\(\int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx\) [106]

Optimal result

Integrand size = 24, antiderivative size = 129 \[ \int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx=\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\left (b^2-4 a c\right ) \sqrt {x} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2} \sqrt {a x+b x^3+c x^5}} \]

[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)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1932, 1928, 1121, 635, 212} \[ \int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx=\frac {\left (b+2 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{8 c \sqrt {x}}-\frac {\sqrt {x} \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2} \sqrt {a x+b x^3+c x^5}} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 1121

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 1928

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 1932

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*} \text {integral}& = \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 ) \text {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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.84 \[ \int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx=\frac {\sqrt {x \left (a+b x^2+c x^4\right )} \left (\sqrt {c} \left (b+2 c x^2\right )+\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a}-\sqrt {a+b x^2+c x^4}}\right )}{\sqrt {a+b x^2+c x^4}}\right )}{8 c^{3/2} \sqrt {x}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {\left (2 c \,x^{2}+b \right ) \left (c \,x^{4}+b \,x^{2}+a \right ) \sqrt {x}}{8 c \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {x}}{16 c^{\frac {3}{2}} \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}\)	\(123\)
default	\(\frac {\sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}\, \left (4 c^{\frac {3}{2}} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}+4 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) a c -\ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) b^{2}+2 b \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}\right )}{16 c^{\frac {3}{2}} \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}\)	\(157\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.80 \[ \int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{5} + 8 \, b c x^{3} + 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {c} \sqrt {x} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) - 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c^{2} x^{2} + b c\right )} \sqrt {x}}{32 \, c^{2} x}, \frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {-c} \sqrt {x}}{2 \, {\left (c^{2} x^{5} + b c x^{3} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c^{2} x^{2} + b c\right )} \sqrt {x}}{16 \, c^{2} x}\right ] \]

[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)]

Sympy [F]

\[ \int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx=\int \sqrt {x} \sqrt {x \left (a + b x^{2} + c x^{4}\right )}\, dx \]

[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)

Maxima [F]

\[ \int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx=\int { \sqrt {c x^{5} + b x^{3} + a x} \sqrt {x} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.94 \[ \int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx=\frac {1}{8} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, x^{2} + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {3}{2}}} - \frac {b^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 4 \, a c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 2 \, \sqrt {a} b \sqrt {c}}{16 \, c^{\frac {3}{2}}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {x} \sqrt {a x+b x^3+c x^5} \, dx=\int \sqrt {x}\,\sqrt {c\,x^5+b\,x^3+a\,x} \,d x \]

[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)