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

Optimal. Leaf size=177 \[ -\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a x+b x^3+c x^5}}{128 c^2 \sqrt{x}}+\frac{3 \sqrt{x} \left (b^2-4 a c\right )^2 \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{256 c^{5/2} \sqrt{a x+b x^3+c x^5}}+\frac{\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}} \]

[Out]

(-3*(b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a*x + b*x^3 + c*x^5])/(128*c^2*Sqrt[x]) + ((b + 2*c*x^2)*(a*x + b*x^3 + c
*x^5)^(3/2))/(16*c*x^(3/2)) + (3*(b^2 - 4*a*c)^2*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])])/(256*c^(5/2)*Sqrt[a*x + b*x^3 + c*x^5])

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Rubi [A]  time = 0.138344, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a x+b x^3+c x^5}}{128 c^2 \sqrt{x}}+\frac{3 \sqrt{x} \left (b^2-4 a c\right )^2 \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{256 c^{5/2} \sqrt{a x+b x^3+c x^5}}+\frac{\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a*x + b*x^3 + c*x^5])/(128*c^2*Sqrt[x]) + ((b + 2*c*x^2)*(a*x + b*x^3 + c
*x^5)^(3/2))/(16*c*x^(3/2)) + (3*(b^2 - 4*a*c)^2*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])])/(256*c^(5/2)*Sqrt[a*x + b*x^3 + c*x^5])

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]

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{\left (a x+b x^3+c x^5\right )^{3/2}}{\sqrt{x}} \, dx &=\frac{\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}-\frac{\left (3 \left (b^2-4 a c\right )\right ) \int \sqrt{x} \sqrt{a x+b x^3+c x^5} \, dx}{16 c}\\ &=-\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a x+b x^3+c x^5}}{128 c^2 \sqrt{x}}+\frac{\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac{\left (3 \left (b^2-4 a c\right )^2\right ) \int \frac{x^{3/2}}{\sqrt{a x+b x^3+c x^5}} \, dx}{128 c^2}\\ &=-\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a x+b x^3+c x^5}}{128 c^2 \sqrt{x}}+\frac{\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac{\left (3 \left (b^2-4 a c\right )^2 \sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{x}{\sqrt{a+b x^2+c x^4}} \, dx}{128 c^2 \sqrt{a x+b x^3+c x^5}}\\ &=-\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a x+b x^3+c x^5}}{128 c^2 \sqrt{x}}+\frac{\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac{\left (3 \left (b^2-4 a c\right )^2 \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 )}{256 c^2 \sqrt{a x+b x^3+c x^5}}\\ &=-\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a x+b x^3+c x^5}}{128 c^2 \sqrt{x}}+\frac{\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac{\left (3 \left (b^2-4 a c\right )^2 \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 )}{128 c^2 \sqrt{a x+b x^3+c x^5}}\\ &=-\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a x+b x^3+c x^5}}{128 c^2 \sqrt{x}}+\frac{\left (b+2 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{16 c x^{3/2}}+\frac{3 \left (b^2-4 a c\right )^2 \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 )}{256 c^{5/2} \sqrt{a x+b x^3+c x^5}}\\ \end{align*}

Mathematica [A]  time = 0.113543, size = 152, normalized size = 0.86 \[ \frac{\sqrt{x} \sqrt{a+b x^2+c x^4} \left (2 \sqrt{c} \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4} \left (4 c \left (5 a+2 c x^4\right )-3 b^2+8 b c x^2\right )+3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )\right )}{256 c^{5/2} \sqrt{x \left (a+b x^2+c x^4\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*(2*Sqrt[c]*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4]*(-3*b^2 + 8*b*c*x^2 + 4*c*(5
*a + 2*c*x^4)) + 3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])]))/(256*c^(5/2)*S
qrt[x*(a + b*x^2 + c*x^4)])

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Maple [A]  time = 0.02, size = 295, normalized size = 1.7 \begin{align*}{\frac{1}{256}\sqrt{x \left ( c{x}^{4}+b{x}^{2}+a \right ) } \left ( 32\,{x}^{6}{c}^{7/2}\sqrt{c{x}^{4}+b{x}^{2}+a}+48\,{x}^{4}b{c}^{5/2}\sqrt{c{x}^{4}+b{x}^{2}+a}+80\,{x}^{2}a{c}^{5/2}\sqrt{c{x}^{4}+b{x}^{2}+a}+4\,{x}^{2}{b}^{2}{c}^{3/2}\sqrt{c{x}^{4}+b{x}^{2}+a}+48\,\ln \left ( 1/2\,{\frac{2\,c{x}^{2}+2\,\sqrt{c{x}^{4}+b{x}^{2}+a}\sqrt{c}+b}{\sqrt{c}}} \right ){a}^{2}{c}^{2}-24\,\ln \left ( 1/2\,{\frac{2\,c{x}^{2}+2\,\sqrt{c{x}^{4}+b{x}^{2}+a}\sqrt{c}+b}{\sqrt{c}}} \right ) a{b}^{2}c+3\,\ln \left ( 1/2\,{\frac{2\,c{x}^{2}+2\,\sqrt{c{x}^{4}+b{x}^{2}+a}\sqrt{c}+b}{\sqrt{c}}} \right ){b}^{4}+40\,ab{c}^{3/2}\sqrt{c{x}^{4}+b{x}^{2}+a}-6\,{b}^{3}\sqrt{c}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(x*(c*x^4+b*x^2+a))^(1/2)/c^(5/2)*(32*x^6*c^(7/2)*(c*x^4+b*x^2+a)^(1/2)+48*x^4*b*c^(5/2)*(c*x^4+b*x^2+a)
^(1/2)+80*x^2*a*c^(5/2)*(c*x^4+b*x^2+a)^(1/2)+4*x^2*b^2*c^(3/2)*(c*x^4+b*x^2+a)^(1/2)+48*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^2*c^2-24*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*b^2*c+3*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^4+40*a*b*c^(3/2)*(c*x^4+b*x^2+a)^(1/2)
-6*b^3*c^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^(1/2)/(c*x^4+b*x^2+a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.51134, size = 768, normalized size = 4.34 \begin{align*} \left [\frac{3 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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 \,{\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \,{\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt{c x^{5} + b x^{3} + a x} \sqrt{x}}{512 \, c^{3} x}, -\frac{3 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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 \,{\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \,{\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt{c x^{5} + b x^{3} + a x} \sqrt{x}}{256 \, c^{3} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/512*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*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*(16*c^4*x^6 + 24*b*c^3*x^4 - 3*b^3*c + 20*a*b*c^2 + 2*(
b^2*c^2 + 20*a*c^3)*x^2)*sqrt(c*x^5 + b*x^3 + a*x)*sqrt(x))/(c^3*x), -1/256*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*
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*(16*c^4*x^6 + 24*b*c^3*x^4 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2 + 20*a*c^3)*x^2)*sqrt(c*x^5 + b*x^3 + a*x)*sq
rt(x))/(c^3*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.5184, size = 624, normalized size = 3.53 \begin{align*} \frac{1}{16} \,{\left (2 \, \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 )}{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}}{c^{\frac{3}{2}}}\right )} a + \frac{1}{96} \,{\left (2 \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \,{\left (4 \, x^{2} + \frac{b}{c}\right )} x^{2} - \frac{3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac{3 \,{\left (b^{3} - 4 \, a b 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 )}{c^{\frac{5}{2}}} + \frac{3 \, b^{3} \log \left ({\left | -b + 2 \, \sqrt{a} \sqrt{c} \right |}\right ) - 12 \, a b c \log \left ({\left | -b + 2 \, \sqrt{a} \sqrt{c} \right |}\right ) + 6 \, \sqrt{a} b^{2} \sqrt{c} - 16 \, a^{\frac{3}{2}} c^{\frac{3}{2}}}{c^{\frac{5}{2}}}\right )} b + \frac{1}{384} \,{\left (\sqrt{c x^{4} + b x^{2} + a}{\left (2 \,{\left (4 \,{\left (6 \, x^{2} + \frac{b}{c}\right )} x^{2} - \frac{5 \, b^{2} c^{7} - 12 \, a c^{8}}{c^{9}}\right )} x^{2} + \frac{15 \, b^{3} c^{6} - 52 \, a b c^{7}}{c^{9}}\right )} - \frac{15 \, \sqrt{a} b^{3} - 52 \, a^{\frac{3}{2}} b c}{c^{3}}\right )} c + \frac{{\left (5 \, b^{4} c^{6} - 24 \, a b^{2} c^{7} + 16 \, a^{2} c^{8}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{17}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*sqrt(c*x^4 + b*x^2 + a)*(2*x^2 + b/c) + (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) - (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))*a + 1/96*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*x^2 + b/c)*x^2 - (3*b^2 - 8*a*c)/c^2) -
3*(b^3 - 4*a*b*c)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(5/2) + (3*b^3*log(abs(-b
+ 2*sqrt(a)*sqrt(c))) - 12*a*b*c*log(abs(-b + 2*sqrt(a)*sqrt(c))) + 6*sqrt(a)*b^2*sqrt(c) - 16*a^(3/2)*c^(3/2
))/c^(5/2))*b + 1/384*(sqrt(c*x^4 + b*x^2 + a)*(2*(4*(6*x^2 + b/c)*x^2 - (5*b^2*c^7 - 12*a*c^8)/c^9)*x^2 + (15
*b^3*c^6 - 52*a*b*c^7)/c^9) - (15*sqrt(a)*b^3 - 52*a^(3/2)*b*c)/c^3)*c + 1/256*(5*b^4*c^6 - 24*a*b^2*c^7 + 16*
a^2*c^8)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(17/2)