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

Optimal. Leaf size=198 \[ \frac {\left (b^2-4 a c\right )^2 \left (-4 a B c-12 A b c+7 b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{1024 c^{9/2}}-\frac {\left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a B c-12 A b c+7 b^2 B\right )}{512 c^4}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2} \left (-4 a B c-12 A b c+7 b^2 B\right )}{192 c^3}-\frac {\left (a+b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2} \]

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Rubi [A]  time = 0.09, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {779, 612, 621, 206} \begin {gather*} \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2} \left (-4 a B c-12 A b c+7 b^2 B\right )}{192 c^3}-\frac {\left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a B c-12 A b c+7 b^2 B\right )}{512 c^4}+\frac {\left (b^2-4 a c\right )^2 \left (-4 a B c-12 A b c+7 b^2 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{1024 c^{9/2}}-\frac {\left (a+b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)*(7*b^2*B - 12*A*b*c - 4*a*B*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(512*c^4) + ((7*b^2*B - 12*A
*b*c - 4*a*B*c)*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(192*c^3) - ((7*b*B - 12*A*c - 10*B*c*x)*(a + b*x + c*x^2
)^(5/2))/(60*c^2) + ((b^2 - 4*a*c)^2*(7*b^2*B - 12*A*b*c - 4*a*B*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*
x + c*x^2])])/(1024*c^(9/2))

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]

Rubi steps

\begin {align*} \int x (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx &=-\frac {(7 b B-12 A c-10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {\left (7 b^2 B-12 A b c-4 a B c\right ) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{24 c^2}\\ &=\frac {\left (7 b^2 B-12 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^3}-\frac {(7 b B-12 A c-10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c^2}-\frac {\left (\left (b^2-4 a c\right ) \left (7 b^2 B-12 A b c-4 a B c\right )\right ) \int \sqrt {a+b x+c x^2} \, dx}{128 c^3}\\ &=-\frac {\left (b^2-4 a c\right ) \left (7 b^2 B-12 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{512 c^4}+\frac {\left (7 b^2 B-12 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^3}-\frac {(7 b B-12 A c-10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {\left (\left (b^2-4 a c\right )^2 \left (7 b^2 B-12 A b c-4 a B c\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{1024 c^4}\\ &=-\frac {\left (b^2-4 a c\right ) \left (7 b^2 B-12 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{512 c^4}+\frac {\left (7 b^2 B-12 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^3}-\frac {(7 b B-12 A c-10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {\left (\left (b^2-4 a c\right )^2 \left (7 b^2 B-12 A b c-4 a B c\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^4}\\ &=-\frac {\left (b^2-4 a c\right ) \left (7 b^2 B-12 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{512 c^4}+\frac {\left (7 b^2 B-12 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^3}-\frac {(7 b B-12 A c-10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {\left (b^2-4 a c\right )^2 \left (7 b^2 B-12 A b c-4 a B c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{1024 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 156, normalized size = 0.79 \begin {gather*} \frac {\frac {5 \left (-4 a B c-12 A b c+7 b^2 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 )}{256 c^{5/2}}+(a+x (b+c x))^{5/2} (2 c (6 A+5 B x)-7 b B)}{60 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-7*b*B + 2*c*(6*A + 5*B*x))*(a + x*(b + c*x))^(5/2) + (5*(7*b^2*B - 12*A*b*c - 4*a*B*c)*(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*S
qrt[c]*Sqrt[a + x*(b + c*x)])]))/(256*c^(5/2)))/(60*c^2)

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IntegrateAlgebraic [A]  time = 1.26, size = 326, normalized size = 1.65 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (1536 a^2 A c^3-1296 a^2 b B c^2+480 a^2 B c^3 x-1200 a A b^2 c^2+672 a A b c^3 x+3072 a A c^4 x^2+760 a b^3 B c-432 a b^2 B c^2 x+288 a b B c^3 x^2+2240 a B c^4 x^3+180 A b^4 c-120 A b^3 c^2 x+96 A b^2 c^3 x^2+2112 A b c^4 x^3+1536 A c^5 x^4-105 b^5 B+70 b^4 B c x-56 b^3 B c^2 x^2+48 b^2 B c^3 x^3+1664 b B c^4 x^4+1280 B c^5 x^5\right )}{7680 c^4}+\frac {\left (64 a^3 B c^3+192 a^2 A b c^3-144 a^2 b^2 B c^2-96 a A b^3 c^2+60 a b^4 B c+12 A b^5 c-7 b^6 B\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{1024 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-105*b^5*B + 180*A*b^4*c + 760*a*b^3*B*c - 1200*a*A*b^2*c^2 - 1296*a^2*b*B*c^2 + 1536*
a^2*A*c^3 + 70*b^4*B*c*x - 120*A*b^3*c^2*x - 432*a*b^2*B*c^2*x + 672*a*A*b*c^3*x + 480*a^2*B*c^3*x - 56*b^3*B*
c^2*x^2 + 96*A*b^2*c^3*x^2 + 288*a*b*B*c^3*x^2 + 3072*a*A*c^4*x^2 + 48*b^2*B*c^3*x^3 + 2112*A*b*c^4*x^3 + 2240
*a*B*c^4*x^3 + 1664*b*B*c^4*x^4 + 1536*A*c^5*x^4 + 1280*B*c^5*x^5))/(7680*c^4) + ((-7*b^6*B + 12*A*b^5*c + 60*
a*b^4*B*c - 96*a*A*b^3*c^2 - 144*a^2*b^2*B*c^2 + 192*a^2*A*b*c^3 + 64*a^3*B*c^3)*Log[b + 2*c*x - 2*Sqrt[c]*Sqr
t[a + b*x + c*x^2]])/(1024*c^(9/2))

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fricas [A]  time = 0.53, size = 669, normalized size = 3.38 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{6} - 64 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} c^{3} + 48 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} c^{2} - 12 \, {\left (5 \, B a b^{4} + A b^{5}\right )} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (1280 \, B c^{6} x^{5} - 105 \, B b^{5} c + 1536 \, A a^{2} c^{4} + 128 \, {\left (13 \, B b c^{5} + 12 \, A c^{6}\right )} x^{4} - 48 \, {\left (27 \, B a^{2} b + 25 \, A a b^{2}\right )} c^{3} + 16 \, {\left (3 \, B b^{2} c^{4} + 4 \, {\left (35 \, B a + 33 \, A b\right )} c^{5}\right )} x^{3} + 20 \, {\left (38 \, B a b^{3} + 9 \, A b^{4}\right )} c^{2} - 8 \, {\left (7 \, B b^{3} c^{3} - 384 \, A a c^{5} - 12 \, {\left (3 \, B a b + A b^{2}\right )} c^{4}\right )} x^{2} + 2 \, {\left (35 \, B b^{4} c^{2} + 48 \, {\left (5 \, B a^{2} + 7 \, A a b\right )} c^{4} - 12 \, {\left (18 \, B a b^{2} + 5 \, A b^{3}\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{30720 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{6} - 64 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} c^{3} + 48 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} c^{2} - 12 \, {\left (5 \, B a b^{4} + A b^{5}\right )} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (1280 \, B c^{6} x^{5} - 105 \, B b^{5} c + 1536 \, A a^{2} c^{4} + 128 \, {\left (13 \, B b c^{5} + 12 \, A c^{6}\right )} x^{4} - 48 \, {\left (27 \, B a^{2} b + 25 \, A a b^{2}\right )} c^{3} + 16 \, {\left (3 \, B b^{2} c^{4} + 4 \, {\left (35 \, B a + 33 \, A b\right )} c^{5}\right )} x^{3} + 20 \, {\left (38 \, B a b^{3} + 9 \, A b^{4}\right )} c^{2} - 8 \, {\left (7 \, B b^{3} c^{3} - 384 \, A a c^{5} - 12 \, {\left (3 \, B a b + A b^{2}\right )} c^{4}\right )} x^{2} + 2 \, {\left (35 \, B b^{4} c^{2} + 48 \, {\left (5 \, B a^{2} + 7 \, A a b\right )} c^{4} - 12 \, {\left (18 \, B a b^{2} + 5 \, A b^{3}\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{15360 \, c^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/30720*(15*(7*B*b^6 - 64*(B*a^3 + 3*A*a^2*b)*c^3 + 48*(3*B*a^2*b^2 + 2*A*a*b^3)*c^2 - 12*(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 - 105*B*b^5*c + 1536*A*a^2*c^4 + 128*(13*B*b*c^5 + 12*A*c^6)*x^4 - 48*(27*B*a^2*b + 25*A*a*b^2)*c^3 +
16*(3*B*b^2*c^4 + 4*(35*B*a + 33*A*b)*c^5)*x^3 + 20*(38*B*a*b^3 + 9*A*b^4)*c^2 - 8*(7*B*b^3*c^3 - 384*A*a*c^5
- 12*(3*B*a*b + A*b^2)*c^4)*x^2 + 2*(35*B*b^4*c^2 + 48*(5*B*a^2 + 7*A*a*b)*c^4 - 12*(18*B*a*b^2 + 5*A*b^3)*c^
3)*x)*sqrt(c*x^2 + b*x + a))/c^5, -1/15360*(15*(7*B*b^6 - 64*(B*a^3 + 3*A*a^2*b)*c^3 + 48*(3*B*a^2*b^2 + 2*A*a
*b^3)*c^2 - 12*(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 - 105*B*b^5*c + 1536*A*a^2*c^4 + 128*(13*B*b*c^5 + 12*A*c^6)*x^4 - 48*(27*
B*a^2*b + 25*A*a*b^2)*c^3 + 16*(3*B*b^2*c^4 + 4*(35*B*a + 33*A*b)*c^5)*x^3 + 20*(38*B*a*b^3 + 9*A*b^4)*c^2 - 8
*(7*B*b^3*c^3 - 384*A*a*c^5 - 12*(3*B*a*b + A*b^2)*c^4)*x^2 + 2*(35*B*b^4*c^2 + 48*(5*B*a^2 + 7*A*a*b)*c^4 - 1
2*(18*B*a*b^2 + 5*A*b^3)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^5]

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giac [A]  time = 0.27, size = 332, normalized size = 1.68 \begin {gather*} \frac {1}{7680} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B c x + \frac {13 \, B b c^{5} + 12 \, A c^{6}}{c^{5}}\right )} x + \frac {3 \, B b^{2} c^{4} + 140 \, B a c^{5} + 132 \, A b c^{5}}{c^{5}}\right )} x - \frac {7 \, B b^{3} c^{3} - 36 \, B a b c^{4} - 12 \, A b^{2} c^{4} - 384 \, A a c^{5}}{c^{5}}\right )} x + \frac {35 \, B b^{4} c^{2} - 216 \, B a b^{2} c^{3} - 60 \, A b^{3} c^{3} + 240 \, B a^{2} c^{4} + 336 \, A a b c^{4}}{c^{5}}\right )} x - \frac {105 \, B b^{5} c - 760 \, B a b^{3} c^{2} - 180 \, A b^{4} c^{2} + 1296 \, B a^{2} b c^{3} + 1200 \, A a b^{2} c^{3} - 1536 \, A a^{2} c^{4}}{c^{5}}\right )} - \frac {{\left (7 \, B b^{6} - 60 \, B a b^{4} c - 12 \, A b^{5} c + 144 \, B a^{2} b^{2} c^{2} + 96 \, 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 {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7680*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(10*B*c*x + (13*B*b*c^5 + 12*A*c^6)/c^5)*x + (3*B*b^2*c^4 + 140*B*a*c
^5 + 132*A*b*c^5)/c^5)*x - (7*B*b^3*c^3 - 36*B*a*b*c^4 - 12*A*b^2*c^4 - 384*A*a*c^5)/c^5)*x + (35*B*b^4*c^2 -
216*B*a*b^2*c^3 - 60*A*b^3*c^3 + 240*B*a^2*c^4 + 336*A*a*b*c^4)/c^5)*x - (105*B*b^5*c - 760*B*a*b^3*c^2 - 180*
A*b^4*c^2 + 1296*B*a^2*b*c^3 + 1200*A*a*b^2*c^3 - 1536*A*a^2*c^4)/c^5) - 1/1024*(7*B*b^6 - 60*B*a*b^4*c - 12*A
*b^5*c + 144*B*a^2*b^2*c^2 + 96*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^(9/2)

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maple [B]  time = 0.05, size = 644, normalized size = 3.25 \begin {gather*} -\frac {3 A \,a^{2} b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {3}{2}}}+\frac {3 A a \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32 c^{\frac {5}{2}}}-\frac {3 A \,b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {7}{2}}}-\frac {B \,a^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {3}{2}}}+\frac {9 B \,a^{2} b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{64 c^{\frac {5}{2}}}-\frac {15 B a \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {7}{2}}}+\frac {7 B \,b^{6} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{1024 c^{\frac {9}{2}}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A a b x}{16 c}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{3} x}{64 c^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, B \,a^{2} x}{16 c}+\frac {\sqrt {c \,x^{2}+b x +a}\, B a \,b^{2} x}{8 c^{2}}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, B \,b^{4} x}{256 c^{3}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A a \,b^{2}}{32 c^{2}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{4}}{128 c^{3}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b x}{8 c}-\frac {\sqrt {c \,x^{2}+b x +a}\, B \,a^{2} b}{32 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, B a \,b^{3}}{16 c^{3}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B a x}{24 c}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, B \,b^{5}}{512 c^{4}}+\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{2} x}{96 c^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{2}}{16 c^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B a b}{48 c^{2}}+\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{3}}{192 c^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B x}{6 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A}{5 c}-\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B b}{60 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16*A*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2-1/8*A*b/c*x*(c*x^2+b*x+a)^(3/2)+3/64*A*b^3/c
^2*(c*x^2+b*x+a)^(1/2)*x-1/16*B*a^2/c*(c*x^2+b*x+a)^(1/2)*x-1/32*B*a^2/c^2*(c*x^2+b*x+a)^(1/2)*b-1/48*B*a/c^2*
(c*x^2+b*x+a)^(3/2)*b-15/256*B*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/24*B*a/c*x*(c*x^2+b
*x+a)^(3/2)+9/64*B*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2-7/256*B*b^4/c^3*(c*x^2+b*x+a)^(
1/2)*x+1/16*B*b^3/c^3*(c*x^2+b*x+a)^(1/2)*a+7/96*B*b^2/c^2*x*(c*x^2+b*x+a)^(3/2)+1/8*B*b^2/c^2*(c*x^2+b*x+a)^(
1/2)*x*a-3/16*A*b/c*(c*x^2+b*x+a)^(1/2)*x*a+7/192*B*b^3/c^3*(c*x^2+b*x+a)^(3/2)+1/6*B*x*(c*x^2+b*x+a)^(5/2)/c-
7/60*B*b/c^2*(c*x^2+b*x+a)^(5/2)-7/512*B*b^5/c^4*(c*x^2+b*x+a)^(1/2)+7/1024*B*b^6/c^(9/2)*ln((c*x+1/2*b)/c^(1/
2)+(c*x^2+b*x+a)^(1/2))-1/16*B*a^3/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/16*A*b^2/c^2*(c*x^2+b
*x+a)^(3/2)+3/128*A*b^4/c^3*(c*x^2+b*x+a)^(1/2)-3/256*A*b^5/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2)
)-3/32*A*b^2/c^2*(c*x^2+b*x+a)^(1/2)*a+3/32*A*b^3/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+1/5*A*
(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*(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 = -0.01 \begin {gather*} \int x\,\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*(A + B*x)*(a + b*x + c*x^2)^(3/2),x)

[Out]

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

[Out]

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

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