3.1562 \(\int \frac{(b+2 c x) (a+b x+c x^2)^{3/2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=303 \[ \frac{\sqrt{a+b x+c x^2} \left (-4 c e (5 b d-a e)+5 b^2 e^2-4 c e x (2 c d-b e)+16 c^2 d^2\right )}{2 e^4}-\frac{(2 c d-b e) \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{4 \sqrt{c} e^5}+\frac{\sqrt{a e^2-b d e+c d^2} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{2 e^5}+\frac{\left (a+b x+c x^2\right )^{3/2} (-3 b e+8 c d+2 c e x)}{3 e^2 (d+e x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.480885, antiderivative size = 303, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {812, 814, 843, 621, 206, 724} \[ \frac{\sqrt{a+b x+c x^2} \left (-4 c e (5 b d-a e)+5 b^2 e^2-4 c e x (2 c d-b e)+16 c^2 d^2\right )}{2 e^4}-\frac{(2 c d-b e) \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{4 \sqrt{c} e^5}+\frac{\sqrt{a e^2-b d e+c d^2} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{2 e^5}+\frac{\left (a+b x+c x^2\right )^{3/2} (-3 b e+8 c d+2 c e x)}{3 e^2 (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 812

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

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

Mathematica [A]  time = 1.08962, size = 395, normalized size = 1.3 \[ \frac{\frac{(a+x (b+c x))^{3/2} \left (c e (-2 a e+11 b d-3 b e x)-3 b^2 e^2+c^2 \left (6 d e x-8 d^2\right )\right )}{3 e^2}+\frac{-2 c^2 e \sqrt{a+x (b+c x)} \left (e (a e-b d)+c d^2\right ) \left (4 c e (a e-5 b d+b e x)+5 b^2 e^2+8 c^2 d (2 d-e x)\right )+2 c^2 \left (4 c e (a e-4 b d)+3 b^2 e^2+16 c^2 d^2\right ) \left (e (a e-b d)+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )+c^{3/2} (2 c d-b e) \left (e (a e-b d)+c d^2\right ) \left (4 c e (3 a e-4 b d)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{4 c^2 e^5}+\frac{(a+x (b+c x))^{5/2} (b e-2 c d)}{d+e x}}{e (b d-a e)-c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.013, size = 6898, normalized size = 22.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out