3.911 \(\int x^3 (A+B x) \sqrt{a+b x+c x^2} \, dx\)

Optimal. Leaf size=280 \[ \frac{(b+2 c x) \sqrt{a+b x+c x^2} \left (16 a^2 B c^2+48 a A b c^2-56 a b^2 B c-28 A b^3 c+21 b^4 B\right )}{512 c^5}-\frac{\left (b^2-4 a c\right ) \left (16 a^2 B c^2+48 a A b c^2-56 a b^2 B c-28 A b^3 c+21 b^4 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{1024 c^{11/2}}-\frac{\left (a+b x+c x^2\right )^{3/2} \left (-6 c x \left (-20 a B c-28 A b c+21 b^2 B\right )+128 a A c^2-196 a b B c-140 A b^2 c+105 b^3 B\right )}{960 c^4}-\frac{x^2 \left (a+b x+c x^2\right )^{3/2} (3 b B-4 A c)}{20 c^2}+\frac{B x^3 \left (a+b x+c x^2\right )^{3/2}}{6 c} \]

[Out]

((21*b^4*B - 28*A*b^3*c - 56*a*b^2*B*c + 48*a*A*b*c^2 + 16*a^2*B*c^2)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(512*
c^5) - ((3*b*B - 4*A*c)*x^2*(a + b*x + c*x^2)^(3/2))/(20*c^2) + (B*x^3*(a + b*x + c*x^2)^(3/2))/(6*c) - ((105*
b^3*B - 140*A*b^2*c - 196*a*b*B*c + 128*a*A*c^2 - 6*c*(21*b^2*B - 28*A*b*c - 20*a*B*c)*x)*(a + b*x + c*x^2)^(3
/2))/(960*c^4) - ((b^2 - 4*a*c)*(21*b^4*B - 28*A*b^3*c - 56*a*b^2*B*c + 48*a*A*b*c^2 + 16*a^2*B*c^2)*ArcTanh[(
b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(1024*c^(11/2))

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Rubi [A]  time = 0.323161, antiderivative size = 280, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {832, 779, 612, 621, 206} \[ \frac{(b+2 c x) \sqrt{a+b x+c x^2} \left (16 a^2 B c^2+48 a A b c^2-56 a b^2 B c-28 A b^3 c+21 b^4 B\right )}{512 c^5}-\frac{\left (b^2-4 a c\right ) \left (16 a^2 B c^2+48 a A b c^2-56 a b^2 B c-28 A b^3 c+21 b^4 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{1024 c^{11/2}}-\frac{\left (a+b x+c x^2\right )^{3/2} \left (-6 c x \left (-20 a B c-28 A b c+21 b^2 B\right )+128 a A c^2-196 a b B c-140 A b^2 c+105 b^3 B\right )}{960 c^4}-\frac{x^2 \left (a+b x+c x^2\right )^{3/2} (3 b B-4 A c)}{20 c^2}+\frac{B x^3 \left (a+b x+c x^2\right )^{3/2}}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(A + B*x)*Sqrt[a + b*x + c*x^2],x]

[Out]

((21*b^4*B - 28*A*b^3*c - 56*a*b^2*B*c + 48*a*A*b*c^2 + 16*a^2*B*c^2)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(512*
c^5) - ((3*b*B - 4*A*c)*x^2*(a + b*x + c*x^2)^(3/2))/(20*c^2) + (B*x^3*(a + b*x + c*x^2)^(3/2))/(6*c) - ((105*
b^3*B - 140*A*b^2*c - 196*a*b*B*c + 128*a*A*c^2 - 6*c*(21*b^2*B - 28*A*b*c - 20*a*B*c)*x)*(a + b*x + c*x^2)^(3
/2))/(960*c^4) - ((b^2 - 4*a*c)*(21*b^4*B - 28*A*b^3*c - 56*a*b^2*B*c + 48*a*A*b*c^2 + 16*a^2*B*c^2)*ArcTanh[(
b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(1024*c^(11/2))

Rule 832

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

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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 x^3 (A+B x) \sqrt{a+b x+c x^2} \, dx &=\frac{B x^3 \left (a+b x+c x^2\right )^{3/2}}{6 c}+\frac{\int x^2 \left (-3 a B-\frac{3}{2} (3 b B-4 A c) x\right ) \sqrt{a+b x+c x^2} \, dx}{6 c}\\ &=-\frac{(3 b B-4 A c) x^2 \left (a+b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (a+b x+c x^2\right )^{3/2}}{6 c}+\frac{\int x \left (3 a (3 b B-4 A c)+\frac{3}{4} \left (21 b^2 B-28 A b c-20 a B c\right ) x\right ) \sqrt{a+b x+c x^2} \, dx}{30 c^2}\\ &=-\frac{(3 b B-4 A c) x^2 \left (a+b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac{\left (105 b^3 B-140 A b^2 c-196 a b B c+128 a A c^2-6 c \left (21 b^2 B-28 A b c-20 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{960 c^4}+\frac{\left (21 b^4 B-28 A b^3 c-56 a b^2 B c+48 a A b c^2+16 a^2 B c^2\right ) \int \sqrt{a+b x+c x^2} \, dx}{128 c^4}\\ &=\frac{\left (21 b^4 B-28 A b^3 c-56 a b^2 B c+48 a A b c^2+16 a^2 B c^2\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{512 c^5}-\frac{(3 b B-4 A c) x^2 \left (a+b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac{\left (105 b^3 B-140 A b^2 c-196 a b B c+128 a A c^2-6 c \left (21 b^2 B-28 A b c-20 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{960 c^4}-\frac{\left (\left (b^2-4 a c\right ) \left (21 b^4 B-28 A b^3 c-56 a b^2 B c+48 a A b c^2+16 a^2 B c^2\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{1024 c^5}\\ &=\frac{\left (21 b^4 B-28 A b^3 c-56 a b^2 B c+48 a A b c^2+16 a^2 B c^2\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{512 c^5}-\frac{(3 b B-4 A c) x^2 \left (a+b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac{\left (105 b^3 B-140 A b^2 c-196 a b B c+128 a A c^2-6 c \left (21 b^2 B-28 A b c-20 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{960 c^4}-\frac{\left (\left (b^2-4 a c\right ) \left (21 b^4 B-28 A b^3 c-56 a b^2 B c+48 a A b c^2+16 a^2 B c^2\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 )}{512 c^5}\\ &=\frac{\left (21 b^4 B-28 A b^3 c-56 a b^2 B c+48 a A b c^2+16 a^2 B c^2\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{512 c^5}-\frac{(3 b B-4 A c) x^2 \left (a+b x+c x^2\right )^{3/2}}{20 c^2}+\frac{B x^3 \left (a+b x+c x^2\right )^{3/2}}{6 c}-\frac{\left (105 b^3 B-140 A b^2 c-196 a b B c+128 a A c^2-6 c \left (21 b^2 B-28 A b c-20 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{960 c^4}-\frac{\left (b^2-4 a c\right ) \left (21 b^4 B-28 A b^3 c-56 a b^2 B c+48 a A b c^2+16 a^2 B c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{1024 c^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.381738, size = 241, normalized size = 0.86 \[ \frac{\frac{3 \left (16 a^2 B c^2+48 a A b c^2-56 a b^2 B c-28 A b^3 c+21 b^4 B\right ) \left (2 \sqrt{c} (b+2 c x) \sqrt{a+x (b+c x)}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right )}{512 c^{9/2}}+\frac{(a+x (b+c x))^{3/2} \left (28 b c (7 a B-6 A c x)-8 a c^2 (16 A+15 B x)+14 b^2 c (10 A+9 B x)-105 b^3 B\right )}{160 c^3}+\frac{3 x^2 (a+x (b+c x))^{3/2} (4 A c-3 b B)}{10 c}+B x^3 (a+x (b+c x))^{3/2}}{6 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(A + B*x)*Sqrt[a + b*x + c*x^2],x]

[Out]

((3*(-3*b*B + 4*A*c)*x^2*(a + x*(b + c*x))^(3/2))/(10*c) + B*x^3*(a + x*(b + c*x))^(3/2) + ((a + x*(b + c*x))^
(3/2)*(-105*b^3*B + 14*b^2*c*(10*A + 9*B*x) - 8*a*c^2*(16*A + 15*B*x) + 28*b*c*(7*a*B - 6*A*c*x)))/(160*c^3) +
(3*(21*b^4*B - 28*A*b^3*c - 56*a*b^2*B*c + 48*a*A*b*c^2 + 16*a^2*B*c^2)*(2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b
+ c*x)] - (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/(512*c^(9/2)))/(6*c)

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Maple [B]  time = 0.01, size = 671, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)*(c*x^2+b*x+a)^(1/2),x)

[Out]

35/256*B*b^4/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-7/64*B*b^3/c^4*a*(c*x^2+b*x+a)^(1/2)-15/64*
B*b^2/c^(7/2)*a^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+49/240*B*b/c^3*a*(c*x^2+b*x+a)^(3/2)-1/8*B*a/c^2
*x*(c*x^2+b*x+a)^(3/2)+1/16*B*a^2/c^2*(c*x^2+b*x+a)^(1/2)*x+21/160*B*b^2/c^3*x*(c*x^2+b*x+a)^(3/2)+21/256*B*b^
4/c^4*(c*x^2+b*x+a)^(1/2)*x+3/16*A*b/c^(5/2)*a^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/32*B*a^2/c^3*(c
*x^2+b*x+a)^(1/2)*b-3/20*B*b/c^2*x^2*(c*x^2+b*x+a)^(3/2)-5/32*A*b^3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+
a)^(1/2))*a+3/32*A*b^2/c^3*a*(c*x^2+b*x+a)^(1/2)-7/64*A*b^3/c^3*(c*x^2+b*x+a)^(1/2)*x-7/40*A*b/c^2*x*(c*x^2+b*
x+a)^(3/2)+1/6*B*x^3*(c*x^2+b*x+a)^(3/2)/c+1/5*A*x^2*(c*x^2+b*x+a)^(3/2)/c+7/48*A*b^2/c^3*(c*x^2+b*x+a)^(3/2)-
7/128*A*b^4/c^4*(c*x^2+b*x+a)^(1/2)+7/256*A*b^5/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-2/15*A*a/c
^2*(c*x^2+b*x+a)^(3/2)+1/16*B*a^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-21/1024*B*b^6/c^(11/2)*l
n((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-7/64*B*b^3/c^4*(c*x^2+b*x+a)^(3/2)+21/512*B*b^5/c^5*(c*x^2+b*x+a)^(
1/2)-7/32*B*b^2/c^3*a*(c*x^2+b*x+a)^(1/2)*x+3/16*A*b/c^2*a*(c*x^2+b*x+a)^(1/2)*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(x^3*(B*x+A)*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.11467, size = 1573, normalized size = 5.62 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/30720*(15*(21*B*b^6 - 64*(B*a^3 + 3*A*a^2*b)*c^3 + 80*(3*B*a^2*b^2 + 2*A*a*b^3)*c^2 - 28*(5*B*a*b^4 + A*b^
5)*c)*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*(1280*
B*c^6*x^5 + 315*B*b^5*c - 1024*A*a^2*c^4 + 128*(B*b*c^5 + 12*A*c^6)*x^4 + 16*(113*B*a^2*b + 115*A*a*b^2)*c^3 -
16*(9*B*b^2*c^4 - 4*(5*B*a + 3*A*b)*c^5)*x^3 - 420*(4*B*a*b^3 + A*b^4)*c^2 + 8*(21*B*b^3*c^3 + 64*A*a*c^5 - 4
*(17*B*a*b + 7*A*b^2)*c^4)*x^2 - 2*(105*B*b^4*c^2 + 16*(15*B*a^2 + 29*A*a*b)*c^4 - 28*(16*B*a*b^2 + 5*A*b^3)*c
^3)*x)*sqrt(c*x^2 + b*x + a))/c^6, 1/15360*(15*(21*B*b^6 - 64*(B*a^3 + 3*A*a^2*b)*c^3 + 80*(3*B*a^2*b^2 + 2*A*
a*b^3)*c^2 - 28*(5*B*a*b^4 + A*b^5)*c)*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*(1280*B*c^6*x^5 + 315*B*b^5*c - 1024*A*a^2*c^4 + 128*(B*b*c^5 + 12*A*c^6)*x^4 + 16*(113*B
*a^2*b + 115*A*a*b^2)*c^3 - 16*(9*B*b^2*c^4 - 4*(5*B*a + 3*A*b)*c^5)*x^3 - 420*(4*B*a*b^3 + A*b^4)*c^2 + 8*(21
*B*b^3*c^3 + 64*A*a*c^5 - 4*(17*B*a*b + 7*A*b^2)*c^4)*x^2 - 2*(105*B*b^4*c^2 + 16*(15*B*a^2 + 29*A*a*b)*c^4 -
28*(16*B*a*b^2 + 5*A*b^3)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^6]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (A + B x\right ) \sqrt{a + b x + c x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x+A)*(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral(x**3*(A + B*x)*sqrt(a + b*x + c*x**2), x)

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Giac [A]  time = 1.40775, size = 436, normalized size = 1.56 \begin{align*} \frac{1}{7680} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (10 \, B x + \frac{B b c^{4} + 12 \, A c^{5}}{c^{5}}\right )} x - \frac{9 \, B b^{2} c^{3} - 20 \, B a c^{4} - 12 \, A b c^{4}}{c^{5}}\right )} x + \frac{21 \, B b^{3} c^{2} - 68 \, B a b c^{3} - 28 \, A b^{2} c^{3} + 64 \, A a c^{4}}{c^{5}}\right )} x - \frac{105 \, B b^{4} c - 448 \, B a b^{2} c^{2} - 140 \, A b^{3} c^{2} + 240 \, B a^{2} c^{3} + 464 \, A a b c^{3}}{c^{5}}\right )} x + \frac{315 \, B b^{5} - 1680 \, B a b^{3} c - 420 \, A b^{4} c + 1808 \, B a^{2} b c^{2} + 1840 \, A a b^{2} c^{2} - 1024 \, A a^{2} c^{3}}{c^{5}}\right )} + \frac{{\left (21 \, B b^{6} - 140 \, B a b^{4} c - 28 \, A b^{5} c + 240 \, B a^{2} b^{2} c^{2} + 160 \, A a b^{3} c^{2} - 64 \, B a^{3} c^{3} - 192 \, A a^{2} b c^{3}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{1024 \, c^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/7680*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(10*B*x + (B*b*c^4 + 12*A*c^5)/c^5)*x - (9*B*b^2*c^3 - 20*B*a*c^4 - 1
2*A*b*c^4)/c^5)*x + (21*B*b^3*c^2 - 68*B*a*b*c^3 - 28*A*b^2*c^3 + 64*A*a*c^4)/c^5)*x - (105*B*b^4*c - 448*B*a*
b^2*c^2 - 140*A*b^3*c^2 + 240*B*a^2*c^3 + 464*A*a*b*c^3)/c^5)*x + (315*B*b^5 - 1680*B*a*b^3*c - 420*A*b^4*c +
1808*B*a^2*b*c^2 + 1840*A*a*b^2*c^2 - 1024*A*a^2*c^3)/c^5) + 1/1024*(21*B*b^6 - 140*B*a*b^4*c - 28*A*b^5*c + 2
40*B*a^2*b^2*c^2 + 160*A*a*b^3*c^2 - 64*B*a^3*c^3 - 192*A*a^2*b*c^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))*sqrt(c) - b))/c^(11/2)