\(\int x \sqrt [4]{b x^3+a x^4} \, dx\) [1658]

Optimal result

Integrand size = 17, antiderivative size = 112 \[ \int x \sqrt [4]{b x^3+a x^4} \, dx=\frac {\left (-7 b^2+4 a b x+32 a^2 x^2\right ) \sqrt [4]{b x^3+a x^4}}{96 a^2}-\frac {7 b^3 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{64 a^{11/4}}+\frac {7 b^3 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{64 a^{11/4}} \]

[Out]

1/96*(32*a^2*x^2+4*a*b*x-7*b^2)*(a*x^4+b*x^3)^(1/4)/a^2-7/64*b^3*arctan(a^(1/4)*x/(a*x^4+b*x^3)^(1/4))/a^(11/4
)+7/64*b^3*arctanh(a^(1/4)*x/(a*x^4+b*x^3)^(1/4))/a^(11/4)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.75, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {2046, 2049, 2057, 65, 338, 304, 209, 212} \[ \int x \sqrt [4]{b x^3+a x^4} \, dx=-\frac {7 b^3 x^{9/4} (a x+b)^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{64 a^{11/4} \left (a x^4+b x^3\right )^{3/4}}+\frac {7 b^3 x^{9/4} (a x+b)^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{64 a^{11/4} \left (a x^4+b x^3\right )^{3/4}}-\frac {7 b^2 \sqrt [4]{a x^4+b x^3}}{96 a^2}+\frac {b x \sqrt [4]{a x^4+b x^3}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{a x^4+b x^3} \]

[In]

Int[x*(b*x^3 + a*x^4)^(1/4),x]

[Out]

(-7*b^2*(b*x^3 + a*x^4)^(1/4))/(96*a^2) + (b*x*(b*x^3 + a*x^4)^(1/4))/(24*a) + (x^2*(b*x^3 + a*x^4)^(1/4))/3 -
(7*b^3*x^(9/4)*(b + a*x)^(3/4)*ArcTan[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(64*a^(11/4)*(b*x^3 + a*x^4)^(3/4))
+ (7*b^3*x^(9/4)*(b + a*x)^(3/4)*ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(64*a^(11/4)*(b*x^3 + a*x^4)^(3/
4))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

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

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 2046

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*(m + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {1}{12} b \int \frac {x^4}{\left (b x^3+a x^4\right )^{3/4}} \, dx \\ & = \frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}-\frac {\left (7 b^2\right ) \int \frac {x^3}{\left (b x^3+a x^4\right )^{3/4}} \, dx}{96 a} \\ & = -\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {\left (7 b^3\right ) \int \frac {x^2}{\left (b x^3+a x^4\right )^{3/4}} \, dx}{128 a^2} \\ & = -\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {\left (7 b^3 x^{9/4} (b+a x)^{3/4}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{128 a^2 \left (b x^3+a x^4\right )^{3/4}} \\ & = -\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {\left (7 b^3 x^{9/4} (b+a x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{32 a^2 \left (b x^3+a x^4\right )^{3/4}} \\ & = -\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {\left (7 b^3 x^{9/4} (b+a x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{32 a^2 \left (b x^3+a x^4\right )^{3/4}} \\ & = -\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {\left (7 b^3 x^{9/4} (b+a x)^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{5/2} \left (b x^3+a x^4\right )^{3/4}}-\frac {\left (7 b^3 x^{9/4} (b+a x)^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{5/2} \left (b x^3+a x^4\right )^{3/4}} \\ & = -\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}-\frac {7 b^3 x^{9/4} (b+a x)^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{11/4} \left (b x^3+a x^4\right )^{3/4}}+\frac {7 b^3 x^{9/4} (b+a x)^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{11/4} \left (b x^3+a x^4\right )^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.15 \[ \int x \sqrt [4]{b x^3+a x^4} \, dx=\frac {x^{9/4} (b+a x)^{3/4} \left (2 a^{3/4} x^{3/4} \sqrt [4]{b+a x} \left (-7 b^2+4 a b x+32 a^2 x^2\right )-21 b^3 \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )+21 b^3 \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )\right )}{192 a^{11/4} \left (x^3 (b+a x)\right )^{3/4}} \]

[In]

Integrate[x*(b*x^3 + a*x^4)^(1/4),x]

[Out]

(x^(9/4)*(b + a*x)^(3/4)*(2*a^(3/4)*x^(3/4)*(b + a*x)^(1/4)*(-7*b^2 + 4*a*b*x + 32*a^2*x^2) - 21*b^3*ArcTan[(a
^(1/4)*x^(1/4))/(b + a*x)^(1/4)] + 21*b^3*ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)]))/(192*a^(11/4)*(x^3*(b +
a*x))^(3/4))

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.17

method	result	size
pseudoelliptic	\(\frac {128 \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}} a^{\frac {11}{4}} x^{2}+16 a^{\frac {7}{4}} b x \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}-28 b^{2} \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}} a^{\frac {3}{4}}+21 \ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}\right ) b^{3}+42 \arctan \left (\frac {\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b^{3}}{384 a^{\frac {11}{4}}}\)	\(131\)

[In]

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

[Out]

1/384/a^(11/4)*(128*(x^3*(a*x+b))^(1/4)*a^(11/4)*x^2+16*a^(7/4)*b*x*(x^3*(a*x+b))^(1/4)-28*b^2*(x^3*(a*x+b))^(
1/4)*a^(3/4)+21*ln((a^(1/4)*x+(x^3*(a*x+b))^(1/4))/(-a^(1/4)*x+(x^3*(a*x+b))^(1/4)))*b^3+42*arctan(1/a^(1/4)/x
*(x^3*(a*x+b))^(1/4))*b^3)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.26 \[ \int x \sqrt [4]{b x^3+a x^4} \, dx=\frac {21 \, a^{2} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} \log \left (\frac {7 \, {\left (a^{3} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{3}\right )}}{x}\right ) - 21 \, a^{2} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (a^{3} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{3}\right )}}{x}\right ) - 21 i \, a^{2} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (i \, a^{3} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{3}\right )}}{x}\right ) + 21 i \, a^{2} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (-i \, a^{3} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{3}\right )}}{x}\right ) + 4 \, {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} {\left (32 \, a^{2} x^{2} + 4 \, a b x - 7 \, b^{2}\right )}}{384 \, a^{2}} \]

[In]

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

[Out]

1/384*(21*a^2*(b^12/a^11)^(1/4)*log(7*(a^3*(b^12/a^11)^(1/4)*x + (a*x^4 + b*x^3)^(1/4)*b^3)/x) - 21*a^2*(b^12/
a^11)^(1/4)*log(-7*(a^3*(b^12/a^11)^(1/4)*x - (a*x^4 + b*x^3)^(1/4)*b^3)/x) - 21*I*a^2*(b^12/a^11)^(1/4)*log(-
7*(I*a^3*(b^12/a^11)^(1/4)*x - (a*x^4 + b*x^3)^(1/4)*b^3)/x) + 21*I*a^2*(b^12/a^11)^(1/4)*log(-7*(-I*a^3*(b^12
/a^11)^(1/4)*x - (a*x^4 + b*x^3)^(1/4)*b^3)/x) + 4*(a*x^4 + b*x^3)^(1/4)*(32*a^2*x^2 + 4*a*b*x - 7*b^2))/a^2

Sympy [F]

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

[In]

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

[Out]

Integral(x*(x**3*(a*x + b))**(1/4), x)

Maxima [F]

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

[In]

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

[Out]

integrate((a*x^4 + b*x^3)^(1/4)*x, x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (92) = 184\).

Time = 0.32 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.33 \[ \int x \sqrt [4]{b x^3+a x^4} \, dx=\frac {\frac {42 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{4} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{3}} + \frac {42 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{4} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{3}} + \frac {21 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{4} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{a^{3}} + \frac {21 \, \sqrt {2} b^{4} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{\left (-a\right )^{\frac {3}{4}} a^{2}} - \frac {8 \, {\left (7 \, {\left (a + \frac {b}{x}\right )}^{\frac {9}{4}} b^{4} - 18 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{4}} a b^{4} - 21 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} a^{2} b^{4}\right )} x^{3}}{a^{2} b^{3}}}{768 \, b} \]

[In]

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

[Out]

1/768*(42*sqrt(2)*(-a)^(1/4)*b^4*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x)^(1/4))/(-a)^(1/4))/a^3 +
42*sqrt(2)*(-a)^(1/4)*b^4*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/a^3 + 21*s
qrt(2)*(-a)^(1/4)*b^4*log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/a^3 + 21*sqrt(2)*b^4*
log(-sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/((-a)^(3/4)*a^2) - 8*(7*(a + b/x)^(9/4)*b^
4 - 18*(a + b/x)^(5/4)*a*b^4 - 21*(a + b/x)^(1/4)*a^2*b^4)*x^3/(a^2*b^3))/b

Mupad [F(-1)]

Timed out. \[ \int x \sqrt [4]{b x^3+a x^4} \, dx=\int x\,{\left (a\,x^4+b\,x^3\right )}^{1/4} \,d x \]

[In]

int(x*(a*x^4 + b*x^3)^(1/4),x)

[Out]

int(x*(a*x^4 + b*x^3)^(1/4), x)