### 3.2360 $$\int \frac{(a+b x+c x^2)^{5/2}}{(d+e x)^2} \, dx$$

Optimal. Leaf size=388 $\frac{5 \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{3/2} e^6}-\frac{5 \sqrt{a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{64 c e^5}-\frac{5 (2 c d-b e) \left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{2 e^6}-\frac{5 \left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{24 e^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{e (d+e x)}$

[Out]

(-5*(64*c^3*d^3 - b^3*e^3 + 4*b*c*e^2*(12*b*d - 11*a*e) - 16*c^2*d*e*(7*b*d - 4*a*e) - 2*c*e*(16*c^2*d^2 + b^2
*e^2 - 4*c*e*(4*b*d - 3*a*e))*x)*Sqrt[a + b*x + c*x^2])/(64*c*e^5) - (5*(8*c*d - 7*b*e - 6*c*e*x)*(a + b*x + c
*x^2)^(3/2))/(24*e^3) - (a + b*x + c*x^2)^(5/2)/(e*(d + e*x)) + (5*(128*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(2*b*d
- 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*c^2*e^2*(3*b^2*d^2 - 4*a*b*d*e + a^2*e^2))*ArcTanh[(b + 2*c*x)/(
2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(3/2)*e^6) - (5*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)^(3/2)*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])])/(2*e^6)

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Rubi [A]  time = 0.663218, antiderivative size = 388, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.273, Rules used = {732, 814, 843, 621, 206, 724} $\frac{5 \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{3/2} e^6}-\frac{5 \sqrt{a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{64 c e^5}-\frac{5 (2 c d-b e) \left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{2 e^6}-\frac{5 \left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{24 e^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{e (d+e x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(d + e*x)^2,x]

[Out]

(-5*(64*c^3*d^3 - b^3*e^3 + 4*b*c*e^2*(12*b*d - 11*a*e) - 16*c^2*d*e*(7*b*d - 4*a*e) - 2*c*e*(16*c^2*d^2 + b^2
*e^2 - 4*c*e*(4*b*d - 3*a*e))*x)*Sqrt[a + b*x + c*x^2])/(64*c*e^5) - (5*(8*c*d - 7*b*e - 6*c*e*x)*(a + b*x + c
*x^2)^(3/2))/(24*e^3) - (a + b*x + c*x^2)^(5/2)/(e*(d + e*x)) + (5*(128*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(2*b*d
- 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*c^2*e^2*(3*b^2*d^2 - 4*a*b*d*e + a^2*e^2))*ArcTanh[(b + 2*c*x)/(
2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(3/2)*e^6) - (5*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)^(3/2)*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])])/(2*e^6)

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 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 )^{5/2}}{(d+e x)^2} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{e (d+e x)}+\frac{5 \int \frac{(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx}{2 e}\\ &=-\frac{5 (8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{24 e^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{e (d+e x)}-\frac{5 \int \frac{\left (c \left (7 b^2 d e+4 a c d e-8 b \left (c d^2+a e^2\right )\right )-c \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{d+e x} \, dx}{16 c e^3}\\ &=-\frac{5 \left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c e^5}-\frac{5 (8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{24 e^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{e (d+e x)}+\frac{5 \int \frac{\frac{1}{2} c \left (d \left (4 b c d-b^2 e-4 a c e\right ) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )+4 c e (b d-2 a e) \left (7 b^2 d e+4 a c d e-8 b \left (c d^2+a e^2\right )\right )\right )+\frac{1}{2} c \left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{64 c^2 e^5}\\ &=-\frac{5 \left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c e^5}-\frac{5 (8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{24 e^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{e (d+e x)}-\frac{\left (5 (2 c d-b e) \left (c d^2-b d e+a e^2\right )^2\right ) \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{2 e^6}+\frac{\left (5 \left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{128 c e^6}\\ &=-\frac{5 \left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c e^5}-\frac{5 (8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{24 e^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{e (d+e x)}+\frac{\left (5 (2 c d-b e) \left (c d^2-b d e+a e^2\right )^2\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 )}{e^6}+\frac{\left (5 \left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\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 )}{64 c e^6}\\ &=-\frac{5 \left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{64 c e^5}-\frac{5 (8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{24 e^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{e (d+e x)}+\frac{5 \left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{3/2} e^6}-\frac{5 (2 c d-b e) \left (c d^2-b d e+a e^2\right )^{3/2} \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 )}{2 e^6}\\ \end{align*}

Mathematica [A]  time = 1.17093, size = 370, normalized size = 0.95 $\frac{5 \left (\left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )+2 \sqrt{c} \left (e \sqrt{a+x (b+c x)} \left (8 c^2 e (a e (3 e x-8 d)+2 b d (7 d-2 e x))+2 b c e^2 (22 a e-24 b d+b e x)+b^3 e^3+32 c^3 d^2 (e x-2 d)\right )+32 c (2 c d-b e) \left (e (a e-b d)+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )\right )\right )}{128 c^{3/2} e^6}+\frac{5 (a+x (b+c x))^{3/2} (7 b e-8 c d+6 c e x)}{24 e^3}-\frac{(a+x (b+c x))^{5/2}}{e (d+e x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(d + e*x)^2,x]

[Out]

(5*(-8*c*d + 7*b*e + 6*c*e*x)*(a + x*(b + c*x))^(3/2))/(24*e^3) - (a + x*(b + c*x))^(5/2)/(e*(d + e*x)) + (5*(
(128*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(2*b*d - 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*c^2*e^2*(3*b^2*d^2 -
4*a*b*d*e + a^2*e^2))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + 2*Sqrt[c]*(e*Sqrt[a + x*(b + c*
x)]*(b^3*e^3 + 32*c^3*d^2*(-2*d + e*x) + 2*b*c*e^2*(-24*b*d + 22*a*e + b*e*x) + 8*c^2*e*(2*b*d*(7*d - 2*e*x) +
a*e*(-8*d + 3*e*x))) + 32*c*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))^(3/2)*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)])])))/(128*c^(3/2)*e^6)

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Maple [B]  time = 0.233, size = 6711, normalized size = 17.3 \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)^(5/2)/(e*x+d)^2,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)^(5/2)/(e*x+d)^2,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)^(5/2)/(e*x+d)^2,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)**(5/2)/(e*x+d)**2,x)

[Out]

Timed out

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Giac [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)^(5/2)/(e*x+d)^2,x, algorithm="giac")

[Out]

Timed out