### 3.13 $$\int x (b x+c x^2)^{3/2} \, dx$$

Optimal. Leaf size=110 $\frac{3 b^3 (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}-\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}-\frac{b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{\left (b x+c x^2\right )^{5/2}}{5 c}$

[Out]

(3*b^3*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) - (b*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(16*c^2) + (b*x + c*x^2)
^(5/2)/(5*c) - (3*b^5*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(7/2))

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Rubi [A]  time = 0.0350046, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.267, Rules used = {640, 612, 620, 206} $\frac{3 b^3 (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}-\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}-\frac{b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{\left (b x+c x^2\right )^{5/2}}{5 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*(b*x + c*x^2)^(3/2),x]

[Out]

(3*b^3*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) - (b*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(16*c^2) + (b*x + c*x^2)
^(5/2)/(5*c) - (3*b^5*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(7/2))

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

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 x \left (b x+c x^2\right )^{3/2} \, dx &=\frac{\left (b x+c x^2\right )^{5/2}}{5 c}-\frac{b \int \left (b x+c x^2\right )^{3/2} \, dx}{2 c}\\ &=-\frac{b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{\left (b x+c x^2\right )^{5/2}}{5 c}+\frac{\left (3 b^3\right ) \int \sqrt{b x+c x^2} \, dx}{32 c^2}\\ &=\frac{3 b^3 (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}-\frac{b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{\left (b x+c x^2\right )^{5/2}}{5 c}-\frac{\left (3 b^5\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{256 c^3}\\ &=\frac{3 b^3 (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}-\frac{b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{\left (b x+c x^2\right )^{5/2}}{5 c}-\frac{\left (3 b^5\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{128 c^3}\\ &=\frac{3 b^3 (b+2 c x) \sqrt{b x+c x^2}}{128 c^3}-\frac{b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac{\left (b x+c x^2\right )^{5/2}}{5 c}-\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.152296, size = 109, normalized size = 0.99 $\frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (8 b^2 c^2 x^2-10 b^3 c x+15 b^4+176 b c^3 x^3+128 c^4 x^4\right )-\frac{15 b^{9/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{640 c^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^4 - 10*b^3*c*x + 8*b^2*c^2*x^2 + 176*b*c^3*x^3 + 128*c^4*x^4) - (15*b^(9/2)*
ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(640*c^(7/2))

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Maple [A]  time = 0.045, size = 126, normalized size = 1.2 \begin{align*}{\frac{1}{5\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{bx}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}}{16\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{b}^{3}x}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,{b}^{4}}{128\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{3\,{b}^{5}}{256}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(c*x^2+b*x)^(5/2)/c-1/8*b/c*x*(c*x^2+b*x)^(3/2)-1/16*b^2/c^2*(c*x^2+b*x)^(3/2)+3/64*b^3/c^2*(c*x^2+b*x)^(1
/2)*x+3/128*b^4/c^3*(c*x^2+b*x)^(1/2)-3/256*b^5/c^(7/2)*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(x*(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.95037, size = 451, normalized size = 4.1 \begin{align*} \left [\frac{15 \, b^{5} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 8 \, b^{2} c^{3} x^{2} - 10 \, b^{3} c^{2} x + 15 \, b^{4} c\right )} \sqrt{c x^{2} + b x}}{1280 \, c^{4}}, \frac{15 \, b^{5} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 8 \, b^{2} c^{3} x^{2} - 10 \, b^{3} c^{2} x + 15 \, b^{4} c\right )} \sqrt{c x^{2} + b x}}{640 \, c^{4}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1280*(15*b^5*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(128*c^5*x^4 + 176*b*c^4*x^3 + 8*b^2*
c^3*x^2 - 10*b^3*c^2*x + 15*b^4*c)*sqrt(c*x^2 + b*x))/c^4, 1/640*(15*b^5*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqr
t(-c)/(c*x)) + (128*c^5*x^4 + 176*b*c^4*x^3 + 8*b^2*c^3*x^2 - 10*b^3*c^2*x + 15*b^4*c)*sqrt(c*x^2 + b*x))/c^4]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x*(x*(b + c*x))**(3/2), x)

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Giac [A]  time = 1.33394, size = 128, normalized size = 1.16 \begin{align*} \frac{3 \, b^{5} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{7}{2}}} + \frac{1}{640} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \, c x + 11 \, b\right )} x + \frac{b^{2}}{c}\right )} x - \frac{5 \, b^{3}}{c^{2}}\right )} x + \frac{15 \, b^{4}}{c^{3}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/256*b^5*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2) + 1/640*sqrt(c*x^2 + b*x)*(2*(4*(2*
(8*c*x + 11*b)*x + b^2/c)*x - 5*b^3/c^2)*x + 15*b^4/c^3)