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

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

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Rubi [A]  time = 0.09, antiderivative size = 101, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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])

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 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]

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*}

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Mathematica [A]  time = 0.10, size = 115, normalized size = 1.14 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (3 b^{7/2} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )+\sqrt {c} x \sqrt {\frac {c x^2}{b}+1} \left (-3 b^3+2 b^2 c x^2+24 b c^2 x^4+16 c^3 x^6\right )\right )}{128 c^{5/2} x \sqrt {\frac {c x^2}{b}+1}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.29, size = 98, normalized size = 0.97 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-3 b^3+2 b^2 c x^2+24 b c^2 x^4+16 c^3 x^6\right )}{128 c^2}-\frac {3 b^4 \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right )}{256 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-3*b^3 + 2*b^2*c*x^2 + 24*b*c^2*x^4 + 16*c^3*x^6))/(128*c^2) - (3*b^4*Log[b + 2*c*x^2 -
2*Sqrt[c]*Sqrt[b*x^2 + c*x^4]])/(256*c^(5/2))

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fricas [A]  time = 1.14, size = 189, normalized size = 1.87 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.19, size = 99, normalized size = 0.98 \begin {gather*} -\frac {3 \, b^{4} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\relax (x)}{128 \, c^{\frac {5}{2}}} + \frac {3 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{256 \, c^{\frac {5}{2}}} + \frac {1}{128} \, {\left (2 \, {\left (4 \, {\left (2 \, c x^{2} \mathrm {sgn}\relax (x) + 3 \, b \mathrm {sgn}\relax (x)\right )} x^{2} + \frac {b^{2} \mathrm {sgn}\relax (x)}{c}\right )} x^{2} - \frac {3 \, b^{3} \mathrm {sgn}\relax (x)}{c^{2}}\right )} \sqrt {c x^{2} + b} x \end {gather*}

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

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maple [A]  time = 0.01, size = 122, normalized size = 1.21 \begin {gather*} \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+3 \sqrt {c \,x^{2}+b}\, b^{3} \sqrt {c}\, x +16 \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} x^{3}+2 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} \sqrt {c}\, x -8 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \sqrt {c}\, x \right )}{128 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} x^{3}} \end {gather*}

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*x^2+b)^(3/2)*b^2*
c^(1/2)*x+3*(c*x^2+b)^(1/2)*b^3*c^(1/2)*x+3*b^4*ln(c^(1/2)*x+(c*x^2+b)^(1/2)))/x^3/(c*x^2+b)^(3/2)/c^(5/2)

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maxima [A]  time = 1.47, size = 118, normalized size = 1.17 \begin {gather*} \frac {1}{8} \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{64 \, c} + \frac {3 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{256 \, c^{\frac {5}{2}}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{128 \, c^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{16 \, c} \end {gather*}

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]

1/8*(c*x^4 + b*x^2)^(3/2)*x^2 - 3/64*sqrt(c*x^4 + b*x^2)*b^2*x^2/c + 3/256*b^4*log(2*c*x^2 + b + 2*sqrt(c*x^4
+ b*x^2)*sqrt(c))/c^(5/2) - 3/128*sqrt(c*x^4 + b*x^2)*b^3/c^2 + 1/16*(c*x^4 + b*x^2)^(3/2)*b/c

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mupad [B]  time = 4.44, size = 99, normalized size = 0.98 \begin {gather*} \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}\,\left (c\,x^2+\frac {b}{2}\right )}{8\,c}-\frac {3\,b^2\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{8\,c^{3/2}}\right )}{32\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^2 + c*x^4)^(3/2),x)

[Out]

((b*x^2 + c*x^4)^(3/2)*(b/2 + c*x^2))/(8*c) - (3*b^2*((b/(4*c) + x^2/2)*(b*x^2 + c*x^4)^(1/2) - (b^2*log((b/2
+ c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(8*c^(3/2))))/(32*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}\, dx \end {gather*}

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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