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

Optimal. Leaf size=155 \[ -\frac {1}{2} a^{3/2} \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )-\frac {b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{3/2}}+\frac {\left (8 a c+b^2+2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c}+\frac {1}{6} \left (a+b x^2+c x^4\right )^{3/2} \]

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Rubi [A]  time = 0.18, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1114, 734, 814, 843, 621, 206, 724} \begin {gather*} -\frac {1}{2} a^{3/2} \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )-\frac {b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{3/2}}+\frac {\left (8 a c+b^2+2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c}+\frac {1}{6} \left (a+b x^2+c x^4\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^(3/2)/x,x]

[Out]

((b^2 + 8*a*c + 2*b*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*c) + (a + b*x^2 + c*x^4)^(3/2)/6 - (a^(3/2)*ArcTanh[(2
*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/2 - (b*(b^2 - 12*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[
a + b*x^2 + c*x^4])])/(32*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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 734

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/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{6} \left (a+b x^2+c x^4\right )^{3/2}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {(-2 a-b x) \sqrt {a+b x+c x^2}}{x} \, dx,x,x^2\right )\\ &=\frac {\left (b^2+8 a c+2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c}+\frac {1}{6} \left (a+b x^2+c x^4\right )^{3/2}+\frac {\operatorname {Subst}\left (\int \frac {8 a^2 c-\frac {1}{2} b \left (b^2-12 a c\right ) x}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{16 c}\\ &=\frac {\left (b^2+8 a c+2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c}+\frac {1}{6} \left (a+b x^2+c x^4\right )^{3/2}+\frac {1}{2} a^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )-\frac {\left (b \left (b^2-12 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{32 c}\\ &=\frac {\left (b^2+8 a c+2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c}+\frac {1}{6} \left (a+b x^2+c x^4\right )^{3/2}-a^2 \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^2}{\sqrt {a+b x^2+c x^4}}\right )-\frac {\left (b \left (b^2-12 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x^2}{\sqrt {a+b x^2+c x^4}}\right )}{16 c}\\ &=\frac {\left (b^2+8 a c+2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c}+\frac {1}{6} \left (a+b x^2+c x^4\right )^{3/2}-\frac {1}{2} a^{3/2} \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )-\frac {b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 143, normalized size = 0.92 \begin {gather*} \frac {1}{96} \left (-48 a^{3/2} \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )-\frac {3 b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{c^{3/2}}+\frac {2 \sqrt {a+b x^2+c x^4} \left (8 c \left (4 a+c x^4\right )+3 b^2+14 b c x^2\right )}{c}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[a + b*x^2 + c*x^4]*(3*b^2 + 14*b*c*x^2 + 8*c*(4*a + c*x^4)))/c - 48*a^(3/2)*ArcTanh[(2*a + b*x^2)/(2*
Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])] - (3*b*(b^2 - 12*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^
4])])/c^(3/2))/96

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IntegrateAlgebraic [A]  time = 0.62, size = 148, normalized size = 0.95 \begin {gather*} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}-\frac {\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )+\frac {\left (b^3-12 a b c\right ) \log \left (-2 c^{3/2} \sqrt {a+b x^2+c x^4}+b c+2 c^2 x^2\right )}{32 c^{3/2}}+\frac {\sqrt {a+b x^2+c x^4} \left (32 a c+3 b^2+14 b c x^2+8 c^2 x^4\right )}{48 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(3*b^2 + 32*a*c + 14*b*c*x^2 + 8*c^2*x^4))/(48*c) + a^(3/2)*ArcTanh[(Sqrt[c]*x^2)/Sqr
t[a] - Sqrt[a + b*x^2 + c*x^4]/Sqrt[a]] + ((b^3 - 12*a*b*c)*Log[b*c + 2*c^2*x^2 - 2*c^(3/2)*Sqrt[a + b*x^2 + c
*x^4]])/(32*c^(3/2))

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fricas [A]  time = 1.46, size = 727, normalized size = 4.69 \begin {gather*} \left [\frac {48 \, a^{\frac {3}{2}} c^{2} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, c^{3} x^{4} + 14 \, b c^{2} x^{2} + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, c^{2}}, \frac {24 \, a^{\frac {3}{2}} c^{2} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, {\left (8 \, c^{3} x^{4} + 14 \, b c^{2} x^{2} + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, c^{2}}, \frac {96 \, \sqrt {-a} a c^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) - 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, c^{3} x^{4} + 14 \, b c^{2} x^{2} + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, c^{2}}, \frac {48 \, \sqrt {-a} a c^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, {\left (8 \, c^{3} x^{4} + 14 \, b c^{2} x^{2} + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/192*(48*a^(3/2)*c^2*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) +
8*a^2)/x^4) - 3*(b^3 - 12*a*b*c)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^
2 + b)*sqrt(c) - 4*a*c) + 4*(8*c^3*x^4 + 14*b*c^2*x^2 + 3*b^2*c + 32*a*c^2)*sqrt(c*x^4 + b*x^2 + a))/c^2, 1/96
*(24*a^(3/2)*c^2*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2
)/x^4) + 3*(b^3 - 12*a*b*c)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*
x^2 + a*c)) + 2*(8*c^3*x^4 + 14*b*c^2*x^2 + 3*b^2*c + 32*a*c^2)*sqrt(c*x^4 + b*x^2 + a))/c^2, 1/192*(96*sqrt(-
a)*a*c^2*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) - 3*(b^3 - 12*a*
b*c)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) + 4*(
8*c^3*x^4 + 14*b*c^2*x^2 + 3*b^2*c + 32*a*c^2)*sqrt(c*x^4 + b*x^2 + a))/c^2, 1/96*(48*sqrt(-a)*a*c^2*arctan(1/
2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + 3*(b^3 - 12*a*b*c)*sqrt(-c)*arct
an(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*(8*c^3*x^4 + 14*b*c^2*x^2
+ 3*b^2*c + 32*a*c^2)*sqrt(c*x^4 + b*x^2 + a))/c^2]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a sub
stitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perha
ps be purged.index.cc index_m operator + Error: Bad Argument Value

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maple [A]  time = 0.02, size = 192, normalized size = 1.24 \begin {gather*} \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, c \,x^{4}}{6}+\frac {7 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b \,x^{2}}{24}-\frac {a^{\frac {3}{2}} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{2}+\frac {3 a b \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 \sqrt {c}}-\frac {b^{3} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {3}{2}}}+\frac {2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a}{3}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2}}{16 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*c*x^4*(c*x^4+b*x^2+a)^(1/2)+7/24*b*x^2*(c*x^4+b*x^2+a)^(1/2)+1/16/c*b^2*(c*x^4+b*x^2+a)^(1/2)-1/32/c^(3/2)
*b^3*ln((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+3/8*a*b*ln((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))/c
^(1/2)+2/3*a*(c*x^4+b*x^2+a)^(1/2)-1/2*a^(3/2)*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)^(3/2)/x,x)

[Out]

int((a + 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 \begin {gather*} \int \frac {\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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