### 3.300 $$\int \frac{(b x+c x^2)^{3/2}}{(d+e x)^2} \, dx$$

Optimal. Leaf size=198 $\frac{3 \left (b^2 e^2-8 b c d e+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 \sqrt{c} e^4}-\frac{3 \sqrt{b x+c x^2} (-3 b e+4 c d-2 c e x)}{4 e^3}-\frac{3 \sqrt{d} \sqrt{c d-b e} (2 c d-b e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{2 e^4}-\frac{\left (b x+c x^2\right )^{3/2}}{e (d+e x)}$

[Out]

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

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Rubi [A]  time = 0.234135, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {732, 814, 843, 620, 206, 724} $\frac{3 \left (b^2 e^2-8 b c d e+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 \sqrt{c} e^4}-\frac{3 \sqrt{b x+c x^2} (-3 b e+4 c d-2 c e x)}{4 e^3}-\frac{3 \sqrt{d} \sqrt{c d-b e} (2 c d-b e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{2 e^4}-\frac{\left (b x+c x^2\right )^{3/2}}{e (d+e x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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

Mathematica [A]  time = 1.04538, size = 218, normalized size = 1.1 $\frac{\sqrt{x (b+c x)} \left (-\frac{12 \sqrt{d} \left (b^2 e^2-3 b c d e+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{\sqrt{b+c x} \sqrt{b e-c d}}+\frac{3 \left (b^2 e^2-8 b c d e+8 c^2 d^2\right ) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{b} \sqrt{c} \sqrt{\frac{c x}{b}+1}}+\frac{e \sqrt{x} \left (b e (9 d+5 e x)-2 c \left (6 d^2+3 d e x-e^2 x^2\right )\right )}{d+e x}\right )}{4 e^4 \sqrt{x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*((e*Sqrt[x]*(b*e*(9*d + 5*e*x) - 2*c*(6*d^2 + 3*d*e*x - e^2*x^2)))/(d + e*x) + (3*(8*c^2*d^
2 - 8*b*c*d*e + b^2*e^2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[b]*Sqrt[c]*Sqrt[1 + (c*x)/b]) - (12*Sqrt[d]
*(2*c^2*d^2 - 3*b*c*d*e + b^2*e^2)*ArcTan[(Sqrt[-(c*d) + b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/(Sqrt[-(c*d)
+ b*e]*Sqrt[b + c*x])))/(4*e^4*Sqrt[x])

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Maple [B]  time = 0.216, size = 1569, normalized size = 7.9 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/(b*e-c*d)/(d/e+x)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(5/2)-1/d/(b*e-c*d)*(c*(d/e+x)^2+(b*
e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(3/2)*b+1/e/(b*e-c*d)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^
(3/2)*c-27/8/e^2*d/(b*e-c*d)*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*
e-c*d)/e^2)^(1/2))*c^(1/2)*b^2+6/e^3*d^2/(b*e-c*d)*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+x)^2+(b*e-
2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))*c^(3/2)*b-6/e^3*d^2/(b*e-c*d)/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-
c*d)/e^2+(b*e-2*c*d)/e*(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^
(1/2))/(d/e+x))*b^2*c+15/2/e^4*d^3/(b*e-c*d)/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/
e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/(d/e+x))*b*c^2+3/2/
e/(b*e-c*d)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*x*b*c-3/2/e^2*d/(b*e-c*d)*(c*(d/e+x)^2+(
b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*x*c^2-21/4/e^2*d/(b*e-c*d)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b
*e-c*d)/e^2)^(1/2)*b*c+3/8/e/(b*e-c*d)*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/
e+x)-d*(b*e-c*d)/e^2)^(1/2))/c^(1/2)*b^3-3/e^4*d^3/(b*e-c*d)*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+
x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))*c^(5/2)+3/2/e^2*d/(b*e-c*d)/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2
*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*
d)/e^2)^(1/2))/(d/e+x))*b^3-3/e^5*d^4/(b*e-c*d)/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*
(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/(d/e+x))*c^3+9/4
/e/(b*e-c*d)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*b^2+3/e^3*d^2/(b*e-c*d)*(c*(d/e+x)^2+(b
*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*c^2-c/d/(b*e-c*d)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^
2)^(3/2)*x

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.96756, size = 2187, normalized size = 11.05 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*(8*c^2*d^3 - 8*b*c*d^2*e + b^2*d*e^2 + (8*c^2*d^2*e - 8*b*c*d*e^2 + b^2*e^3)*x)*sqrt(c)*log(2*c*x + b
+ 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 12*(2*c^2*d^2 - b*c*d*e + (2*c^2*d*e - b*c*e^2)*x)*sqrt(c*d^2 - b*d*e)*log((b
*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) + 2*(2*c^2*e^3*x^2 - 12*c^2*d^2*e +
9*b*c*d*e^2 - (6*c^2*d*e^2 - 5*b*c*e^3)*x)*sqrt(c*x^2 + b*x))/(c*e^5*x + c*d*e^4), -1/8*(24*(2*c^2*d^2 - b*c*
d*e + (2*c^2*d*e - b*c*e^2)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c*d - b*e
)*x)) - 3*(8*c^2*d^3 - 8*b*c*d^2*e + b^2*d*e^2 + (8*c^2*d^2*e - 8*b*c*d*e^2 + b^2*e^3)*x)*sqrt(c)*log(2*c*x +
b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(2*c^2*e^3*x^2 - 12*c^2*d^2*e + 9*b*c*d*e^2 - (6*c^2*d*e^2 - 5*b*c*e^3)*x
)*sqrt(c*x^2 + b*x))/(c*e^5*x + c*d*e^4), -1/4*(3*(8*c^2*d^3 - 8*b*c*d^2*e + b^2*d*e^2 + (8*c^2*d^2*e - 8*b*c*
d*e^2 + b^2*e^3)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + 6*(2*c^2*d^2 - b*c*d*e + (2*c^2*d*e -
b*c*e^2)*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d
)) - (2*c^2*e^3*x^2 - 12*c^2*d^2*e + 9*b*c*d*e^2 - (6*c^2*d*e^2 - 5*b*c*e^3)*x)*sqrt(c*x^2 + b*x))/(c*e^5*x +
c*d*e^4), -1/4*(12*(2*c^2*d^2 - b*c*d*e + (2*c^2*d*e - b*c*e^2)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 +
b*d*e)*sqrt(c*x^2 + b*x)/((c*d - b*e)*x)) + 3*(8*c^2*d^3 - 8*b*c*d^2*e + b^2*d*e^2 + (8*c^2*d^2*e - 8*b*c*d*e^
2 + b^2*e^3)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2*c^2*e^3*x^2 - 12*c^2*d^2*e + 9*b*c*d*e^
2 - (6*c^2*d*e^2 - 5*b*c*e^3)*x)*sqrt(c*x^2 + b*x))/(c*e^5*x + c*d*e^4)]

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

[Out]

Timed out