3.239 \(\int x (b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=101 \[ -\frac{3 b^2 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{128 c^2}+\frac{3 b^4 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{128 c^{5/2}}+\frac{\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{16 c} \]

[Out]

(-3*b^2*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(128*c^2) + ((b + 2*c*x^2)*(b*x^2 + c*x^4)^(3/2))/(16*c) + (3*b^4*A
rcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(128*c^(5/2))

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Rubi [A]  time = 0.0922218, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2013, 612, 620, 206} \[ -\frac{3 b^2 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{128 c^2}+\frac{3 b^4 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{128 c^{5/2}}+\frac{\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{16 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(-3*b^2*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(128*c^2) + ((b + 2*c*x^2)*(b*x^2 + c*x^4)^(3/2))/(16*c) + (3*b^4*A
rcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(128*c^(5/2))

Rule 2013

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[(a*x^Simplify[j/n]
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && IntegerQ[Simplify[j
/n]] && EqQ[Simplify[m - n + 1], 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 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^2+c x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{16 c}-\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \sqrt{b x+c x^2} \, dx,x,x^2\right )}{32 c}\\ &=-\frac{3 b^2 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{128 c^2}+\frac{\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{16 c}+\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{256 c^2}\\ &=-\frac{3 b^2 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{128 c^2}+\frac{\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{16 c}+\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{b x^2+c x^4}}\right )}{128 c^2}\\ &=-\frac{3 b^2 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{128 c^2}+\frac{\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{16 c}+\frac{3 b^4 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{128 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.101972, size = 115, normalized size = 1.14 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (\sqrt{c} x \sqrt{\frac{c x^2}{b}+1} \left (2 b^2 c x^2-3 b^3+24 b c^2 x^4+16 c^3 x^6\right )+3 b^{7/2} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )\right )}{128 c^{5/2} x \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/b]*(-3*b^3 + 2*b^2*c*x^2 + 24*b*c^2*x^4 + 16*c^3*x^6) + 3*b
^(7/2)*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(128*c^(5/2)*x*Sqrt[1 + (c*x^2)/b])

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Maple [A]  time = 0.048, size = 122, normalized size = 1.2 \begin{align*}{\frac{1}{128\,{x}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 16\,{x}^{3} \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{3/2}-8\,\sqrt{c} \left ( c{x}^{2}+b \right ) ^{5/2}xb+2\,\sqrt{c} \left ( c{x}^{2}+b \right ) ^{3/2}x{b}^{2}+3\,\sqrt{c}\sqrt{c{x}^{2}+b}x{b}^{3}+3\,\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{4} \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*x^4+b*x^2)^(3/2),x)

[Out]

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

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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^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.34129, size = 423, normalized size = 4.19 \begin{align*} \left [\frac{3 \, b^{4} \sqrt{c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) + 2 \,{\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} - 3 \, b^{3} c\right )} \sqrt{c x^{4} + b x^{2}}}{256 \, c^{3}}, -\frac{3 \, b^{4} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) -{\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} - 3 \, b^{3} c\right )} \sqrt{c x^{4} + b x^{2}}}{128 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/256*(3*b^4*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*(16*c^4*x^6 + 24*b*c^3*x^4 + 2*b^2
*c^2*x^2 - 3*b^3*c)*sqrt(c*x^4 + b*x^2))/c^3, -1/128*(3*b^4*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^
2 + b)) - (16*c^4*x^6 + 24*b*c^3*x^4 + 2*b^2*c^2*x^2 - 3*b^3*c)*sqrt(c*x^4 + b*x^2))/c^3]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x**4+b*x**2)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.28781, size = 134, normalized size = 1.33 \begin{align*} -\frac{3 \, b^{4} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + b} \right |}\right ) \mathrm{sgn}\left (x\right )}{128 \, c^{\frac{5}{2}}} + \frac{3 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm{sgn}\left (x\right )}{256 \, c^{\frac{5}{2}}} + \frac{1}{128} \,{\left (2 \,{\left (4 \,{\left (2 \, c x^{2} \mathrm{sgn}\left (x\right ) + 3 \, b \mathrm{sgn}\left (x\right )\right )} x^{2} + \frac{b^{2} \mathrm{sgn}\left (x\right )}{c}\right )} x^{2} - \frac{3 \, b^{3} \mathrm{sgn}\left (x\right )}{c^{2}}\right )} \sqrt{c x^{2} + b} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

-3/128*b^4*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))*sgn(x)/c^(5/2) + 3/256*b^4*log(abs(b))*sgn(x)/c^(5/2) + 1/12
8*(2*(4*(2*c*x^2*sgn(x) + 3*b*sgn(x))*x^2 + b^2*sgn(x)/c)*x^2 - 3*b^3*sgn(x)/c^2)*sqrt(c*x^2 + b)*x