3.9.96 \(\int \frac {x^3 (A+B x)}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=189 \[ -\frac {2 \left (x \left (24 a^2 B c^2+8 a A b c^2-22 a b^2 B c+3 b^4 B\right )+a \left (16 a A c^2-20 a b B c+3 b^3 B\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {B \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 189, 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, 777, 621, 206} \begin {gather*} -\frac {2 \left (x \left (24 a^2 B c^2+8 a A b c^2-22 a b^2 B c+3 b^4 B\right )+a \left (16 a A c^2-20 a b B c+3 b^3 B\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {B \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*x^2*(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*(a*(
3*b^3*B - 20*a*b*B*c + 16*a*A*c^2) + (3*b^4*B - 22*a*b^2*B*c + 8*a*A*b*c^2 + 24*a^2*B*c^2)*x))/(3*c^2*(b^2 - 4
*a*c)^2*Sqrt[a + b*x + c*x^2]) + (B*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(5/2)

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 777

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((2
*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x)*(a + b*x + c*x^2)^
(p + 1))/(c*(p + 1)*(b^2 - 4*a*c)), 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))/(c*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N
eQ[b^2 - 4*a*c, 0] && LtQ[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 )^{5/2}} \, 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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \int \frac {x \left (2 a (b B-2 A c)+\frac {3}{2} B \left (b^2-4 a 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^2 \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 \left (a \left (3 b^3 B-20 a b B c+16 a A c^2\right )+\left (3 b^4 B-22 a b^2 B c+8 a A b c^2+24 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 {B \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{c^2}\\ &=-\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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (a \left (3 b^3 B-20 a b B c+16 a A c^2\right )+\left (3 b^4 B-22 a b^2 B c+8 a A b c^2+24 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 {(2 B) \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^2}\\ &=-\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 )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (a \left (3 b^3 B-20 a b B c+16 a A c^2\right )+\left (3 b^4 B-22 a b^2 B c+8 a A b c^2+24 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 {B \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 201, normalized size = 1.06 \begin {gather*} \frac {B \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{c^{5/2}}-\frac {2 \left (4 a^3 c (4 A c-5 b B+6 B c x)+a^2 \left (24 A b c^2 x+8 c^3 x^2 (3 A+4 B x)+3 b^3 B-42 b^2 B c x\right )+2 a b x \left (b c^2 x (3 A-14 B x)+6 A c^3 x^2+3 b^3 B-9 b^2 B c x\right )+b^3 x^2 \left (-A c^2 x+3 b^2 B+4 b B c x\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(4*a^3*c*(-5*b*B + 4*A*c + 6*B*c*x) + b^3*x^2*(3*b^2*B + 4*b*B*c*x - A*c^2*x) + 2*a*b*x*(3*b^3*B - 9*b^2*B
*c*x + 6*A*c^3*x^2 + b*c^2*x*(3*A - 14*B*x)) + a^2*(3*b^3*B - 42*b^2*B*c*x + 24*A*b*c^2*x + 8*c^3*x^2*(3*A + 4
*B*x))))/(3*c^2*(b^2 - 4*a*c)^2*(a + x*(b + c*x))^(3/2)) + (B*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c
*x)])])/c^(5/2)

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IntegrateAlgebraic [A]  time = 1.73, size = 246, normalized size = 1.30 \begin {gather*} -\frac {2 \left (16 a^3 A c^2-20 a^3 b B c+24 a^3 B c^2 x+24 a^2 A b c^2 x+24 a^2 A c^3 x^2+3 a^2 b^3 B-42 a^2 b^2 B c x+32 a^2 B c^3 x^3+6 a A b^2 c^2 x^2+12 a A b c^3 x^3+6 a b^4 B x-18 a b^3 B c x^2-28 a b^2 B c^2 x^3-A b^3 c^2 x^3+3 b^5 B x^2+4 b^4 B c x^3\right )}{3 c^2 \left (4 a c-b^2\right )^2 \left (a+b x+c x^2\right )^{3/2}}-\frac {B \log \left (-2 c^{5/2} \sqrt {a+b x+c x^2}+b c^2+2 c^3 x\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^3*(A + B*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(3*a^2*b^3*B - 20*a^3*b*B*c + 16*a^3*A*c^2 + 6*a*b^4*B*x - 42*a^2*b^2*B*c*x + 24*a^2*A*b*c^2*x + 24*a^3*B*
c^2*x + 3*b^5*B*x^2 - 18*a*b^3*B*c*x^2 + 6*a*A*b^2*c^2*x^2 + 24*a^2*A*c^3*x^2 + 4*b^4*B*c*x^3 - A*b^3*c^2*x^3
- 28*a*b^2*B*c^2*x^3 + 12*a*A*b*c^3*x^3 + 32*a^2*B*c^3*x^3))/(3*c^2*(-b^2 + 4*a*c)^2*(a + b*x + c*x^2)^(3/2))
- (B*Log[b*c^2 + 2*c^3*x - 2*c^(5/2)*Sqrt[a + b*x + c*x^2]])/c^(5/2)

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fricas [B]  time = 1.49, size = 1061, normalized size = 5.61 \begin {gather*} \left [\frac {3 \, {\left (B a^{2} b^{4} - 8 \, B a^{3} b^{2} c + 16 \, B a^{4} c^{2} + {\left (B b^{4} c^{2} - 8 \, B a b^{2} c^{3} + 16 \, B a^{2} c^{4}\right )} x^{4} + 2 \, {\left (B b^{5} c - 8 \, B a b^{3} c^{2} + 16 \, B a^{2} b c^{3}\right )} x^{3} + {\left (B b^{6} - 6 \, B a b^{4} c + 32 \, B a^{3} c^{3}\right )} x^{2} + 2 \, {\left (B a b^{5} - 8 \, B a^{2} b^{3} c + 16 \, B a^{3} b c^{2}\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 (3 \, B a^{2} b^{3} c - 20 \, B a^{3} b c^{2} + 16 \, A a^{3} c^{3} + {\left (4 \, B b^{4} c^{2} + 4 \, {\left (8 \, B a^{2} + 3 \, A a b\right )} c^{4} - {\left (28 \, B a b^{2} + A b^{3}\right )} c^{3}\right )} x^{3} + 3 \, {\left (B b^{5} c - 6 \, B a b^{3} c^{2} + 2 \, A a b^{2} c^{3} + 8 \, A a^{2} c^{4}\right )} x^{2} + 6 \, {\left (B a b^{4} c - 7 \, B a^{2} b^{2} c^{2} + 4 \, {\left (B a^{3} + A a^{2} b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{6 \, {\left (a^{2} b^{4} c^{3} - 8 \, a^{3} b^{2} c^{4} + 16 \, a^{4} c^{5} + {\left (b^{4} c^{5} - 8 \, a b^{2} c^{6} + 16 \, a^{2} c^{7}\right )} x^{4} + 2 \, {\left (b^{5} c^{4} - 8 \, a b^{3} c^{5} + 16 \, a^{2} b c^{6}\right )} x^{3} + {\left (b^{6} c^{3} - 6 \, a b^{4} c^{4} + 32 \, a^{3} c^{6}\right )} x^{2} + 2 \, {\left (a b^{5} c^{3} - 8 \, a^{2} b^{3} c^{4} + 16 \, a^{3} b c^{5}\right )} x\right )}}, -\frac {3 \, {\left (B a^{2} b^{4} - 8 \, B a^{3} b^{2} c + 16 \, B a^{4} c^{2} + {\left (B b^{4} c^{2} - 8 \, B a b^{2} c^{3} + 16 \, B a^{2} c^{4}\right )} x^{4} + 2 \, {\left (B b^{5} c - 8 \, B a b^{3} c^{2} + 16 \, B a^{2} b c^{3}\right )} x^{3} + {\left (B b^{6} - 6 \, B a b^{4} c + 32 \, B a^{3} c^{3}\right )} x^{2} + 2 \, {\left (B a b^{5} - 8 \, B a^{2} b^{3} c + 16 \, B a^{3} b c^{2}\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 (3 \, B a^{2} b^{3} c - 20 \, B a^{3} b c^{2} + 16 \, A a^{3} c^{3} + {\left (4 \, B b^{4} c^{2} + 4 \, {\left (8 \, B a^{2} + 3 \, A a b\right )} c^{4} - {\left (28 \, B a b^{2} + A b^{3}\right )} c^{3}\right )} x^{3} + 3 \, {\left (B b^{5} c - 6 \, B a b^{3} c^{2} + 2 \, A a b^{2} c^{3} + 8 \, A a^{2} c^{4}\right )} x^{2} + 6 \, {\left (B a b^{4} c - 7 \, B a^{2} b^{2} c^{2} + 4 \, {\left (B a^{3} + A a^{2} b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (a^{2} b^{4} c^{3} - 8 \, a^{3} b^{2} c^{4} + 16 \, a^{4} c^{5} + {\left (b^{4} c^{5} - 8 \, a b^{2} c^{6} + 16 \, a^{2} c^{7}\right )} x^{4} + 2 \, {\left (b^{5} c^{4} - 8 \, a b^{3} c^{5} + 16 \, a^{2} b c^{6}\right )} x^{3} + {\left (b^{6} c^{3} - 6 \, a b^{4} c^{4} + 32 \, a^{3} c^{6}\right )} x^{2} + 2 \, {\left (a b^{5} c^{3} - 8 \, a^{2} b^{3} c^{4} + 16 \, a^{3} 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)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(B*a^2*b^4 - 8*B*a^3*b^2*c + 16*B*a^4*c^2 + (B*b^4*c^2 - 8*B*a*b^2*c^3 + 16*B*a^2*c^4)*x^4 + 2*(B*b^5*
c - 8*B*a*b^3*c^2 + 16*B*a^2*b*c^3)*x^3 + (B*b^6 - 6*B*a*b^4*c + 32*B*a^3*c^3)*x^2 + 2*(B*a*b^5 - 8*B*a^2*b^3*
c + 16*B*a^3*b*c^2)*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*(3*B*a^2*b^3*c - 20*B*a^3*b*c^2 + 16*A*a^3*c^3 + (4*B*b^4*c^2 + 4*(8*B*a^2 + 3*A*a*b)*c^4 - (28*B*a
*b^2 + A*b^3)*c^3)*x^3 + 3*(B*b^5*c - 6*B*a*b^3*c^2 + 2*A*a*b^2*c^3 + 8*A*a^2*c^4)*x^2 + 6*(B*a*b^4*c - 7*B*a^
2*b^2*c^2 + 4*(B*a^3 + A*a^2*b)*c^3)*x)*sqrt(c*x^2 + b*x + a))/(a^2*b^4*c^3 - 8*a^3*b^2*c^4 + 16*a^4*c^5 + (b^
4*c^5 - 8*a*b^2*c^6 + 16*a^2*c^7)*x^4 + 2*(b^5*c^4 - 8*a*b^3*c^5 + 16*a^2*b*c^6)*x^3 + (b^6*c^3 - 6*a*b^4*c^4
+ 32*a^3*c^6)*x^2 + 2*(a*b^5*c^3 - 8*a^2*b^3*c^4 + 16*a^3*b*c^5)*x), -1/3*(3*(B*a^2*b^4 - 8*B*a^3*b^2*c + 16*B
*a^4*c^2 + (B*b^4*c^2 - 8*B*a*b^2*c^3 + 16*B*a^2*c^4)*x^4 + 2*(B*b^5*c - 8*B*a*b^3*c^2 + 16*B*a^2*b*c^3)*x^3 +
(B*b^6 - 6*B*a*b^4*c + 32*B*a^3*c^3)*x^2 + 2*(B*a*b^5 - 8*B*a^2*b^3*c + 16*B*a^3*b*c^2)*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*(3*B*a^2*b^3*c - 20*B*a^3*b*c^2 + 16
*A*a^3*c^3 + (4*B*b^4*c^2 + 4*(8*B*a^2 + 3*A*a*b)*c^4 - (28*B*a*b^2 + A*b^3)*c^3)*x^3 + 3*(B*b^5*c - 6*B*a*b^3
*c^2 + 2*A*a*b^2*c^3 + 8*A*a^2*c^4)*x^2 + 6*(B*a*b^4*c - 7*B*a^2*b^2*c^2 + 4*(B*a^3 + A*a^2*b)*c^3)*x)*sqrt(c*
x^2 + b*x + a))/(a^2*b^4*c^3 - 8*a^3*b^2*c^4 + 16*a^4*c^5 + (b^4*c^5 - 8*a*b^2*c^6 + 16*a^2*c^7)*x^4 + 2*(b^5*
c^4 - 8*a*b^3*c^5 + 16*a^2*b*c^6)*x^3 + (b^6*c^3 - 6*a*b^4*c^4 + 32*a^3*c^6)*x^2 + 2*(a*b^5*c^3 - 8*a^2*b^3*c^
4 + 16*a^3*b*c^5)*x)]

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giac [A]  time = 0.28, size = 314, normalized size = 1.66 \begin {gather*} -\frac {2 \, {\left ({\left ({\left (\frac {{\left (4 \, B b^{4} c - 28 \, B a b^{2} c^{2} - A b^{3} c^{2} + 32 \, B a^{2} c^{3} + 12 \, A a b c^{3}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac {3 \, {\left (B b^{5} - 6 \, B a b^{3} c + 2 \, A a b^{2} c^{2} + 8 \, A a^{2} c^{3}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac {6 \, {\left (B a b^{4} - 7 \, B a^{2} b^{2} c + 4 \, B a^{3} c^{2} + 4 \, A a^{2} b c^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac {3 \, B a^{2} b^{3} - 20 \, B a^{3} b c + 16 \, A a^{3} c^{2}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} - \frac {B \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {5}{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)^(5/2),x, algorithm="giac")

[Out]

-2/3*((((4*B*b^4*c - 28*B*a*b^2*c^2 - A*b^3*c^2 + 32*B*a^2*c^3 + 12*A*a*b*c^3)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a
^2*c^4) + 3*(B*b^5 - 6*B*a*b^3*c + 2*A*a*b^2*c^2 + 8*A*a^2*c^3)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + 6*(B
*a*b^4 - 7*B*a^2*b^2*c + 4*B*a^3*c^2 + 4*A*a^2*b*c^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + (3*B*a^2*b^3 -
20*B*a^3*b*c + 16*A*a^3*c^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2) - B*log(abs(-2*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)

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maple [B]  time = 0.06, size = 860, normalized size = 4.55 \begin {gather*} -\frac {8 A a b x}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}+\frac {2 A \,b^{3} x}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, c}+\frac {4 B a \,b^{2} x}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, c}-\frac {B \,b^{4} x}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {4 A a \,b^{2}}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, c}-\frac {A a b x}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c}+\frac {A \,b^{4}}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {A \,b^{3} x}{12 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{2}}+\frac {2 B a \,b^{3}}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {B a \,b^{2} x}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{2}}-\frac {B \,b^{5}}{6 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, c^{3}}-\frac {B \,b^{4} x}{24 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{3}}-\frac {B \,x^{3}}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c}-\frac {A a \,b^{2}}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{2}}+\frac {A \,b^{4}}{24 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{3}}-\frac {A \,x^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c}+\frac {B a \,b^{3}}{4 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{3}}-\frac {B \,b^{5}}{48 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{4}}+\frac {B \,b^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {B b \,x^{2}}{2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{2}}-\frac {A b x}{4 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{2}}+\frac {B \,b^{3}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {B \,b^{2} x}{8 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{3}}-\frac {2 A a}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{2}}+\frac {A \,b^{2}}{24 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{3}}+\frac {B a b}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{3}}-\frac {B \,b^{3}}{48 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{4}}-\frac {B x}{\sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {B \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {5}{2}}}+\frac {B b}{2 \sqrt {c \,x^{2}+b x +a}\, c^{3}} \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)^(5/2),x)

[Out]

1/2*B/c^3*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-1/4*A*b/c^2*x/(c*x^2+b*x+a)^(3/2)-4*A*b^2/c*a/(4*a*c-b^2)^2/(c*x
^2+b*x+a)^(1/2)+2/3*A*b^3/c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x+4*B*b^2/c*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*
x-A*b/c*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+1/2*B*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+1/2*B/c^3*b/(c*x
^2+b*x+a)^(1/2)-1/3*B*x^3/c/(c*x^2+b*x+a)^(3/2)-1/48*B*b^3/c^4/(c*x^2+b*x+a)^(3/2)-B/c^2*x/(c*x^2+b*x+a)^(1/2)
-A*x^2/c/(c*x^2+b*x+a)^(3/2)-2/3*A*a/c^2/(c*x^2+b*x+a)^(3/2)+1/24*A*b^2/c^3/(c*x^2+b*x+a)^(3/2)+B/c^(5/2)*ln((
c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/12*A*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+B/c^2*b^2/(4*a*c-b^2)
/(c*x^2+b*x+a)^(1/2)*x+1/4*B*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+2*B*b^3/c^2*a/(4*a*c-b^2)^2/(c*x^2+b*x+
a)^(1/2)+1/8*B*b^2/c^3*x/(c*x^2+b*x+a)^(3/2)+1/2*B*b/c^2*x^2/(c*x^2+b*x+a)^(3/2)+1/24*A*b^4/c^3/(4*a*c-b^2)/(c
*x^2+b*x+a)^(3/2)+1/3*A*b^4/c^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)-1/2*A*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3
/2)-8*A*b*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-1/24*B*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x-1/3*B*b^4/c^2
/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-1/48*B*b^5/c^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)-1/6*B*b^5/c^3/(4*a*c-b^2)^
2/(c*x^2+b*x+a)^(1/2)+1/3*B*b/c^3*a/(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^3*(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 = -0.01 \begin {gather*} \int \frac {x^3\,\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^3*(A + B*x))/(a + b*x + c*x^2)^(5/2),x)

[Out]

int((x^3*(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**3*(B*x+A)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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