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

Optimal. Leaf size=356 \[ \frac {3 \left (b^2-4 a c\right )^2 \left (16 a^2 B c^2+64 a A b c^2-72 a b^2 B c-48 A b^3 c+33 b^4 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32768 c^{13/2}}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} \left (16 a^2 B c^2+64 a A b c^2-72 a b^2 B c-48 A b^3 c+33 b^4 B\right )}{16384 c^6}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2} \left (16 a^2 B c^2+64 a A b c^2-72 a b^2 B c-48 A b^3 c+33 b^4 B\right )}{2048 c^5}-\frac {\left (a+b x+c x^2\right )^{5/2} \left (-10 c x \left (-28 a B c-48 A b c+33 b^2 B\right )+256 a A c^2-372 a b B c-336 A b^2 c+231 b^3 B\right )}{4480 c^4}-\frac {x^2 \left (a+b x+c x^2\right )^{5/2} (11 b B-16 A c)}{112 c^2}+\frac {B x^3 \left (a+b x+c x^2\right )^{5/2}}{8 c} \]

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Rubi [A]  time = 0.40, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \begin {gather*} \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2} \left (16 a^2 B c^2+64 a A b c^2-72 a b^2 B c-48 A b^3 c+33 b^4 B\right )}{2048 c^5}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} \left (16 a^2 B c^2+64 a A b c^2-72 a b^2 B c-48 A b^3 c+33 b^4 B\right )}{16384 c^6}+\frac {3 \left (b^2-4 a c\right )^2 \left (16 a^2 B c^2+64 a A b c^2-72 a b^2 B c-48 A b^3 c+33 b^4 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32768 c^{13/2}}-\frac {\left (a+b x+c x^2\right )^{5/2} \left (-10 c x \left (-28 a B c-48 A b c+33 b^2 B\right )+256 a A c^2-372 a b B c-336 A b^2 c+231 b^3 B\right )}{4480 c^4}-\frac {x^2 \left (a+b x+c x^2\right )^{5/2} (11 b B-16 A c)}{112 c^2}+\frac {B x^3 \left (a+b x+c x^2\right )^{5/2}}{8 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b^2 - 4*a*c)*(33*b^4*B - 48*A*b^3*c - 72*a*b^2*B*c + 64*a*A*b*c^2 + 16*a^2*B*c^2)*(b + 2*c*x)*Sqrt[a + b*
x + c*x^2])/(16384*c^6) + ((33*b^4*B - 48*A*b^3*c - 72*a*b^2*B*c + 64*a*A*b*c^2 + 16*a^2*B*c^2)*(b + 2*c*x)*(a
+ b*x + c*x^2)^(3/2))/(2048*c^5) - ((11*b*B - 16*A*c)*x^2*(a + b*x + c*x^2)^(5/2))/(112*c^2) + (B*x^3*(a + b*
x + c*x^2)^(5/2))/(8*c) - ((231*b^3*B - 336*A*b^2*c - 372*a*b*B*c + 256*a*A*c^2 - 10*c*(33*b^2*B - 48*A*b*c -
28*a*B*c)*x)*(a + b*x + c*x^2)^(5/2))/(4480*c^4) + (3*(b^2 - 4*a*c)^2*(33*b^4*B - 48*A*b^3*c - 72*a*b^2*B*c +
64*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])])/(32768*c^(13/2))

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 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 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 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])

Rubi steps

\begin {align*} \int x^3 (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac {B x^3 \left (a+b x+c x^2\right )^{5/2}}{8 c}+\frac {\int x^2 \left (-3 a B-\frac {1}{2} (11 b B-16 A c) x\right ) \left (a+b x+c x^2\right )^{3/2} \, dx}{8 c}\\ &=-\frac {(11 b B-16 A c) x^2 \left (a+b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (a+b x+c x^2\right )^{5/2}}{8 c}+\frac {\int x \left (a (11 b B-16 A c)+\frac {3}{4} \left (33 b^2 B-48 A b c-28 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2} \, dx}{56 c^2}\\ &=-\frac {(11 b B-16 A c) x^2 \left (a+b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (a+b x+c x^2\right )^{5/2}}{8 c}-\frac {\left (231 b^3 B-336 A b^2 c-372 a b B c+256 a A c^2-10 c \left (33 b^2 B-48 A b c-28 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{5/2}}{4480 c^4}+\frac {\left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 a A b c^2+16 a^2 B c^2\right ) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{256 c^4}\\ &=\frac {\left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 a A b c^2+16 a^2 B c^2\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{2048 c^5}-\frac {(11 b B-16 A c) x^2 \left (a+b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (a+b x+c x^2\right )^{5/2}}{8 c}-\frac {\left (231 b^3 B-336 A b^2 c-372 a b B c+256 a A c^2-10 c \left (33 b^2 B-48 A b c-28 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{5/2}}{4480 c^4}-\frac {\left (3 \left (b^2-4 a c\right ) \left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 a A b c^2+16 a^2 B c^2\right )\right ) \int \sqrt {a+b x+c x^2} \, dx}{4096 c^5}\\ &=-\frac {3 \left (b^2-4 a c\right ) \left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 a A b c^2+16 a^2 B c^2\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{16384 c^6}+\frac {\left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 a A b c^2+16 a^2 B c^2\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{2048 c^5}-\frac {(11 b B-16 A c) x^2 \left (a+b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (a+b x+c x^2\right )^{5/2}}{8 c}-\frac {\left (231 b^3 B-336 A b^2 c-372 a b B c+256 a A c^2-10 c \left (33 b^2 B-48 A b c-28 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{5/2}}{4480 c^4}+\frac {\left (3 \left (b^2-4 a c\right )^2 \left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 a A b c^2+16 a^2 B c^2\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{32768 c^6}\\ &=-\frac {3 \left (b^2-4 a c\right ) \left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 a A b c^2+16 a^2 B c^2\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{16384 c^6}+\frac {\left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 a A b c^2+16 a^2 B c^2\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{2048 c^5}-\frac {(11 b B-16 A c) x^2 \left (a+b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (a+b x+c x^2\right )^{5/2}}{8 c}-\frac {\left (231 b^3 B-336 A b^2 c-372 a b B c+256 a A c^2-10 c \left (33 b^2 B-48 A b c-28 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{5/2}}{4480 c^4}+\frac {\left (3 \left (b^2-4 a c\right )^2 \left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 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 )}{16384 c^6}\\ &=-\frac {3 \left (b^2-4 a c\right ) \left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 a A b c^2+16 a^2 B c^2\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{16384 c^6}+\frac {\left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 a A b c^2+16 a^2 B c^2\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{2048 c^5}-\frac {(11 b B-16 A c) x^2 \left (a+b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (a+b x+c x^2\right )^{5/2}}{8 c}-\frac {\left (231 b^3 B-336 A b^2 c-372 a b B c+256 a A c^2-10 c \left (33 b^2 B-48 A b c-28 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{5/2}}{4480 c^4}+\frac {3 \left (b^2-4 a c\right )^2 \left (33 b^4 B-48 A b^3 c-72 a b^2 B c+64 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 )}{32768 c^{13/2}}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 267, normalized size = 0.75 \begin {gather*} \frac {\frac {\left (16 a^2 B c^2+64 a A b c^2-72 a b^2 B c-48 A b^3 c+33 b^4 B\right ) \left (2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)} \left (4 c \left (5 a+2 c x^2\right )-3 b^2+8 b c x\right )+3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{4096 c^{11/2}}+\frac {(a+x (b+c x))^{5/2} \left (12 b c (31 a B-40 A c x)-8 a c^2 (32 A+35 B x)+6 b^2 c (56 A+55 B x)-231 b^3 B\right )}{560 c^3}+\frac {x^2 (a+x (b+c x))^{5/2} (16 A c-11 b B)}{14 c}+B x^3 (a+x (b+c x))^{5/2}}{8 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-11*b*B + 16*A*c)*x^2*(a + x*(b + c*x))^(5/2))/(14*c) + B*x^3*(a + x*(b + c*x))^(5/2) + ((a + x*(b + c*x))^
(5/2)*(-231*b^3*B - 8*a*c^2*(32*A + 35*B*x) + 6*b^2*c*(56*A + 55*B*x) + 12*b*c*(31*a*B - 40*A*c*x)))/(560*c^3)
+ ((33*b^4*B - 48*A*b^3*c - 72*a*b^2*B*c + 64*a*A*b*c^2 + 16*a^2*B*c^2)*(2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b
+ c*x)]*(-3*b^2 + 8*b*c*x + 4*c*(5*a + 2*c*x^2)) + 3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x
*(b + c*x)])]))/(4096*c^(11/2)))/(8*c)

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IntegrateAlgebraic [A]  time = 2.50, size = 535, normalized size = 1.50 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-32768 a^3 A c^4+58816 a^3 b B c^3-13440 a^3 B c^4 x+87808 a^2 A b^2 c^3-37376 a^2 A b c^4 x+16384 a^2 A c^5 x^2-81648 a^2 b^3 B c^2+37792 a^2 b^2 B c^3 x-19328 a^2 b B c^4 x^2+8960 a^2 B c^5 x^3-40320 a A b^4 c^2+23296 a A b^3 c^3 x-15872 a A b^2 c^4 x^2+11264 a A b c^5 x^3+131072 a A c^6 x^4+30660 a b^5 B c-17976 a b^4 B c^2 x+12480 a b^3 B c^3 x^2-9088 a b^2 B c^4 x^3+6656 a b B c^5 x^4+107520 a B c^6 x^5+5040 A b^6 c-3360 A b^5 c^2 x+2688 A b^4 c^3 x^2-2304 A b^3 c^4 x^3+2048 A b^2 c^5 x^4+102400 A b c^6 x^5+81920 A c^7 x^6-3465 b^7 B+2310 b^6 B c x-1848 b^5 B c^2 x^2+1584 b^4 B c^3 x^3-1408 b^3 B c^4 x^4+1280 b^2 B c^5 x^5+87040 b B c^6 x^6+71680 B c^7 x^7\right )}{573440 c^6}-\frac {3 \left (256 a^4 B c^4+1024 a^3 A b c^4-1280 a^3 b^2 B c^3-1280 a^2 A b^3 c^3+1120 a^2 b^4 B c^2+448 a A b^5 c^2-336 a b^6 B c-48 A b^7 c+33 b^8 B\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{32768 c^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-3465*b^7*B + 5040*A*b^6*c + 30660*a*b^5*B*c - 40320*a*A*b^4*c^2 - 81648*a^2*b^3*B*c^2
+ 87808*a^2*A*b^2*c^3 + 58816*a^3*b*B*c^3 - 32768*a^3*A*c^4 + 2310*b^6*B*c*x - 3360*A*b^5*c^2*x - 17976*a*b^4
*B*c^2*x + 23296*a*A*b^3*c^3*x + 37792*a^2*b^2*B*c^3*x - 37376*a^2*A*b*c^4*x - 13440*a^3*B*c^4*x - 1848*b^5*B*
c^2*x^2 + 2688*A*b^4*c^3*x^2 + 12480*a*b^3*B*c^3*x^2 - 15872*a*A*b^2*c^4*x^2 - 19328*a^2*b*B*c^4*x^2 + 16384*a
^2*A*c^5*x^2 + 1584*b^4*B*c^3*x^3 - 2304*A*b^3*c^4*x^3 - 9088*a*b^2*B*c^4*x^3 + 11264*a*A*b*c^5*x^3 + 8960*a^2
*B*c^5*x^3 - 1408*b^3*B*c^4*x^4 + 2048*A*b^2*c^5*x^4 + 6656*a*b*B*c^5*x^4 + 131072*a*A*c^6*x^4 + 1280*b^2*B*c^
5*x^5 + 102400*A*b*c^6*x^5 + 107520*a*B*c^6*x^5 + 87040*b*B*c^6*x^6 + 81920*A*c^7*x^6 + 71680*B*c^7*x^7))/(573
440*c^6) - (3*(33*b^8*B - 48*A*b^7*c - 336*a*b^6*B*c + 448*a*A*b^5*c^2 + 1120*a^2*b^4*B*c^2 - 1280*a^2*A*b^3*c
^3 - 1280*a^3*b^2*B*c^3 + 1024*a^3*A*b*c^4 + 256*a^4*B*c^4)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/
(32768*c^(13/2))

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fricas [A]  time = 0.62, size = 1037, normalized size = 2.91

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2293760*(105*(33*B*b^8 + 256*(B*a^4 + 4*A*a^3*b)*c^4 - 1280*(B*a^3*b^2 + A*a^2*b^3)*c^3 + 224*(5*B*a^2*b^4
+ 2*A*a*b^5)*c^2 - 48*(7*B*a*b^6 + A*b^7)*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*(71680*B*c^8*x^7 - 3465*B*b^7*c - 32768*A*a^3*c^5 + 5120*(17*B*b*c^7 + 16*A*c
^8)*x^6 + 1280*(B*b^2*c^6 + 4*(21*B*a + 20*A*b)*c^7)*x^5 + 64*(919*B*a^3*b + 1372*A*a^2*b^2)*c^4 - 128*(11*B*b
^3*c^5 - 1024*A*a*c^7 - 4*(13*B*a*b + 4*A*b^2)*c^6)*x^4 - 1008*(81*B*a^2*b^3 + 40*A*a*b^4)*c^3 + 16*(99*B*b^4*
c^4 + 16*(35*B*a^2 + 44*A*a*b)*c^6 - 8*(71*B*a*b^2 + 18*A*b^3)*c^5)*x^3 + 420*(73*B*a*b^5 + 12*A*b^6)*c^2 - 8*
(231*B*b^5*c^3 - 2048*A*a^2*c^6 + 16*(151*B*a^2*b + 124*A*a*b^2)*c^5 - 24*(65*B*a*b^3 + 14*A*b^4)*c^4)*x^2 + 2
*(1155*B*b^6*c^2 - 64*(105*B*a^3 + 292*A*a^2*b)*c^5 + 16*(1181*B*a^2*b^2 + 728*A*a*b^3)*c^4 - 84*(107*B*a*b^4
+ 20*A*b^5)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^7, -1/1146880*(105*(33*B*b^8 + 256*(B*a^4 + 4*A*a^3*b)*c^4 - 1280
*(B*a^3*b^2 + A*a^2*b^3)*c^3 + 224*(5*B*a^2*b^4 + 2*A*a*b^5)*c^2 - 48*(7*B*a*b^6 + A*b^7)*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*(71680*B*c^8*x^7 - 3465*B*b^7*c - 3
2768*A*a^3*c^5 + 5120*(17*B*b*c^7 + 16*A*c^8)*x^6 + 1280*(B*b^2*c^6 + 4*(21*B*a + 20*A*b)*c^7)*x^5 + 64*(919*B
*a^3*b + 1372*A*a^2*b^2)*c^4 - 128*(11*B*b^3*c^5 - 1024*A*a*c^7 - 4*(13*B*a*b + 4*A*b^2)*c^6)*x^4 - 1008*(81*B
*a^2*b^3 + 40*A*a*b^4)*c^3 + 16*(99*B*b^4*c^4 + 16*(35*B*a^2 + 44*A*a*b)*c^6 - 8*(71*B*a*b^2 + 18*A*b^3)*c^5)*
x^3 + 420*(73*B*a*b^5 + 12*A*b^6)*c^2 - 8*(231*B*b^5*c^3 - 2048*A*a^2*c^6 + 16*(151*B*a^2*b + 124*A*a*b^2)*c^5
- 24*(65*B*a*b^3 + 14*A*b^4)*c^4)*x^2 + 2*(1155*B*b^6*c^2 - 64*(105*B*a^3 + 292*A*a^2*b)*c^5 + 16*(1181*B*a^2
*b^2 + 728*A*a*b^3)*c^4 - 84*(107*B*a*b^4 + 20*A*b^5)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^7]

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giac [A]  time = 0.27, size = 524, normalized size = 1.47 \begin {gather*} \frac {1}{573440} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, {\left (4 \, {\left (14 \, B c x + \frac {17 \, B b c^{7} + 16 \, A c^{8}}{c^{7}}\right )} x + \frac {B b^{2} c^{6} + 84 \, B a c^{7} + 80 \, A b c^{7}}{c^{7}}\right )} x - \frac {11 \, B b^{3} c^{5} - 52 \, B a b c^{6} - 16 \, A b^{2} c^{6} - 1024 \, A a c^{7}}{c^{7}}\right )} x + \frac {99 \, B b^{4} c^{4} - 568 \, B a b^{2} c^{5} - 144 \, A b^{3} c^{5} + 560 \, B a^{2} c^{6} + 704 \, A a b c^{6}}{c^{7}}\right )} x - \frac {231 \, B b^{5} c^{3} - 1560 \, B a b^{3} c^{4} - 336 \, A b^{4} c^{4} + 2416 \, B a^{2} b c^{5} + 1984 \, A a b^{2} c^{5} - 2048 \, A a^{2} c^{6}}{c^{7}}\right )} x + \frac {1155 \, B b^{6} c^{2} - 8988 \, B a b^{4} c^{3} - 1680 \, A b^{5} c^{3} + 18896 \, B a^{2} b^{2} c^{4} + 11648 \, A a b^{3} c^{4} - 6720 \, B a^{3} c^{5} - 18688 \, A a^{2} b c^{5}}{c^{7}}\right )} x - \frac {3465 \, B b^{7} c - 30660 \, B a b^{5} c^{2} - 5040 \, A b^{6} c^{2} + 81648 \, B a^{2} b^{3} c^{3} + 40320 \, A a b^{4} c^{3} - 58816 \, B a^{3} b c^{4} - 87808 \, A a^{2} b^{2} c^{4} + 32768 \, A a^{3} c^{5}}{c^{7}}\right )} - \frac {3 \, {\left (33 \, B b^{8} - 336 \, B a b^{6} c - 48 \, A b^{7} c + 1120 \, B a^{2} b^{4} c^{2} + 448 \, A a b^{5} c^{2} - 1280 \, B a^{3} b^{2} c^{3} - 1280 \, A a^{2} b^{3} c^{3} + 256 \, B a^{4} c^{4} + 1024 \, A a^{3} b c^{4}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{32768 \, c^{\frac {13}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/573440*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(10*(4*(14*B*c*x + (17*B*b*c^7 + 16*A*c^8)/c^7)*x + (B*b^2*c^6 + 84
*B*a*c^7 + 80*A*b*c^7)/c^7)*x - (11*B*b^3*c^5 - 52*B*a*b*c^6 - 16*A*b^2*c^6 - 1024*A*a*c^7)/c^7)*x + (99*B*b^4
*c^4 - 568*B*a*b^2*c^5 - 144*A*b^3*c^5 + 560*B*a^2*c^6 + 704*A*a*b*c^6)/c^7)*x - (231*B*b^5*c^3 - 1560*B*a*b^3
*c^4 - 336*A*b^4*c^4 + 2416*B*a^2*b*c^5 + 1984*A*a*b^2*c^5 - 2048*A*a^2*c^6)/c^7)*x + (1155*B*b^6*c^2 - 8988*B
*a*b^4*c^3 - 1680*A*b^5*c^3 + 18896*B*a^2*b^2*c^4 + 11648*A*a*b^3*c^4 - 6720*B*a^3*c^5 - 18688*A*a^2*b*c^5)/c^
7)*x - (3465*B*b^7*c - 30660*B*a*b^5*c^2 - 5040*A*b^6*c^2 + 81648*B*a^2*b^3*c^3 + 40320*A*a*b^4*c^3 - 58816*B*
a^3*b*c^4 - 87808*A*a^2*b^2*c^4 + 32768*A*a^3*c^5)/c^7) - 3/32768*(33*B*b^8 - 336*B*a*b^6*c - 48*A*b^7*c + 112
0*B*a^2*b^4*c^2 + 448*A*a*b^5*c^2 - 1280*B*a^3*b^2*c^3 - 1280*A*a^2*b^3*c^3 + 256*B*a^4*c^4 + 1024*A*a^3*b*c^4
)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(13/2)

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maple [B]  time = 0.06, size = 1061, normalized size = 2.98

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-57/512*B*b^2/c^3*a^2*(c*x^2+b*x+a)^(1/2)*x-9/128*B*b^2/c^3*a*(c*x^2+b*x+a)^(3/2)*x+153/2048*B*b^4/c^4*(c*x^2+
b*x+a)^(1/2)*x*a-3/32*A*b^3/c^3*(c*x^2+b*x+a)^(1/2)*x*a+1/16*A*b/c^2*a*(c*x^2+b*x+a)^(3/2)*x+3/32*A*b/c^2*a^2*
(c*x^2+b*x+a)^(1/2)*x-99/8192*B*b^6/c^5*(c*x^2+b*x+a)^(1/2)*x+105/1024*B*b^4/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c
*x^2+b*x+a)^(1/2))*a^2-11/112*B*b/c^2*x^2*(c*x^2+b*x+a)^(5/2)-57/1024*B*b^3/c^4*a^2*(c*x^2+b*x+a)^(1/2)+21/512
*A*b^5/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+3/32*A*b/c^(5/2)*a^3*ln((c*x+1/2*b)/c^(1/2)+(c*x^
2+b*x+a)^(1/2))+3/256*B*a^3/c^3*(c*x^2+b*x+a)^(1/2)*b-15/128*B*b^2/c^(7/2)*a^3*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b
*x+a)^(1/2))+33/448*B*b^2/c^3*x*(c*x^2+b*x+a)^(5/2)-63/2048*B*b^6/c^(11/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a
)^(1/2))*a+3/64*A*b^2/c^3*a^2*(c*x^2+b*x+a)^(1/2)-3/28*A*b/c^2*x*(c*x^2+b*x+a)^(5/2)-3/64*A*b^3/c^3*(c*x^2+b*x
+a)^(3/2)*x+9/512*A*b^5/c^4*(c*x^2+b*x+a)^(1/2)*x-3/64*A*b^4/c^4*(c*x^2+b*x+a)^(1/2)*a+1/32*A*b^2/c^3*a*(c*x^2
+b*x+a)^(3/2)-15/128*A*b^3/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2+3/128*B*a^3/c^2*(c*x^2+b*x+
a)^(1/2)*x+33/1024*B*b^4/c^4*(c*x^2+b*x+a)^(3/2)*x+153/4096*B*b^5/c^5*(c*x^2+b*x+a)^(1/2)*a-9/256*B*b^3/c^4*a*
(c*x^2+b*x+a)^(3/2)+93/1120*B*b/c^3*a*(c*x^2+b*x+a)^(5/2)-1/16*B*a/c^2*x*(c*x^2+b*x+a)^(5/2)+1/64*B*a^2/c^2*(c
*x^2+b*x+a)^(3/2)*x+1/128*B*a^2/c^3*(c*x^2+b*x+a)^(3/2)*b+33/2048*B*b^5/c^5*(c*x^2+b*x+a)^(3/2)-99/16384*B*b^7
/c^6*(c*x^2+b*x+a)^(1/2)-33/640*B*b^3/c^4*(c*x^2+b*x+a)^(5/2)+99/32768*B*b^8/c^(13/2)*ln((c*x+1/2*b)/c^(1/2)+(
c*x^2+b*x+a)^(1/2))+3/128*B*a^4/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/7*A*x^2*(c*x^2+b*x+a)^(5
/2)/c+9/1024*A*b^6/c^5*(c*x^2+b*x+a)^(1/2)+3/40*A*b^2/c^3*(c*x^2+b*x+a)^(5/2)-3/128*A*b^4/c^4*(c*x^2+b*x+a)^(3
/2)-9/2048*A*b^7/c^(11/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-2/35*A*a/c^2*(c*x^2+b*x+a)^(5/2)+1/8*B*x
^3*(c*x^2+b*x+a)^(5/2)/c

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^3\,\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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