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

Optimal. Leaf size=112 \[ \frac {3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2}}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {612, 621, 206} \begin {gather*} -\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2}}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(3/2),x]

[Out]

(-3*(b^2 - 4*a*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(64*c^2) + ((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(8*c) +
(3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(5/2))

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]

Rubi steps

\begin {align*} \int \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \int \sqrt {a+b x+c x^2} \, dx}{16 c}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}+\frac {\left (3 \left (b^2-4 a c\right )^2\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{128 c^2}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}+\frac {\left (3 \left (b^2-4 a c\right )^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^2}\\ &=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}+\frac {3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.34, size = 114, normalized size = 1.02 \begin {gather*} \frac {\sqrt {a+x (b+c x)} \left (2 (b+2 c x) \left (4 c \left (5 a+2 c x^2\right )-3 b^2+8 b c x\right )-\frac {3 \left (b^2-4 a c\right )^{3/2} \sin ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}}}\right )}{128 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[a + x*(b + c*x)]*(2*(b + 2*c*x)*(-3*b^2 + 8*b*c*x + 4*c*(5*a + 2*c*x^2)) - (3*(b^2 - 4*a*c)^(3/2)*ArcSin
[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]))/(128*c^2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.45, size = 120, normalized size = 1.07 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (20 a b c+40 a c^2 x-3 b^3+2 b^2 c x+24 b c^2 x^2+16 c^3 x^3\right )}{64 c^2}-\frac {3 \left (16 a^2 c^2-8 a b^2 c+b^4\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{128 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-3*b^3 + 20*a*b*c + 2*b^2*c*x + 40*a*c^2*x + 24*b*c^2*x^2 + 16*c^3*x^3))/(64*c^2) - (3
*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(128*c^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 277, normalized size = 2.47 \begin {gather*} \left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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 (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{256 \, c^{3}}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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 (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{128 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/256*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*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*(16*c^4*x^3 + 24*b*c^3*x^2 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2 + 20*a*c^3)*x)*sqr
t(c*x^2 + b*x + a))/c^3, -1/128*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*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*(16*c^4*x^3 + 24*b*c^3*x^2 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2
+ 20*a*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^3]

________________________________________________________________________________________

giac [A]  time = 0.22, size = 123, normalized size = 1.10 \begin {gather*} \frac {1}{64} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, c x + 3 \, b\right )} x + \frac {b^{2} c^{2} + 20 \, a c^{3}}{c^{3}}\right )} x - \frac {3 \, b^{3} c - 20 \, a b c^{2}}{c^{3}}\right )} - \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} 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 {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*sqrt(c*x^2 + b*x + a)*(2*(4*(2*c*x + 3*b)*x + (b^2*c^2 + 20*a*c^3)/c^3)*x - (3*b^3*c - 20*a*b*c^2)/c^3) -
3/128*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)

________________________________________________________________________________________

maple [B]  time = 0.05, size = 201, normalized size = 1.79 \begin {gather*} \frac {3 a^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}-\frac {3 a \,b^{2} \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 b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {5}{2}}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, a x}{8}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, b^{2} x}{32 c}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, a b}{16 c}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, b^{3}}{64 c^{2}}+\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c+3/8*(c*x^2+b*x+a)^(1/2)*x*a-3/32/c*(c*x^2+b*x+a)^(1/2)*x*b^2+3/16/c*(c*x^2
+b*x+a)^(1/2)*b*a-3/64/c^2*(c*x^2+b*x+a)^(1/2)*b^3+3/8/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2
-3/16/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a*b^2+3/128/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*
x+a)^(1/2))*b^4

________________________________________________________________________________________

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((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?

________________________________________________________________________________________

mupad [B]  time = 0.17, size = 103, normalized size = 0.92 \begin {gather*} \frac {\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )\,\left (3\,a\,c-\frac {3\,b^2}{4}\right )}{4\,c}+\frac {\left (\frac {b}{2}+c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(3/2),x)

[Out]

(((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))
/(2*c^(3/2)))*(3*a*c - (3*b^2)/4))/(4*c) + ((b/2 + c*x)*(a + b*x + c*x^2)^(3/2))/(4*c)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \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((c*x**2+b*x+a)**(3/2),x)

[Out]

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

________________________________________________________________________________________