3.13.0 ∫ 4 √ ________ −bx3 + ax4 x(d+cx+x2) dx [1271]

Integrand size = 30, antiderivative size = 92 \[ \int \frac {\sqrt [4]{-b x^3+a x^4}}{x \left (d+c x+x^2\right )} \, dx=b \text {RootSum}\left [b^2+a b c+a^2 d-b c \text {$\#$1}^4-2 a d \text {$\#$1}^4+d \text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [4]{-b x^3+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-b c-2 a d+2 d \text {$\#$1}^4}\&\right ] \]

Rubi [A] (veriﬁed)

Time = 0.35 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.00, number of steps used = 9, number of rules used = 5, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.167, Rules used = {2081, 925, 129, 525, 524} \[ \int \frac {\sqrt [4]{-b x^3+a x^4}}{x \left (d+c x+x^2\right )} \, dx=-\frac {8 \sqrt [4]{a x^4-b x^3} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {2 x}{c-\sqrt {c^2-4 d}},\frac {a x}{b}\right )}{3 \left (-c \sqrt {c^2-4 d}+c^2-4 d\right ) \sqrt [4]{1-\frac {a x}{b}}}-\frac {8 \sqrt [4]{a x^4-b x^3} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {2 x}{c+\sqrt {c^2-4 d}},\frac {a x}{b}\right )}{3 \left (c \sqrt {c^2-4 d}+c^2-4 d\right ) \sqrt [4]{1-\frac {a x}{b}}} \]

(-8*(-(b*x^3) + a*x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, (-2*x)/(c - Sqrt[c^2 - 4*d]), (a*x)/b])/(3*(c^2 - c*S

qrt[c^2 - 4*d] - 4*d)*(1 - (a*x)/b)^(1/4)) - (8*(-(b*x^3) + a*x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, (-2*x)/(c

+ Sqrt[c^2 - 4*d]), (a*x)/b])/(3*(c^2 + c*Sqrt[c^2 - 4*d] - 4*d)*(1 - (a*x)/b)^(1/4))

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]

}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*

((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free

Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar

t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x

] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && !(IntegerQ[

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> In

t[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x

] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[m] && !IntegerQ[n]

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart

[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] && !IntegerQ[p

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [4]{-b x^3+a x^4} \int \frac {\sqrt [4]{-b+a x}}{\sqrt [4]{x} \left (d+c x+x^2\right )} \, dx}{x^{3/4} \sqrt [4]{-b+a x}} \\ & = \frac {\sqrt [4]{-b x^3+a x^4} \int \left (\frac {2 \sqrt [4]{-b+a x}}{\sqrt {c^2-4 d} \sqrt [4]{x} \left (c-\sqrt {c^2-4 d}+2 x\right )}-\frac {2 \sqrt [4]{-b+a x}}{\sqrt {c^2-4 d} \sqrt [4]{x} \left (c+\sqrt {c^2-4 d}+2 x\right )}\right ) \, dx}{x^{3/4} \sqrt [4]{-b+a x}} \\ & = \frac {\left (2 \sqrt [4]{-b x^3+a x^4}\right ) \int \frac {\sqrt [4]{-b+a x}}{\sqrt [4]{x} \left (c-\sqrt {c^2-4 d}+2 x\right )} \, dx}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}-\frac {\left (2 \sqrt [4]{-b x^3+a x^4}\right ) \int \frac {\sqrt [4]{-b+a x}}{\sqrt [4]{x} \left (c+\sqrt {c^2-4 d}+2 x\right )} \, dx}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}} \\ & = \frac {\left (8 \sqrt [4]{-b x^3+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{-b+a x^4}}{c-\sqrt {c^2-4 d}+2 x^4} \, dx,x,\sqrt [4]{x}\right )}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}-\frac {\left (8 \sqrt [4]{-b x^3+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{-b+a x^4}}{c+\sqrt {c^2-4 d}+2 x^4} \, dx,x,\sqrt [4]{x}\right )}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}} \\ & = \frac {\left (8 \sqrt [4]{-b x^3+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{1-\frac {a x^4}{b}}}{c-\sqrt {c^2-4 d}+2 x^4} \, dx,x,\sqrt [4]{x}\right )}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{1-\frac {a x}{b}}}-\frac {\left (8 \sqrt [4]{-b x^3+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{1-\frac {a x^4}{b}}}{c+\sqrt {c^2-4 d}+2 x^4} \, dx,x,\sqrt [4]{x}\right )}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{1-\frac {a x}{b}}} \\ & = -\frac {8 \sqrt [4]{-b x^3+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {2 x}{c-\sqrt {c^2-4 d}},\frac {a x}{b}\right )}{3 \left (c^2-c \sqrt {c^2-4 d}-4 d\right ) \sqrt [4]{1-\frac {a x}{b}}}-\frac {8 \sqrt [4]{-b x^3+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {2 x}{c+\sqrt {c^2-4 d}},\frac {a x}{b}\right )}{3 \left (c^2+c \sqrt {c^2-4 d}-4 d\right ) \sqrt [4]{1-\frac {a x}{b}}} \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 0.49 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.37 \[ \int \frac {\sqrt [4]{-b x^3+a x^4}}{x \left (d+c x+x^2\right )} \, dx=\frac {b \sqrt [4]{x^3 (-b+a x)} \text {RootSum}\left [b^2+a b c+a^2 d-b c \text {$\#$1}^4-2 a d \text {$\#$1}^4+d \text {$\#$1}^8\&,\frac {-\log \left (\sqrt [4]{x}\right ) \text {$\#$1}+\log \left (\sqrt [4]{-b+a x}-\sqrt [4]{x} \text {$\#$1}\right ) \text {$\#$1}}{-b c-2 a d+2 d \text {$\#$1}^4}\&\right ]}{x^{3/4} \sqrt [4]{-b+a x}} \]

(b*(x^3*(-b + a*x))^(1/4)*RootSum[b^2 + a*b*c + a^2*d - b*c*#1^4 - 2*a*d*#1^4 + d*#1^8 & , (-(Log[x^(1/4)]*#1)

+ Log[(-b + a*x)^(1/4) - x^(1/4)*#1]*#1)/(-(b*c) - 2*a*d + 2*d*#1^4) & ])/(x^(3/4)*(-b + a*x)^(1/4))

Maple [N/A] (veriﬁed)

Time = 2.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.87


method	result	size

pseudoelliptic	$b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{8}+\left (-2 a d -b c \right ) \textit {\_Z}^{4}+a^{2} d +a b c +b^{2}\right )}{\sum }\left (-\frac {\textit {\_R} \ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (a x -b \right )\right )^{\frac {1}{4}}}{x}\right )}{-2 \textit {\_R}^{4} d +2 a d +b c}\right )\right )$	$80$

b*sum(-_R*ln((-_R*x+(x^3*(a*x-b))^(1/4))/x)/(-2*_R^4*d+2*a*d+b*c),_R=RootOf(d*_Z^8+(-2*a*d-b*c)*_Z^4+a^2*d+a*b

Fricas [C] (veriﬁcation not implemented)

Time = 0.34 (sec) , antiderivative size = 2253, normalized size of antiderivative = 24.49 \[ \int \frac {\sqrt [4]{-b x^3+a x^4}}{x \left (d+c x+x^2\right )} \, dx=\text {Too large to display} \]

-1/2*sqrt(2)*sqrt(sqrt(2)*sqrt((b*c + 2*a*d + (c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48

*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3)))*log((sqrt(2)*(c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2/(c^6*d^

2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5))*x*sqrt(sqrt(2)*sqrt((b*c + 2*a*d + (c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^

2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3))) + 2*(a*x^4 - b*x^3)^(1/4)*b)/x

) + 1/2*sqrt(2)*sqrt(sqrt(2)*sqrt((b*c + 2*a*d + (c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 +

48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3)))*log(-(sqrt(2)*(c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2/(c^

6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5))*x*sqrt(sqrt(2)*sqrt((b*c + 2*a*d + (c^4*d - 8*c^2*d^2 + 16*d^3)*sqr

t(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3))) - 2*(a*x^4 - b*x^3)^(1/4)*

b)/x) - 1/2*sqrt(2)*sqrt(-sqrt(2)*sqrt((b*c + 2*a*d + (c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*

d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3)))*log((sqrt(2)*(c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2

/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5))*x*sqrt(-sqrt(2)*sqrt((b*c + 2*a*d + (c^4*d - 8*c^2*d^2 + 16*d^3

)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3))) + 2*(a*x^4 - b*x^3)^(

1/4)*b)/x) + 1/2*sqrt(2)*sqrt(-sqrt(2)*sqrt((b*c + 2*a*d + (c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2/(c^6*d^2 - 12

*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3)))*log(-(sqrt(2)*(c^4*d - 8*c^2*d^2 + 16*d^3)*sq

rt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5))*x*sqrt(-sqrt(2)*sqrt((b*c + 2*a*d + (c^4*d - 8*c^2*d^2 +

16*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3))) - 2*(a*x^4 - b*

x^3)^(1/4)*b)/x) + 1/2*sqrt(2)*sqrt(sqrt(2)*sqrt((b*c + 2*a*d - (c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2/(c^6*d^2

- 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3)))*log((sqrt(2)*(c^4*d - 8*c^2*d^2 + 16*d^3

)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5))*x*sqrt(sqrt(2)*sqrt((b*c + 2*a*d - (c^4*d - 8*c^2*d^2

+ 16*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3))) + 2*(a*x^4 -

b*x^3)^(1/4)*b)/x) - 1/2*sqrt(2)*sqrt(sqrt(2)*sqrt((b*c + 2*a*d - (c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2/(c^6*

d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3)))*log(-(sqrt(2)*(c^4*d - 8*c^2*d^2 + 16

*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5))*x*sqrt(sqrt(2)*sqrt((b*c + 2*a*d - (c^4*d - 8*c^2

*d^2 + 16*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3))) - 2*(a*x

^4 - b*x^3)^(1/4)*b)/x) + 1/2*sqrt(2)*sqrt(-sqrt(2)*sqrt((b*c + 2*a*d - (c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2/

(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3)))*log((sqrt(2)*(c^4*d - 8*c^2*d^2

+ 16*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5))*x*sqrt(-sqrt(2)*sqrt((b*c + 2*a*d - (c^4*d -

8*c^2*d^2 + 16*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3))) + 2

*(a*x^4 - b*x^3)^(1/4)*b)/x) - 1/2*sqrt(2)*sqrt(-sqrt(2)*sqrt((b*c + 2*a*d - (c^4*d - 8*c^2*d^2 + 16*d^3)*sqrt

(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3)))*log(-(sqrt(2)*(c^4*d - 8*c^

2*d^2 + 16*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5))*x*sqrt(-sqrt(2)*sqrt((b*c + 2*a*d - (c^

4*d - 8*c^2*d^2 + 16*d^3)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))/(c^4*d - 8*c^2*d^2 + 16*d^3)

Sympy [N/A]

Time = 1.72 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.24 \[ \int \frac {\sqrt [4]{-b x^3+a x^4}}{x \left (d+c x+x^2\right )} \, dx=\int \frac {\sqrt [4]{x^{3} \left (a x - b\right )}}{x \left (c x + d + x^{2}\right )}\, dx \]

Maxima [N/A]

Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt [4]{-b x^3+a x^4}}{x \left (d+c x+x^2\right )} \, dx=\int { \frac {{\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}}}{{\left (c x + x^{2} + d\right )} x} \,d x } \]

Giac [N/A]

Time = 12.55 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt [4]{-b x^3+a x^4}}{x \left (d+c x+x^2\right )} \, dx=\int { \frac {{\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}}}{{\left (c x + x^{2} + d\right )} x} \,d x } \]

Mupad [N/A]

Time = 5.76 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt [4]{-b x^3+a x^4}}{x \left (d+c x+x^2\right )} \, dx=\int \frac {{\left (a\,x^4-b\,x^3\right )}^{1/4}}{x\,\left (x^2+c\,x+d\right )} \,d x \]

\(\int \frac {\sqrt [4]{-b x^3+a x^4}}{x (d+c x+x^2)} \, dx\) [1271]

Optimal result