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

Optimal. Leaf size=197 \[ \frac {3 \left (-4 a B c-4 A b c+5 b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{7/2}}-\frac {2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2} \left (-2 c x \left (-12 a B c-4 A b c+5 b^2 B\right )+32 a A c^2-52 a b B c-12 A b^2 c+15 b^3 B\right )}{4 c^3 \left (b^2-4 a c\right )} \]

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Rubi [A]  time = 0.14, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {818, 779, 621, 206} \begin {gather*} -\frac {\sqrt {a+b x+c x^2} \left (-2 c x \left (-12 a B c-4 A b c+5 b^2 B\right )+32 a A c^2-52 a b B c-12 A b^2 c+15 b^3 B\right )}{4 c^3 \left (b^2-4 a c\right )}+\frac {3 \left (-4 a B c-4 A b c+5 b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{7/2}}-\frac {2 x^2 \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^3*(A + B*x))/(a + b*x + c*x^2)^(3/2),x]

[Out]

(-2*x^2*(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]) - ((15*b^3*B
- 12*A*b^2*c - 52*a*b*B*c + 32*a*A*c^2 - 2*c*(5*b^2*B - 4*A*b*c - 12*a*B*c)*x)*Sqrt[a + b*x + c*x^2])/(4*c^3*(
b^2 - 4*a*c)) + (3*(5*b^2*B - 4*A*b*c - 4*a*B*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*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 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])

Rubi steps

\begin {align*} \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 x^2 \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 \left (2 a (b B-2 A c)+\frac {1}{2} \left (5 b^2 B-4 A b c-12 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^2 \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 (15 b^3 B-12 A b^2 c-52 a b B c+32 a A c^2-2 c \left (5 b^2 B-4 A b c-12 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac {\left (3 \left (5 b^2 B-4 A b c-4 a B c\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 c^3}\\ &=-\frac {2 x^2 \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 (15 b^3 B-12 A b^2 c-52 a b B c+32 a A c^2-2 c \left (5 b^2 B-4 A b c-12 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac {\left (3 \left (5 b^2 B-4 A b c-4 a B c\right )\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 )}{4 c^3}\\ &=-\frac {2 x^2 \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 (15 b^3 B-12 A b^2 c-52 a b B c+32 a A c^2-2 c \left (5 b^2 B-4 A b c-12 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac {3 \left (5 b^2 B-4 A b c-4 a B c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 221, normalized size = 1.12 \begin {gather*} \frac {2 \sqrt {c} \left (4 a^2 c (8 A c-13 b B+6 B c x)+a \left (-2 b^2 c (6 A+31 B x)-20 b c^2 x (B x-2 A)+8 c^3 x^2 (2 A+B x)+15 b^3 B\right )+b^2 x \left (b (5 B c x-12 A c)-2 c^2 x (2 A+B x)+15 b^2 B\right )\right )-3 \left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \left (-4 a B c-4 A b c+5 b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{8 c^{7/2} \left (4 a c-b^2\right ) \sqrt {a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*(4*a^2*c*(-13*b*B + 8*A*c + 6*B*c*x) + a*(15*b^3*B - 20*b*c^2*x*(-2*A + B*x) + 8*c^3*x^2*(2*A + B*x
) - 2*b^2*c*(6*A + 31*B*x)) + b^2*x*(15*b^2*B - 2*c^2*x*(2*A + B*x) + b*(-12*A*c + 5*B*c*x))) - 3*(b^2 - 4*a*c
)*(5*b^2*B - 4*A*b*c - 4*a*B*c)*Sqrt[a + x*(b + c*x)]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/
(8*c^(7/2)*(-b^2 + 4*a*c)*Sqrt[a + x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.97, size = 235, normalized size = 1.19 \begin {gather*} \frac {32 a^2 A c^2-52 a^2 b B c+24 a^2 B c^2 x-12 a A b^2 c+40 a A b c^2 x+16 a A c^3 x^2+15 a b^3 B-62 a b^2 B c x-20 a b B c^2 x^2+8 a B c^3 x^3-12 A b^3 c x-4 A b^2 c^2 x^2+15 b^4 B x+5 b^3 B c x^2-2 b^2 B c^2 x^3}{4 c^3 \left (4 a c-b^2\right ) \sqrt {a+b x+c x^2}}-\frac {3 \left (-4 a B c-4 A b c+5 b^2 B\right ) \log \left (-2 c^{7/2} \sqrt {a+b x+c x^2}+b c^3+2 c^4 x\right )}{8 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a*b^3*B - 12*a*A*b^2*c - 52*a^2*b*B*c + 32*a^2*A*c^2 + 15*b^4*B*x - 12*A*b^3*c*x - 62*a*b^2*B*c*x + 40*a*A
*b*c^2*x + 24*a^2*B*c^2*x + 5*b^3*B*c*x^2 - 4*A*b^2*c^2*x^2 - 20*a*b*B*c^2*x^2 + 16*a*A*c^3*x^2 - 2*b^2*B*c^2*
x^3 + 8*a*B*c^3*x^3)/(4*c^3*(-b^2 + 4*a*c)*Sqrt[a + b*x + c*x^2]) - (3*(5*b^2*B - 4*A*b*c - 4*a*B*c)*Log[b*c^3
+ 2*c^4*x - 2*c^(7/2)*Sqrt[a + b*x + c*x^2]])/(8*c^(7/2))

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fricas [B]  time = 0.71, size = 793, normalized size = 4.03 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B a b^{4} + 16 \, {\left (B a^{3} + A a^{2} b\right )} c^{2} + {\left (5 \, B b^{4} c + 16 \, {\left (B a^{2} + A a b\right )} c^{3} - 4 \, {\left (6 \, B a b^{2} + A b^{3}\right )} c^{2}\right )} x^{2} - 4 \, {\left (6 \, B a^{2} b^{2} + A a b^{3}\right )} c + {\left (5 \, B b^{5} + 16 \, {\left (B a^{2} b + A a b^{2}\right )} c^{2} - 4 \, {\left (6 \, B a b^{3} + A b^{4}\right )} c\right )} x\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (15 \, B a b^{3} c + 32 \, A a^{2} c^{3} - 2 \, {\left (B b^{2} c^{3} - 4 \, B a c^{4}\right )} x^{3} - 4 \, {\left (13 \, B a^{2} b + 3 \, A a b^{2}\right )} c^{2} + {\left (5 \, B b^{3} c^{2} + 16 \, A a c^{4} - 4 \, {\left (5 \, B a b + A b^{2}\right )} c^{3}\right )} x^{2} + {\left (15 \, B b^{4} c + 8 \, {\left (3 \, B a^{2} + 5 \, A a b\right )} c^{3} - 2 \, {\left (31 \, B a b^{2} + 6 \, A b^{3}\right )} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{16 \, {\left (a b^{2} c^{4} - 4 \, a^{2} c^{5} + {\left (b^{2} c^{5} - 4 \, a c^{6}\right )} x^{2} + {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} x\right )}}, -\frac {3 \, {\left (5 \, B a b^{4} + 16 \, {\left (B a^{3} + A a^{2} b\right )} c^{2} + {\left (5 \, B b^{4} c + 16 \, {\left (B a^{2} + A a b\right )} c^{3} - 4 \, {\left (6 \, B a b^{2} + A b^{3}\right )} c^{2}\right )} x^{2} - 4 \, {\left (6 \, B a^{2} b^{2} + A a b^{3}\right )} c + {\left (5 \, B b^{5} + 16 \, {\left (B a^{2} b + A a b^{2}\right )} c^{2} - 4 \, {\left (6 \, B a b^{3} + A b^{4}\right )} c\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (15 \, B a b^{3} c + 32 \, A a^{2} c^{3} - 2 \, {\left (B b^{2} c^{3} - 4 \, B a c^{4}\right )} x^{3} - 4 \, {\left (13 \, B a^{2} b + 3 \, A a b^{2}\right )} c^{2} + {\left (5 \, B b^{3} c^{2} + 16 \, A a c^{4} - 4 \, {\left (5 \, B a b + A b^{2}\right )} c^{3}\right )} x^{2} + {\left (15 \, B b^{4} c + 8 \, {\left (3 \, B a^{2} + 5 \, A a b\right )} c^{3} - 2 \, {\left (31 \, B a b^{2} + 6 \, A b^{3}\right )} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{8 \, {\left (a b^{2} c^{4} - 4 \, a^{2} c^{5} + {\left (b^{2} c^{5} - 4 \, a c^{6}\right )} x^{2} + {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(3*(5*B*a*b^4 + 16*(B*a^3 + A*a^2*b)*c^2 + (5*B*b^4*c + 16*(B*a^2 + A*a*b)*c^3 - 4*(6*B*a*b^2 + A*b^3)*
c^2)*x^2 - 4*(6*B*a^2*b^2 + A*a*b^3)*c + (5*B*b^5 + 16*(B*a^2*b + A*a*b^2)*c^2 - 4*(6*B*a*b^3 + A*b^4)*c)*x)*s
qrt(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*b^3*c
+ 32*A*a^2*c^3 - 2*(B*b^2*c^3 - 4*B*a*c^4)*x^3 - 4*(13*B*a^2*b + 3*A*a*b^2)*c^2 + (5*B*b^3*c^2 + 16*A*a*c^4 -
4*(5*B*a*b + A*b^2)*c^3)*x^2 + (15*B*b^4*c + 8*(3*B*a^2 + 5*A*a*b)*c^3 - 2*(31*B*a*b^2 + 6*A*b^3)*c^2)*x)*sqr
t(c*x^2 + b*x + a))/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x^2 + (b^3*c^4 - 4*a*b*c^5)*x), -1/8*(3*(5*B*
a*b^4 + 16*(B*a^3 + A*a^2*b)*c^2 + (5*B*b^4*c + 16*(B*a^2 + A*a*b)*c^3 - 4*(6*B*a*b^2 + A*b^3)*c^2)*x^2 - 4*(6
*B*a^2*b^2 + A*a*b^3)*c + (5*B*b^5 + 16*(B*a^2*b + A*a*b^2)*c^2 - 4*(6*B*a*b^3 + A*b^4)*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*b^3*c + 32*A*a^2*c^3 - 2*(
B*b^2*c^3 - 4*B*a*c^4)*x^3 - 4*(13*B*a^2*b + 3*A*a*b^2)*c^2 + (5*B*b^3*c^2 + 16*A*a*c^4 - 4*(5*B*a*b + A*b^2)*
c^3)*x^2 + (15*B*b^4*c + 8*(3*B*a^2 + 5*A*a*b)*c^3 - 2*(31*B*a*b^2 + 6*A*b^3)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(
a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x^2 + (b^3*c^4 - 4*a*b*c^5)*x)]

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giac [A]  time = 0.25, size = 268, normalized size = 1.36 \begin {gather*} \frac {{\left ({\left (\frac {2 \, {\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} x}{b^{2} c^{3} - 4 \, a c^{4}} - \frac {5 \, B b^{3} c - 20 \, B a b c^{2} - 4 \, A b^{2} c^{2} + 16 \, A a c^{3}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac {15 \, B b^{4} - 62 \, B a b^{2} c - 12 \, A b^{3} c + 24 \, B a^{2} c^{2} + 40 \, A a b c^{2}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac {15 \, B a b^{3} - 52 \, B a^{2} b c - 12 \, A a b^{2} c + 32 \, A a^{2} c^{2}}{b^{2} c^{3} - 4 \, a c^{4}}}{4 \, \sqrt {c x^{2} + b x + a}} - \frac {3 \, {\left (5 \, B b^{2} - 4 \, B a c - 4 \, A b c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{8 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(((2*(B*b^2*c^2 - 4*B*a*c^3)*x/(b^2*c^3 - 4*a*c^4) - (5*B*b^3*c - 20*B*a*b*c^2 - 4*A*b^2*c^2 + 16*A*a*c^3)
/(b^2*c^3 - 4*a*c^4))*x - (15*B*b^4 - 62*B*a*b^2*c - 12*A*b^3*c + 24*B*a^2*c^2 + 40*A*a*b*c^2)/(b^2*c^3 - 4*a*
c^4))*x - (15*B*a*b^3 - 52*B*a^2*b*c - 12*A*a*b^2*c + 32*A*a^2*c^2)/(b^2*c^3 - 4*a*c^4))/sqrt(c*x^2 + b*x + a)
- 3/8*(5*B*b^2 - 4*B*a*c - 4*A*b*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 = 576, normalized size = 2.92 \begin {gather*} \frac {4 A a b x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {3 A \,b^{3} x}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {13 B a \,b^{2} x}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {15 B \,b^{4} x}{8 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {B \,x^{3}}{2 \sqrt {c \,x^{2}+b x +a}\, c}+\frac {2 A a \,b^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 A \,b^{4}}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {A \,x^{2}}{\sqrt {c \,x^{2}+b x +a}\, c}-\frac {13 B a \,b^{3}}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {15 B \,b^{5}}{16 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{4}}-\frac {5 B b \,x^{2}}{4 \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {3 A b x}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {3 B a x}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {15 B \,b^{2} x}{8 \sqrt {c \,x^{2}+b x +a}\, c^{3}}-\frac {3 A b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {5}{2}}}-\frac {3 B 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 B \,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 {2 A a}{\sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 A \,b^{2}}{4 \sqrt {c \,x^{2}+b x +a}\, c^{3}}-\frac {13 B a b}{4 \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {15 B \,b^{3}}{16 \sqrt {c \,x^{2}+b x +a}\, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*x^3/c/(c*x^2+b*x+a)^(1/2)-5/4*B*b/c^2*x^2/(c*x^2+b*x+a)^(1/2)-15/8*B*b^2/c^3*x/(c*x^2+b*x+a)^(1/2)+15/16
*B*b^3/c^4/(c*x^2+b*x+a)^(1/2)+15/8*B*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+15/16*B*b^5/c^4/(4*a*c-b^2)/(c
*x^2+b*x+a)^(1/2)+15/8*B*b^2/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-13/4*B*b/c^3*a/(c*x^2+b*x+a)^
(1/2)-13/2*B*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-13/4*B*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+3/2*
B*a/c^2*x/(c*x^2+b*x+a)^(1/2)-3/2*B*a/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+A*x^2/c/(c*x^2+b*x+a
)^(1/2)+3/2*A*b/c^2*x/(c*x^2+b*x+a)^(1/2)-3/4*A*b^2/c^3/(c*x^2+b*x+a)^(1/2)-3/2*A*b^3/c^2/(4*a*c-b^2)/(c*x^2+b
*x+a)^(1/2)*x-3/4*A*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-3/2*A*b/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+
a)^(1/2))+2*A*a/c^2/(c*x^2+b*x+a)^(1/2)+4*A*a/c*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+2*A*a/c^2*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^3*(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 = -0.01 \begin {gather*} \int \frac {x^3\,\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^3*(A + B*x))/(a + b*x + c*x^2)^(3/2),x)

[Out]

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

[Out]

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

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