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

Optimal. Leaf size=148 \[ \frac {3 b^2 c \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^3}}\right )}{16 a^{7/4}}-\frac {3 b^2 c \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^3}}\right )}{16 a^{7/4}}+\frac {\sqrt [4]{a x^4-b x^3} \left (-128 a^3 d x^2-32 a^2 b d x+180 a b^2 c x^4+160 a b^2 d-45 b^3 c x^3\right )}{360 a b^2 x^3} \]

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Rubi [A]  time = 0.44, antiderivative size = 239, normalized size of antiderivative = 1.61, number of steps used = 12, number of rules used = 11, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {2052, 2004, 2024, 2032, 63, 331, 298, 203, 206, 2016, 2014} \begin {gather*} \frac {3 b^2 c x^{9/4} (a x-b)^{3/4} \tan ^{-1}\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} \tanh ^{-1}\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} \end {gather*}

Antiderivative was successfully verified.

[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 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 2004

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 2014

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] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

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 2024

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 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 2052

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

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Mathematica [C]  time = 0.28, size = 237, normalized size = 1.60 \begin {gather*} -\frac {4 \sqrt [4]{x^3 (a x-b)} \left (4 a^6 d x^2 \sqrt [4]{1-\frac {a x}{b}}+a^5 b d x \sqrt [4]{1-\frac {a x}{b}}-5 a^4 b^2 d \sqrt [4]{1-\frac {a x}{b}}-24 a^2 b^4 c x^2 \sqrt [4]{1-\frac {a x}{b}}+5 b^6 c \, _2F_1\left (-\frac {17}{4},-\frac {9}{4};-\frac {5}{4};\frac {a x}{b}\right )-20 b^6 c \, _2F_1\left (-\frac {13}{4},-\frac {9}{4};-\frac {5}{4};\frac {a x}{b}\right )+30 b^6 c \, _2F_1\left (-\frac {9}{4},-\frac {9}{4};-\frac {5}{4};\frac {a x}{b}\right )-15 b^6 c \sqrt [4]{1-\frac {a x}{b}}+39 a b^5 c x \sqrt [4]{1-\frac {a x}{b}}\right )}{45 a^4 b^2 x^3 \sqrt [4]{1-\frac {a x}{b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(x^3*(-b + a*x))^(1/4)*(-15*b^6*c*(1 - (a*x)/b)^(1/4) - 5*a^4*b^2*d*(1 - (a*x)/b)^(1/4) + 39*a*b^5*c*x*(1
- (a*x)/b)^(1/4) + a^5*b*d*x*(1 - (a*x)/b)^(1/4) - 24*a^2*b^4*c*x^2*(1 - (a*x)/b)^(1/4) + 4*a^6*d*x^2*(1 - (a*
x)/b)^(1/4) + 5*b^6*c*Hypergeometric2F1[-17/4, -9/4, -5/4, (a*x)/b] - 20*b^6*c*Hypergeometric2F1[-13/4, -9/4,
-5/4, (a*x)/b] + 30*b^6*c*Hypergeometric2F1[-9/4, -9/4, -5/4, (a*x)/b]))/(45*a^4*b^2*x^3*(1 - (a*x)/b)^(1/4))

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IntegrateAlgebraic [A]  time = 0.83, size = 148, normalized size = 1.00 \begin {gather*} \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 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^3+a x^4}}\right )}{16 a^{7/4}}-\frac {3 b^2 c \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^3+a x^4}}\right )}{16 a^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(b*x^3) + a*x^4)^(1/4)*(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))/(360*
a*b^2*x^3) + (3*b^2*c*ArcTan[(a^(1/4)*x)/(-(b*x^3) + a*x^4)^(1/4)])/(16*a^(7/4)) - (3*b^2*c*ArcTanh[(a^(1/4)*x
)/(-(b*x^3) + a*x^4)^(1/4)])/(16*a^(7/4))

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fricas [B]  time = 0.50, size = 334, normalized size = 2.26 \begin {gather*} \frac {540 \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{2} x^{3} \arctan \left (-\frac {\left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {3}{4}} {\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} a^{5} b^{2} c - \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {3}{4}} a^{5} x \sqrt {\frac {\sqrt {a x^{4} - b x^{3}} b^{4} c^{2} + \sqrt {\frac {b^{8} c^{4}}{a^{7}}} a^{4} x^{2}}{x^{2}}}}{b^{8} c^{4} 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 ) + 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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1440*(540*(b^8*c^4/a^7)^(1/4)*a*b^2*x^3*arctan(-((b^8*c^4/a^7)^(3/4)*(a*x^4 - b*x^3)^(1/4)*a^5*b^2*c - (b^8*
c^4/a^7)^(3/4)*a^5*x*sqrt((sqrt(a*x^4 - b*x^3)*b^4*c^2 + sqrt(b^8*c^4/a^7)*a^4*x^2)/x^2))/(b^8*c^4*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) + 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)

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giac [B]  time = 1.26, size = 296, normalized size = 2.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.85, size = 111, normalized size = 0.75 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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^{4} - d\right )}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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