\(\int \frac {\sqrt [4]{-b x^3+a x^4} (-d+c x^4)}{x^4} \, dx\) [2061]

Optimal result

Integrand size = 29, antiderivative size = 148 \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\frac {\sqrt [4]{-b x^3+a x^4} \left (160 a b^2 d-32 a^2 b d x-128 a^3 d x^2-45 b^3 c x^3+180 a b^2 c x^4\right )}{360 a b^2 x^3}+\frac {3 b^2 c \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^3+a x^4}}\right )}{16 a^{7/4}}-\frac {3 b^2 c \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^3+a x^4}}\right )}{16 a^{7/4}} \]

[Out]

1/360*(a*x^4-b*x^3)^(1/4)*(180*a*b^2*c*x^4-45*b^3*c*x^3-128*a^3*d*x^2-32*a^2*b*d*x+160*a*b^2*d)/a/b^2/x^3+3/16
*b^2*c*arctan(a^(1/4)*x/(a*x^4-b*x^3)^(1/4))/a^(7/4)-3/16*b^2*c*arctanh(a^(1/4)*x/(a*x^4-b*x^3)^(1/4))/a^(7/4)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.61, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {2077, 2029, 2049, 2057, 65, 338, 304, 209, 212, 2041, 2039} \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\frac {3 b^2 c x^{9/4} (a x-b)^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x-b}}\right )}{16 a^{7/4} \left (a x^4-b x^3\right )^{3/4}}-\frac {3 b^2 c x^{9/4} (a x-b)^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x-b}}\right )}{16 a^{7/4} \left (a x^4-b x^3\right )^{3/4}}-\frac {16 a d \left (a x^4-b x^3\right )^{5/4}}{45 b^2 x^5}+\frac {1}{2} c x \sqrt [4]{a x^4-b x^3}-\frac {b c \sqrt [4]{a x^4-b x^3}}{8 a}-\frac {4 d \left (a x^4-b x^3\right )^{5/4}}{9 b x^6} \]

[In]

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

[Out]

-1/8*(b*c*(-(b*x^3) + a*x^4)^(1/4))/a + (c*x*(-(b*x^3) + a*x^4)^(1/4))/2 - (4*d*(-(b*x^3) + a*x^4)^(5/4))/(9*b
*x^6) - (16*a*d*(-(b*x^3) + a*x^4)^(5/4))/(45*b^2*x^5) + (3*b^2*c*x^(9/4)*(-b + a*x)^(3/4)*ArcTan[(a^(1/4)*x^(
1/4))/(-b + a*x)^(1/4)])/(16*a^(7/4)*(-(b*x^3) + a*x^4)^(3/4)) - (3*b^2*c*x^(9/4)*(-b + a*x)^(3/4)*ArcTanh[(a^
(1/4)*x^(1/4))/(-b + a*x)^(1/4)])/(16*a^(7/4)*(-(b*x^3) + a*x^4)^(3/4))

Rule 65

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 209

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

Rule 212

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

Rule 304

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 338

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 2029

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(n*p + 1)), x] + Dist[a
*(n - j)*(p/(n*p + 1)), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]

Rule 2039

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2041

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[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 2077

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \int \left (c \sqrt [4]{-b x^3+a x^4}-\frac {d \sqrt [4]{-b x^3+a x^4}}{x^4}\right ) \, dx \\ & = c \int \sqrt [4]{-b x^3+a x^4} \, dx-d \int \frac {\sqrt [4]{-b x^3+a x^4}}{x^4} \, dx \\ & = \frac {1}{2} c x \sqrt [4]{-b x^3+a x^4}-\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{9 b x^6}-\frac {1}{8} (b c) \int \frac {x^3}{\left (-b x^3+a x^4\right )^{3/4}} \, dx-\frac {(4 a d) \int \frac {\sqrt [4]{-b x^3+a x^4}}{x^3} \, dx}{9 b} \\ & = -\frac {b c \sqrt [4]{-b x^3+a x^4}}{8 a}+\frac {1}{2} c x \sqrt [4]{-b x^3+a x^4}-\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{9 b x^6}-\frac {16 a d \left (-b x^3+a x^4\right )^{5/4}}{45 b^2 x^5}-\frac {\left (3 b^2 c\right ) \int \frac {x^2}{\left (-b x^3+a x^4\right )^{3/4}} \, dx}{32 a} \\ & = -\frac {b c \sqrt [4]{-b x^3+a x^4}}{8 a}+\frac {1}{2} c x \sqrt [4]{-b x^3+a x^4}-\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{9 b x^6}-\frac {16 a d \left (-b x^3+a x^4\right )^{5/4}}{45 b^2 x^5}-\frac {\left (3 b^2 c x^{9/4} (-b+a x)^{3/4}\right ) \int \frac {1}{\sqrt [4]{x} (-b+a x)^{3/4}} \, dx}{32 a \left (-b x^3+a x^4\right )^{3/4}} \\ & = -\frac {b c \sqrt [4]{-b x^3+a x^4}}{8 a}+\frac {1}{2} c x \sqrt [4]{-b x^3+a x^4}-\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{9 b x^6}-\frac {16 a d \left (-b x^3+a x^4\right )^{5/4}}{45 b^2 x^5}-\frac {\left (3 b^2 c x^{9/4} (-b+a x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (-b+a x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{8 a \left (-b x^3+a x^4\right )^{3/4}} \\ & = -\frac {b c \sqrt [4]{-b x^3+a x^4}}{8 a}+\frac {1}{2} c x \sqrt [4]{-b x^3+a x^4}-\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{9 b x^6}-\frac {16 a d \left (-b x^3+a x^4\right )^{5/4}}{45 b^2 x^5}-\frac {\left (3 b^2 c x^{9/4} (-b+a x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{8 a \left (-b x^3+a x^4\right )^{3/4}} \\ & = -\frac {b c \sqrt [4]{-b x^3+a x^4}}{8 a}+\frac {1}{2} c x \sqrt [4]{-b x^3+a x^4}-\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{9 b x^6}-\frac {16 a d \left (-b x^3+a x^4\right )^{5/4}}{45 b^2 x^5}-\frac {\left (3 b^2 c x^{9/4} (-b+a x)^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{16 a^{3/2} \left (-b x^3+a x^4\right )^{3/4}}+\frac {\left (3 b^2 c x^{9/4} (-b+a x)^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{16 a^{3/2} \left (-b x^3+a x^4\right )^{3/4}} \\ & = -\frac {b c \sqrt [4]{-b x^3+a x^4}}{8 a}+\frac {1}{2} c x \sqrt [4]{-b x^3+a x^4}-\frac {4 d \left (-b x^3+a x^4\right )^{5/4}}{9 b x^6}-\frac {16 a d \left (-b x^3+a x^4\right )^{5/4}}{45 b^2 x^5}+\frac {3 b^2 c x^{9/4} (-b+a x)^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{16 a^{7/4} \left (-b x^3+a x^4\right )^{3/4}}-\frac {3 b^2 c x^{9/4} (-b+a x)^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )}{16 a^{7/4} \left (-b x^3+a x^4\right )^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\frac {(-b+a x)^{3/4} \left (2 a^{3/4} \sqrt [4]{-b+a x} \left (-32 a^2 b d x-128 a^3 d x^2-45 b^3 c x^3+20 a b^2 \left (8 d+9 c x^4\right )\right )+135 b^4 c x^{9/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )-135 b^4 c x^{9/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )\right )}{720 a^{7/4} b^2 \left (x^3 (-b+a x)\right )^{3/4}} \]

[In]

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

[Out]

((-b + a*x)^(3/4)*(2*a^(3/4)*(-b + a*x)^(1/4)*(-32*a^2*b*d*x - 128*a^3*d*x^2 - 45*b^3*c*x^3 + 20*a*b^2*(8*d +
9*c*x^4)) + 135*b^4*c*x^(9/4)*ArcTan[(a^(1/4)*x^(1/4))/(-b + a*x)^(1/4)] - 135*b^4*c*x^(9/4)*ArcTanh[(a^(1/4)*
x^(1/4))/(-b + a*x)^(1/4)]))/(720*a^(7/4)*b^2*(x^3*(-b + a*x))^(3/4))

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.07

method	result	size
pseudoelliptic	\(\frac {-\frac {3 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (x^{3} \left (a x -b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (x^{3} \left (a x -b \right )\right )^{\frac {1}{4}}}\right ) b^{4} c \,x^{3}}{16}-\frac {3 \arctan \left (\frac {\left (x^{3} \left (a x -b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b^{4} c \,x^{3}}{8}+\left (b^{2} \left (c \,x^{4}+\frac {8 d}{9}\right ) a^{\frac {7}{4}}-\frac {8 x \left (\frac {45 b^{3} c \,x^{2} a^{\frac {3}{4}}}{32}+a^{\frac {11}{4}} b d +4 a^{\frac {15}{4}} d x \right )}{45}\right ) \left (x^{3} \left (a x -b \right )\right )^{\frac {1}{4}}}{2 a^{\frac {7}{4}} x^{3} b^{2}}\)	\(159\)

[In]

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

[Out]

1/2*(-3/16*ln((-a^(1/4)*x-(x^3*(a*x-b))^(1/4))/(a^(1/4)*x-(x^3*(a*x-b))^(1/4)))*b^4*c*x^3-3/8*arctan(1/a^(1/4)
/x*(x^3*(a*x-b))^(1/4))*b^4*c*x^3+(b^2*(c*x^4+8/9*d)*a^(7/4)-8/45*x*(45/32*b^3*c*x^2*a^(3/4)+a^(11/4)*b*d+4*a^
(15/4)*d*x))*(x^3*(a*x-b))^(1/4))/a^(7/4)/x^3/b^2

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.24 \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=-\frac {135 \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{2} x^{3} \log \left (\frac {3 \, {\left ({\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} b^{2} c + \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a^{2} x\right )}}{x}\right ) + 135 i \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{2} x^{3} \log \left (\frac {3 \, {\left ({\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} b^{2} c + i \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a^{2} x\right )}}{x}\right ) - 135 i \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{2} x^{3} \log \left (\frac {3 \, {\left ({\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} b^{2} c - i \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a^{2} x\right )}}{x}\right ) - 135 \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{2} x^{3} \log \left (\frac {3 \, {\left ({\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} b^{2} c - \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a^{2} x\right )}}{x}\right ) - 4 \, {\left (180 \, a b^{2} c x^{4} - 45 \, b^{3} c x^{3} - 128 \, a^{3} d x^{2} - 32 \, a^{2} b d x + 160 \, a b^{2} d\right )} {\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}}}{1440 \, a b^{2} x^{3}} \]

[In]

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

[Out]

-1/1440*(135*(b^8*c^4/a^7)^(1/4)*a*b^2*x^3*log(3*((a*x^4 - b*x^3)^(1/4)*b^2*c + (b^8*c^4/a^7)^(1/4)*a^2*x)/x)
+ 135*I*(b^8*c^4/a^7)^(1/4)*a*b^2*x^3*log(3*((a*x^4 - b*x^3)^(1/4)*b^2*c + I*(b^8*c^4/a^7)^(1/4)*a^2*x)/x) - 1
35*I*(b^8*c^4/a^7)^(1/4)*a*b^2*x^3*log(3*((a*x^4 - b*x^3)^(1/4)*b^2*c - I*(b^8*c^4/a^7)^(1/4)*a^2*x)/x) - 135*
(b^8*c^4/a^7)^(1/4)*a*b^2*x^3*log(3*((a*x^4 - b*x^3)^(1/4)*b^2*c - (b^8*c^4/a^7)^(1/4)*a^2*x)/x) - 4*(180*a*b^
2*c*x^4 - 45*b^3*c*x^3 - 128*a^3*d*x^2 - 32*a^2*b*d*x + 160*a*b^2*d)*(a*x^4 - b*x^3)^(1/4))/(a*b^2*x^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (128) = 256\).

Time = 0.31 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.00 \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\frac {\frac {270 \, \sqrt {2} b^{3} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {270 \, \sqrt {2} b^{3} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {135 \, \sqrt {2} b^{3} c \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x}}\right )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {135 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} c \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x}}\right )}{a^{2}} + \frac {360 \, {\left ({\left (a - \frac {b}{x}\right )}^{\frac {5}{4}} b^{3} c + 3 \, {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}} a b^{3} c\right )} x^{2}}{a b^{2}} + \frac {256 \, {\left (5 \, {\left (a - \frac {b}{x}\right )}^{\frac {9}{4}} b^{8} d - 9 \, {\left (a - \frac {b}{x}\right )}^{\frac {5}{4}} a b^{8} d\right )}}{b^{9}}}{2880 \, b} \]

[In]

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

[Out]

1/2880*(270*sqrt(2)*b^3*c*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a - b/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*
a) + 270*sqrt(2)*b^3*c*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a - b/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a)
+ 135*sqrt(2)*b^3*c*log(sqrt(2)*(-a)^(1/4)*(a - b/x)^(1/4) + sqrt(-a) + sqrt(a - b/x))/((-a)^(3/4)*a) + 135*s
qrt(2)*(-a)^(1/4)*b^3*c*log(-sqrt(2)*(-a)^(1/4)*(a - b/x)^(1/4) + sqrt(-a) + sqrt(a - b/x))/a^2 + 360*((a - b/
x)^(5/4)*b^3*c + 3*(a - b/x)^(1/4)*a*b^3*c)*x^2/(a*b^2) + 256*(5*(a - b/x)^(9/4)*b^8*d - 9*(a - b/x)^(5/4)*a*b
^8*d)/b^9)/b

Mupad [B] (verification not implemented)

Time = 6.70 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\frac {4\,d\,{\left (a\,x^4-b\,x^3\right )}^{1/4}}{9\,x^3}-\frac {16\,a^2\,d\,{\left (a\,x^4-b\,x^3\right )}^{1/4}}{45\,b^2\,x}+\frac {4\,c\,x\,{\left (a\,x^4-b\,x^3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {7}{4};\ \frac {11}{4};\ \frac {a\,x}{b}\right )}{7\,{\left (1-\frac {a\,x}{b}\right )}^{1/4}}-\frac {4\,a\,d\,{\left (a\,x^4-b\,x^3\right )}^{1/4}}{45\,b\,x^2} \]

[In]

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

[Out]

(4*d*(a*x^4 - b*x^3)^(1/4))/(9*x^3) - (16*a^2*d*(a*x^4 - b*x^3)^(1/4))/(45*b^2*x) + (4*c*x*(a*x^4 - b*x^3)^(1/
4)*hypergeom([-1/4, 7/4], 11/4, (a*x)/b))/(7*(1 - (a*x)/b)^(1/4)) - (4*a*d*(a*x^4 - b*x^3)^(1/4))/(45*b*x^2)