### 3.12 $$\int x^2 (b x+c x^2)^{3/2} \, dx$$

Optimal. Leaf size=134 $-\frac{7 b^4 (b+2 c x) \sqrt{b x+c x^2}}{512 c^4}+\frac{7 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^3}+\frac{7 b^6 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{512 c^{9/2}}-\frac{7 b \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac{x \left (b x+c x^2\right )^{5/2}}{6 c}$

[Out]

(-7*b^4*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(512*c^4) + (7*b^2*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(192*c^3) - (7*b*(b
*x + c*x^2)^(5/2))/(60*c^2) + (x*(b*x + c*x^2)^(5/2))/(6*c) + (7*b^6*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(
512*c^(9/2))

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Rubi [A]  time = 0.0551097, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.294, Rules used = {670, 640, 612, 620, 206} $-\frac{7 b^4 (b+2 c x) \sqrt{b x+c x^2}}{512 c^4}+\frac{7 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^3}+\frac{7 b^6 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{512 c^{9/2}}-\frac{7 b \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac{x \left (b x+c x^2\right )^{5/2}}{6 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2*(b*x + c*x^2)^(3/2),x]

[Out]

(-7*b^4*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(512*c^4) + (7*b^2*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(192*c^3) - (7*b*(b
*x + c*x^2)^(5/2))/(60*c^2) + (x*(b*x + c*x^2)^(5/2))/(6*c) + (7*b^6*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(
512*c^(9/2))

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

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^2 \left (b x+c x^2\right )^{3/2} \, dx &=\frac{x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac{(7 b) \int x \left (b x+c x^2\right )^{3/2} \, dx}{12 c}\\ &=-\frac{7 b \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac{x \left (b x+c x^2\right )^{5/2}}{6 c}+\frac{\left (7 b^2\right ) \int \left (b x+c x^2\right )^{3/2} \, dx}{24 c^2}\\ &=\frac{7 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^3}-\frac{7 b \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac{x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac{\left (7 b^4\right ) \int \sqrt{b x+c x^2} \, dx}{128 c^3}\\ &=-\frac{7 b^4 (b+2 c x) \sqrt{b x+c x^2}}{512 c^4}+\frac{7 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^3}-\frac{7 b \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac{x \left (b x+c x^2\right )^{5/2}}{6 c}+\frac{\left (7 b^6\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{1024 c^4}\\ &=-\frac{7 b^4 (b+2 c x) \sqrt{b x+c x^2}}{512 c^4}+\frac{7 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^3}-\frac{7 b \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac{x \left (b x+c x^2\right )^{5/2}}{6 c}+\frac{\left (7 b^6\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{512 c^4}\\ &=-\frac{7 b^4 (b+2 c x) \sqrt{b x+c x^2}}{512 c^4}+\frac{7 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^3}-\frac{7 b \left (b x+c x^2\right )^{5/2}}{60 c^2}+\frac{x \left (b x+c x^2\right )^{5/2}}{6 c}+\frac{7 b^6 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{512 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.167331, size = 120, normalized size = 0.9 $\frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-56 b^3 c^2 x^2+48 b^2 c^3 x^3+70 b^4 c x-105 b^5+1664 b c^4 x^4+1280 c^5 x^5\right )+\frac{105 b^{11/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{7680 c^{9/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2*(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^5 + 70*b^4*c*x - 56*b^3*c^2*x^2 + 48*b^2*c^3*x^3 + 1664*b*c^4*x^4 + 1280*c
^5*x^5) + (105*b^(11/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(7680*c^(9/2))

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Maple [A]  time = 0.048, size = 146, normalized size = 1.1 \begin{align*}{\frac{x}{6\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{7\,b}{60\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{7\,{b}^{2}x}{96\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{b}^{3}}{192\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{b}^{4}x}{256\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{7\,{b}^{5}}{512\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{7\,{b}^{6}}{1024}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{9}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x*(c*x^2+b*x)^(5/2)/c-7/60*b*(c*x^2+b*x)^(5/2)/c^2+7/96*b^2/c^2*(c*x^2+b*x)^(3/2)*x+7/192*b^3/c^3*(c*x^2+b
*x)^(3/2)-7/256*b^4/c^3*(c*x^2+b*x)^(1/2)*x-7/512*b^5/c^4*(c*x^2+b*x)^(1/2)+7/1024*b^6/c^(9/2)*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^2*(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94666, size = 514, normalized size = 3.84 \begin{align*} \left [\frac{105 \, b^{6} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (1280 \, c^{6} x^{5} + 1664 \, b c^{5} x^{4} + 48 \, b^{2} c^{4} x^{3} - 56 \, b^{3} c^{3} x^{2} + 70 \, b^{4} c^{2} x - 105 \, b^{5} c\right )} \sqrt{c x^{2} + b x}}{15360 \, c^{5}}, -\frac{105 \, b^{6} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (1280 \, c^{6} x^{5} + 1664 \, b c^{5} x^{4} + 48 \, b^{2} c^{4} x^{3} - 56 \, b^{3} c^{3} x^{2} + 70 \, b^{4} c^{2} x - 105 \, b^{5} c\right )} \sqrt{c x^{2} + b x}}{7680 \, c^{5}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15360*(105*b^6*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(1280*c^6*x^5 + 1664*b*c^5*x^4 + 48
*b^2*c^4*x^3 - 56*b^3*c^3*x^2 + 70*b^4*c^2*x - 105*b^5*c)*sqrt(c*x^2 + b*x))/c^5, -1/7680*(105*b^6*sqrt(-c)*ar
ctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (1280*c^6*x^5 + 1664*b*c^5*x^4 + 48*b^2*c^4*x^3 - 56*b^3*c^3*x^2 + 70
*b^4*c^2*x - 105*b^5*c)*sqrt(c*x^2 + b*x))/c^5]

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

[Out]

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

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Giac [A]  time = 1.34954, size = 146, normalized size = 1.09 \begin{align*} -\frac{7 \, b^{6} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{1024 \, c^{\frac{9}{2}}} + \frac{1}{7680} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (10 \, c x + 13 \, b\right )} x + \frac{3 \, b^{2}}{c}\right )} x - \frac{7 \, b^{3}}{c^{2}}\right )} x + \frac{35 \, b^{4}}{c^{3}}\right )} x - \frac{105 \, b^{5}}{c^{4}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/1024*b^6*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2) + 1/7680*sqrt(c*x^2 + b*x)*(2*(4*
(2*(8*(10*c*x + 13*b)*x + 3*b^2/c)*x - 7*b^3/c^2)*x + 35*b^4/c^3)*x - 105*b^5/c^4)