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

Optimal. Leaf size=206 \[ \frac {\left (48 a^2 B c^2+96 a A b c^2-120 a b^2 B c-40 A b^3 c+35 b^4 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{9/2}}-\frac {\sqrt {a+b x+c x^2} \left (-2 c x \left (-36 a B c-40 A b c+35 b^2 B\right )+128 a A c^2-220 a b B c-120 A b^2 c+105 b^3 B\right )}{192 c^4}-\frac {x^2 \sqrt {a+b x+c x^2} (7 b B-8 A c)}{24 c^2}+\frac {B x^3 \sqrt {a+b x+c x^2}}{4 c} \]

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Rubi [A]  time = 0.24, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {832, 779, 621, 206} \begin {gather*} \frac {\left (48 a^2 B c^2+96 a A b c^2-120 a b^2 B c-40 A b^3 c+35 b^4 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{9/2}}-\frac {\sqrt {a+b x+c x^2} \left (-2 c x \left (-36 a B c-40 A b c+35 b^2 B\right )+128 a A c^2-220 a b B c-120 A b^2 c+105 b^3 B\right )}{192 c^4}-\frac {x^2 \sqrt {a+b x+c x^2} (7 b B-8 A c)}{24 c^2}+\frac {B x^3 \sqrt {a+b x+c x^2}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((7*b*B - 8*A*c)*x^2*Sqrt[a + b*x + c*x^2])/(24*c^2) + (B*x^3*Sqrt[a + b*x + c*x^2])/(4*c) - ((105*b^3*B - 12
0*A*b^2*c - 220*a*b*B*c + 128*a*A*c^2 - 2*c*(35*b^2*B - 40*A*b*c - 36*a*B*c)*x)*Sqrt[a + b*x + c*x^2])/(192*c^
4) + ((35*b^4*B - 40*A*b^3*c - 120*a*b^2*B*c + 96*a*A*b*c^2 + 48*a^2*B*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr
t[a + b*x + c*x^2])])/(128*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 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 \frac {x^3 (A+B x)}{\sqrt {a+b x+c x^2}} \, dx &=\frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\int \frac {x^2 \left (-3 a B-\frac {1}{2} (7 b B-8 A c) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{4 c}\\ &=-\frac {(7 b B-8 A c) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\int \frac {x \left (a (7 b B-8 A c)+\frac {1}{4} \left (35 b^2 B-40 A b c-36 a B c\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{12 c^2}\\ &=-\frac {(7 b B-8 A c) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (105 b^3 B-120 A b^2 c-220 a b B c+128 a A c^2-2 c \left (35 b^2 B-40 A b c-36 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{192 c^4}+\frac {\left (35 b^4 B-40 A b^3 c-120 a b^2 B c+96 a A b c^2+48 a^2 B c^2\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{128 c^4}\\ &=-\frac {(7 b B-8 A c) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (105 b^3 B-120 A b^2 c-220 a b B c+128 a A c^2-2 c \left (35 b^2 B-40 A b c-36 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{192 c^4}+\frac {\left (35 b^4 B-40 A b^3 c-120 a b^2 B c+96 a A b c^2+48 a^2 B c^2\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 )}{64 c^4}\\ &=-\frac {(7 b B-8 A c) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (105 b^3 B-120 A b^2 c-220 a b B c+128 a A c^2-2 c \left (35 b^2 B-40 A b c-36 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{192 c^4}+\frac {\left (35 b^4 B-40 A b^3 c-120 a b^2 B c+96 a A b c^2+48 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 )}{128 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 169, normalized size = 0.82 \begin {gather*} \frac {\left (48 a^2 B c^2+96 a A b c^2-120 a b^2 B c-40 A b^3 c+35 b^4 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{128 c^{9/2}}+\frac {\sqrt {a+x (b+c x)} \left (4 b c (55 a B-2 c x (10 A+7 B x))+8 c^2 \left (-16 a A-9 a B x+8 A c x^2+6 B c x^3\right )+10 b^2 c (12 A+7 B x)-105 b^3 B\right )}{192 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + x*(b + c*x)]*(-105*b^3*B + 10*b^2*c*(12*A + 7*B*x) + 8*c^2*(-16*a*A - 9*a*B*x + 8*A*c*x^2 + 6*B*c*x^
3) + 4*b*c*(55*a*B - 2*c*x*(10*A + 7*B*x))))/(192*c^4) + ((35*b^4*B - 40*A*b^3*c - 120*a*b^2*B*c + 96*a*A*b*c^
2 + 48*a^2*B*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(128*c^(9/2))

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IntegrateAlgebraic [A]  time = 0.68, size = 183, normalized size = 0.89 \begin {gather*} \frac {\left (-48 a^2 B c^2-96 a A b c^2+120 a b^2 B c+40 A b^3 c-35 b^4 B\right ) \log \left (-2 c^{9/2} \sqrt {a+b x+c x^2}+b c^4+2 c^5 x\right )}{128 c^{9/2}}+\frac {\sqrt {a+b x+c x^2} \left (-128 a A c^2+220 a b B c-72 a B c^2 x+120 A b^2 c-80 A b c^2 x+64 A c^3 x^2-105 b^3 B+70 b^2 B c x-56 b B c^2 x^2+48 B c^3 x^3\right )}{192 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-105*b^3*B + 120*A*b^2*c + 220*a*b*B*c - 128*a*A*c^2 + 70*b^2*B*c*x - 80*A*b*c^2*x - 7
2*a*B*c^2*x - 56*b*B*c^2*x^2 + 64*A*c^3*x^2 + 48*B*c^3*x^3))/(192*c^4) + ((-35*b^4*B + 40*A*b^3*c + 120*a*b^2*
B*c - 96*a*A*b*c^2 - 48*a^2*B*c^2)*Log[b*c^4 + 2*c^5*x - 2*c^(9/2)*Sqrt[a + b*x + c*x^2]])/(128*c^(9/2))

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fricas [A]  time = 0.52, size = 395, normalized size = 1.92 \begin {gather*} \left [\frac {3 \, {\left (35 \, B b^{4} + 48 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 40 \, {\left (3 \, B a b^{2} + A b^{3}\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 (48 \, B c^{4} x^{3} - 105 \, B b^{3} c - 128 \, A a c^{3} + 20 \, {\left (11 \, B a b + 6 \, A b^{2}\right )} c^{2} - 8 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (35 \, B b^{2} c^{2} - 4 \, {\left (9 \, B a + 10 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{5}}, -\frac {3 \, {\left (35 \, B b^{4} + 48 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 40 \, {\left (3 \, B a b^{2} + A b^{3}\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 (48 \, B c^{4} x^{3} - 105 \, B b^{3} c - 128 \, A a c^{3} + 20 \, {\left (11 \, B a b + 6 \, A b^{2}\right )} c^{2} - 8 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (35 \, B b^{2} c^{2} - 4 \, {\left (9 \, B a + 10 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{5}}\right ] \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)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(3*(35*B*b^4 + 48*(B*a^2 + 2*A*a*b)*c^2 - 40*(3*B*a*b^2 + A*b^3)*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*(48*B*c^4*x^3 - 105*B*b^3*c - 128*A*a*c^3 + 20*
(11*B*a*b + 6*A*b^2)*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^2 + 2*(35*B*b^2*c^2 - 4*(9*B*a + 10*A*b)*c^3)*x)*sqrt(c*x
^2 + b*x + a))/c^5, -1/384*(3*(35*B*b^4 + 48*(B*a^2 + 2*A*a*b)*c^2 - 40*(3*B*a*b^2 + A*b^3)*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*(48*B*c^4*x^3 - 105*B*b^3*c - 128
*A*a*c^3 + 20*(11*B*a*b + 6*A*b^2)*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^2 + 2*(35*B*b^2*c^2 - 4*(9*B*a + 10*A*b)*c^
3)*x)*sqrt(c*x^2 + b*x + a))/c^5]

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giac [A]  time = 0.27, size = 183, normalized size = 0.89 \begin {gather*} \frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (\frac {6 \, B x}{c} - \frac {7 \, B b c^{2} - 8 \, A c^{3}}{c^{4}}\right )} x + \frac {35 \, B b^{2} c - 36 \, B a c^{2} - 40 \, A b c^{2}}{c^{4}}\right )} x - \frac {105 \, B b^{3} - 220 \, B a b c - 120 \, A b^{2} c + 128 \, A a c^{2}}{c^{4}}\right )} - \frac {{\left (35 \, B b^{4} - 120 \, B a b^{2} c - 40 \, A b^{3} c + 48 \, B a^{2} c^{2} + 96 \, A a b c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {9}{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)^(1/2),x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*B*x/c - (7*B*b*c^2 - 8*A*c^3)/c^4)*x + (35*B*b^2*c - 36*B*a*c^2 - 40*A*b*
c^2)/c^4)*x - (105*B*b^3 - 220*B*a*b*c - 120*A*b^2*c + 128*A*a*c^2)/c^4) - 1/128*(35*B*b^4 - 120*B*a*b^2*c - 4
0*A*b^3*c + 48*B*a^2*c^2 + 96*A*a*b*c^2)*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 = 379, normalized size = 1.84 \begin {gather*} \frac {\sqrt {c \,x^{2}+b x +a}\, B \,x^{3}}{4 c}+\frac {\sqrt {c \,x^{2}+b x +a}\, A \,x^{2}}{3 c}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, B b \,x^{2}}{24 c^{2}}+\frac {3 A a b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {5}{2}}}-\frac {5 A \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {7}{2}}}+\frac {3 B \,a^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {5}{2}}}-\frac {15 B a \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {7}{2}}}+\frac {35 B \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {9}{2}}}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, A b x}{12 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, B a x}{8 c^{2}}+\frac {35 \sqrt {c \,x^{2}+b x +a}\, B \,b^{2} x}{96 c^{3}}-\frac {2 \sqrt {c \,x^{2}+b x +a}\, A a}{3 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2}}{8 c^{3}}+\frac {55 \sqrt {c \,x^{2}+b x +a}\, B a b}{48 c^{3}}-\frac {35 \sqrt {c \,x^{2}+b x +a}\, B \,b^{3}}{64 c^{4}} \end {gather*}

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]

1/4*B*x^3*(c*x^2+b*x+a)^(1/2)/c-7/24*B*b/c^2*x^2*(c*x^2+b*x+a)^(1/2)+35/96*B*b^2/c^3*x*(c*x^2+b*x+a)^(1/2)-35/
64*B*b^3/c^4*(c*x^2+b*x+a)^(1/2)+35/128*B*b^4/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-15/16*B*b^2/
c^(7/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+55/48*B*b/c^3*a*(c*x^2+b*x+a)^(1/2)-3/8*B*a/c^2*x*(c*x^2
+b*x+a)^(1/2)+3/8*B*a^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/3*A*x^2/c*(c*x^2+b*x+a)^(1/2)-5/
12*A*b/c^2*x*(c*x^2+b*x+a)^(1/2)+5/8*A*b^2/c^3*(c*x^2+b*x+a)^(1/2)-5/16*A*b^3/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(
c*x^2+b*x+a)^(1/2))+3/4*A*b/c^(5/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-2/3*A*a/c^2*(c*x^2+b*x+a)^(1
/2)

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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)^(1/2),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 zero or nonzero?

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (A + B x\right )}{\sqrt {a + b x + c x^{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)**(1/2),x)

[Out]

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

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