3.25.5 \(\int \frac {\sqrt [4]{b x^3+a x^4} (-d+c x^8)}{x^8} \, dx\)

Optimal. Leaf size=193 \[ \frac {3 b^2 c \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{16 a^{7/4}}-\frac {3 b^2 c \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{16 a^{7/4}}+\frac {\sqrt [4]{a x^4+b x^3} \left (-262144 a^7 d x^6+65536 a^6 b d x^5-40960 a^5 b^2 d x^4+30720 a^4 b^3 d x^3-24960 a^3 b^4 d x^2+21216 a^2 b^5 d x+1392300 a b^6 c x^8+445536 a b^6 d+348075 b^7 c x^7\right )}{2784600 a b^6 x^7} \]

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Rubi [A]  time = 0.61, antiderivative size = 341, normalized size of antiderivative = 1.77, number of steps used = 16, number of rules used = 11, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, 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 {32768 a^5 d \left (a x^4+b x^3\right )^{5/4}}{348075 b^6 x^5}+\frac {8192 a^4 d \left (a x^4+b x^3\right )^{5/4}}{69615 b^5 x^6}-\frac {1024 a^3 d \left (a x^4+b x^3\right )^{5/4}}{7735 b^4 x^7}+\frac {256 a^2 d \left (a x^4+b x^3\right )^{5/4}}{1785 b^3 x^8}-\frac {16 a d \left (a x^4+b x^3\right )^{5/4}}{105 b^2 x^9}+\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}}{25 b x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((b*x^3 + a*x^4)^(1/4)*(-d + c*x^8))/x^8,x]

[Out]

(b*c*(b*x^3 + a*x^4)^(1/4))/(8*a) + (c*x*(b*x^3 + a*x^4)^(1/4))/2 + (4*d*(b*x^3 + a*x^4)^(5/4))/(25*b*x^10) -
(16*a*d*(b*x^3 + a*x^4)^(5/4))/(105*b^2*x^9) + (256*a^2*d*(b*x^3 + a*x^4)^(5/4))/(1785*b^3*x^8) - (1024*a^3*d*
(b*x^3 + a*x^4)^(5/4))/(7735*b^4*x^7) + (8192*a^4*d*(b*x^3 + a*x^4)^(5/4))/(69615*b^5*x^6) - (32768*a^5*d*(b*x
^3 + a*x^4)^(5/4))/(348075*b^6*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^8\right )}{x^8} \, dx &=\int \left (c \sqrt [4]{b x^3+a x^4}-\frac {d \sqrt [4]{b x^3+a x^4}}{x^8}\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^8} \, 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}}{25 b x^{10}}+\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^7} \, dx}{5 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}}{25 b x^{10}}-\frac {16 a d \left (b x^3+a x^4\right )^{5/4}}{105 b^2 x^9}-\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 {\left (64 a^2 d\right ) \int \frac {\sqrt [4]{b x^3+a x^4}}{x^6} \, dx}{105 b^2}\\ &=\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}}{25 b x^{10}}-\frac {16 a d \left (b x^3+a x^4\right )^{5/4}}{105 b^2 x^9}+\frac {256 a^2 d \left (b x^3+a x^4\right )^{5/4}}{1785 b^3 x^8}+\frac {\left (256 a^3 d\right ) \int \frac {\sqrt [4]{b x^3+a x^4}}{x^5} \, dx}{595 b^3}-\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}}{25 b x^{10}}-\frac {16 a d \left (b x^3+a x^4\right )^{5/4}}{105 b^2 x^9}+\frac {256 a^2 d \left (b x^3+a x^4\right )^{5/4}}{1785 b^3 x^8}-\frac {1024 a^3 d \left (b x^3+a x^4\right )^{5/4}}{7735 b^4 x^7}-\frac {\left (2048 a^4 d\right ) \int \frac {\sqrt [4]{b x^3+a x^4}}{x^4} \, dx}{7735 b^4}-\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}}{25 b x^{10}}-\frac {16 a d \left (b x^3+a x^4\right )^{5/4}}{105 b^2 x^9}+\frac {256 a^2 d \left (b x^3+a x^4\right )^{5/4}}{1785 b^3 x^8}-\frac {1024 a^3 d \left (b x^3+a x^4\right )^{5/4}}{7735 b^4 x^7}+\frac {8192 a^4 d \left (b x^3+a x^4\right )^{5/4}}{69615 b^5 x^6}+\frac {\left (8192 a^5 d\right ) \int \frac {\sqrt [4]{b x^3+a x^4}}{x^3} \, dx}{69615 b^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}}{25 b x^{10}}-\frac {16 a d \left (b x^3+a x^4\right )^{5/4}}{105 b^2 x^9}+\frac {256 a^2 d \left (b x^3+a x^4\right )^{5/4}}{1785 b^3 x^8}-\frac {1024 a^3 d \left (b x^3+a x^4\right )^{5/4}}{7735 b^4 x^7}+\frac {8192 a^4 d \left (b x^3+a x^4\right )^{5/4}}{69615 b^5 x^6}-\frac {32768 a^5 d \left (b x^3+a x^4\right )^{5/4}}{348075 b^6 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}}{25 b x^{10}}-\frac {16 a d \left (b x^3+a x^4\right )^{5/4}}{105 b^2 x^9}+\frac {256 a^2 d \left (b x^3+a x^4\right )^{5/4}}{1785 b^3 x^8}-\frac {1024 a^3 d \left (b x^3+a x^4\right )^{5/4}}{7735 b^4 x^7}+\frac {8192 a^4 d \left (b x^3+a x^4\right )^{5/4}}{69615 b^5 x^6}-\frac {32768 a^5 d \left (b x^3+a x^4\right )^{5/4}}{348075 b^6 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 = 0.44, size = 424, normalized size = 2.20 \begin {gather*} \frac {4 \sqrt [4]{x^3 (a x+b)} \left (-8192 a^{14} d x^6 \sqrt [4]{\frac {a x}{b}+1}+2048 a^{13} b d x^5 \sqrt [4]{\frac {a x}{b}+1}-1280 a^{12} b^2 d x^4 \sqrt [4]{\frac {a x}{b}+1}+960 a^{11} b^3 d x^3 \sqrt [4]{\frac {a x}{b}+1}-780 a^{10} b^4 d x^2 \sqrt [4]{\frac {a x}{b}+1}+663 a^9 b^5 d x \sqrt [4]{\frac {a x}{b}+1}+13923 a^8 b^6 d \sqrt [4]{\frac {a x}{b}+1}+1092048 a^6 b^8 c x^6 \sqrt [4]{\frac {a x}{b}+1}+4600038 a^5 b^9 c x^5 \sqrt [4]{\frac {a x}{b}+1}+8698470 a^4 b^{10} c x^4 \sqrt [4]{\frac {a x}{b}+1}+9313560 a^3 b^{11} c x^3 \sqrt [4]{\frac {a x}{b}+1}+5833620 a^2 b^{12} c x^2 \sqrt [4]{\frac {a x}{b}+1}-13923 b^{14} c \, _2F_1\left (-\frac {33}{4},-\frac {25}{4};-\frac {21}{4};-\frac {a x}{b}\right )+111384 b^{14} c \, _2F_1\left (-\frac {29}{4},-\frac {25}{4};-\frac {21}{4};-\frac {a x}{b}\right )-389844 b^{14} c \, _2F_1\left (-\frac {25}{4},-\frac {25}{4};-\frac {21}{4};-\frac {a x}{b}\right )+292383 b^{14} c \sqrt [4]{\frac {a x}{b}+1}+2002923 a b^{13} c x \sqrt [4]{\frac {a x}{b}+1}\right )}{348075 a^8 b^6 x^7 \sqrt [4]{\frac {a x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((b*x^3 + a*x^4)^(1/4)*(-d + c*x^8))/x^8,x]

[Out]

(4*(x^3*(b + a*x))^(1/4)*(292383*b^14*c*(1 + (a*x)/b)^(1/4) + 13923*a^8*b^6*d*(1 + (a*x)/b)^(1/4) + 2002923*a*
b^13*c*x*(1 + (a*x)/b)^(1/4) + 663*a^9*b^5*d*x*(1 + (a*x)/b)^(1/4) + 5833620*a^2*b^12*c*x^2*(1 + (a*x)/b)^(1/4
) - 780*a^10*b^4*d*x^2*(1 + (a*x)/b)^(1/4) + 9313560*a^3*b^11*c*x^3*(1 + (a*x)/b)^(1/4) + 960*a^11*b^3*d*x^3*(
1 + (a*x)/b)^(1/4) + 8698470*a^4*b^10*c*x^4*(1 + (a*x)/b)^(1/4) - 1280*a^12*b^2*d*x^4*(1 + (a*x)/b)^(1/4) + 46
00038*a^5*b^9*c*x^5*(1 + (a*x)/b)^(1/4) + 2048*a^13*b*d*x^5*(1 + (a*x)/b)^(1/4) + 1092048*a^6*b^8*c*x^6*(1 + (
a*x)/b)^(1/4) - 8192*a^14*d*x^6*(1 + (a*x)/b)^(1/4) - 13923*b^14*c*Hypergeometric2F1[-33/4, -25/4, -21/4, -((a
*x)/b)] + 111384*b^14*c*Hypergeometric2F1[-29/4, -25/4, -21/4, -((a*x)/b)] - 389844*b^14*c*Hypergeometric2F1[-
25/4, -25/4, -21/4, -((a*x)/b)]))/(348075*a^8*b^6*x^7*(1 + (a*x)/b)^(1/4))

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IntegrateAlgebraic [A]  time = 2.62, size = 193, normalized size = 1.00 \begin {gather*} \frac {\sqrt [4]{b x^3+a x^4} \left (445536 a b^6 d+21216 a^2 b^5 d x-24960 a^3 b^4 d x^2+30720 a^4 b^3 d x^3-40960 a^5 b^2 d x^4+65536 a^6 b d x^5-262144 a^7 d x^6+348075 b^7 c x^7+1392300 a b^6 c x^8\right )}{2784600 a b^6 x^7}+\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^8))/x^8,x]

[Out]

((b*x^3 + a*x^4)^(1/4)*(445536*a*b^6*d + 21216*a^2*b^5*d*x - 24960*a^3*b^4*d*x^2 + 30720*a^4*b^3*d*x^3 - 40960
*a^5*b^2*d*x^4 + 65536*a^6*b*d*x^5 - 262144*a^7*d*x^6 + 348075*b^7*c*x^7 + 1392300*a*b^6*c*x^8))/(2784600*a*b^
6*x^7) + (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 = 0.51, size = 377, normalized size = 1.95 \begin {gather*} \frac {4176900 \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{6} x^{7} \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 ) - 1044225 \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{6} x^{7} \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 ) + 1044225 \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{6} x^{7} \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 (1392300 \, a b^{6} c x^{8} + 348075 \, b^{7} c x^{7} - 262144 \, a^{7} d x^{6} + 65536 \, a^{6} b d x^{5} - 40960 \, a^{5} b^{2} d x^{4} + 30720 \, a^{4} b^{3} d x^{3} - 24960 \, a^{3} b^{4} d x^{2} + 21216 \, a^{2} b^{5} d x + 445536 \, a b^{6} d\right )} {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{11138400 \, a b^{6} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4+b*x^3)^(1/4)*(c*x^8-d)/x^8,x, algorithm="fricas")

[Out]

1/11138400*(4176900*(b^8*c^4/a^7)^(1/4)*a*b^6*x^7*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))
- 1044225*(b^8*c^4/a^7)^(1/4)*a*b^6*x^7*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) +
1044225*(b^8*c^4/a^7)^(1/4)*a*b^6*x^7*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*(
1392300*a*b^6*c*x^8 + 348075*b^7*c*x^7 - 262144*a^7*d*x^6 + 65536*a^6*b*d*x^5 - 40960*a^5*b^2*d*x^4 + 30720*a^
4*b^3*d*x^3 - 24960*a^3*b^4*d*x^2 + 21216*a^2*b^5*d*x + 445536*a*b^6*d)*(a*x^4 + b*x^3)^(1/4))/(a*b^6*x^7)

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giac [B]  time = 0.67, size = 358, normalized size = 1.85 \begin {gather*} \frac {\frac {2088450 \, \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 {2088450 \, \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 {1044225 \, \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 {1044225 \, \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 {2784600 \, {\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 (13923 \, {\left (a + \frac {b}{x}\right )}^{\frac {25}{4}} b^{120} d - 82875 \, {\left (a + \frac {b}{x}\right )}^{\frac {21}{4}} a b^{120} d + 204750 \, {\left (a + \frac {b}{x}\right )}^{\frac {17}{4}} a^{2} b^{120} d - 267750 \, {\left (a + \frac {b}{x}\right )}^{\frac {13}{4}} a^{3} b^{120} d + 193375 \, {\left (a + \frac {b}{x}\right )}^{\frac {9}{4}} a^{4} b^{120} d - 69615 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{4}} a^{5} b^{120} d\right )}}{b^{125}}}{22276800 \, 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^8-d)/x^8,x, algorithm="giac")

[Out]

1/22276800*(2088450*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) + 2088450*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) + 1044225*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) + 1044225*sqrt(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 + 2784600*((a + b/x)^(5/4)*b^3*c + 3*(a + b/x)^(1/4)*a*b^3*c)*x^2/(a*b^2) + 256*(13923*(a + b/x)^(25/4)*
b^120*d - 82875*(a + b/x)^(21/4)*a*b^120*d + 204750*(a + b/x)^(17/4)*a^2*b^120*d - 267750*(a + b/x)^(13/4)*a^3
*b^120*d + 193375*(a + b/x)^(9/4)*a^4*b^120*d - 69615*(a + b/x)^(5/4)*a^5*b^120*d)/b^125)/b

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}+b \,x^{3}\right )^{\frac {1}{4}} \left (c \,x^{8}-d \right )}{x^{8}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4+b*x^3)^(1/4)*(c*x^8-d)/x^8,x)

[Out]

int((a*x^4+b*x^3)^(1/4)*(c*x^8-d)/x^8,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c x^{8} - d\right )} {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{x^{8}}\,{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^8-d)/x^8,x, algorithm="maxima")

[Out]

integrate((c*x^8 - d)*(a*x^4 + b*x^3)^(1/4)/x^8, x)

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mupad [B]  time = 4.81, size = 207, normalized size = 1.07 \begin {gather*} \frac {4\,d\,{\left (a\,x^4+b\,x^3\right )}^{1/4}}{25\,x^7}-\frac {16\,a^2\,d\,{\left (a\,x^4+b\,x^3\right )}^{1/4}}{1785\,b^2\,x^5}+\frac {256\,a^3\,d\,{\left (a\,x^4+b\,x^3\right )}^{1/4}}{23205\,b^3\,x^4}-\frac {1024\,a^4\,d\,{\left (a\,x^4+b\,x^3\right )}^{1/4}}{69615\,b^4\,x^3}+\frac {8192\,a^5\,d\,{\left (a\,x^4+b\,x^3\right )}^{1/4}}{348075\,b^5\,x^2}-\frac {32768\,a^6\,d\,{\left (a\,x^4+b\,x^3\right )}^{1/4}}{348075\,b^6\,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 (\frac {a\,x}{b}+1\right )}^{1/4}}+\frac {4\,a\,d\,{\left (a\,x^4+b\,x^3\right )}^{1/4}}{525\,b\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((d - c*x^8)*(a*x^4 + b*x^3)^(1/4))/x^8,x)

[Out]

(4*d*(a*x^4 + b*x^3)^(1/4))/(25*x^7) - (16*a^2*d*(a*x^4 + b*x^3)^(1/4))/(1785*b^2*x^5) + (256*a^3*d*(a*x^4 + b
*x^3)^(1/4))/(23205*b^3*x^4) - (1024*a^4*d*(a*x^4 + b*x^3)^(1/4))/(69615*b^4*x^3) + (8192*a^5*d*(a*x^4 + b*x^3
)^(1/4))/(348075*b^5*x^2) - (32768*a^6*d*(a*x^4 + b*x^3)^(1/4))/(348075*b^6*x) + (4*c*x*(a*x^4 + b*x^3)^(1/4)*
hypergeom([-1/4, 7/4], 11/4, -(a*x)/b))/(7*((a*x)/b + 1)^(1/4)) + (4*a*d*(a*x^4 + b*x^3)^(1/4))/(525*b*x^6)

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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^{8} - d\right )}{x^{8}}\, 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**8-d)/x**8,x)

[Out]

Integral((x**3*(a*x + b))**(1/4)*(c*x**8 - d)/x**8, x)

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