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

Optimal. Leaf size=153 \[ \frac {3 e \left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 c^{5/2}}+\frac {9 e^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{2 c^2}+\frac {3 e^2 (d+e x) \sqrt {a+b x+c x^2}}{c}-\frac {2 (d+e x)^3}{\sqrt {a+b x+c x^2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {768, 742, 640, 621, 206} \begin {gather*} \frac {3 e \left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 c^{5/2}}+\frac {9 e^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{2 c^2}+\frac {3 e^2 (d+e x) \sqrt {a+b x+c x^2}}{c}-\frac {2 (d+e x)^3}{\sqrt {a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^3)/Sqrt[a + b*x + c*x^2] + (9*e^2*(2*c*d - b*e)*Sqrt[a + b*x + c*x^2])/(2*c^2) + (3*e^2*(d + e*x
)*Sqrt[a + b*x + c*x^2])/c + (3*e*(8*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(2*b*d + a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]
*Sqrt[a + b*x + c*x^2])])/(4*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 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 742

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

Rule 768

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))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(b+2 c x) (d+e x)^3}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 (d+e x)^3}{\sqrt {a+b x+c x^2}}+(6 e) \int \frac {(d+e x)^2}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 (d+e x)^3}{\sqrt {a+b x+c x^2}}+\frac {3 e^2 (d+e x) \sqrt {a+b x+c x^2}}{c}+\frac {(3 e) \int \frac {\frac {1}{2} \left (4 c d^2-e (b d+2 a e)\right )+\frac {3}{2} e (2 c d-b e) x}{\sqrt {a+b x+c x^2}} \, dx}{c}\\ &=-\frac {2 (d+e x)^3}{\sqrt {a+b x+c x^2}}+\frac {9 e^2 (2 c d-b e) \sqrt {a+b x+c x^2}}{2 c^2}+\frac {3 e^2 (d+e x) \sqrt {a+b x+c x^2}}{c}+\frac {\left (3 e \left (8 c^2 d^2+3 b^2 e^2-4 c e (2 b d+a e)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{4 c^2}\\ &=-\frac {2 (d+e x)^3}{\sqrt {a+b x+c x^2}}+\frac {9 e^2 (2 c d-b e) \sqrt {a+b x+c x^2}}{2 c^2}+\frac {3 e^2 (d+e x) \sqrt {a+b x+c x^2}}{c}+\frac {\left (3 e \left (8 c^2 d^2+3 b^2 e^2-4 c e (2 b d+a e)\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 )}{2 c^2}\\ &=-\frac {2 (d+e x)^3}{\sqrt {a+b x+c x^2}}+\frac {9 e^2 (2 c d-b e) \sqrt {a+b x+c x^2}}{2 c^2}+\frac {3 e^2 (d+e x) \sqrt {a+b x+c x^2}}{c}+\frac {3 e \left (8 c^2 d^2+3 b^2 e^2-4 c e (2 b d+a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 164, normalized size = 1.07 \begin {gather*} \frac {3 e \left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{4 c^{5/2}}+\frac {3 c e^2 (2 a (4 d+e x)+b x (8 d-e x))-9 b e^3 (a+b x)-2 c^2 \left (2 d^3+6 d^2 e x-6 d e^2 x^2-e^3 x^3\right )}{2 c^2 \sqrt {a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*b*e^3*(a + b*x) - 2*c^2*(2*d^3 + 6*d^2*e*x - 6*d*e^2*x^2 - e^3*x^3) + 3*c*e^2*(b*x*(8*d - e*x) + 2*a*(4*d
+ e*x)))/(2*c^2*Sqrt[a + x*(b + c*x)]) + (3*e*(8*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(2*b*d + a*e))*ArcTanh[(b + 2*c*x
)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(4*c^(5/2))

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IntegrateAlgebraic [A]  time = 2.35, size = 191, normalized size = 1.25 \begin {gather*} \frac {-9 a b e^3+24 a c d e^2+6 a c e^3 x-9 b^2 e^3 x+24 b c d e^2 x-3 b c e^3 x^2-4 c^2 d^3-12 c^2 d^2 e x+12 c^2 d e^2 x^2+2 c^2 e^3 x^3}{2 c^2 \sqrt {a+b x+c x^2}}-\frac {3 \left (-4 a c e^3+3 b^2 e^3-8 b c d e^2+8 c^2 d^2 e\right ) \log \left (-2 c^{5/2} \sqrt {a+b x+c x^2}+b c^2+2 c^3 x\right )}{4 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*c^2*d^3 + 24*a*c*d*e^2 - 9*a*b*e^3 - 12*c^2*d^2*e*x + 24*b*c*d*e^2*x - 9*b^2*e^3*x + 6*a*c*e^3*x + 12*c^2*
d*e^2*x^2 - 3*b*c*e^3*x^2 + 2*c^2*e^3*x^3)/(2*c^2*Sqrt[a + b*x + c*x^2]) - (3*(8*c^2*d^2*e - 8*b*c*d*e^2 + 3*b
^2*e^3 - 4*a*c*e^3)*Log[b*c^2 + 2*c^3*x - 2*c^(5/2)*Sqrt[a + b*x + c*x^2]])/(4*c^(5/2))

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fricas [B]  time = 0.77, size = 615, normalized size = 4.02 \begin {gather*} \left [-\frac {3 \, {\left (8 \, a c^{2} d^{2} e - 8 \, a b c d e^{2} + {\left (3 \, a b^{2} - 4 \, a^{2} c\right )} e^{3} + {\left (8 \, c^{3} d^{2} e - 8 \, b c^{2} d e^{2} + {\left (3 \, b^{2} c - 4 \, a c^{2}\right )} e^{3}\right )} x^{2} + {\left (8 \, b c^{2} d^{2} e - 8 \, b^{2} c d e^{2} + {\left (3 \, b^{3} - 4 \, a b 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 \, c^{3} e^{3} x^{3} - 4 \, c^{3} d^{3} + 24 \, a c^{2} d e^{2} - 9 \, a b c e^{3} + 3 \, {\left (4 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{2} - 3 \, {\left (4 \, c^{3} d^{2} e - 8 \, b c^{2} d e^{2} + {\left (3 \, b^{2} c - 2 \, a c^{2}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{8 \, {\left (c^{4} x^{2} + b c^{3} x + a c^{3}\right )}}, -\frac {3 \, {\left (8 \, a c^{2} d^{2} e - 8 \, a b c d e^{2} + {\left (3 \, a b^{2} - 4 \, a^{2} c\right )} e^{3} + {\left (8 \, c^{3} d^{2} e - 8 \, b c^{2} d e^{2} + {\left (3 \, b^{2} c - 4 \, a c^{2}\right )} e^{3}\right )} x^{2} + {\left (8 \, b c^{2} d^{2} e - 8 \, b^{2} c d e^{2} + {\left (3 \, b^{3} - 4 \, a b 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 \, c^{3} e^{3} x^{3} - 4 \, c^{3} d^{3} + 24 \, a c^{2} d e^{2} - 9 \, a b c e^{3} + 3 \, {\left (4 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{2} - 3 \, {\left (4 \, c^{3} d^{2} e - 8 \, b c^{2} d e^{2} + {\left (3 \, b^{2} c - 2 \, a c^{2}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{4 \, {\left (c^{4} x^{2} + b c^{3} x + a c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*(8*a*c^2*d^2*e - 8*a*b*c*d*e^2 + (3*a*b^2 - 4*a^2*c)*e^3 + (8*c^3*d^2*e - 8*b*c^2*d*e^2 + (3*b^2*c -
4*a*c^2)*e^3)*x^2 + (8*b*c^2*d^2*e - 8*b^2*c*d*e^2 + (3*b^3 - 4*a*b*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*c^3*e^3*x^3 - 4*c^3*d^3 + 24*a*c^2*d*e^2
- 9*a*b*c*e^3 + 3*(4*c^3*d*e^2 - b*c^2*e^3)*x^2 - 3*(4*c^3*d^2*e - 8*b*c^2*d*e^2 + (3*b^2*c - 2*a*c^2)*e^3)*x
)*sqrt(c*x^2 + b*x + a))/(c^4*x^2 + b*c^3*x + a*c^3), -1/4*(3*(8*a*c^2*d^2*e - 8*a*b*c*d*e^2 + (3*a*b^2 - 4*a^
2*c)*e^3 + (8*c^3*d^2*e - 8*b*c^2*d*e^2 + (3*b^2*c - 4*a*c^2)*e^3)*x^2 + (8*b*c^2*d^2*e - 8*b^2*c*d*e^2 + (3*b
^3 - 4*a*b*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*c^3*e^3*x^3 - 4*c^3*d^3 + 24*a*c^2*d*e^2 - 9*a*b*c*e^3 + 3*(4*c^3*d*e^2 - b*c^2*e^3)*x^2 - 3*(4*c^3*d^2
*e - 8*b*c^2*d*e^2 + (3*b^2*c - 2*a*c^2)*e^3)*x)*sqrt(c*x^2 + b*x + a))/(c^4*x^2 + b*c^3*x + a*c^3)]

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giac [B]  time = 0.32, size = 347, normalized size = 2.27 \begin {gather*} \frac {{\left ({\left (\frac {2 \, {\left (b^{2} c^{2} e^{3} - 4 \, a c^{3} e^{3}\right )} x}{b^{2} c^{2} - 4 \, a c^{3}} + \frac {3 \, {\left (4 \, b^{2} c^{2} d e^{2} - 16 \, a c^{3} d e^{2} - b^{3} c e^{3} + 4 \, a b c^{2} e^{3}\right )}}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x - \frac {3 \, {\left (4 \, b^{2} c^{2} d^{2} e - 16 \, a c^{3} d^{2} e - 8 \, b^{3} c d e^{2} + 32 \, a b c^{2} d e^{2} + 3 \, b^{4} e^{3} - 14 \, a b^{2} c e^{3} + 8 \, a^{2} c^{2} e^{3}\right )}}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x - \frac {4 \, b^{2} c^{2} d^{3} - 16 \, a c^{3} d^{3} - 24 \, a b^{2} c d e^{2} + 96 \, a^{2} c^{2} d e^{2} + 9 \, a b^{3} e^{3} - 36 \, a^{2} b c e^{3}}{b^{2} c^{2} - 4 \, a c^{3}}}{2 \, \sqrt {c x^{2} + b x + a}} - \frac {3 \, {\left (8 \, c^{2} d^{2} e - 8 \, b c d e^{2} + 3 \, b^{2} e^{3} - 4 \, a c e^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{4 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(((2*(b^2*c^2*e^3 - 4*a*c^3*e^3)*x/(b^2*c^2 - 4*a*c^3) + 3*(4*b^2*c^2*d*e^2 - 16*a*c^3*d*e^2 - b^3*c*e^3 +
4*a*b*c^2*e^3)/(b^2*c^2 - 4*a*c^3))*x - 3*(4*b^2*c^2*d^2*e - 16*a*c^3*d^2*e - 8*b^3*c*d*e^2 + 32*a*b*c^2*d*e^
2 + 3*b^4*e^3 - 14*a*b^2*c*e^3 + 8*a^2*c^2*e^3)/(b^2*c^2 - 4*a*c^3))*x - (4*b^2*c^2*d^3 - 16*a*c^3*d^3 - 24*a*
b^2*c*d*e^2 + 96*a^2*c^2*d*e^2 + 9*a*b^3*e^3 - 36*a^2*b*c*e^3)/(b^2*c^2 - 4*a*c^3))/sqrt(c*x^2 + b*x + a) - 3/
4*(8*c^2*d^2*e - 8*b*c*d*e^2 + 3*b^2*e^3 - 4*a*c*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 = 0.06, size = 788, normalized size = 5.15 \begin {gather*} -\frac {9 a \,b^{2} e^{3} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {24 a b d \,e^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {9 b^{4} e^{3} x}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {6 b^{3} d \,e^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {4 b c \,d^{3} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {e^{3} x^{3}}{\sqrt {c \,x^{2}+b x +a}}-\frac {9 a \,b^{3} e^{3}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {12 a \,b^{2} d \,e^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {9 b^{5} e^{3}}{8 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}-\frac {3 b^{4} d \,e^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {2 b^{2} d^{3}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {3 b \,e^{3} x^{2}}{2 \sqrt {c \,x^{2}+b x +a}\, c}+\frac {6 d \,e^{2} x^{2}}{\sqrt {c \,x^{2}+b x +a}}+\frac {3 a \,e^{3} x}{\sqrt {c \,x^{2}+b x +a}\, c}-\frac {9 b^{2} e^{3} x}{4 \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {6 b d \,e^{2} x}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {2 \left (2 c x +b \right ) b \,d^{3}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {6 d^{2} e x}{\sqrt {c \,x^{2}+b x +a}}-\frac {3 a \,e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {9 b^{2} e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {5}{2}}}-\frac {6 b 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 {6 d^{2} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}-\frac {9 a b \,e^{3}}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {12 a d \,e^{2}}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {9 b^{3} e^{3}}{8 \sqrt {c \,x^{2}+b x +a}\, c^{3}}-\frac {3 b^{2} d \,e^{2}}{\sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {2 d^{3}}{\sqrt {c \,x^{2}+b x +a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

24*a*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*d*e^2+12*a/c*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d*e^2-6*b^3/c/(4*a*c
-b^2)/(c*x^2+b*x+a)^(1/2)*x*d*e^2-2/(c*x^2+b*x+a)^(1/2)*d^3+9/4/c^(5/2)*e^3*b^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+
b*x+a)^(1/2))+6/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d^2*e-6*x/(c*x^2+b*x+a)^(1/2)*d^2*e-2*b^2/
(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d^3+6*x^2/(c*x^2+b*x+a)^(1/2)*d*e^2+9/8/c^3*e^3*b^3/(c*x^2+b*x+a)^(1/2)-3/c^(3
/2)*e^3*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-9/2/c^2*e^3*b^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-3*b^2/
c^2/(c*x^2+b*x+a)^(1/2)*d*e^2-3/2/c*e^3*b*x^2/(c*x^2+b*x+a)^(1/2)-9/4/c^2*e^3*b^2*x/(c*x^2+b*x+a)^(1/2)+e^3*x^
3/(c*x^2+b*x+a)^(1/2)-9/c*e^3*b^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+3/c*e^3*a*x/(c*x^2+b*x+a)^(1/2)+12*a/c/(
c*x^2+b*x+a)^(1/2)*d*e^2-4*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*c*d^3-9/2/c^2*e^3*b*a/(c*x^2+b*x+a)^(1/2)+2*b*d
^3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-6*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*e^2+9/8
/c^3*e^3*b^5/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-3*b^4/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d*e^2+6*b/c*x/(c*x^2+b*
x+a)^(1/2)*d*e^2+9/4/c^2*e^3*b^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x

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

[Out]

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

[Out]

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

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