3.2346 \(\int (a+b x+c x^2)^{3/2} \, dx\)

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

[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))

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Rubi [A]  time = 0.0331895, antiderivative size = 112, normalized size of antiderivative = 1., 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} \[ -\frac{3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{64 c^2}+\frac{3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{5/2}}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c} \]

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 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 \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 = 0.355323, size = 114, normalized size = 1.02 \[ \frac{\sqrt{a+x (b+c x)} \left (2 (b+2 c x) \left (4 c \left (5 a+2 c x^2\right )-3 b^2+8 b c x\right )-\frac{3 \left (b^2-4 a c\right )^{3/2} \sin ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}\right )}{128 c^2} \]

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)

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Maple [B]  time = 0.044, size = 201, normalized size = 1.8 \begin{align*}{\frac{2\,cx+b}{8\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,ax}{8}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,{b}^{2}x}{32\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,ab}{16\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,{b}^{3}}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{a}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{3\,{b}^{2}a}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{3\,{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.73169, size = 657, normalized size = 5.87 \begin{align*} \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{align*}

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]

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

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)

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Giac [A]  time = 1.13939, size = 166, normalized size = 1.48 \begin{align*} \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{align*}

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)