3.2.10 \(\int \frac {\sqrt {x}}{(a x+b x^3+c x^5)^{3/2}} \, dx\)

Optimal. Leaf size=103 \[ \frac {\sqrt {x} \left (-2 a c+b^2+b c x^2\right )}{a \left (b^2-4 a c\right ) \sqrt {a x+b x^3+c x^5}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x+b x^3+c x^5}}\right )}{2 a^{3/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1922, 1913, 206} \begin {gather*} \frac {\sqrt {x} \left (-2 a c+b^2+b c x^2\right )}{a \left (b^2-4 a c\right ) \sqrt {a x+b x^3+c x^5}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x+b x^3+c x^5}}\right )}{2 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a*x + b*x^3 + c*x^5)^(3/2),x]

[Out]

(Sqrt[x]*(b^2 - 2*a*c + b*c*x^2))/(a*(b^2 - 4*a*c)*Sqrt[a*x + b*x^3 + c*x^5]) - ArcTanh[(Sqrt[x]*(2*a + b*x^2)
)/(2*Sqrt[a]*Sqrt[a*x + b*x^3 + c*x^5])]/(2*a^(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 1913

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-2/(n - q), Sub
st[Int[1/(4*a - x^2), x], x, (x^(m + 1)*(2*a + b*x^(n - q)))/Sqrt[a*x^q + b*x^n + c*x^r]], x] /; FreeQ[{a, b,
c, m, n, q, r}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && NeQ[b^2 - 4*a*c, 0] && EqQ[m, q/2 - 1]

Rule 1922

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> -Simp[(x^(m - q + 1
)*(b^2 - 2*a*c + b*c*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(a*(n - q)*(p + 1)*(b^2 - 4*a*c)), x]
 + Dist[(2*a*c - b^2*(p + 2))/(a*(p + 1)*(b^2 - 4*a*c)), Int[x^(m - q)*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1)
, x], x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && I
GtQ[n, 0] && LtQ[p, -1] && RationalQ[m, p, q] && EqQ[m + p*q + 1, -((n - q)*(2*p + 3))]

Rubi steps

\begin {align*} \int \frac {\sqrt {x}}{\left (a x+b x^3+c x^5\right )^{3/2}} \, dx &=\frac {\sqrt {x} \left (b^2-2 a c+b c x^2\right )}{a \left (b^2-4 a c\right ) \sqrt {a x+b x^3+c x^5}}+\frac {\int \frac {1}{\sqrt {x} \sqrt {a x+b x^3+c x^5}} \, dx}{a}\\ &=\frac {\sqrt {x} \left (b^2-2 a c+b c x^2\right )}{a \left (b^2-4 a c\right ) \sqrt {a x+b x^3+c x^5}}-\frac {\operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {\sqrt {x} \left (2 a+b x^2\right )}{\sqrt {a x+b x^3+c x^5}}\right )}{a}\\ &=\frac {\sqrt {x} \left (b^2-2 a c+b c x^2\right )}{a \left (b^2-4 a c\right ) \sqrt {a x+b x^3+c x^5}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x+b x^3+c x^5}}\right )}{2 a^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 126, normalized size = 1.22 \begin {gather*} \frac {\sqrt {x} \left (\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )-2 \sqrt {a} \left (-2 a c+b^2+b c x^2\right )\right )}{2 a^{3/2} \left (4 a c-b^2\right ) \sqrt {x \left (a+b x^2+c x^4\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(a*x + b*x^3 + c*x^5)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 1.84, size = 123, normalized size = 1.19 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {c} x^{5/2}-\sqrt {a x+b x^3+c x^5}}\right )}{a^{3/2}}+\frac {\sqrt {a x+b x^3+c x^5} \left (2 a c-b^2-b c x^2\right )}{a \sqrt {x} \left (4 a c-b^2\right ) \left (a+b x^2+c x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[x]/(a*x + b*x^3 + c*x^5)^(3/2),x]

[Out]

((-b^2 + 2*a*c - b*c*x^2)*Sqrt[a*x + b*x^3 + c*x^5])/(a*(-b^2 + 4*a*c)*Sqrt[x]*(a + b*x^2 + c*x^4)) + ArcTanh[
(Sqrt[a]*Sqrt[x])/(Sqrt[c]*x^(5/2) - Sqrt[a*x + b*x^3 + c*x^5])]/a^(3/2)

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fricas [B]  time = 1.56, size = 424, normalized size = 4.12 \begin {gather*} \left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{5} + {\left (b^{3} - 4 \, a b c\right )} x^{3} + {\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{5} + 8 \, a b x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} \sqrt {x}}{x^{5}}\right ) + 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (a b c x^{2} + a b^{2} - 2 \, a^{2} c\right )} \sqrt {x}}{4 \, {\left ({\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{5} + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{3} + {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x\right )}}, \frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{5} + {\left (b^{3} - 4 \, a b c\right )} x^{3} + {\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a} \sqrt {x}}{2 \, {\left (a c x^{5} + a b x^{3} + a^{2} x\right )}}\right ) + 2 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (a b c x^{2} + a b^{2} - 2 \, a^{2} c\right )} \sqrt {x}}{2 \, {\left ({\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{5} + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{3} + {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(((b^2*c - 4*a*c^2)*x^5 + (b^3 - 4*a*b*c)*x^3 + (a*b^2 - 4*a^2*c)*x)*sqrt(a)*log(-((b^2 + 4*a*c)*x^5 + 8*
a*b*x^3 + 8*a^2*x - 4*sqrt(c*x^5 + b*x^3 + a*x)*(b*x^2 + 2*a)*sqrt(a)*sqrt(x))/x^5) + 4*sqrt(c*x^5 + b*x^3 + a
*x)*(a*b*c*x^2 + a*b^2 - 2*a^2*c)*sqrt(x))/((a^2*b^2*c - 4*a^3*c^2)*x^5 + (a^2*b^3 - 4*a^3*b*c)*x^3 + (a^3*b^2
 - 4*a^4*c)*x), 1/2*(((b^2*c - 4*a*c^2)*x^5 + (b^3 - 4*a*b*c)*x^3 + (a*b^2 - 4*a^2*c)*x)*sqrt(-a)*arctan(1/2*s
qrt(c*x^5 + b*x^3 + a*x)*(b*x^2 + 2*a)*sqrt(-a)*sqrt(x)/(a*c*x^5 + a*b*x^3 + a^2*x)) + 2*sqrt(c*x^5 + b*x^3 +
a*x)*(a*b*c*x^2 + a*b^2 - 2*a^2*c)*sqrt(x))/((a^2*b^2*c - 4*a^3*c^2)*x^5 + (a^2*b^3 - 4*a^3*b*c)*x^3 + (a^3*b^
2 - 4*a^4*c)*x)]

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giac [B]  time = 0.66, size = 193, normalized size = 1.87 \begin {gather*} \frac {\frac {a b c x^{2}}{a^{2} b^{2} - 4 \, a^{3} c} + \frac {a b^{2} - 2 \, a^{2} c}{a^{2} b^{2} - 4 \, a^{3} c}}{\sqrt {c x^{4} + b x^{2} + a}} - \frac {a b^{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) - 4 \, a^{2} c \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + \sqrt {-a} \sqrt {a} b^{2} - 2 \, \sqrt {-a} a^{\frac {3}{2}} c}{\sqrt {-a} a^{2} b^{2} - 4 \, \sqrt {-a} a^{3} c} + \frac {\arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="giac")

[Out]

(a*b*c*x^2/(a^2*b^2 - 4*a^3*c) + (a*b^2 - 2*a^2*c)/(a^2*b^2 - 4*a^3*c))/sqrt(c*x^4 + b*x^2 + a) - (a*b^2*arcta
n(sqrt(a)/sqrt(-a)) - 4*a^2*c*arctan(sqrt(a)/sqrt(-a)) + sqrt(-a)*sqrt(a)*b^2 - 2*sqrt(-a)*a^(3/2)*c)/(sqrt(-a
)*a^2*b^2 - 4*sqrt(-a)*a^3*c) + arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a)

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maple [B]  time = 0.02, size = 179, normalized size = 1.74 \begin {gather*} -\frac {\sqrt {\left (c \,x^{4}+b \,x^{2}+a \right ) x}\, \left (2 \sqrt {a}\, b c \,x^{2}+4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )-\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )-4 a^{\frac {3}{2}} c +2 \sqrt {a}\, b^{2}\right )}{2 \left (c \,x^{4}+b \,x^{2}+a \right ) \left (4 a c -b^{2}\right ) a^{\frac {3}{2}} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(c*x^5+b*x^3+a*x)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x}}{{\left (c x^{5} + b x^{3} + a x\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x)/(c*x^5 + b*x^3 + a*x)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x}}{{\left (c\,x^5+b\,x^3+a\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(a*x + b*x^3 + c*x^5)^(3/2),x)

[Out]

int(x^(1/2)/(a*x + b*x^3 + c*x^5)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x}}{\left (x \left (a + b x^{2} + c x^{4}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/(c*x**5+b*x**3+a*x)**(3/2),x)

[Out]

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

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