### 3.2350 $$\int \frac{(a+b x+c x^2)^{3/2}}{(d+e x)^4} \, dx$$

Optimal. Leaf size=307 $-\frac{\sqrt{a+b x+c x^2} \left (e x \left (-4 c e (3 b d-2 a e)+b^2 e^2+12 c^2 d^2\right )-2 c d e (3 b d-2 a e)-b e^2 (b d-2 a e)+8 c^2 d^3\right )}{8 e^3 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}-\frac{(2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{16 e^4 \left (a e^2-b d e+c d^2\right )^{3/2}}+\frac{c^{3/2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{e^4}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^3}$

[Out]

-((8*c^2*d^3 - b*e^2*(b*d - 2*a*e) - 2*c*d*e*(3*b*d - 2*a*e) + e*(12*c^2*d^2 + b^2*e^2 - 4*c*e*(3*b*d - 2*a*e)
)*x)*Sqrt[a + b*x + c*x^2])/(8*e^3*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) - (a + b*x + c*x^2)^(3/2)/(3*e*(d + e*
x)^3) + (c^(3/2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/e^4 - ((2*c*d - b*e)*(8*c^2*d^2 - b^2
*e^2 - 4*c*e*(2*b*d - 3*a*e))*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a +
b*x + c*x^2])])/(16*e^4*(c*d^2 - b*d*e + a*e^2)^(3/2))

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Rubi [A]  time = 0.355388, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.273, Rules used = {732, 810, 843, 621, 206, 724} $-\frac{\sqrt{a+b x+c x^2} \left (e x \left (-4 c e (3 b d-2 a e)+b^2 e^2+12 c^2 d^2\right )-2 c d e (3 b d-2 a e)-b e^2 (b d-2 a e)+8 c^2 d^3\right )}{8 e^3 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}-\frac{(2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{16 e^4 \left (a e^2-b d e+c d^2\right )^{3/2}}+\frac{c^{3/2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{e^4}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(d + e*x)^4,x]

[Out]

-((8*c^2*d^3 - b*e^2*(b*d - 2*a*e) - 2*c*d*e*(3*b*d - 2*a*e) + e*(12*c^2*d^2 + b^2*e^2 - 4*c*e*(3*b*d - 2*a*e)
)*x)*Sqrt[a + b*x + c*x^2])/(8*e^3*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) - (a + b*x + c*x^2)^(3/2)/(3*e*(d + e*
x)^3) + (c^(3/2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/e^4 - ((2*c*d - b*e)*(8*c^2*d^2 - b^2
*e^2 - 4*c*e*(2*b*d - 3*a*e))*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a +
b*x + c*x^2])])/(16*e^4*(c*d^2 - b*d*e + a*e^2)^(3/2))

Rule 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 810

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*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 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 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 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{3/2}}{(d+e x)^4} \, dx &=-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^3}+\frac{\int \frac{(b+2 c x) \sqrt{a+b x+c x^2}}{(d+e x)^3} \, dx}{2 e}\\ &=-\frac{\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{8 e^3 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^3}-\frac{\int \frac{\frac{1}{2} \left (6 b^2 c d e+8 a c^2 d e+b^3 e^2-4 b c \left (2 c d^2+3 a e^2\right )\right )-8 c^2 \left (c d^2-b d e+a e^2\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{8 e^3 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{8 e^3 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^3}+\frac{c^2 \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{e^4}-\frac{\left (8 c^2 d \left (c d^2-b d e+a e^2\right )+\frac{1}{2} e \left (6 b^2 c d e+8 a c^2 d e+b^3 e^2-4 b c \left (2 c d^2+3 a e^2\right )\right )\right ) \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{8 e^4 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{8 e^3 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^3}+\frac{\left (2 c^2\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 )}{e^4}+\frac{\left (8 c^2 d \left (c d^2-b d e+a e^2\right )+\frac{1}{2} e \left (6 b^2 c d e+8 a c^2 d e+b^3 e^2-4 b c \left (2 c d^2+3 a e^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac{-b d+2 a e-(2 c d-b e) x}{\sqrt{a+b x+c x^2}}\right )}{4 e^4 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{8 e^3 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^3}+\frac{c^{3/2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{e^4}-\frac{(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) \tanh ^{-1}\left (\frac{b d-2 a e+(2 c d-b e) x}{2 \sqrt{c d^2-b d e+a e^2} \sqrt{a+b x+c x^2}}\right )}{16 e^4 \left (c d^2-b d e+a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.71053, size = 421, normalized size = 1.37 $\frac{\frac{3 \left (\frac{2 (a+x (b+c x))^{3/2} \left (4 c e (b d-2 a e)+b^2 e^2-4 c^2 d^2\right )}{d+e x}+\frac{2 \sqrt{a+x (b+c x)} \left (2 c^2 e (2 a e (2 e x-3 d)+b d (7 d-2 e x))-b c e^2 (-10 a e+5 b d+b e x)-b^3 e^3+4 c^3 d^2 (e x-2 d)\right )}{e^2}+\frac{(2 c d-b e) \left (4 c e (3 a e-2 b d)-b^2 e^2+8 c^2 d^2\right ) \sqrt{e (a e-b d)+c d^2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )+16 c^{3/2} \left (e (a e-b d)+c d^2\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{e^3}\right )}{8 \left (e (a e-b d)+c d^2\right )^2}+\frac{3 (a+x (b+c x))^{3/2} (2 c d-b e)}{2 (d+e x)^2 \left (e (a e-b d)+c d^2\right )}-\frac{2 (a+x (b+c x))^{3/2}}{(d+e x)^3}}{6 e}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(d + e*x)^4,x]

[Out]

((-2*(a + x*(b + c*x))^(3/2))/(d + e*x)^3 + (3*(2*c*d - b*e)*(a + x*(b + c*x))^(3/2))/(2*(c*d^2 + e*(-(b*d) +
a*e))*(d + e*x)^2) + (3*((2*(-4*c^2*d^2 + b^2*e^2 + 4*c*e*(b*d - 2*a*e))*(a + x*(b + c*x))^(3/2))/(d + e*x) +
(2*Sqrt[a + x*(b + c*x)]*(-(b^3*e^3) + 4*c^3*d^2*(-2*d + e*x) - b*c*e^2*(5*b*d - 10*a*e + b*e*x) + 2*c^2*e*(b*
d*(7*d - 2*e*x) + 2*a*e*(-3*d + 2*e*x))))/e^2 + (16*c^(3/2)*(c*d^2 + e*(-(b*d) + a*e))^2*ArcTanh[(b + 2*c*x)/(
2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + (2*c*d - b*e)*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*(8*c^2*d^2 - b^2*e^2 + 4*c*e*
(-2*b*d + 3*a*e))*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x + b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b +
c*x)])])/e^3))/(8*(c*d^2 + e*(-(b*d) + a*e))^2))/(6*e)

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Maple [B]  time = 0.246, size = 10401, normalized size = 33.9 \begin{align*} \text{output too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(3/2)/(e*x+d)^4,x)

[Out]

result too large to display

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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((c*x^2+b*x+a)^(3/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(3/2)/(e*x+d)**4,x)

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: TypeError