\(\int \frac {\sqrt [4]{b x^3+a x^4}}{x^2} \, dx\) [1089]

Optimal result

Integrand size = 19, antiderivative size = 81 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x^2} \, dx=-\frac {4 \sqrt [4]{b x^3+a x^4}}{x}-2 \sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )+2 \sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.67, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2045, 2057, 65, 338, 304, 209, 212} \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x^2} \, dx=-\frac {2 \sqrt [4]{a} x^{9/4} (a x+b)^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{\left (a x^4+b x^3\right )^{3/4}}+\frac {2 \sqrt [4]{a} x^{9/4} (a x+b)^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{\left (a x^4+b x^3\right )^{3/4}}-\frac {4 \sqrt [4]{a x^4+b x^3}}{x} \]

[In]

Int[(b*x^3 + a*x^4)^(1/4)/x^2,x]

[Out]

(-4*(b*x^3 + a*x^4)^(1/4))/x - (2*a^(1/4)*x^(9/4)*(b + a*x)^(3/4)*ArcTan[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(
b*x^3 + a*x^4)^(3/4) + (2*a^(1/4)*x^(9/4)*(b + a*x)^(3/4)*ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(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 2045

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 + j*p + 1))), x] - Dist[b*p*((n - j)/(c^n*(m + j*p + 1))), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*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 {4 \sqrt [4]{b x^3+a x^4}}{x}+a \int \frac {x^2}{\left (b x^3+a x^4\right )^{3/4}} \, dx \\ & = -\frac {4 \sqrt [4]{b x^3+a x^4}}{x}+\frac {\left (a x^{9/4} (b+a x)^{3/4}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{\left (b x^3+a x^4\right )^{3/4}} \\ & = -\frac {4 \sqrt [4]{b x^3+a x^4}}{x}+\frac {\left (4 a 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 )}{\left (b x^3+a x^4\right )^{3/4}} \\ & = -\frac {4 \sqrt [4]{b x^3+a x^4}}{x}+\frac {\left (4 a 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 )}{\left (b x^3+a x^4\right )^{3/4}} \\ & = -\frac {4 \sqrt [4]{b x^3+a x^4}}{x}+\frac {\left (2 \sqrt {a} 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 )}{\left (b x^3+a x^4\right )^{3/4}}-\frac {\left (2 \sqrt {a} 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 )}{\left (b x^3+a x^4\right )^{3/4}} \\ & = -\frac {4 \sqrt [4]{b x^3+a x^4}}{x}-\frac {2 \sqrt [4]{a} x^{9/4} (b+a x)^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{\left (b x^3+a x^4\right )^{3/4}}+\frac {2 \sqrt [4]{a} x^{9/4} (b+a x)^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{\left (b x^3+a x^4\right )^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.28 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x^2} \, dx=-\frac {2 x^2 (b+a x)^{3/4} \left (2 \sqrt [4]{b+a x}+\sqrt [4]{a} \sqrt [4]{x} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )-\sqrt [4]{a} \sqrt [4]{x} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )\right )}{\left (x^3 (b+a x)\right )^{3/4}} \]

[In]

Integrate[(b*x^3 + a*x^4)^(1/4)/x^2,x]

[Out]

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

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(\frac {x \left (\ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )\right ) a^{\frac {1}{4}}-4 \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{x}\)	\(85\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.78 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x^2} \, dx=\frac {a^{\frac {1}{4}} x \log \left (\frac {a^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - a^{\frac {1}{4}} x \log \left (-\frac {a^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + i \, a^{\frac {1}{4}} x \log \left (\frac {i \, a^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - i \, a^{\frac {1}{4}} x \log \left (\frac {-i \, a^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 4 \, {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{x} \]

[In]

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

[Out]

(a^(1/4)*x*log((a^(1/4)*x + (a*x^4 + b*x^3)^(1/4))/x) - a^(1/4)*x*log(-(a^(1/4)*x - (a*x^4 + b*x^3)^(1/4))/x)
+ I*a^(1/4)*x*log((I*a^(1/4)*x + (a*x^4 + b*x^3)^(1/4))/x) - I*a^(1/4)*x*log((-I*a^(1/4)*x + (a*x^4 + b*x^3)^(
1/4))/x) - 4*(a*x^4 + b*x^3)^(1/4))/x

Sympy [F]

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

[In]

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

[Out]

Integral((x**3*(a*x + b))**(1/4)/x**2, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (67) = 134\).

Time = 0.28 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.26 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x^2} \, dx=\sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right ) - \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right ) - 4 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x^2} \, dx=-\frac {4\,{\left (a\,x^4+b\,x^3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ -\frac {a\,x}{b}\right )}{x\,{\left (\frac {a\,x}{b}+1\right )}^{1/4}} \]

[In]

int((a*x^4 + b*x^3)^(1/4)/x^2,x)

[Out]

-(4*(a*x^4 + b*x^3)^(1/4)*hypergeom([-1/4, -1/4], 3/4, -(a*x)/b))/(x*((a*x)/b + 1)^(1/4))