3.3.39 \(\int x (b x^2+c x^4)^{3/2} \, dx\) [239]

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

[Out]

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

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Rubi [A]
time = 0.06, 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 = {2038, 626, 634, 212} \begin {gather*} \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} \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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 626

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 634

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 2038

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} \text {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 ) \text {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 ) \text {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 ) \text {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 = 0.09, size = 113, normalized size = 1.12 \begin {gather*} \frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (-3 b^3+2 b^2 c x^2+24 b c^2 x^4+16 c^3 x^6\right )-3 b^4 \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{128 c^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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Maple [A]
time = 0.10, size = 122, normalized size = 1.21

method	result	size
risch	\(-\frac {\left (-16 c^{3} x^{6}-24 b \,c^{2} x^{4}-2 b^{2} c \,x^{2}+3 b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{128 c^{2}}+\frac {3 b^{4} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{128 c^{\frac {5}{2}} x \sqrt {c \,x^{2}+b}}\)	\(101\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (16 x^{3} \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}}-8 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {5}{2}} b x +2 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x +3 \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{3} x +3 \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) b^{4}\right )}{128 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}}}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*b*x+2*c^(1/2)*(c*x^2+b)^(3
/2)*b^2*x+3*c^(1/2)*(c*x^2+b)^(1/2)*b^3*x+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 [A]
time = 0.30, 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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Fricas [A]
time = 0.37, 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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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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Giac [A]
time = 3.04, 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}\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 {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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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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