### 3.298 $$\int (b x+c x^2)^{3/2} \, dx$$

Optimal. Leaf size=89 $-\frac{3 b^2 (b+2 c x) \sqrt{b x+c x^2}}{64 c^2}+\frac{3 b^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{5/2}}+\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}$

[Out]

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

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Rubi [A]  time = 0.0252396, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.231, Rules used = {612, 620, 206} $-\frac{3 b^2 (b+2 c x) \sqrt{b x+c x^2}}{64 c^2}+\frac{3 b^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{5/2}}+\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*x + c*x^2)^(3/2),x]

[Out]

(-3*b^2*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^2) + ((b + 2*c*x)*(b*x + c*x^2)^(3/2))/(8*c) + (3*b^4*ArcTanh[(Sq
rt[c]*x)/Sqrt[b*x + c*x^2]])/(64*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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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

Mathematica [A]  time = 0.0876433, size = 98, normalized size = 1.1 $\frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (2 b^2 c x-3 b^3+24 b c^2 x^2+16 c^3 x^3\right )+\frac{3 b^{7/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{64 c^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-3*b^3 + 2*b^2*c*x + 24*b*c^2*x^2 + 16*c^3*x^3) + (3*b^(7/2)*ArcSinh[(Sqrt[c]*Sqr
t[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(64*c^(5/2))

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Maple [A]  time = 0.051, size = 95, normalized size = 1.1 \begin{align*}{\frac{2\,cx+b}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{b}^{2}x}{32\,c}\sqrt{c{x}^{2}+bx}}-{\frac{3\,{b}^{3}}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.36783, size = 393, normalized size = 4.42 \begin{align*} \left [\frac{3 \, b^{4} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} + 2 \, b^{2} c^{2} x - 3 \, b^{3} c\right )} \sqrt{c x^{2} + b x}}{128 \, c^{3}}, -\frac{3 \, b^{4} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} + 2 \, b^{2} c^{2} x - 3 \, b^{3} c\right )} \sqrt{c x^{2} + b x}}{64 \, c^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/128*(3*b^4*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(16*c^4*x^3 + 24*b*c^3*x^2 + 2*b^2*c^2*
x - 3*b^3*c)*sqrt(c*x^2 + b*x))/c^3, -1/64*(3*b^4*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (16*c^4*
x^3 + 24*b*c^3*x^2 + 2*b^2*c^2*x - 3*b^3*c)*sqrt(c*x^2 + b*x))/c^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (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)**(3/2),x)

[Out]

Integral((b*x + c*x**2)**(3/2), x)

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Giac [A]  time = 1.28646, size = 112, normalized size = 1.26 \begin{align*} -\frac{3 \, b^{4} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{128 \, c^{\frac{5}{2}}} + \frac{1}{64} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \, c x + 3 \, b\right )} x + \frac{b^{2}}{c}\right )} x - \frac{3 \, b^{3}}{c^{2}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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