3.2383 \(\int \frac {(d+e x)^3}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=177 \[ \frac {e \sqrt {a+b x+c x^2} \left (-2 c e (4 a e+3 b d)+3 b^2 e^2+2 c e x (2 c d-b e)+8 c^2 d^2\right )}{c^2 \left (b^2-4 a c\right )}-\frac {2 (d+e x)^2 (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{5/2}} \]

[Out]

3/2*e^2*(-b*e+2*c*d)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)-2*(e*x+d)^2*(b*d-2*a*e+(-b*e+2
*c*d)*x)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(1/2)+e*(8*c^2*d^2+3*b^2*e^2-2*c*e*(4*a*e+3*b*d)+2*c*e*(-b*e+2*c*d)*x)*(c*
x^2+b*x+a)^(1/2)/c^2/(-4*a*c+b^2)

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Rubi [A]  time = 0.18, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {738, 779, 621, 206} \[ \frac {e \sqrt {a+b x+c x^2} \left (-2 c e (4 a e+3 b d)+3 b^2 e^2+2 c e x (2 c d-b e)+8 c^2 d^2\right )}{c^2 \left (b^2-4 a c\right )}-\frac {2 (d+e x)^2 (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a + b*x + c*x^2)^(3/2),x]

[Out]

(-2*(d + e*x)^2*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (e*(8*c^2*d^2 + 3*b^2
*e^2 - 2*c*e*(3*b*d + 4*a*e) + 2*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(c^2*(b^2 - 4*a*c)) + (3*e^2*(2*c
*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(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 738

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 + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

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]

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {(d+e x) (-2 e (b d-2 a e)-2 e (2 c d-b e) x)}{\sqrt {a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=-\frac {2 (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {e \left (8 c^2 d^2+3 b^2 e^2-2 c e (3 b d+4 a e)+2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{c^2 \left (b^2-4 a c\right )}+\frac {\left (3 e^2 (2 c d-b e)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 c^2}\\ &=-\frac {2 (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {e \left (8 c^2 d^2+3 b^2 e^2-2 c e (3 b d+4 a e)+2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{c^2 \left (b^2-4 a c\right )}+\frac {\left (3 e^2 (2 c d-b e)\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 )}{c^2}\\ &=-\frac {2 (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {e \left (8 c^2 d^2+3 b^2 e^2-2 c e (3 b d+4 a e)+2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{c^2 \left (b^2-4 a c\right )}+\frac {3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 196, normalized size = 1.11 \[ \frac {\frac {2 \sqrt {c} \left (4 c \left (2 a^2 e^3+a c e \left (-3 d^2-3 d e x+e^2 x^2\right )+c^2 d^3 x\right )-b^2 e^2 (3 a e+c x (e x-6 d))+2 b c \left (a e^2 (3 d+5 e x)+c d^2 (d-3 e x)\right )-3 b^3 e^3 x\right )}{\sqrt {a+x (b+c x)}}+3 e^2 \left (b^2-4 a c\right ) (b e-2 c d) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{2 c^{5/2} \left (4 a c-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[c]*(-3*b^3*e^3*x - b^2*e^2*(3*a*e + c*x*(-6*d + e*x)) + 2*b*c*(c*d^2*(d - 3*e*x) + a*e^2*(3*d + 5*e*x
)) + 4*c*(2*a^2*e^3 + c^2*d^3*x + a*c*e*(-3*d^2 - 3*d*e*x + e^2*x^2))))/Sqrt[a + x*(b + c*x)] + 3*(b^2 - 4*a*c
)*e^2*(-2*c*d + b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(2*c^(5/2)*(-b^2 + 4*a*c))

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fricas [B]  time = 1.42, size = 755, normalized size = 4.27 \[ \left [-\frac {3 \, {\left (2 \, {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} d e^{2} - {\left (a b^{3} - 4 \, a^{2} b c\right )} e^{3} + {\left (2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d e^{2} - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} x^{2} + {\left (2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} - {\left (b^{4} - 4 \, a b^{2} c\right )} e^{3}\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 (2 \, b c^{3} d^{3} - 12 \, a c^{3} d^{2} e + 6 \, a b c^{2} d e^{2} - {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{3} x^{2} - {\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} e^{3} + {\left (4 \, c^{4} d^{3} - 6 \, b c^{3} d^{2} e + 6 \, {\left (b^{2} c^{2} - 2 \, a c^{3}\right )} d e^{2} - {\left (3 \, b^{3} c - 10 \, a b c^{2}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{4 \, {\left (a b^{2} c^{3} - 4 \, a^{2} c^{4} + {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{2} + {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x\right )}}, -\frac {3 \, {\left (2 \, {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} d e^{2} - {\left (a b^{3} - 4 \, a^{2} b c\right )} e^{3} + {\left (2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d e^{2} - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} x^{2} + {\left (2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} - {\left (b^{4} - 4 \, a b^{2} c\right )} e^{3}\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 (2 \, b c^{3} d^{3} - 12 \, a c^{3} d^{2} e + 6 \, a b c^{2} d e^{2} - {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{3} x^{2} - {\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} e^{3} + {\left (4 \, c^{4} d^{3} - 6 \, b c^{3} d^{2} e + 6 \, {\left (b^{2} c^{2} - 2 \, a c^{3}\right )} d e^{2} - {\left (3 \, b^{3} c - 10 \, a b c^{2}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (a b^{2} c^{3} - 4 \, a^{2} c^{4} + {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{2} + {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(3*(2*(a*b^2*c - 4*a^2*c^2)*d*e^2 - (a*b^3 - 4*a^2*b*c)*e^3 + (2*(b^2*c^2 - 4*a*c^3)*d*e^2 - (b^3*c - 4*
a*b*c^2)*e^3)*x^2 + (2*(b^3*c - 4*a*b*c^2)*d*e^2 - (b^4 - 4*a*b^2*c)*e^3)*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*(2*b*c^3*d^3 - 12*a*c^3*d^2*e + 6*a*b*c^2*d*e
^2 - (b^2*c^2 - 4*a*c^3)*e^3*x^2 - (3*a*b^2*c - 8*a^2*c^2)*e^3 + (4*c^4*d^3 - 6*b*c^3*d^2*e + 6*(b^2*c^2 - 2*a
*c^3)*d*e^2 - (3*b^3*c - 10*a*b*c^2)*e^3)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^3 - 4*a^2*c^4 + (b^2*c^4 - 4*a*c^
5)*x^2 + (b^3*c^3 - 4*a*b*c^4)*x), -1/2*(3*(2*(a*b^2*c - 4*a^2*c^2)*d*e^2 - (a*b^3 - 4*a^2*b*c)*e^3 + (2*(b^2*
c^2 - 4*a*c^3)*d*e^2 - (b^3*c - 4*a*b*c^2)*e^3)*x^2 + (2*(b^3*c - 4*a*b*c^2)*d*e^2 - (b^4 - 4*a*b^2*c)*e^3)*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*(2*b*c^3*d^3 - 12
*a*c^3*d^2*e + 6*a*b*c^2*d*e^2 - (b^2*c^2 - 4*a*c^3)*e^3*x^2 - (3*a*b^2*c - 8*a^2*c^2)*e^3 + (4*c^4*d^3 - 6*b*
c^3*d^2*e + 6*(b^2*c^2 - 2*a*c^3)*d*e^2 - (3*b^3*c - 10*a*b*c^2)*e^3)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^3 - 4
*a^2*c^4 + (b^2*c^4 - 4*a*c^5)*x^2 + (b^3*c^3 - 4*a*b*c^4)*x)]

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giac [A]  time = 0.29, size = 233, normalized size = 1.32 \[ \frac {{\left (\frac {{\left (b^{2} c e^{3} - 4 \, a c^{2} e^{3}\right )} x}{b^{2} c^{2} - 4 \, a c^{3}} - \frac {4 \, c^{3} d^{3} - 6 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} - 12 \, a c^{2} d e^{2} - 3 \, b^{3} e^{3} + 10 \, a b c e^{3}}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x - \frac {2 \, b c^{2} d^{3} - 12 \, a c^{2} d^{2} e + 6 \, a b c d e^{2} - 3 \, a b^{2} e^{3} + 8 \, a^{2} c e^{3}}{b^{2} c^{2} - 4 \, a c^{3}}}{\sqrt {c x^{2} + b x + a}} - \frac {3 \, {\left (2 \, c d e^{2} - b e^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((b^2*c*e^3 - 4*a*c^2*e^3)*x/(b^2*c^2 - 4*a*c^3) - (4*c^3*d^3 - 6*b*c^2*d^2*e + 6*b^2*c*d*e^2 - 12*a*c^2*d*e^
2 - 3*b^3*e^3 + 10*a*b*c*e^3)/(b^2*c^2 - 4*a*c^3))*x - (2*b*c^2*d^3 - 12*a*c^2*d^2*e + 6*a*b*c*d*e^2 - 3*a*b^2
*e^3 + 8*a^2*c*e^3)/(b^2*c^2 - 4*a*c^3))/sqrt(c*x^2 + b*x + a) - 3/2*(2*c*d*e^2 - b*e^3)*log(abs(-2*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)

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maple [B]  time = 0.09, size = 541, normalized size = 3.06 \[ \frac {4 a b \,e^{3} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {3 b^{3} e^{3} x}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {3 b^{2} d \,e^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {6 b \,d^{2} e x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {2 a \,b^{2} e^{3}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 b^{4} e^{3}}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {3 b^{3} d \,e^{2}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 b^{2} d^{2} e}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {e^{3} x^{2}}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {3 b \,e^{3} x}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 d \,e^{2} x}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {2 \left (2 c x +b \right ) d^{3}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {3 b \,e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {5}{2}}}+\frac {3 d \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {2 a \,e^{3}}{\sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 b^{2} e^{3}}{4 \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {3 b d \,e^{2}}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 d^{2} e}{\sqrt {c \,x^{2}+b x +a}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^2+b*x+a)^(3/2),x)

[Out]

e^3*x^2/c/(c*x^2+b*x+a)^(1/2)+3/2*e^3*b/c^2*x/(c*x^2+b*x+a)^(1/2)-3/4*e^3*b^2/c^3/(c*x^2+b*x+a)^(1/2)-3/2*e^3*
b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-3/4*e^3*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-3/2*e^3*b/c^(5/2)*ln
((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+2*e^3*a/c^2/(c*x^2+b*x+a)^(1/2)+4*e^3*a/c*b/(4*a*c-b^2)/(c*x^2+b*x+a
)^(1/2)*x+2*e^3*a/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-3*d*e^2*x/c/(c*x^2+b*x+a)^(1/2)+3/2*d*e^2*b/c^2/(c*x
^2+b*x+a)^(1/2)+3*d*e^2*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+3/2*d*e^2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1
/2)+3*d*e^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3*d^2*e/c/(c*x^2+b*x+a)^(1/2)-6*d^2*e*b/(4*a*c
-b^2)/(c*x^2+b*x+a)^(1/2)*x-3*d^2*e*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+2*d^3*(2*c*x+b)/(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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(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 \[ \int \frac {{\left (d+e\,x\right )}^3}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + b*x + c*x^2)^(3/2),x)

[Out]

int((d + e*x)^3/(a + b*x + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{3}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((d + e*x)**3/(a + b*x + c*x**2)**(3/2), x)

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