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

Optimal. Leaf size=111 \[ \frac {(b c-4 a d) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^3}}\right )}{2 a^{3/4}}+\frac {(4 a d-b c) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^3}}\right )}{2 a^{3/4}}+\frac {\sqrt [4]{a x^4-b x^3} (c x-4 d)}{x} \]

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Rubi [A]  time = 0.28, antiderivative size = 198, normalized size of antiderivative = 1.78, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2038, 2021, 2032, 63, 331, 298, 203, 206} \begin {gather*} \frac {x^{9/4} (a x-b)^{3/4} (b c-4 a d) \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x-b}}\right )}{2 a^{3/4} \left (a x^4-b x^3\right )^{3/4}}-\frac {x^{9/4} (a x-b)^{3/4} (b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x-b}}\right )}{2 a^{3/4} \left (a x^4-b x^3\right )^{3/4}}+\frac {\sqrt [4]{a x^4-b x^3} (b c-4 a d)}{b}+\frac {4 d \left (a x^4-b x^3\right )^{5/4}}{b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - 4*a*d)*(-(b*x^3) + a*x^4)^(1/4))/b + (4*d*(-(b*x^3) + a*x^4)^(5/4))/(b*x^4) + ((b*c - 4*a*d)*x^(9/4)*(
-b + a*x)^(3/4)*ArcTan[(a^(1/4)*x^(1/4))/(-b + a*x)^(1/4)])/(2*a^(3/4)*(-(b*x^3) + a*x^4)^(3/4)) - ((b*c - 4*a
*d)*x^(9/4)*(-b + a*x)^(3/4)*ArcTanh[(a^(1/4)*x^(1/4))/(-b + a*x)^(1/4)])/(2*a^(3/4)*(-(b*x^3) + a*x^4)^(3/4))

Rule 63

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 203

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

Rule 206

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

Rule 298

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 331

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 2021

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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]

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {(d+c x) \sqrt [4]{-b x^3+a x^4}}{x^2} \, dx &=\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{b x^4}+\frac {\left (4 \left (\frac {b c}{4}-a d\right )\right ) \int \frac {\sqrt [4]{-b x^3+a x^4}}{x} \, dx}{b}\\ &=\frac {(b c-4 a d) \sqrt [4]{-b x^3+a x^4}}{b}+\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{b x^4}+\frac {1}{4} (-b c+4 a d) \int \frac {x^2}{\left (-b x^3+a x^4\right )^{3/4}} \, dx\\ &=\frac {(b c-4 a d) \sqrt [4]{-b x^3+a x^4}}{b}+\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{b x^4}+\frac {\left ((-b c+4 a d) x^{9/4} (-b+a x)^{3/4}\right ) \int \frac {1}{\sqrt [4]{x} (-b+a x)^{3/4}} \, dx}{4 \left (-b x^3+a x^4\right )^{3/4}}\\ &=\frac {(b c-4 a d) \sqrt [4]{-b x^3+a x^4}}{b}+\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{b x^4}+\frac {\left ((-b c+4 a d) x^{9/4} (-b+a x)^{3/4}\right ) \operatorname {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 {(b c-4 a d) \sqrt [4]{-b x^3+a x^4}}{b}+\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{b x^4}+\frac {\left ((-b c+4 a d) x^{9/4} (-b+a x)^{3/4}\right ) \operatorname {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 {(b c-4 a d) \sqrt [4]{-b x^3+a x^4}}{b}+\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{b x^4}+\frac {\left ((-b c+4 a d) x^{9/4} (-b+a x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{2 \sqrt {a} \left (-b x^3+a x^4\right )^{3/4}}-\frac {\left ((-b c+4 a d) x^{9/4} (-b+a x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{2 \sqrt {a} \left (-b x^3+a x^4\right )^{3/4}}\\ &=\frac {(b c-4 a d) \sqrt [4]{-b x^3+a x^4}}{b}+\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{b x^4}+\frac {(b c-4 a d) x^{9/4} (-b+a x)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{2 a^{3/4} \left (-b x^3+a x^4\right )^{3/4}}-\frac {(b c-4 a d) x^{9/4} (-b+a x)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{2 a^{3/4} \left (-b x^3+a x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 87, normalized size = 0.78 \begin {gather*} \frac {4 \sqrt [4]{x^3 (a x-b)} \left (x (b c-4 a d) \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};\frac {a x}{b}\right )-3 d (b-a x) \sqrt [4]{1-\frac {a x}{b}}\right )}{3 b x \sqrt [4]{1-\frac {a x}{b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(x^3*(-b + a*x))^(1/4)*(-3*d*(b - a*x)*(1 - (a*x)/b)^(1/4) + (b*c - 4*a*d)*x*Hypergeometric2F1[-1/4, 3/4, 7
/4, (a*x)/b]))/(3*b*x*(1 - (a*x)/b)^(1/4))

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IntegrateAlgebraic [A]  time = 0.66, size = 111, normalized size = 1.00 \begin {gather*} \frac {(-4 d+c x) \sqrt [4]{-b x^3+a x^4}}{x}+\frac {(b c-4 a d) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^3+a x^4}}\right )}{2 a^{3/4}}+\frac {(-b c+4 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^3+a x^4}}\right )}{2 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.66, size = 693, normalized size = 6.24 \begin {gather*} -\frac {4 \, x \left (\frac {b^{4} c^{4} - 16 \, a b^{3} c^{3} d + 96 \, a^{2} b^{2} c^{2} d^{2} - 256 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {a^{2} x \sqrt {\frac {a^{2} x^{2} \sqrt {\frac {b^{4} c^{4} - 16 \, a b^{3} c^{3} d + 96 \, a^{2} b^{2} c^{2} d^{2} - 256 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{3}}} + \sqrt {a x^{4} - b x^{3}} {\left (b^{2} c^{2} - 8 \, a b c d + 16 \, a^{2} d^{2}\right )}}{x^{2}}} \left (\frac {b^{4} c^{4} - 16 \, a b^{3} c^{3} d + 96 \, a^{2} b^{2} c^{2} d^{2} - 256 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{3}}\right )^{\frac {3}{4}} + {\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} {\left (a^{2} b c - 4 \, a^{3} d\right )} \left (\frac {b^{4} c^{4} - 16 \, a b^{3} c^{3} d + 96 \, a^{2} b^{2} c^{2} d^{2} - 256 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{3}}\right )^{\frac {3}{4}}}{{\left (b^{4} c^{4} - 16 \, a b^{3} c^{3} d + 96 \, a^{2} b^{2} c^{2} d^{2} - 256 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}\right )} x}\right ) + x \left (\frac {b^{4} c^{4} - 16 \, a b^{3} c^{3} d + 96 \, a^{2} b^{2} c^{2} d^{2} - 256 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {a x \left (\frac {b^{4} c^{4} - 16 \, a b^{3} c^{3} d + 96 \, a^{2} b^{2} c^{2} d^{2} - 256 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{3}}\right )^{\frac {1}{4}} + {\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} {\left (b c - 4 \, a d\right )}}{x}\right ) - x \left (\frac {b^{4} c^{4} - 16 \, a b^{3} c^{3} d + 96 \, a^{2} b^{2} c^{2} d^{2} - 256 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (\frac {a x \left (\frac {b^{4} c^{4} - 16 \, a b^{3} c^{3} d + 96 \, a^{2} b^{2} c^{2} d^{2} - 256 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{3}}\right )^{\frac {1}{4}} - {\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} {\left (b c - 4 \, a d\right )}}{x}\right ) - 4 \, {\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} {\left (c x - 4 \, d\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*x*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2 - 256*a^3*b*c*d^3 + 256*a^4*d^4)/a^3)^(1/4)*arctan((
a^2*x*sqrt((a^2*x^2*sqrt((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2 - 256*a^3*b*c*d^3 + 256*a^4*d^4)/a^3)
+ sqrt(a*x^4 - b*x^3)*(b^2*c^2 - 8*a*b*c*d + 16*a^2*d^2))/x^2)*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2
- 256*a^3*b*c*d^3 + 256*a^4*d^4)/a^3)^(3/4) + (a*x^4 - b*x^3)^(1/4)*(a^2*b*c - 4*a^3*d)*((b^4*c^4 - 16*a*b^3*
c^3*d + 96*a^2*b^2*c^2*d^2 - 256*a^3*b*c*d^3 + 256*a^4*d^4)/a^3)^(3/4))/((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^
2*c^2*d^2 - 256*a^3*b*c*d^3 + 256*a^4*d^4)*x)) + x*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2 - 256*a^3*b
*c*d^3 + 256*a^4*d^4)/a^3)^(1/4)*log(-(a*x*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2 - 256*a^3*b*c*d^3 +
256*a^4*d^4)/a^3)^(1/4) + (a*x^4 - b*x^3)^(1/4)*(b*c - 4*a*d))/x) - x*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2
*c^2*d^2 - 256*a^3*b*c*d^3 + 256*a^4*d^4)/a^3)^(1/4)*log((a*x*((b^4*c^4 - 16*a*b^3*c^3*d + 96*a^2*b^2*c^2*d^2
- 256*a^3*b*c*d^3 + 256*a^4*d^4)/a^3)^(1/4) - (a*x^4 - b*x^3)^(1/4)*(b*c - 4*a*d))/x) - 4*(a*x^4 - b*x^3)^(1/4
)*(c*x - 4*d))/x

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giac [B]  time = 0.22, size = 257, normalized size = 2.32 \begin {gather*} \frac {8 \, {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}} b c x - 32 \, {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}} b d + \frac {2 \, \sqrt {2} {\left (b^{2} c - 4 \, a b d\right )} \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 )}{\left (-a\right )^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} {\left (b^{2} c - 4 \, a b d\right )} \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 )}{\left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} {\left (b^{2} c - 4 \, a b d\right )} \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}}} - \frac {\sqrt {2} {\left (b^{2} c - 4 \, a b d\right )} \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}}}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*(a - b/x)^(1/4)*b*c*x - 32*(a - b/x)^(1/4)*b*d + 2*sqrt(2)*(b^2*c - 4*a*b*d)*arctan(1/2*sqrt(2)*(sqrt(2
)*(-a)^(1/4) + 2*(a - b/x)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 2*sqrt(2)*(b^2*c - 4*a*b*d)*arctan(-1/2*sqrt(2)*(sq
rt(2)*(-a)^(1/4) - 2*(a - b/x)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + sqrt(2)*(b^2*c - 4*a*b*d)*log(sqrt(2)*(-a)^(1/4
)*(a - b/x)^(1/4) + sqrt(-a) + sqrt(a - b/x))/(-a)^(3/4) - sqrt(2)*(b^2*c - 4*a*b*d)*log(-sqrt(2)*(-a)^(1/4)*(
a - b/x)^(1/4) + sqrt(-a) + sqrt(a - b/x))/(-a)^(3/4))/b

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (c x +d \right ) \left (a \,x^{4}-b \,x^{3}\right )^{\frac {1}{4}}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 - b*x^3)^(1/4)*(c*x + d)/x^2, 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}\,\left (d+c\,x\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((a*x^4 - b*x^3)^(1/4)*(d + c*x))/x^2, 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 )} \left (c x + d\right )}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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