3.13.72 \(\int \frac {\sqrt [4]{-b x^3+a x^4}}{x (d+c x+x^2)} \, dx\) [1272]

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

[Out]

Unintegrable

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Rubi [A]
time = 0.46, antiderivative size = 184, normalized size of antiderivative = 2.00, number of steps used = 9, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2081, 925, 129, 525, 524} \begin {gather*} -\frac {8 \sqrt [4]{a x^4-b x^3} F_1\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} F_1\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}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-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))

Rule 129

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
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 524

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
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

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[
p] || GtQ[a, 0])

Rule 925

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]

Rule 2081

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
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{-b x^3+a x^4}}{x \left (d+c x+x^2\right )} \, dx &=\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 (\left (2 b+a \left (c-\sqrt {c^2-4 d}\right )\right ) \sqrt [4]{-b x^3+a x^4}\right ) \int \frac {1}{\sqrt [4]{x} \left (c-\sqrt {c^2-4 d}+2 x\right ) (-b+a x)^{3/4}} \, dx}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}+\frac {\left (\left (2 b+a \left (c+\sqrt {c^2-4 d}\right )\right ) \sqrt [4]{-b x^3+a x^4}\right ) \int \frac {1}{\sqrt [4]{x} \left (c+\sqrt {c^2-4 d}+2 x\right ) (-b+a x)^{3/4}} \, dx}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}\\ &=-\frac {\left (4 \left (2 b+a \left (c-\sqrt {c^2-4 d}\right )\right ) \sqrt [4]{-b x^3+a x^4}\right ) \text {Subst}\left (\int \frac {x^2}{c-\sqrt {c^2-4 d}-\left (2 b+a \left (c-\sqrt {c^2-4 d}\right )\right ) x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}+\frac {\left (4 \left (2 b+a \left (c+\sqrt {c^2-4 d}\right )\right ) \sqrt [4]{-b x^3+a x^4}\right ) \text {Subst}\left (\int \frac {x^2}{c+\sqrt {c^2-4 d}-\left (2 b+a \left (c+\sqrt {c^2-4 d}\right )\right ) x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}\\ &=-\frac {\left (2 \sqrt {2 b+a \left (c-\sqrt {c^2-4 d}\right )} \sqrt [4]{-b x^3+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c-\sqrt {c^2-4 d}}-\sqrt {2 b+a c-a \sqrt {c^2-4 d}} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}+\frac {\left (2 \sqrt {2 b+a \left (c-\sqrt {c^2-4 d}\right )} \sqrt [4]{-b x^3+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c-\sqrt {c^2-4 d}}+\sqrt {2 b+a c-a \sqrt {c^2-4 d}} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}+\frac {\left (2 \sqrt {2 b+a \left (c+\sqrt {c^2-4 d}\right )} \sqrt [4]{-b x^3+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\sqrt {c^2-4 d}}-\sqrt {2 b+a c+a \sqrt {c^2-4 d}} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}-\frac {\left (2 \sqrt {2 b+a \left (c+\sqrt {c^2-4 d}\right )} \sqrt [4]{-b x^3+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\sqrt {c^2-4 d}}+\sqrt {2 b+a c+a \sqrt {c^2-4 d}} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{\sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}\\ &=\frac {2 \sqrt [4]{2 b+a \left (c-\sqrt {c^2-4 d}\right )} \sqrt [4]{-b x^3+a x^4} \tan ^{-1}\left (\frac {\sqrt [4]{2 b+a \left (c-\sqrt {c^2-4 d}\right )} \sqrt [4]{x}}{\sqrt [4]{c-\sqrt {c^2-4 d}} \sqrt [4]{-b+a x}}\right )}{\sqrt [4]{c-\sqrt {c^2-4 d}} \sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}-\frac {2 \sqrt [4]{2 b+a \left (c+\sqrt {c^2-4 d}\right )} \sqrt [4]{-b x^3+a x^4} \tan ^{-1}\left (\frac {\sqrt [4]{2 b+a \left (c+\sqrt {c^2-4 d}\right )} \sqrt [4]{x}}{\sqrt [4]{c+\sqrt {c^2-4 d}} \sqrt [4]{-b+a x}}\right )}{\sqrt [4]{c+\sqrt {c^2-4 d}} \sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}-\frac {2 \sqrt [4]{2 b+a \left (c-\sqrt {c^2-4 d}\right )} \sqrt [4]{-b x^3+a x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{2 b+a \left (c-\sqrt {c^2-4 d}\right )} \sqrt [4]{x}}{\sqrt [4]{c-\sqrt {c^2-4 d}} \sqrt [4]{-b+a x}}\right )}{\sqrt [4]{c-\sqrt {c^2-4 d}} \sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}+\frac {2 \sqrt [4]{2 b+a \left (c+\sqrt {c^2-4 d}\right )} \sqrt [4]{-b x^3+a x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{2 b+a \left (c+\sqrt {c^2-4 d}\right )} \sqrt [4]{x}}{\sqrt [4]{c+\sqrt {c^2-4 d}} \sqrt [4]{-b+a x}}\right )}{\sqrt [4]{c+\sqrt {c^2-4 d}} \sqrt {c^2-4 d} x^{3/4} \sqrt [4]{-b+a x}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 126, normalized size = 1.37 \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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))

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}-b \,x^{3}\right )^{\frac {1}{4}}}{x \left (c x +x^{2}+d \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 3 vs. order 1.
time = 0.56, size = 2783, normalized size = 30.25 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)))*arctan(-1/8*sqrt(2)*(sqrt(2)*(b^3*c^4 - 8*b^3*c^2*d + 16*b^3
*d^2 - (b^2*c^7*d - 128*a*b*d^5 + 32*(3*a*b*c^2 - 2*b^2*c)*d^4 - 24*(a*b*c^4 - 2*b^2*c^3)*d^3 + 2*(a*b*c^6 - 6
*b^2*c^5)*d^2)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))*(a*x^4 - b*x^3)^(1/4)*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 + 1
6*d^3)) + ((b*c^7*d - 128*a*d^5 + 32*(3*a*c^2 - 2*b*c)*d^4 - 24*(a*c^4 - 2*b*c^3)*d^3 + 2*(a*c^6 - 6*b*c^5)*d^
2)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5))*x - (b^2*c^4 - 8*b^2*c^2*d + 16*b^2*d^2)*x)*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))*sqrt((sqrt(2)*(b^2*c^2 - 4*b^2*d)*x^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*sqrt(a*x^4 - b*x^3)*b^2)/x
^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)))/((a*b^3*c + a^2*b^2*d + b^4)*x)) + 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)))*arctan(1/8*sqrt(2)*(sqrt(2)*(b^3*c^4 - 8*b^3*c^2*d + 16*b^3*d^2 + (b^2*c^7*d - 128*a*b*d^5 +
32*(3*a*b*c^2 - 2*b^2*c)*d^4 - 24*(a*b*c^4 - 2*b^2*c^3)*d^3 + 2*(a*b*c^6 - 6*b^2*c^5)*d^2)*sqrt(b^2/(c^6*d^2
- 12*c^4*d^3 + 48*c^2*d^4 - 64*d^5)))*(a*x^4 - b*x^3)^(1/4)*sqrt((b*c + 2*a*d - (c^4*d - 8*c^2*d^2 + 16*d^3)*s
qrt(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)) - ((b*c^7*d - 128*a*d^5 +
32*(3*a*c^2 - 2*b*c)*d^4 - 24*(a*c^4 - 2*b*c^3)*d^3 + 2*(a*c^6 - 6*b*c^5)*d^2)*sqrt(b^2/(c^6*d^2 - 12*c^4*d^3
+ 48*c^2*d^4 - 64*d^5))*x + (b^2*c^4 - 8*b^2*c^2*d + 16*b^2*d^2)*x)*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))*sqrt((sqrt(2)*(b
^2*c^2 - 4*b^2*d)*x^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*sqrt(a*x^4 - b*x^3)*b^2)/x^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)))/((a*b^3*c + a^2*b^2*d + b^4)*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((s
qrt(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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{x^{3} \left (a x - b\right )}}{x \left (c x + d + x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a\,x^4-b\,x^3\right )}^{1/4}}{x\,\left (x^2+c\,x+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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