\(\int \frac {\sqrt {a x+b x^3+c x^5}}{x^{3/2}} \, dx\) [108]

Optimal result

Integrand size = 24, antiderivative size = 194 \[ \int \frac {\sqrt {a x+b x^3+c x^5}}{x^{3/2}} \, dx=\frac {\sqrt {a x+b x^3+c x^5}}{2 \sqrt {x}}-\frac {\sqrt {a} \sqrt {x} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {a x+b x^3+c x^5}}+\frac {b \sqrt {x} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {c} \sqrt {a x+b x^3+c x^5}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1935, 1967, 1265, 857, 635, 212, 738} \[ \int \frac {\sqrt {a x+b x^3+c x^5}}{x^{3/2}} \, dx=-\frac {\sqrt {a} \sqrt {x} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {a x+b x^3+c x^5}}+\frac {b \sqrt {x} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {c} \sqrt {a x+b x^3+c x^5}}+\frac {\sqrt {a x+b x^3+c x^5}}{2 \sqrt {x}} \]

[In]

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

[Out]

Sqrt[a*x + b*x^3 + c*x^5]/(2*Sqrt[x]) - (Sqrt[a]*Sqrt[x]*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*Sqrt[a*x + b*x^3 + c*x^5]) + (b*Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*ArcTanh[(b +
2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4*Sqrt[c]*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 738

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 857

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 1265

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

Rule 1935

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a*
x^q + b*x^n + c*x^(2*n - q))^p/(m + p*(2*n - q) + 1)), x] + Dist[(n - q)*(p/(m + p*(2*n - q) + 1)), Int[x^(m +
q)*(2*a + b*x^(n - 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] && IGtQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] && Gt
Q[m + p*q + 1, -(n - q)] && NeQ[m + p*(2*n - q) + 1, 0]

Rule 1967

Int[((x_)^(m_.)*((A_) + (B_.)*(x_)^(j_.)))/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Sym
bol] :> 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)*((A + B*x^(n - q))/Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]), x], x] /; FreeQ[{a, b, c, A, B, m, n, q}, x
] && EqQ[j, n - q] && EqQ[r, 2*n - q] && PosQ[n - q] && (EqQ[m, 1/2] || EqQ[m, -2^(-1)]) && EqQ[n, 3] && EqQ[q
, 1]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {a x+b x^3+c x^5}}{x^{3/2}} \, dx=\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (2 \sqrt {c} \sqrt {a+b x^2+c x^4}+4 \sqrt {a} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )-b \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )\right )}{4 \sqrt {c} \sqrt {x \left (a+b x^2+c x^4\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.70

method	result	size
default	\(-\frac {\sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}\, \left (2 \sqrt {a}\, \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right ) \sqrt {c}-b \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right )-2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}\right )}{4 \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}}\)	\(136\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 666, normalized size of antiderivative = 3.43 \[ \int \frac {\sqrt {a x+b x^3+c x^5}}{x^{3/2}} \, dx=\left [\frac {b \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{5} + 8 \, b c x^{3} + 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {c} \sqrt {x} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 2 \, \sqrt {a} c x \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{5} + 8 \, a b x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} \sqrt {x}}{x^{5}}\right ) + 4 \, \sqrt {c x^{5} + b x^{3} + a x} c \sqrt {x}}{8 \, c x}, -\frac {b \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {-c} \sqrt {x}}{2 \, {\left (c^{2} x^{5} + b c x^{3} + a c x\right )}}\right ) - \sqrt {a} c x \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{5} + 8 \, a b x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} \sqrt {x}}{x^{5}}\right ) - 2 \, \sqrt {c x^{5} + b x^{3} + a x} c \sqrt {x}}{4 \, c x}, \frac {4 \, \sqrt {-a} c x \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a} \sqrt {x}}{2 \, {\left (a c x^{5} + a b x^{3} + a^{2} x\right )}}\right ) + b \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{5} + 8 \, b c x^{3} + 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {c} \sqrt {x} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, \sqrt {c x^{5} + b x^{3} + a x} c \sqrt {x}}{8 \, c x}, \frac {2 \, \sqrt {-a} c x \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a} \sqrt {x}}{2 \, {\left (a c x^{5} + a b x^{3} + a^{2} x\right )}}\right ) - b \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {-c} \sqrt {x}}{2 \, {\left (c^{2} x^{5} + b c x^{3} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{5} + b x^{3} + a x} c \sqrt {x}}{4 \, c x}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt {a x+b x^3+c x^5}}{x^{3/2}} \, dx=\int { \frac {\sqrt {c x^{5} + b x^{3} + a x}}{x^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a x+b x^3+c x^5}}{x^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a x+b x^3+c x^5}}{x^{3/2}} \, dx=\int \frac {\sqrt {c\,x^5+b\,x^3+a\,x}}{x^{3/2}} \,d x \]

[In]

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

[Out]

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