\(\int \frac {a+b x}{x (-d+c x) \sqrt [4]{-x^3+x^4}} \, dx\) [2574]

Optimal result

Integrand size = 31, antiderivative size = 221 \[ \int \frac {a+b x}{x (-d+c x) \sqrt [4]{-x^3+x^4}} \, dx=-\frac {4 a \left (-x^3+x^4\right )^{3/4}}{3 d x^3}+\frac {\sqrt {2} (a c+b d) \arctan \left (\frac {\sqrt {2} \sqrt [4]{c-d} \sqrt [4]{d} x \sqrt [4]{-x^3+x^4}}{\sqrt {c-d} x^2-\sqrt {d} \sqrt {-x^3+x^4}}\right )}{\sqrt [4]{c-d} d^{7/4}}-\frac {\sqrt {2} (a c+b d) \text {arctanh}\left (\frac {\sqrt {c-d} x^2+\sqrt {d} \sqrt {-x^3+x^4}}{\sqrt {2} \sqrt [4]{c-d} \sqrt [4]{d} x \sqrt [4]{-x^3+x^4}}\right )}{\sqrt [4]{c-d} d^{7/4}} \]

[Out]

-4/3*a*(x^4-x^3)^(3/4)/d/x^3+2^(1/2)*(a*c+b*d)*arctan(2^(1/2)*(c-d)^(1/4)*d^(1/4)*x*(x^4-x^3)^(1/4)/((c-d)^(1/
2)*x^2-d^(1/2)*(x^4-x^3)^(1/2)))/(c-d)^(1/4)/d^(7/4)-2^(1/2)*(a*c+b*d)*arctanh(1/2*((c-d)^(1/2)*x^2+d^(1/2)*(x
^4-x^3)^(1/2))*2^(1/2)/(c-d)^(1/4)/d^(1/4)/x/(x^4-x^3)^(1/4))/(c-d)^(1/4)/d^(7/4)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2081, 160, 12, 95, 218, 214, 211} \[ \int \frac {a+b x}{x (-d+c x) \sqrt [4]{-x^3+x^4}} \, dx=-\frac {2 \sqrt [4]{x-1} x^{3/4} (a c+b d) \arctan \left (\frac {\sqrt [4]{x} \sqrt [4]{d-c}}{\sqrt [4]{d} \sqrt [4]{x-1}}\right )}{d^{7/4} \sqrt [4]{x^4-x^3} \sqrt [4]{d-c}}-\frac {2 \sqrt [4]{x-1} x^{3/4} (a c+b d) \text {arctanh}\left (\frac {\sqrt [4]{x} \sqrt [4]{d-c}}{\sqrt [4]{d} \sqrt [4]{x-1}}\right )}{d^{7/4} \sqrt [4]{x^4-x^3} \sqrt [4]{d-c}}+\frac {4 a (1-x)}{3 d \sqrt [4]{x^4-x^3}} \]

[In]

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

[Out]

(4*a*(1 - x))/(3*d*(-x^3 + x^4)^(1/4)) - (2*(a*c + b*d)*(-1 + x)^(1/4)*x^(3/4)*ArcTan[((-c + d)^(1/4)*x^(1/4))
/(d^(1/4)*(-1 + x)^(1/4))])/(d^(7/4)*(-c + d)^(1/4)*(-x^3 + x^4)^(1/4)) - (2*(a*c + b*d)*(-1 + x)^(1/4)*x^(3/4
)*ArcTanh[((-c + d)^(1/4)*x^(1/4))/(d^(1/4)*(-1 + x)^(1/4))])/(d^(7/4)*(-c + d)^(1/4)*(-x^3 + x^4)^(1/4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 160

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1])))

Rule 211

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

Rule 214

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

Rule 218

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

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

Mathematica [A] (verified)

Time = 6.07 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.04 \[ \int \frac {a+b x}{x (-d+c x) \sqrt [4]{-x^3+x^4}} \, dx=\frac {-4 a \sqrt [4]{c-d} d^{3/4} (-1+x)+3 \sqrt {2} (a c+b d) \sqrt [4]{-1+x} x^{3/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{c-d} \sqrt [4]{d} \sqrt [4]{-1+x} \sqrt [4]{x}}{-\sqrt {d} \sqrt {-1+x}+\sqrt {c-d} \sqrt {x}}\right )-3 \sqrt {2} (a c+b d) \sqrt [4]{-1+x} x^{3/4} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {-1+x}+\sqrt {c-d} \sqrt {x}}{\sqrt {2} \sqrt [4]{c-d} \sqrt [4]{d} \sqrt [4]{-1+x} \sqrt [4]{x}}\right )}{3 \sqrt [4]{c-d} d^{7/4} \sqrt [4]{(-1+x) x^3}} \]

[In]

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

[Out]

(-4*a*(c - d)^(1/4)*d^(3/4)*(-1 + x) + 3*Sqrt[2]*(a*c + b*d)*(-1 + x)^(1/4)*x^(3/4)*ArcTan[(Sqrt[2]*(c - d)^(1
/4)*d^(1/4)*(-1 + x)^(1/4)*x^(1/4))/(-(Sqrt[d]*Sqrt[-1 + x]) + Sqrt[c - d]*Sqrt[x])] - 3*Sqrt[2]*(a*c + b*d)*(
-1 + x)^(1/4)*x^(3/4)*ArcTanh[(Sqrt[d]*Sqrt[-1 + x] + Sqrt[c - d]*Sqrt[x])/(Sqrt[2]*(c - d)^(1/4)*d^(1/4)*(-1
+ x)^(1/4)*x^(1/4))])/(3*(c - d)^(1/4)*d^(7/4)*((-1 + x)*x^3)^(1/4))

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.16

method	result	size
pseudoelliptic	\(-\frac {4 \left (a \left (x^{3} \left (-1+x \right )\right )^{\frac {3}{4}} d \left (\frac {c -d}{d}\right )^{\frac {1}{4}}-\frac {3 \left (a c +b d \right ) x^{3} \left (\ln \left (\frac {-\left (\frac {c -d}{d}\right )^{\frac {1}{4}} \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {c -d}{d}}\, x^{2}+\sqrt {x^{3} \left (-1+x \right )}}{\left (\frac {c -d}{d}\right )^{\frac {1}{4}} \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {c -d}{d}}\, x^{2}+\sqrt {x^{3} \left (-1+x \right )}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}-\left (\frac {c -d}{d}\right )^{\frac {1}{4}} x}{\left (\frac {c -d}{d}\right )^{\frac {1}{4}} x}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}+\left (\frac {c -d}{d}\right )^{\frac {1}{4}} x}{\left (\frac {c -d}{d}\right )^{\frac {1}{4}} x}\right )\right ) \sqrt {2}}{8}\right )}{3 \left (\frac {c -d}{d}\right )^{\frac {1}{4}} d^{2} x^{3}}\)	\(256\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 843, normalized size of antiderivative = 3.81 \[ \int \frac {a+b x}{x (-d+c x) \sqrt [4]{-x^3+x^4}} \, dx=\frac {3 \, d x^{3} \left (-\frac {a^{4} c^{4} + 4 \, a^{3} b c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a b^{3} c d^{3} + b^{4} d^{4}}{c d^{7} - d^{8}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (c d^{5} - d^{6}\right )} x \left (-\frac {a^{4} c^{4} + 4 \, a^{3} b c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a b^{3} c d^{3} + b^{4} d^{4}}{c d^{7} - d^{8}}\right )^{\frac {3}{4}} + {\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} d + 3 \, a b^{2} c d^{2} + b^{3} d^{3}\right )} {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 3 \, d x^{3} \left (-\frac {a^{4} c^{4} + 4 \, a^{3} b c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a b^{3} c d^{3} + b^{4} d^{4}}{c d^{7} - d^{8}}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (c d^{5} - d^{6}\right )} x \left (-\frac {a^{4} c^{4} + 4 \, a^{3} b c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a b^{3} c d^{3} + b^{4} d^{4}}{c d^{7} - d^{8}}\right )^{\frac {3}{4}} - {\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} d + 3 \, a b^{2} c d^{2} + b^{3} d^{3}\right )} {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 3 i \, d x^{3} \left (-\frac {a^{4} c^{4} + 4 \, a^{3} b c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a b^{3} c d^{3} + b^{4} d^{4}}{c d^{7} - d^{8}}\right )^{\frac {1}{4}} \log \left (\frac {i \, {\left (c d^{5} - d^{6}\right )} x \left (-\frac {a^{4} c^{4} + 4 \, a^{3} b c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a b^{3} c d^{3} + b^{4} d^{4}}{c d^{7} - d^{8}}\right )^{\frac {3}{4}} + {\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} d + 3 \, a b^{2} c d^{2} + b^{3} d^{3}\right )} {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 3 i \, d x^{3} \left (-\frac {a^{4} c^{4} + 4 \, a^{3} b c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a b^{3} c d^{3} + b^{4} d^{4}}{c d^{7} - d^{8}}\right )^{\frac {1}{4}} \log \left (\frac {-i \, {\left (c d^{5} - d^{6}\right )} x \left (-\frac {a^{4} c^{4} + 4 \, a^{3} b c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a b^{3} c d^{3} + b^{4} d^{4}}{c d^{7} - d^{8}}\right )^{\frac {3}{4}} + {\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} d + 3 \, a b^{2} c d^{2} + b^{3} d^{3}\right )} {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 4 \, {\left (x^{4} - x^{3}\right )}^{\frac {3}{4}} a}{3 \, d x^{3}} \]

[In]

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

[Out]

1/3*(3*d*x^3*(-(a^4*c^4 + 4*a^3*b*c^3*d + 6*a^2*b^2*c^2*d^2 + 4*a*b^3*c*d^3 + b^4*d^4)/(c*d^7 - d^8))^(1/4)*lo
g(((c*d^5 - d^6)*x*(-(a^4*c^4 + 4*a^3*b*c^3*d + 6*a^2*b^2*c^2*d^2 + 4*a*b^3*c*d^3 + b^4*d^4)/(c*d^7 - d^8))^(3
/4) + (a^3*c^3 + 3*a^2*b*c^2*d + 3*a*b^2*c*d^2 + b^3*d^3)*(x^4 - x^3)^(1/4))/x) - 3*d*x^3*(-(a^4*c^4 + 4*a^3*b
*c^3*d + 6*a^2*b^2*c^2*d^2 + 4*a*b^3*c*d^3 + b^4*d^4)/(c*d^7 - d^8))^(1/4)*log(-((c*d^5 - d^6)*x*(-(a^4*c^4 +
4*a^3*b*c^3*d + 6*a^2*b^2*c^2*d^2 + 4*a*b^3*c*d^3 + b^4*d^4)/(c*d^7 - d^8))^(3/4) - (a^3*c^3 + 3*a^2*b*c^2*d +
3*a*b^2*c*d^2 + b^3*d^3)*(x^4 - x^3)^(1/4))/x) - 3*I*d*x^3*(-(a^4*c^4 + 4*a^3*b*c^3*d + 6*a^2*b^2*c^2*d^2 + 4
*a*b^3*c*d^3 + b^4*d^4)/(c*d^7 - d^8))^(1/4)*log((I*(c*d^5 - d^6)*x*(-(a^4*c^4 + 4*a^3*b*c^3*d + 6*a^2*b^2*c^2
*d^2 + 4*a*b^3*c*d^3 + b^4*d^4)/(c*d^7 - d^8))^(3/4) + (a^3*c^3 + 3*a^2*b*c^2*d + 3*a*b^2*c*d^2 + b^3*d^3)*(x^
4 - x^3)^(1/4))/x) + 3*I*d*x^3*(-(a^4*c^4 + 4*a^3*b*c^3*d + 6*a^2*b^2*c^2*d^2 + 4*a*b^3*c*d^3 + b^4*d^4)/(c*d^
7 - d^8))^(1/4)*log((-I*(c*d^5 - d^6)*x*(-(a^4*c^4 + 4*a^3*b*c^3*d + 6*a^2*b^2*c^2*d^2 + 4*a*b^3*c*d^3 + b^4*d
^4)/(c*d^7 - d^8))^(3/4) + (a^3*c^3 + 3*a^2*b*c^2*d + 3*a*b^2*c*d^2 + b^3*d^3)*(x^4 - x^3)^(1/4))/x) - 4*(x^4
- x^3)^(3/4)*a)/(d*x^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.53 \[ \int \frac {a+b x}{x (-d+c x) \sqrt [4]{-x^3+x^4}} \, dx=-\frac {\sqrt {2} {\left (a c + b d\right )} \log \left (\sqrt {2} \left (\frac {c - d}{d}\right )^{\frac {1}{4}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + \sqrt {\frac {c - d}{d}} + \sqrt {-\frac {1}{x} + 1}\right )}{2 \, {\left (c d^{3} - d^{4}\right )}^{\frac {1}{4}} d} + \frac {{\left ({\left (c d^{3} - d^{4}\right )}^{\frac {3}{4}} a c + {\left (c d^{3} - d^{4}\right )}^{\frac {3}{4}} b d\right )} \log \left (-\sqrt {2} \left (\frac {c - d}{d}\right )^{\frac {1}{4}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + \sqrt {\frac {c - d}{d}} + \sqrt {-\frac {1}{x} + 1}\right )}{\sqrt {2} c d^{4} - \sqrt {2} d^{5}} - \frac {4 \, a {\left (-\frac {1}{x} + 1\right )}^{\frac {3}{4}}}{3 \, d} + \frac {{\left (\sqrt {2} a c + \sqrt {2} b d\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c - d}{d}\right )^{\frac {1}{4}} + 2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c - d}{d}\right )^{\frac {1}{4}}}\right )}{{\left (c d^{3} - d^{4}\right )}^{\frac {1}{4}} d} + \frac {{\left (\sqrt {2} a c + \sqrt {2} b d\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c - d}{d}\right )^{\frac {1}{4}} - 2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c - d}{d}\right )^{\frac {1}{4}}}\right )}{{\left (c d^{3} - d^{4}\right )}^{\frac {1}{4}} d} \]

[In]

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

[Out]

-1/2*sqrt(2)*(a*c + b*d)*log(sqrt(2)*((c - d)/d)^(1/4)*(-1/x + 1)^(1/4) + sqrt((c - d)/d) + sqrt(-1/x + 1))/((
c*d^3 - d^4)^(1/4)*d) + ((c*d^3 - d^4)^(3/4)*a*c + (c*d^3 - d^4)^(3/4)*b*d)*log(-sqrt(2)*((c - d)/d)^(1/4)*(-1
/x + 1)^(1/4) + sqrt((c - d)/d) + sqrt(-1/x + 1))/(sqrt(2)*c*d^4 - sqrt(2)*d^5) - 4/3*a*(-1/x + 1)^(3/4)/d + (
sqrt(2)*a*c + sqrt(2)*b*d)*arctan(1/2*sqrt(2)*(sqrt(2)*((c - d)/d)^(1/4) + 2*(-1/x + 1)^(1/4))/((c - d)/d)^(1/
4))/((c*d^3 - d^4)^(1/4)*d) + (sqrt(2)*a*c + sqrt(2)*b*d)*arctan(-1/2*sqrt(2)*(sqrt(2)*((c - d)/d)^(1/4) - 2*(
-1/x + 1)^(1/4))/((c - d)/d)^(1/4))/((c*d^3 - d^4)^(1/4)*d)

Mupad [F(-1)]

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

[In]

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

[Out]

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