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

Optimal. Leaf size=221 \[ \frac {\sqrt {2} (a c+b d) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} x \sqrt [4]{x^4-x^3} \sqrt [4]{c-d}}{x^2 \sqrt {c-d}-\sqrt {d} \sqrt {x^4-x^3}}\right )}{d^{7/4} \sqrt [4]{c-d}}-\frac {\sqrt {2} (a c+b d) \tanh ^{-1}\left (\frac {x^2 \sqrt {c-d}+\sqrt {d} \sqrt {x^4-x^3}}{\sqrt {2} \sqrt [4]{d} x \sqrt [4]{x^4-x^3} \sqrt [4]{c-d}}\right )}{d^{7/4} \sqrt [4]{c-d}}-\frac {4 a \left (x^4-x^3\right )^{3/4}}{3 d x^3} \]

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Rubi [A]  time = 0.34, antiderivative size = 179, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2056, 155, 12, 93, 212, 208, 205} \begin {gather*} -\frac {2 \sqrt [4]{x-1} x^{3/4} (a c+b d) \tan ^{-1}\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) \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[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 93

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 155

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 205

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

Rule 208

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 2056

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 {a+b x}{x (-d+c x) \sqrt [4]{-x^3+x^4}} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \tan ^{-1}\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} \tanh ^{-1}\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*}

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Mathematica [C]  time = 0.05, size = 69, normalized size = 0.31 \begin {gather*} -\frac {4 \left ((x-1) x^3\right )^{3/4} \left ((a c+b d) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {d-d x}{c x-d x}\right )+a (d-c)\right )}{3 d x^3 (d-c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 3.14, size = 221, normalized size = 1.00 \begin {gather*} -\frac {4 a \left (-x^3+x^4\right )^{3/4}}{3 d x^3}+\frac {\sqrt {2} (a c+b d) \tan ^{-1}\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) \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.69, size = 1021, normalized size = 4.62 \begin {gather*} -\frac {12 \, 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}} \arctan \left (\frac {d^{2} 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 {1}{4}} \sqrt {-\frac {{\left (a^{4} c^{5} d^{3} - b^{4} d^{8} - {\left (a^{4} - 4 \, a^{3} b\right )} c^{4} d^{4} - 2 \, {\left (2 \, a^{3} b - 3 \, a^{2} b^{2}\right )} c^{3} d^{5} - 2 \, {\left (3 \, a^{2} b^{2} - 2 \, a b^{3}\right )} c^{2} d^{6} - {\left (4 \, a b^{3} - b^{4}\right )} c d^{7}\right )} x^{2} \sqrt {-\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}}} - {\left (a^{6} c^{6} + 6 \, a^{5} b c^{5} d + 15 \, a^{4} b^{2} c^{4} d^{2} + 20 \, a^{3} b^{3} c^{3} d^{3} + 15 \, a^{2} b^{4} c^{2} d^{4} + 6 \, a b^{5} c d^{5} + b^{6} d^{6}\right )} \sqrt {x^{4} - x^{3}}}{x^{2}}} - {\left (a^{3} c^{3} d^{2} + 3 \, a^{2} b c^{2} d^{3} + 3 \, a b^{2} c d^{4} + b^{3} d^{5}\right )} {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} \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}}}{{\left (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}\right )} 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 \, 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 ) + 4 \, {\left (x^{4} - x^{3}\right )}^{\frac {3}{4}} a}{3 \, d x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(12*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)*
arctan((d^2*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))^(1/4)*s
qrt(-((a^4*c^5*d^3 - b^4*d^8 - (a^4 - 4*a^3*b)*c^4*d^4 - 2*(2*a^3*b - 3*a^2*b^2)*c^3*d^5 - 2*(3*a^2*b^2 - 2*a*
b^3)*c^2*d^6 - (4*a*b^3 - b^4)*c*d^7)*x^2*sqrt(-(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)) - (a^6*c^6 + 6*a^5*b*c^5*d + 15*a^4*b^2*c^4*d^2 + 20*a^3*b^3*c^3*d^3 + 15*a^2*b^4*c^2
*d^4 + 6*a*b^5*c*d^5 + b^6*d^6)*sqrt(x^4 - x^3))/x^2) - (a^3*c^3*d^2 + 3*a^2*b*c^2*d^3 + 3*a*b^2*c*d^4 + b^3*d
^5)*(x^4 - x^3)^(1/4)*(-(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))/((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)*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*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) + 4*(x^4 - x^3)^(3/4)*a)/(d*x^3)

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giac [B]  time = 0.21, size = 386, normalized size = 1.75 \begin {gather*} -\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 (\sqrt {2} {\left (c d^{3} - d^{4}\right )}^{\frac {3}{4}} a c + \sqrt {2} {\left (c d^{3} - d^{4}\right )}^{\frac {3}{4}} 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 )}{c d^{4} - d^{5}} - \frac {{\left (\sqrt {2} {\left (c d^{3} - d^{4}\right )}^{\frac {3}{4}} a c + \sqrt {2} {\left (c d^{3} - d^{4}\right )}^{\frac {3}{4}} 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 )}{c d^{4} - d^{5}} + \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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) - (sqrt(2)*(c*d^3 - d^4)^(3/4)*a*c + sqrt(2)*(c*d^3 - d^4)^(3/4)*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^4 - d^5) - (sqrt(2)*(c*d^3 - d^4)^(
3/4)*a*c + sqrt(2)*(c*d^3 - d^4)^(3/4)*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^4 - d^5) + ((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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {b x +a}{x \left (c x -d \right ) \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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