 a B c\right ) x\right )}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 x \left (a \left (5 b^3 B-2 A b^2 c-28 a b B c+24 a A c^2\right )+\left (5 b^4 B-2 A b^3 c-32 a b^2 B c+16 a A b c^2+32 a^2 B c^2\right ) x\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {4 \int \frac {\frac {1}{2} a \left (5 b^3 B-2 A b^2 c-28 a b B c+24 a A c^2\right )+\frac {1}{4} \left (15 b^4 B-6 A b^3 c-100 a b^2 B c+40 a A b c^2+128 a^2 B c^2\right ) x}{\sqrt {a+b x+c x^2}} \, dx}{3 c^2 \left (b^2-4 a c\right )^2}\\ &=-\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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 x \left (a \left (5 b^3 B-2 A b^2 c-28 a b B c+24 a A c^2\right )+\left (5 b^4 B-2 A b^3 c-32 a b^2 B c+16 a A b c^2+32 a^2 B c^2\right ) x\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {\left (15 b^4 B-6 A b^3 c-100 a b^2 B c+40 a A b c^2+128 a^2 B c^2\right ) \sqrt {a+b x+c x^2}}{3 c^3 \left (b^2-4 a c\right )^2}-\frac {(5 b B-2 A c) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 c^3}\\ &=-\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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 x \left (a \left (5 b^3 B-2 A b^2 c-28 a b B c+24 a A c^2\right )+\left (5 b^4 B-2 A b^3 c-32 a b^2 B c+16 a A b c^2+32 a^2 B c^2\right ) x\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {\left (15 b^4 B-6 A b^3 c-100 a b^2 B c+40 a A b c^2+128 a^2 B c^2\right ) \sqrt {a+b x+c x^2}}{3 c^3 \left (b^2-4 a c\right )^2}-\frac {(5 b B-2 A c) \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 )}{c^3}\\ &=-\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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 x \left (a \left (5 b^3 B-2 A b^2 c-28 a b B c+24 a A c^2\right )+\left (5 b^4 B-2 A b^3 c-32 a b^2 B c+16 a A b c^2+32 a^2 B c^2\right ) x\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {\left (15 b^4 B-6 A b^3 c-100 a b^2 B c+40 a A b c^2+128 a^2 B c^2\right ) \sqrt {a+b x+c x^2}}{3 c^3 \left (b^2-4 a c\right )^2}-\frac {(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 )}{2 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.79, size = 414, normalized size = 1.45 \begin {gather*} \frac {2 \left (\frac {x^5 \left (A \left (16 a^2 c^2-28 a b^2 c-20 a b c^2 x+7 b^4+7 b^3 c x\right )+2 a B \left (16 a b c+12 a c^2 x-5 b^3-5 b^2 c x\right )\right )}{a \left (4 a c-b^2\right ) \sqrt {a+x (b+c x)}}+\frac {3 a^2 \left (b^2-4 a c\right )^2 (5 b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {c} \sqrt {a+x (b+c x)} \left (128 a^4 B c^2-4 a^3 c \left (-2 b c (5 A+7 B x)+4 c^2 x (3 A+4 B x)+25 b^2 B\right )+a^2 \left (-2 b^3 c (3 A+5 B x)+4 b^2 c^2 x (A+2 B x)+16 b c^3 x^2 (A+B x)+16 c^4 x^3 (2 A+3 B x)+15 b^4 B\right )-4 a b c^3 x^3 (4 A b+10 A c x+5 b B x)+14 A b^3 c^3 x^4\right )}{4 a c^{7/2} \left (4 a c-b^2\right )}+\frac {x^5 \left (A \left (-2 a c+b^2+b c x\right )-a B (b+2 c x)\right )}{(a+x (b+c x))^{3/2}}\right )}{3 a \left (b^2-4 a c\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(2*((x^5*(-(a*B*(b + 2*c*x)) + A*(b^2 - 2*a*c + b*c*x)))/(a + x*(b + c*x))^(3/2) + (x^5*(2*a*B*(-5*b^3 + 16*a*
b*c - 5*b^2*c*x + 12*a*c^2*x) + A*(7*b^4 - 28*a*b^2*c + 16*a^2*c^2 + 7*b^3*c*x - 20*a*b*c^2*x)))/(a*(-b^2 + 4*
a*c)*Sqrt[a + x*(b + c*x)]) + (-2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(128*a^4*B*c^2 + 14*A*b^3*c^3*x^4 - 4*a*b*c^3*
x^3*(4*A*b + 5*b*B*x + 10*A*c*x) + a^2*(15*b^4*B + 16*b*c^3*x^2*(A + B*x) + 4*b^2*c^2*x*(A + 2*B*x) + 16*c^4*x
^3*(2*A + 3*B*x) - 2*b^3*c*(3*A + 5*B*x)) - 4*a^3*c*(25*b^2*B + 4*c^2*x*(3*A + 4*B*x) - 2*b*c*(5*A + 7*B*x)))
+ 3*a^2*(b^2 - 4*a*c)^2*(5*b*B - 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(4*a*c^(7/2)*(
-b^2 + 4*a*c))))/(3*a*(b^2 - 4*a*c))

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IntegrateAlgebraic [A]  time = 2.35, size = 371, normalized size = 1.30 \begin {gather*} \frac {128 a^4 B c^2+40 a^3 A b c^2-48 a^3 A c^3 x-100 a^3 b^2 B c+312 a^3 b B c^2 x+192 a^3 B c^3 x^2-6 a^2 A b^3 c+84 a^2 A b^2 c^2 x-64 a^2 A c^4 x^3+15 a^2 b^4 B-210 a^2 b^3 B c x+48 a^2 b^2 B c^2 x^2+256 a^2 b B c^3 x^3+48 a^2 B c^4 x^4-12 a A b^4 c x+36 a A b^3 c^2 x^2+56 a A b^2 c^3 x^3+30 a b^5 B x-90 a b^4 B c x^2-148 a b^3 B c^2 x^3-24 a b^2 B c^3 x^4-6 A b^5 c x^2-8 A b^4 c^2 x^3+15 b^6 B x^2+20 b^5 B c x^3+3 b^4 B c^2 x^4}{3 c^3 \left (4 a c-b^2\right )^2 \left (a+b x+c x^2\right )^{3/2}}+\frac {(5 b B-2 A c) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{2 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(A + B*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(15*a^2*b^4*B - 6*a^2*A*b^3*c - 100*a^3*b^2*B*c + 40*a^3*A*b*c^2 + 128*a^4*B*c^2 + 30*a*b^5*B*x - 12*a*A*b^4*c
*x - 210*a^2*b^3*B*c*x + 84*a^2*A*b^2*c^2*x + 312*a^3*b*B*c^2*x - 48*a^3*A*c^3*x + 15*b^6*B*x^2 - 6*A*b^5*c*x^
2 - 90*a*b^4*B*c*x^2 + 36*a*A*b^3*c^2*x^2 + 48*a^2*b^2*B*c^2*x^2 + 192*a^3*B*c^3*x^2 + 20*b^5*B*c*x^3 - 8*A*b^
4*c^2*x^3 - 148*a*b^3*B*c^2*x^3 + 56*a*A*b^2*c^3*x^3 + 256*a^2*b*B*c^3*x^3 - 64*a^2*A*c^4*x^3 + 3*b^4*B*c^2*x^
4 - 24*a*b^2*B*c^3*x^4 + 48*a^2*B*c^4*x^4)/(3*c^3*(-b^2 + 4*a*c)^2*(a + b*x + c*x^2)^(3/2)) + ((5*b*B - 2*A*c)
*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(2*c^(7/2))

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fricas [B]  time = 1.48, size = 1601, normalized size = 5.62

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)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(5*B*a^2*b^5 - 32*A*a^4*c^3 + (5*B*b^5*c^2 - 32*A*a^2*c^5 + 16*(5*B*a^2*b + A*a*b^2)*c^4 - 2*(20*B*a
*b^3 + A*b^4)*c^3)*x^4 + 2*(5*B*b^6*c - 32*A*a^2*b*c^4 + 16*(5*B*a^2*b^2 + A*a*b^3)*c^3 - 2*(20*B*a*b^4 + A*b^
5)*c^2)*x^3 + 16*(5*B*a^4*b + A*a^3*b^2)*c^2 + (5*B*b^7 + 12*A*a*b^4*c^2 + 160*B*a^3*b*c^3 - 64*A*a^3*c^4 - 2*
(15*B*a*b^5 + A*b^6)*c)*x^2 - 2*(20*B*a^3*b^3 + A*a^2*b^4)*c + 2*(5*B*a*b^6 - 32*A*a^3*b*c^3 + 16*(5*B*a^3*b^2
+ A*a^2*b^3)*c^2 - 2*(20*B*a^2*b^4 + A*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*(15*B*a^2*b^4*c + 3*(B*b^4*c^3 - 8*B*a*b^2*c^4 + 16*B*a^2*c^5)*x^4 +
8*(16*B*a^4 + 5*A*a^3*b)*c^3 + 4*(5*B*b^5*c^2 - 16*A*a^2*c^5 + 2*(32*B*a^2*b + 7*A*a*b^2)*c^4 - (37*B*a*b^3 +
2*A*b^4)*c^3)*x^3 - 2*(50*B*a^3*b^2 + 3*A*a^2*b^3)*c^2 + 3*(5*B*b^6*c + 64*B*a^3*c^4 + 4*(4*B*a^2*b^2 + 3*A*a*
b^3)*c^3 - 2*(15*B*a*b^4 + A*b^5)*c^2)*x^2 + 6*(5*B*a*b^5*c - 8*A*a^3*c^4 + 2*(26*B*a^3*b + 7*A*a^2*b^2)*c^3 -
(35*B*a^2*b^3 + 2*A*a*b^4)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(a^2*b^4*c^4 - 8*a^3*b^2*c^5 + 16*a^4*c^6 + (b^4*c^
6 - 8*a*b^2*c^7 + 16*a^2*c^8)*x^4 + 2*(b^5*c^5 - 8*a*b^3*c^6 + 16*a^2*b*c^7)*x^3 + (b^6*c^4 - 6*a*b^4*c^5 + 32
*a^3*c^7)*x^2 + 2*(a*b^5*c^4 - 8*a^2*b^3*c^5 + 16*a^3*b*c^6)*x), 1/6*(3*(5*B*a^2*b^5 - 32*A*a^4*c^3 + (5*B*b^5
*c^2 - 32*A*a^2*c^5 + 16*(5*B*a^2*b + A*a*b^2)*c^4 - 2*(20*B*a*b^3 + A*b^4)*c^3)*x^4 + 2*(5*B*b^6*c - 32*A*a^2
*b*c^4 + 16*(5*B*a^2*b^2 + A*a*b^3)*c^3 - 2*(20*B*a*b^4 + A*b^5)*c^2)*x^3 + 16*(5*B*a^4*b + A*a^3*b^2)*c^2 + (
5*B*b^7 + 12*A*a*b^4*c^2 + 160*B*a^3*b*c^3 - 64*A*a^3*c^4 - 2*(15*B*a*b^5 + A*b^6)*c)*x^2 - 2*(20*B*a^3*b^3 +
A*a^2*b^4)*c + 2*(5*B*a*b^6 - 32*A*a^3*b*c^3 + 16*(5*B*a^3*b^2 + A*a^2*b^3)*c^2 - 2*(20*B*a^2*b^4 + A*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*(15*B*a^2*b^4
*c + 3*(B*b^4*c^3 - 8*B*a*b^2*c^4 + 16*B*a^2*c^5)*x^4 + 8*(16*B*a^4 + 5*A*a^3*b)*c^3 + 4*(5*B*b^5*c^2 - 16*A*a
^2*c^5 + 2*(32*B*a^2*b + 7*A*a*b^2)*c^4 - (37*B*a*b^3 + 2*A*b^4)*c^3)*x^3 - 2*(50*B*a^3*b^2 + 3*A*a^2*b^3)*c^2
+ 3*(5*B*b^6*c + 64*B*a^3*c^4 + 4*(4*B*a^2*b^2 + 3*A*a*b^3)*c^3 - 2*(15*B*a*b^4 + A*b^5)*c^2)*x^2 + 6*(5*B*a*
b^5*c - 8*A*a^3*c^4 + 2*(26*B*a^3*b + 7*A*a^2*b^2)*c^3 - (35*B*a^2*b^3 + 2*A*a*b^4)*c^2)*x)*sqrt(c*x^2 + b*x +
a))/(a^2*b^4*c^4 - 8*a^3*b^2*c^5 + 16*a^4*c^6 + (b^4*c^6 - 8*a*b^2*c^7 + 16*a^2*c^8)*x^4 + 2*(b^5*c^5 - 8*a*b
^3*c^6 + 16*a^2*b*c^7)*x^3 + (b^6*c^4 - 6*a*b^4*c^5 + 32*a^3*c^7)*x^2 + 2*(a*b^5*c^4 - 8*a^2*b^3*c^5 + 16*a^3*
b*c^6)*x)]

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giac [A]  time = 0.29, size = 458, normalized size = 1.61 \begin {gather*} \frac {{\left ({\left ({\left (\frac {3 \, {\left (B b^{4} c^{2} - 8 \, B a b^{2} c^{3} + 16 \, B a^{2} c^{4}\right )} x}{b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}} + \frac {4 \, {\left (5 \, B b^{5} c - 37 \, B a b^{3} c^{2} - 2 \, A b^{4} c^{2} + 64 \, B a^{2} b c^{3} + 14 \, A a b^{2} c^{3} - 16 \, A a^{2} c^{4}\right )}}{b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}}\right )} x + \frac {3 \, {\left (5 \, B b^{6} - 30 \, B a b^{4} c - 2 \, A b^{5} c + 16 \, B a^{2} b^{2} c^{2} + 12 \, A a b^{3} c^{2} + 64 \, B a^{3} c^{3}\right )}}{b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}}\right )} x + \frac {6 \, {\left (5 \, B a b^{5} - 35 \, B a^{2} b^{3} c - 2 \, A a b^{4} c + 52 \, B a^{3} b c^{2} + 14 \, A a^{2} b^{2} c^{2} - 8 \, A a^{3} c^{3}\right )}}{b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}}\right )} x + \frac {15 \, B a^{2} b^{4} - 100 \, B a^{3} b^{2} c - 6 \, A a^{2} b^{3} c + 128 \, B a^{4} c^{2} + 40 \, A a^{3} b c^{2}}{b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, B b - 2 \, A c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {7}{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)^(5/2),x, algorithm="giac")

[Out]

1/3*((((3*(B*b^4*c^2 - 8*B*a*b^2*c^3 + 16*B*a^2*c^4)*x/(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5) + 4*(5*B*b^5*c - 3
7*B*a*b^3*c^2 - 2*A*b^4*c^2 + 64*B*a^2*b*c^3 + 14*A*a*b^2*c^3 - 16*A*a^2*c^4)/(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*
c^5))*x + 3*(5*B*b^6 - 30*B*a*b^4*c - 2*A*b^5*c + 16*B*a^2*b^2*c^2 + 12*A*a*b^3*c^2 + 64*B*a^3*c^3)/(b^4*c^3 -
8*a*b^2*c^4 + 16*a^2*c^5))*x + 6*(5*B*a*b^5 - 35*B*a^2*b^3*c - 2*A*a*b^4*c + 52*B*a^3*b*c^2 + 14*A*a^2*b^2*c^
2 - 8*A*a^3*c^3)/(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5))*x + (15*B*a^2*b^4 - 100*B*a^3*b^2*c - 6*A*a^2*b^3*c + 1
28*B*a^4*c^2 + 40*A*a^3*b*c^2)/(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5))/(c*x^2 + b*x + a)^(3/2) + 1/2*(5*B*b - 2*
A*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)

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maple [B]  time = 0.06, size = 1262, normalized size = 4.43

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)/(c*x^2+b*x+a)^(5/2),x)

[Out]

4*B*a^2/c^2*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x-38/3*B*b^3/c^2*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x+1/2*A*b^2
/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+4*A*b^2/c*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x+32*B*a^2/c*b/(4*a*c-b
^2)^2/(c*x^2+b*x+a)^(1/2)*x-19/12*B*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x-5/4*B*b^4/c^4/(4*a*c-b^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))+B*x^4/c/(c*x^2+b*x+a)^(3/2)-1/3*A*x^3/c/
(c*x^2+b*x+a)^(3/2)-1/48*A*b^3/c^4/(c*x^2+b*x+a)^(3/2)-A/c^2*x/(c*x^2+b*x+a)^(1/2)+1/2*A/c^3*b/(c*x^2+b*x+a)^(
1/2)-1/3*A*b^4/c^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-1/24*A*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+5/6*B*
b^5/c^3/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x+B*a/c^3*b*x/(c*x^2+b*x+a)^(3/2)+2*B*a^2/c^3*b^2/(4*a*c-b^2)/(c*x^2
+b*x+a)^(3/2)-5/2*B*b^3/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-19/24*B*b^4/c^4*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2
)-19/3*B*b^4/c^3*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+1/4*A*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+2*A*b^3/c
^2*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+A/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+5/6*B*b/c^2*x^3/(c*x^2+b*x+
a)^(3/2)+8/3*B*a^2/c^3/(c*x^2+b*x+a)^(3/2)+5/96*B*b^4/c^5/(c*x^2+b*x+a)^(3/2)-5/4*B*b^2/c^4/(c*x^2+b*x+a)^(1/2
)-5/2*B*b/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+16*B*a^2/c^2*b^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/
2)+5/48*B*b^5/c^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x-5/4*B*b^2/c^3*x^2/(c*x^2+b*x+a)^(3/2)+4*B*a/c^2*x^2/(c*x^2
+b*x+a)^(3/2)-5/16*B*b^3/c^4*x/(c*x^2+b*x+a)^(3/2)+5/96*B*b^6/c^5/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+5/12*B*b^6/c
^4/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)-B*b^2/c^4*a/(c*x^2+b*x+a)^(3/2)+5/2*B*b/c^3*x/(c*x^2+b*x+a)^(1/2)+1/3*A*b
/c^3*a/(c*x^2+b*x+a)^(3/2)+1/8*A*b^2/c^3*x/(c*x^2+b*x+a)^(3/2)-1/48*A*b^5/c^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)-
1/6*A*b^5/c^3/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+1/2*A/c^3*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+1/2*A*b/c^2*x^2/
(c*x^2+b*x+a)^(3/2)

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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)^(5/2),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 )}^{5/2}} \,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)^(5/2),x)

[Out]

int((x^4*(A + B*x))/(a + b*x + c*x^2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**(5/2),x)

[Out]

Timed out

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