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

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

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Rubi [A]  time = 0.13, antiderivative size = 124, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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]

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*}

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Mathematica [A]  time = 0.10, size = 126, normalized size = 1.02 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {c} x \sqrt {\frac {c x^2}{b}+1} \left (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\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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.34, size = 109, normalized size = 0.88 \begin {gather*} \frac {3 b^5 \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right )}{512 c^{7/2}}+\frac {\sqrt {b x^2+c x^4} \left (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\right )}{1280 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(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))/(1280*c^3) + (3*b^
5*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[b*x^2 + c*x^4]])/(512*c^(7/2))

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fricas [A]  time = 2.61, size = 210, normalized size = 1.69 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.28, size = 115, normalized size = 0.93 \begin {gather*} \frac {3 \, b^{5} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\relax (x)}{256 \, c^{\frac {7}{2}}} - \frac {3 \, b^{5} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{512 \, c^{\frac {7}{2}}} + \frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c x^{2} \mathrm {sgn}\relax (x) + 11 \, b \mathrm {sgn}\relax (x)\right )} x^{2} + \frac {b^{2} \mathrm {sgn}\relax (x)}{c}\right )} x^{2} - \frac {5 \, b^{3} \mathrm {sgn}\relax (x)}{c^{2}}\right )} x^{2} + \frac {15 \, b^{4} \mathrm {sgn}\relax (x)}{c^{3}}\right )} \sqrt {c x^{2} + b} x \end {gather*}

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

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maple [A]  time = 0.01, size = 142, normalized size = 1.15 \begin {gather*} \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (128 \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {5}{2}} x^{5}-15 b^{5} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-15 \sqrt {c \,x^{2}+b}\, b^{4} \sqrt {c}\, x -80 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,c^{\frac {3}{2}} x^{3}-10 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3} \sqrt {c}\, x +40 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} \sqrt {c}\, x \right )}{1280 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {7}{2}} x^{3}} \end {gather*}

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

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maxima [A]  time = 1.39, size = 142, normalized size = 1.15 \begin {gather*} \frac {3 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{128 \, c^{2}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{16 \, c} - \frac {3 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{512 \, c^{\frac {7}{2}}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{256 \, c^{3}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{32 \, c^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}}}{10 \, c} \end {gather*}

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]

3/128*sqrt(c*x^4 + b*x^2)*b^3*x^2/c^2 - 1/16*(c*x^4 + b*x^2)^(3/2)*b*x^2/c - 3/512*b^5*log(2*c*x^2 + b + 2*sqr
t(c*x^4 + b*x^2)*sqrt(c))/c^(7/2) + 3/256*sqrt(c*x^4 + b*x^2)*b^4/c^3 - 1/32*(c*x^4 + b*x^2)^(3/2)*b^2/c^2 + 1
/10*(c*x^4 + b*x^2)^(5/2)/c

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mupad [B]  time = 4.35, size = 134, normalized size = 1.08 \begin {gather*} \frac {{\left (c\,x^4+b\,x^2\right )}^{5/2}}{10\,c}-\frac {b\,\left (\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{4}-\frac {3\,b^2\,\left (\frac {\left (2\,c\,x^2+b\right )\,\sqrt {c\,x^4+b\,x^2}}{4\,c}-\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 )}{16\,c}+\frac {b\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}\right )}{4\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2 + c*x^4)^(3/2),x)

[Out]

(b*x^2 + c*x^4)^(5/2)/(10*c) - (b*((x^2*(b*x^2 + c*x^4)^(3/2))/4 - (3*b^2*(((b + 2*c*x^2)*(b*x^2 + c*x^4)^(1/2
))/(4*c) - (b^2*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(8*c^(3/2))))/(16*c) + (b*(b*x^2 + c*x^4)^
(3/2))/(8*c)))/(4*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \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**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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