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

Optimal. Leaf size=124 \[ \frac{3 b^3 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{256 c^3}-\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{256 c^{7/2}}-\frac{b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (b x^2+c x^4\right )^{5/2}}{10 c} \]

[Out]

(3*b^3*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(256*c^3) - (b*(b + 2*c*x^2)*(b*x^2 + c*x^4)^(3/2))/(32*c^2) + (b*x^
2 + c*x^4)^(5/2)/(10*c) - (3*b^5*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(256*c^(7/2))

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Rubi [A]  time = 0.132281, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2018, 640, 612, 620, 206} \[ \frac{3 b^3 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{256 c^3}-\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{256 c^{7/2}}-\frac{b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (b x^2+c x^4\right )^{5/2}}{10 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(3*b^3*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(256*c^3) - (b*(b + 2*c*x^2)*(b*x^2 + c*x^4)^(3/2))/(32*c^2) + (b*x^
2 + c*x^4)^(5/2)/(10*c) - (3*b^5*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(256*c^(7/2))

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(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]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

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^3 \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{\left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac{b \operatorname{Subst}\left (\int \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac{b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (b x^2+c x^4\right )^{5/2}}{10 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \sqrt{b x+c x^2} \, dx,x,x^2\right )}{64 c^2}\\ &=\frac{3 b^3 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{256 c^3}-\frac{b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac{\left (3 b^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{512 c^3}\\ &=\frac{3 b^3 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{256 c^3}-\frac{b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac{\left (3 b^5\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{b x^2+c x^4}}\right )}{256 c^3}\\ &=\frac{3 b^3 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{256 c^3}-\frac{b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{256 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.101127, size = 126, normalized size = 1.02 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (\sqrt{c} x \sqrt{\frac{c x^2}{b}+1} \left (8 b^2 c^2 x^4-10 b^3 c x^2+15 b^4+176 b c^3 x^6+128 c^4 x^8\right )-15 b^{9/2} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )\right )}{1280 c^{7/2} x \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(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]*(15*b^4 - 10*b^3*c*x^2 + 8*b^2*c^2*x^4 + 176*b*c^3*x^6 +
128*c^4*x^8) - 15*b^(9/2)*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(1280*c^(7/2)*x*Sqrt[1 + (c*x^2)/b])

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Maple [A]  time = 0.052, size = 142, normalized size = 1.2 \begin{align*}{\frac{1}{1280\,{x}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 128\,{x}^{5} \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{5/2}-80\, \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{3/2}{x}^{3}b+40\, \left ( c{x}^{2}+b \right ) ^{5/2}\sqrt{c}x{b}^{2}-10\, \left ( c{x}^{2}+b \right ) ^{3/2}\sqrt{c}x{b}^{3}-15\,\sqrt{c{x}^{2}+b}\sqrt{c}x{b}^{4}-15\,\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{5} \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1280*(c*x^4+b*x^2)^(3/2)*(128*x^5*(c*x^2+b)^(5/2)*c^(5/2)-80*(c*x^2+b)^(5/2)*c^(3/2)*x^3*b+40*(c*x^2+b)^(5/2
)*c^(1/2)*x*b^2-10*(c*x^2+b)^(3/2)*c^(1/2)*x*b^3-15*(c*x^2+b)^(1/2)*c^(1/2)*x*b^4-15*ln(x*c^(1/2)+(c*x^2+b)^(1
/2))*b^5)/x^3/(c*x^2+b)^(3/2)/c^(7/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64961, size = 481, normalized size = 3.88 \begin{align*} \left [\frac{15 \, b^{5} \sqrt{c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) + 2 \,{\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 8 \, b^{2} c^{3} x^{4} - 10 \, b^{3} c^{2} x^{2} + 15 \, b^{4} c\right )} \sqrt{c x^{4} + b x^{2}}}{2560 \, c^{4}}, \frac{15 \, b^{5} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) +{\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 8 \, b^{2} c^{3} x^{4} - 10 \, b^{3} c^{2} x^{2} + 15 \, b^{4} c\right )} \sqrt{c x^{4} + b x^{2}}}{1280 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2560*(15*b^5*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*(128*c^5*x^8 + 176*b*c^4*x^6 + 8
*b^2*c^3*x^4 - 10*b^3*c^2*x^2 + 15*b^4*c)*sqrt(c*x^4 + b*x^2))/c^4, 1/1280*(15*b^5*sqrt(-c)*arctan(sqrt(c*x^4
+ b*x^2)*sqrt(-c)/(c*x^2 + b)) + (128*c^5*x^8 + 176*b*c^4*x^6 + 8*b^2*c^3*x^4 - 10*b^3*c^2*x^2 + 15*b^4*c)*sqr
t(c*x^4 + b*x^2))/c^4]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \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**3*(c*x**4+b*x**2)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.20441, size = 155, normalized size = 1.25 \begin{align*} \frac{3 \, b^{5} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + b} \right |}\right ) \mathrm{sgn}\left (x\right )}{256 \, c^{\frac{7}{2}}} - \frac{3 \, b^{5} \log \left ({\left | b \right |}\right ) \mathrm{sgn}\left (x\right )}{512 \, c^{\frac{7}{2}}} + \frac{1}{1280} \,{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \, c x^{2} \mathrm{sgn}\left (x\right ) + 11 \, b \mathrm{sgn}\left (x\right )\right )} x^{2} + \frac{b^{2} \mathrm{sgn}\left (x\right )}{c}\right )} x^{2} - \frac{5 \, b^{3} \mathrm{sgn}\left (x\right )}{c^{2}}\right )} x^{2} + \frac{15 \, b^{4} \mathrm{sgn}\left (x\right )}{c^{3}}\right )} \sqrt{c x^{2} + b} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/256*b^5*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))*sgn(x)/c^(7/2) - 3/512*b^5*log(abs(b))*sgn(x)/c^(7/2) + 1/128
0*(2*(4*(2*(8*c*x^2*sgn(x) + 11*b*sgn(x))*x^2 + b^2*sgn(x)/c)*x^2 - 5*b^3*sgn(x)/c^2)*x^2 + 15*b^4*sgn(x)/c^3)
*sqrt(c*x^2 + b)*x