3.106 \(\int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx\)

Optimal. Leaf size=129 \[ \frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\sqrt {x} \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2} \sqrt {a x+b x^3+c x^5}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1918, 1914, 1107, 621, 206} \[ \frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\sqrt {x} \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2} \sqrt {a x+b x^3+c x^5}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*Sqrt[a*x + b*x^3 + c*x^5],x]

[Out]

((b + 2*c*x^2)*Sqrt[a*x + b*x^3 + c*x^5])/(8*c*Sqrt[x]) - ((b^2 - 4*a*c)*Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*ArcTa
nh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*c^(3/2)*Sqrt[a*x + b*x^3 + c*x^5])

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 1107

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

Rule 1914

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[(x^(q/2)*Sqrt[a
+ b*x^(n - q) + c*x^(2*(n - q))])/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)], Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

Rule 1918

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

Rubi steps

\begin {align*} \int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx &=\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\left (b^2-4 a c\right ) \int \frac {x^{3/2}}{\sqrt {a x+b x^3+c x^5}} \, dx}{8 c}\\ &=\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\left (\left (b^2-4 a c\right ) \sqrt {x} \sqrt {a+b x^2+c x^4}\right ) \int \frac {x}{\sqrt {a+b x^2+c x^4}} \, dx}{8 c \sqrt {a x+b x^3+c x^5}}\\ &=\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\left (\left (b^2-4 a c\right ) \sqrt {x} \sqrt {a+b x^2+c x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{16 c \sqrt {a x+b x^3+c x^5}}\\ &=\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\left (\left (b^2-4 a c\right ) \sqrt {x} \sqrt {a+b x^2+c x^4}\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 )}{8 c \sqrt {a x+b x^3+c x^5}}\\ &=\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\left (b^2-4 a c\right ) \sqrt {x} \sqrt {a+b x^2+c x^4} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2} \sqrt {a x+b x^3+c x^5}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 126, normalized size = 0.98 \[ \frac {\sqrt {x \left (a+b x^2+c x^4\right )} \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{8 c^{3/2}}\right )}{2 \sqrt {x} \sqrt {a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*Sqrt[a*x + b*x^3 + c*x^5],x]

[Out]

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

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fricas [A]  time = 0.82, size = 232, normalized size = 1.80 \[ \left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{5} + 8 \, b c x^{3} + 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {c} \sqrt {x} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) - 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c^{2} x^{2} + b c\right )} \sqrt {x}}{32 \, c^{2} x}, \frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {-c} \sqrt {x}}{2 \, {\left (c^{2} x^{5} + b c x^{3} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c^{2} x^{2} + b c\right )} \sqrt {x}}{16 \, c^{2} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*((b^2 - 4*a*c)*sqrt(c)*x*log(-(8*c^2*x^5 + 8*b*c*x^3 + 4*sqrt(c*x^5 + b*x^3 + a*x)*(2*c*x^2 + b)*sqrt(c
)*sqrt(x) + (b^2 + 4*a*c)*x)/x) - 4*sqrt(c*x^5 + b*x^3 + a*x)*(2*c^2*x^2 + b*c)*sqrt(x))/(c^2*x), 1/16*((b^2 -
4*a*c)*sqrt(-c)*x*arctan(1/2*sqrt(c*x^5 + b*x^3 + a*x)*(2*c*x^2 + b)*sqrt(-c)*sqrt(x)/(c^2*x^5 + b*c*x^3 + a*
c*x)) + 2*sqrt(c*x^5 + b*x^3 + a*x)*(2*c^2*x^2 + b*c)*sqrt(x))/(c^2*x)]

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giac [A]  time = 0.97, size = 127, normalized size = 0.98 \[ \frac {1}{8} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, x^{2} + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {3}{2}}} - \frac {b^{2} \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 4 \, a c \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 2 \, \sqrt {a} b \sqrt {c}}{16 \, c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 157, normalized size = 1.22 \[ \frac {\sqrt {\left (c \,x^{4}+b \,x^{2}+a \right ) x}\, \left (4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{\frac {3}{2}} x^{2}+4 a c \ln \left (\frac {2 c \,x^{2}+b +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}}{2 \sqrt {c}}\right )-b^{2} \ln \left (\frac {2 c \,x^{2}+b +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}}{2 \sqrt {c}}\right )+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b \sqrt {c}\right )}{16 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{\frac {3}{2}} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((c*x^4+b*x^2+a)*x)^(1/2)/c^(3/2)*(4*x^2*c^(3/2)*(c*x^4+b*x^2+a)^(1/2)+4*ln(1/2*(2*c*x^2+2*(c*x^4+b*x^2+a
)^(1/2)*c^(1/2)+b)/c^(1/2))*a*c-ln(1/2*(2*c*x^2+2*(c*x^4+b*x^2+a)^(1/2)*c^(1/2)+b)/c^(1/2))*b^2+2*b*(c*x^4+b*x
^2+a)^(1/2)*c^(1/2))/x^(1/2)/(c*x^4+b*x^2+a)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{5} + b x^{3} + a x} \sqrt {x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {x}\,\sqrt {c\,x^5+b\,x^3+a\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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