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

Optimal. Leaf size=88 \[ -\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{3/2}}+\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c}+\frac {1}{6} \left (b x^2+c x^4\right )^{3/2} \]

[Out]

1/6*(c*x^4+b*x^2)^(3/2)-1/16*b^3*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(3/2)+1/16*b*(2*c*x^2+b)*(c*x^4+b*
x^2)^(1/2)/c

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Rubi [A]  time = 0.10, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2018, 664, 612, 620, 206} \[ -\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{3/2}}+\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c}+\frac {1}{6} \left (b x^2+c x^4\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x,x]

[Out]

(b*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(16*c) + (b*x^2 + c*x^4)^(3/2)/6 - (b^3*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2
+ c*x^4]])/(16*c^(3/2))

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 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

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 \frac {\left (b x^2+c x^4\right )^{3/2}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (b x+c x^2\right )^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{6} \left (b x^2+c x^4\right )^{3/2}+\frac {1}{4} b \operatorname {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c}+\frac {1}{6} \left (b x^2+c x^4\right )^{3/2}-\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{32 c}\\ &=\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c}+\frac {1}{6} \left (b x^2+c x^4\right )^{3/2}-\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c}\\ &=\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c}+\frac {1}{6} \left (b x^2+c x^4\right )^{3/2}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 104, normalized size = 1.18 \[ \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {c} x \sqrt {\frac {c x^2}{b}+1} \left (3 b^2+14 b c x^2+8 c^2 x^4\right )-3 b^{5/2} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )\right )}{48 c^{3/2} x \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x,x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/b]*(3*b^2 + 14*b*c*x^2 + 8*c^2*x^4) - 3*b^(5/2)*ArcSinh[(Sq
rt[c]*x)/Sqrt[b]]))/(48*c^(3/2)*x*Sqrt[1 + (c*x^2)/b])

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fricas [A]  time = 0.65, size = 166, normalized size = 1.89 \[ \left [\frac {3 \, b^{3} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} x^{4} + 14 \, b c^{2} x^{2} + 3 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{2}}, \frac {3 \, b^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (8 \, c^{3} x^{4} + 14 \, b c^{2} x^{2} + 3 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*b^3*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*(8*c^3*x^4 + 14*b*c^2*x^2 + 3*b^2*c
)*sqrt(c*x^4 + b*x^2))/c^2, 1/48*(3*b^3*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (8*c^3*x^4
+ 14*b*c^2*x^2 + 3*b^2*c)*sqrt(c*x^4 + b*x^2))/c^2]

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giac [A]  time = 0.21, size = 84, normalized size = 0.95 \[ \frac {b^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\relax (x)}{16 \, c^{\frac {3}{2}}} - \frac {b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{32 \, c^{\frac {3}{2}}} + \frac {1}{48} \, {\left (2 \, {\left (4 \, c x^{2} \mathrm {sgn}\relax (x) + 7 \, b \mathrm {sgn}\relax (x)\right )} x^{2} + \frac {3 \, b^{2} \mathrm {sgn}\relax (x)}{c}\right )} \sqrt {c x^{2} + b} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*b^3*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))*sgn(x)/c^(3/2) - 1/32*b^3*log(abs(b))*sgn(x)/c^(3/2) + 1/48*(2
*(4*c*x^2*sgn(x) + 7*b*sgn(x))*x^2 + 3*b^2*sgn(x)/c)*sqrt(c*x^2 + b)*x

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maple [A]  time = 0.01, size = 102, normalized size = 1.16 \[ \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-3 b^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-3 \sqrt {c \,x^{2}+b}\, b^{2} \sqrt {c}\, x -2 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \sqrt {c}\, x +8 \left (c \,x^{2}+b \right )^{\frac {5}{2}} \sqrt {c}\, x \right )}{48 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(c*x^4+b*x^2)^(3/2)*(8*x*(c*x^2+b)^(5/2)*c^(1/2)-2*(c*x^2+b)^(3/2)*b*c^(1/2)*x-3*(c*x^2+b)^(1/2)*b^2*c^(1
/2)*x-3*b^3*ln(c^(1/2)*x+(c*x^2+b)^(1/2)))/x^3/(c*x^2+b)^(3/2)/c^(3/2)

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maxima [A]  time = 1.42, size = 91, normalized size = 1.03 \[ \frac {1}{8} \, \sqrt {c x^{4} + b x^{2}} b x^{2} - \frac {b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{32 \, c^{\frac {3}{2}}} + \frac {1}{6} \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} + \frac {\sqrt {c x^{4} + b x^{2}} b^{2}}{16 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(c*x^4 + b*x^2)*b*x^2 - 1/32*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(3/2) + 1/6*(c*x^4
+ b*x^2)^(3/2) + 1/16*sqrt(c*x^4 + b*x^2)*b^2/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x,x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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