3.1582 \(\int \frac{(b+2 c x) (d+e x)^3}{(a+b x+c x^2)^{3/2}} \, dx\)

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

[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))

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Rubi [A]  time = 0.154252, antiderivative size = 153, normalized size of antiderivative = 1., 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} \[ \frac{3 e \left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{4 c^{5/2}}+\frac{9 e^2 \sqrt{a+b x+c x^2} (2 c d-b e)}{2 c^2}+\frac{3 e^2 (d+e x) \sqrt{a+b x+c x^2}}{c}-\frac{2 (d+e x)^3}{\sqrt{a+b x+c x^2}} \]

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 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]

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 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 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 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])

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*}

Mathematica [A]  time = 0.228671, size = 164, normalized size = 1.07 \[ \frac{3 e \left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{4 c^{5/2}}+\frac{3 c e^2 (2 a (4 d+e x)+b x (8 d-e x))-9 b e^3 (a+b x)-2 c^2 \left (6 d^2 e x+2 d^3-6 d e^2 x^2-e^3 x^3\right )}{2 c^2 \sqrt{a+x (b+c x)}} \]

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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Maple [B]  time = 0.009, size = 788, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

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]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 4.03729, size = 1328, normalized size = 8.68 \begin{align*} \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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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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Giac [B]  time = 1.35625, size = 468, normalized size = 3.06 \begin{align*} \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{align*}

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)