\(\int \frac {x^3 \sqrt {-x+x^4}}{-b+a x^3} \, dx\) [1655]

Optimal result

Integrand size = 26, antiderivative size = 112 \[ \int \frac {x^3 \sqrt {-x+x^4}}{-b+a x^3} \, dx=\frac {x \sqrt {-x+x^4}}{3 a}+\frac {2 \sqrt {a-b} \sqrt {b} \arctan \left (\frac {\sqrt {a-b} x \sqrt {-x+x^4}}{\sqrt {b} (-1+x) \left (1+x+x^2\right )}\right )}{3 a^2}+\frac {(-a+2 b) \text {arctanh}\left (\frac {x^2}{\sqrt {-x+x^4}}\right )}{3 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.33, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {2067, 477, 476, 489, 537, 223, 212, 385, 211} \[ \int \frac {x^3 \sqrt {-x+x^4}}{-b+a x^3} \, dx=\frac {2 \sqrt {b} \sqrt {x^4-x} \sqrt {a-b} \arctan \left (\frac {x^{3/2} \sqrt {a-b}}{\sqrt {b} \sqrt {x^3-1}}\right )}{3 a^2 \sqrt {x^3-1} \sqrt {x}}-\frac {\sqrt {x^4-x} (a-2 b) \text {arctanh}\left (\frac {x^{3/2}}{\sqrt {x^3-1}}\right )}{3 a^2 \sqrt {x^3-1} \sqrt {x}}+\frac {\sqrt {x^4-x} x}{3 a} \]

[In]

Int[(x^3*Sqrt[-x + x^4])/(-b + a*x^3),x]

[Out]

(x*Sqrt[-x + x^4])/(3*a) + (2*Sqrt[a - b]*Sqrt[b]*Sqrt[-x + x^4]*ArcTan[(Sqrt[a - b]*x^(3/2))/(Sqrt[b]*Sqrt[-1
+ x^3])])/(3*a^2*Sqrt[x]*Sqrt[-1 + x^3]) - ((a - 2*b)*Sqrt[-x + x^4]*ArcTanh[x^(3/2)/Sqrt[-1 + x^3]])/(3*a^2*
Sqrt[x]*Sqrt[-1 + x^3])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 476

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 489

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*(m + n*(p + q) + 1))), x] - Dist[e^n/(b*(m + n*(p +
q) + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 1) + (a*d*(m - n + 1) - n*q*(b
*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] &&
GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 2067

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.))^(q_.), x_Symbol]
:> Dist[e^IntPart[m]*(e*x)^FracPart[m]*((a*x^j + b*x^(j + n))^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a
+ b*x^n)^FracPart[p])), Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, j, m, n,
p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-x+x^4} \int \frac {x^{7/2} \sqrt {-1+x^3}}{-b+a x^3} \, dx}{\sqrt {x} \sqrt {-1+x^3}} \\ & = \frac {\left (2 \sqrt {-x+x^4}\right ) \text {Subst}\left (\int \frac {x^8 \sqrt {-1+x^6}}{-b+a x^6} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {-1+x^3}} \\ & = \frac {\left (2 \sqrt {-x+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt {-1+x^2}}{-b+a x^2} \, dx,x,x^{3/2}\right )}{3 \sqrt {x} \sqrt {-1+x^3}} \\ & = \frac {x \sqrt {-x+x^4}}{3 a}-\frac {\sqrt {-x+x^4} \text {Subst}\left (\int \frac {b+(a-2 b) x^2}{\sqrt {-1+x^2} \left (-b+a x^2\right )} \, dx,x,x^{3/2}\right )}{3 a \sqrt {x} \sqrt {-1+x^3}} \\ & = \frac {x \sqrt {-x+x^4}}{3 a}-\frac {\left ((a-2 b) \sqrt {-x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^2}} \, dx,x,x^{3/2}\right )}{3 a^2 \sqrt {x} \sqrt {-1+x^3}}-\frac {\left (2 (a-b) b \sqrt {-x+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^2} \left (-b+a x^2\right )} \, dx,x,x^{3/2}\right )}{3 a^2 \sqrt {x} \sqrt {-1+x^3}} \\ & = \frac {x \sqrt {-x+x^4}}{3 a}-\frac {\left ((a-2 b) \sqrt {-x+x^4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {-1+x^3}}\right )}{3 a^2 \sqrt {x} \sqrt {-1+x^3}}-\frac {\left (2 (a-b) b \sqrt {-x+x^4}\right ) \text {Subst}\left (\int \frac {1}{-b-(a-b) x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {-1+x^3}}\right )}{3 a^2 \sqrt {x} \sqrt {-1+x^3}} \\ & = \frac {x \sqrt {-x+x^4}}{3 a}+\frac {2 \sqrt {a-b} \sqrt {b} \sqrt {-x+x^4} \arctan \left (\frac {\sqrt {a-b} x^{3/2}}{\sqrt {b} \sqrt {-1+x^3}}\right )}{3 a^2 \sqrt {x} \sqrt {-1+x^3}}-\frac {(a-2 b) \sqrt {-x+x^4} \text {arctanh}\left (\frac {x^{3/2}}{\sqrt {-1+x^3}}\right )}{3 a^2 \sqrt {x} \sqrt {-1+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.15 \[ \int \frac {x^3 \sqrt {-x+x^4}}{-b+a x^3} \, dx=\frac {\sqrt {x} \sqrt {-1+x^3} \left (a x^{3/2} \sqrt {-1+x^3}+2 \sqrt {a-b} \sqrt {b} \arctan \left (\frac {b-a x^{3/2} \left (x^{3/2}+\sqrt {-1+x^3}\right )}{\sqrt {a-b} \sqrt {b}}\right )-(a-2 b) \log \left (x^{3/2}+\sqrt {-1+x^3}\right )\right )}{3 a^2 \sqrt {x \left (-1+x^3\right )}} \]

[In]

Integrate[(x^3*Sqrt[-x + x^4])/(-b + a*x^3),x]

[Out]

(Sqrt[x]*Sqrt[-1 + x^3]*(a*x^(3/2)*Sqrt[-1 + x^3] + 2*Sqrt[a - b]*Sqrt[b]*ArcTan[(b - a*x^(3/2)*(x^(3/2) + Sqr
t[-1 + x^3]))/(Sqrt[a - b]*Sqrt[b])] - (a - 2*b)*Log[x^(3/2) + Sqrt[-1 + x^3]]))/(3*a^2*Sqrt[x*(-1 + x^3)])

Maple [A] (verified)

Time = 1.57 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.93

method	result	size
risch	\(\frac {x^{2} \left (x^{3}-1\right )}{3 a \sqrt {x \left (x^{3}-1\right )}}-\frac {\frac {\left (a -2 b \right ) \ln \left (-2 x^{3}-2 x \sqrt {x^{4}-x}+1\right )}{3 a}+\frac {4 \left (a -b \right ) b \arctan \left (\frac {\sqrt {x^{4}-x}\, b}{x^{2} \sqrt {\left (a -b \right ) b}}\right )}{3 a \sqrt {\left (a -b \right ) b}}}{2 a}\)	\(104\)
pseudoelliptic	\(-\frac {2 \left (\left (a -b \right ) b \arctan \left (\frac {\sqrt {x^{4}-x}\, b}{x^{2} \sqrt {\left (a -b \right ) b}}\right )-\frac {\sqrt {\left (a -b \right ) b}\, \left (\left (\frac {a}{2}-b \right ) \ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )+\left (-\frac {a}{2}+b \right ) \ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )+\sqrt {x^{4}-x}\, a x \right )}{2}\right )}{3 \sqrt {\left (a -b \right ) b}\, a^{2}}\)	\(123\)
default	\(\frac {\frac {x \sqrt {x^{4}-x}}{3}+\frac {\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )}{6}-\frac {\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )}{6}}{a}-\frac {b \left (\left (2 a -2 b \right ) \arctan \left (\frac {\sqrt {x^{4}-x}\, b}{x^{2} \sqrt {\left (a -b \right ) b}}\right )+\sqrt {\left (a -b \right ) b}\, \left (\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )\right )\right )}{3 a^{2} \sqrt {\left (a -b \right ) b}}\)	\(160\)
elliptic	\(\text {Expression too large to display}\)	\(665\)

[In]

int(x^3*(x^4-x)^(1/2)/(a*x^3-b),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 1.02 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.21 \[ \int \frac {x^3 \sqrt {-x+x^4}}{-b+a x^3} \, dx=\left [\frac {2 \, \sqrt {x^{4} - x} a x - {\left (a - 2 \, b\right )} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} - x} x + 1\right ) + \sqrt {-a b + b^{2}} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} x^{6} + 2 \, {\left (3 \, a b - 4 \, b^{2}\right )} x^{3} + 4 \, {\left ({\left (a - 2 \, b\right )} x^{4} + b x\right )} \sqrt {x^{4} - x} \sqrt {-a b + b^{2}} + b^{2}}{a^{2} x^{6} - 2 \, a b x^{3} + b^{2}}\right )}{6 \, a^{2}}, \frac {2 \, \sqrt {x^{4} - x} a x - {\left (a - 2 \, b\right )} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} - x} x + 1\right ) + 2 \, \sqrt {a b - b^{2}} \arctan \left (-\frac {2 \, \sqrt {x^{4} - x} \sqrt {a b - b^{2}} x}{{\left (a - 2 \, b\right )} x^{3} + b}\right )}{6 \, a^{2}}\right ] \]

[In]

integrate(x^3*(x^4-x)^(1/2)/(a*x^3-b),x, algorithm="fricas")

[Out]

[1/6*(2*sqrt(x^4 - x)*a*x - (a - 2*b)*log(-2*x^3 - 2*sqrt(x^4 - x)*x + 1) + sqrt(-a*b + b^2)*log(-((a^2 - 8*a*
b + 8*b^2)*x^6 + 2*(3*a*b - 4*b^2)*x^3 + 4*((a - 2*b)*x^4 + b*x)*sqrt(x^4 - x)*sqrt(-a*b + b^2) + b^2)/(a^2*x^
6 - 2*a*b*x^3 + b^2)))/a^2, 1/6*(2*sqrt(x^4 - x)*a*x - (a - 2*b)*log(-2*x^3 - 2*sqrt(x^4 - x)*x + 1) + 2*sqrt(
a*b - b^2)*arctan(-2*sqrt(x^4 - x)*sqrt(a*b - b^2)*x/((a - 2*b)*x^3 + b)))/a^2]

Sympy [F]

\[ \int \frac {x^3 \sqrt {-x+x^4}}{-b+a x^3} \, dx=\int \frac {x^{3} \sqrt {x \left (x - 1\right ) \left (x^{2} + x + 1\right )}}{a x^{3} - b}\, dx \]

[In]

integrate(x**3*(x**4-x)**(1/2)/(a*x**3-b),x)

[Out]

Integral(x**3*sqrt(x*(x - 1)*(x**2 + x + 1))/(a*x**3 - b), x)

Maxima [F]

\[ \int \frac {x^3 \sqrt {-x+x^4}}{-b+a x^3} \, dx=\int { \frac {\sqrt {x^{4} - x} x^{3}}{a x^{3} - b} \,d x } \]

[In]

integrate(x^3*(x^4-x)^(1/2)/(a*x^3-b),x, algorithm="maxima")

[Out]

integrate(sqrt(x^4 - x)*x^3/(a*x^3 - b), x)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89 \[ \int \frac {x^3 \sqrt {-x+x^4}}{-b+a x^3} \, dx=\frac {\sqrt {x^{4} - x} x}{3 \, a} - \frac {{\left (a - 2 \, b\right )} \log \left (\sqrt {-\frac {1}{x^{3}} + 1} + 1\right )}{6 \, a^{2}} + \frac {{\left (a - 2 \, b\right )} \log \left ({\left | \sqrt {-\frac {1}{x^{3}} + 1} - 1 \right |}\right )}{6 \, a^{2}} - \frac {2 \, \sqrt {a b - b^{2}} \arctan \left (\frac {b \sqrt {-\frac {1}{x^{3}} + 1}}{\sqrt {a b - b^{2}}}\right )}{3 \, a^{2}} \]

[In]

integrate(x^3*(x^4-x)^(1/2)/(a*x^3-b),x, algorithm="giac")

[Out]

1/3*sqrt(x^4 - x)*x/a - 1/6*(a - 2*b)*log(sqrt(-1/x^3 + 1) + 1)/a^2 + 1/6*(a - 2*b)*log(abs(sqrt(-1/x^3 + 1) -
1))/a^2 - 2/3*sqrt(a*b - b^2)*arctan(b*sqrt(-1/x^3 + 1)/sqrt(a*b - b^2))/a^2

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \sqrt {-x+x^4}}{-b+a x^3} \, dx=-\int \frac {x^3\,\sqrt {x^4-x}}{b-a\,x^3} \,d x \]

[In]

int(-(x^3*(x^4 - x)^(1/2))/(b - a*x^3),x)

[Out]

-int((x^3*(x^4 - x)^(1/2))/(b - a*x^3), x)