3.25.39 \(\int \frac {x^2 \sqrt [4]{b x^3+a x^4}}{-b+a x} \, dx\)

Optimal. Leaf size=198 \[ -\frac {155 b^3 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{64 a^{15/4}}+\frac {2 \sqrt [4]{2} b^3 \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{a^{15/4}}+\frac {155 b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{64 a^{15/4}}-\frac {2 \sqrt [4]{2} b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{a^{15/4}}+\frac {\left (32 a^2 x^2+52 a b x+101 b^2\right ) \sqrt [4]{a x^4+b x^3}}{96 a^3} \]

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Rubi [A]  time = 0.46, antiderivative size = 339, normalized size of antiderivative = 1.71, number of steps used = 27, number of rules used = 11, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {2042, 101, 157, 50, 63, 331, 298, 203, 206, 105, 93} \begin {gather*} -\frac {155 b^3 \sqrt [4]{a x^4+b x^3} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{64 a^{15/4} x^{3/4} \sqrt [4]{a x+b}}+\frac {2 \sqrt [4]{2} b^3 \sqrt [4]{a x^4+b x^3} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{a^{15/4} x^{3/4} \sqrt [4]{a x+b}}+\frac {155 b^3 \sqrt [4]{a x^4+b x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{64 a^{15/4} x^{3/4} \sqrt [4]{a x+b}}-\frac {2 \sqrt [4]{2} b^3 \sqrt [4]{a x^4+b x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{a^{15/4} x^{3/4} \sqrt [4]{a x+b}}+\frac {101 b^2 \sqrt [4]{a x^4+b x^3}}{96 a^3}+\frac {13 b x \sqrt [4]{a x^4+b x^3}}{24 a^2}+\frac {x^2 \sqrt [4]{a x^4+b x^3}}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(b*x^3 + a*x^4)^(1/4))/(-b + a*x),x]

[Out]

(101*b^2*(b*x^3 + a*x^4)^(1/4))/(96*a^3) + (13*b*x*(b*x^3 + a*x^4)^(1/4))/(24*a^2) + (x^2*(b*x^3 + a*x^4)^(1/4
))/(3*a) - (155*b^3*(b*x^3 + a*x^4)^(1/4)*ArcTan[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(64*a^(15/4)*x^(3/4)*(b +
a*x)^(1/4)) + (2*2^(1/4)*b^3*(b*x^3 + a*x^4)^(1/4)*ArcTan[(2^(1/4)*a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(a^(15/
4)*x^(3/4)*(b + a*x)^(1/4)) + (155*b^3*(b*x^3 + a*x^4)^(1/4)*ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(64*a
^(15/4)*x^(3/4)*(b + a*x)^(1/4)) - (2*2^(1/4)*b^3*(b*x^3 + a*x^4)^(1/4)*ArcTanh[(2^(1/4)*a^(1/4)*x^(1/4))/(b +
a*x)^(1/4)])/(a^(15/4)*x^(3/4)*(b + a*x)^(1/4))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 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 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 2042

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.))^(q_.), x_Symbol]
:> Dist[(e^IntPart[m]*(e*x)^FracPart[m]*(a*x^j + b*x^(j + n))^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a
+ b*x^n)^FracPart[p]), Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, j, m, n,
p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])

Rubi steps

\begin {align*} \int \frac {x^2 \sqrt [4]{b x^3+a x^4}}{-b+a x} \, dx &=\frac {\sqrt [4]{b x^3+a x^4} \int \frac {x^{11/4} \sqrt [4]{b+a x}}{-b+a x} \, dx}{x^{3/4} \sqrt [4]{b+a x}}\\ &=\frac {x^2 \sqrt [4]{b x^3+a x^4}}{3 a}-\frac {\sqrt [4]{b x^3+a x^4} \int \frac {x^{7/4} \left (-\frac {11 b^2}{4}-\frac {13 a b x}{4}\right )}{(-b+a x) (b+a x)^{3/4}} \, dx}{3 a x^{3/4} \sqrt [4]{b+a x}}\\ &=\frac {x^2 \sqrt [4]{b x^3+a x^4}}{3 a}+\frac {\left (13 b \sqrt [4]{b x^3+a x^4}\right ) \int \frac {x^{7/4}}{(b+a x)^{3/4}} \, dx}{12 a x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (2 b^2 \sqrt [4]{b x^3+a x^4}\right ) \int \frac {x^{7/4}}{(-b+a x) (b+a x)^{3/4}} \, dx}{a x^{3/4} \sqrt [4]{b+a x}}\\ &=\frac {13 b x \sqrt [4]{b x^3+a x^4}}{24 a^2}+\frac {x^2 \sqrt [4]{b x^3+a x^4}}{3 a}-\frac {\left (91 b^2 \sqrt [4]{b x^3+a x^4}\right ) \int \frac {x^{3/4}}{(b+a x)^{3/4}} \, dx}{96 a^2 x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (2 b^2 \sqrt [4]{b x^3+a x^4}\right ) \int \frac {x^{3/4}}{(b+a x)^{3/4}} \, dx}{a^2 x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (2 b^3 \sqrt [4]{b x^3+a x^4}\right ) \int \frac {x^{3/4}}{(-b+a x) (b+a x)^{3/4}} \, dx}{a^2 x^{3/4} \sqrt [4]{b+a x}}\\ &=\frac {101 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^3}+\frac {13 b x \sqrt [4]{b x^3+a x^4}}{24 a^2}+\frac {x^2 \sqrt [4]{b x^3+a x^4}}{3 a}+\frac {\left (91 b^3 \sqrt [4]{b x^3+a x^4}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{128 a^3 x^{3/4} \sqrt [4]{b+a x}}-\frac {\left (3 b^3 \sqrt [4]{b x^3+a x^4}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{2 a^3 x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (2 b^3 \sqrt [4]{b x^3+a x^4}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{a^3 x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (2 b^4 \sqrt [4]{b x^3+a x^4}\right ) \int \frac {1}{\sqrt [4]{x} (-b+a x) (b+a x)^{3/4}} \, dx}{a^3 x^{3/4} \sqrt [4]{b+a x}}\\ &=\frac {101 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^3}+\frac {13 b x \sqrt [4]{b x^3+a x^4}}{24 a^2}+\frac {x^2 \sqrt [4]{b x^3+a x^4}}{3 a}+\frac {\left (91 b^3 \sqrt [4]{b x^3+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{32 a^3 x^{3/4} \sqrt [4]{b+a x}}-\frac {\left (6 b^3 \sqrt [4]{b x^3+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{a^3 x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (8 b^3 \sqrt [4]{b x^3+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{a^3 x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (8 b^4 \sqrt [4]{b x^3+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{-b+2 a b x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{a^3 x^{3/4} \sqrt [4]{b+a x}}\\ &=\frac {101 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^3}+\frac {13 b x \sqrt [4]{b x^3+a x^4}}{24 a^2}+\frac {x^2 \sqrt [4]{b x^3+a x^4}}{3 a}-\frac {\left (2 \sqrt {2} b^3 \sqrt [4]{b x^3+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} \sqrt {a} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{a^{7/2} x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (2 \sqrt {2} b^3 \sqrt [4]{b x^3+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} \sqrt {a} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{a^{7/2} x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (91 b^3 \sqrt [4]{b x^3+a x^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 )}{32 a^3 x^{3/4} \sqrt [4]{b+a x}}-\frac {\left (6 b^3 \sqrt [4]{b x^3+a x^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 )}{a^3 x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (8 b^3 \sqrt [4]{b x^3+a x^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 )}{a^3 x^{3/4} \sqrt [4]{b+a x}}\\ &=\frac {101 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^3}+\frac {13 b x \sqrt [4]{b x^3+a x^4}}{24 a^2}+\frac {x^2 \sqrt [4]{b x^3+a x^4}}{3 a}+\frac {2 \sqrt [4]{2} b^3 \sqrt [4]{b x^3+a x^4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{a^{15/4} x^{3/4} \sqrt [4]{b+a x}}-\frac {2 \sqrt [4]{2} b^3 \sqrt [4]{b x^3+a x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{a^{15/4} x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (91 b^3 \sqrt [4]{b x^3+a x^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 )}{64 a^{7/2} x^{3/4} \sqrt [4]{b+a x}}-\frac {\left (91 b^3 \sqrt [4]{b x^3+a x^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 )}{64 a^{7/2} x^{3/4} \sqrt [4]{b+a x}}-\frac {\left (3 b^3 \sqrt [4]{b x^3+a x^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 )}{a^{7/2} x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (3 b^3 \sqrt [4]{b x^3+a x^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 )}{a^{7/2} x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (4 b^3 \sqrt [4]{b x^3+a x^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 )}{a^{7/2} x^{3/4} \sqrt [4]{b+a x}}-\frac {\left (4 b^3 \sqrt [4]{b x^3+a x^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 )}{a^{7/2} x^{3/4} \sqrt [4]{b+a x}}\\ &=\frac {101 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^3}+\frac {13 b x \sqrt [4]{b x^3+a x^4}}{24 a^2}+\frac {x^2 \sqrt [4]{b x^3+a x^4}}{3 a}-\frac {155 b^3 \sqrt [4]{b x^3+a x^4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{15/4} x^{3/4} \sqrt [4]{b+a x}}+\frac {2 \sqrt [4]{2} b^3 \sqrt [4]{b x^3+a x^4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{a^{15/4} x^{3/4} \sqrt [4]{b+a x}}+\frac {155 b^3 \sqrt [4]{b x^3+a x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{15/4} x^{3/4} \sqrt [4]{b+a x}}-\frac {2 \sqrt [4]{2} b^3 \sqrt [4]{b x^3+a x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{a^{15/4} x^{3/4} \sqrt [4]{b+a x}}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 183, normalized size = 0.92 \begin {gather*} \frac {4 x^3 \left (21 a^2 x^2 (a x+b) \, _2F_1\left (-\frac {1}{4},\frac {11}{4};\frac {15}{4};-\frac {a x}{b}\right )+77 b^2 (a x+b) \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-\frac {a x}{b}\right )+77 b^2 \left ((a x+b) \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};-\frac {a x}{b}\right )-2 b \sqrt [4]{\frac {a x}{b}+1} \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {2 a x}{b+a x}\right )\right )+33 a b x (a x+b) \, _2F_1\left (-\frac {1}{4},\frac {7}{4};\frac {11}{4};-\frac {a x}{b}\right )\right )}{231 a^3 \left (x^3 (a x+b)\right )^{3/4} \sqrt [4]{\frac {a x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(b*x^3 + a*x^4)^(1/4))/(-b + a*x),x]

[Out]

(4*x^3*(77*b^2*(b + a*x)*Hypergeometric2F1[-1/4, 3/4, 7/4, -((a*x)/b)] + 33*a*b*x*(b + a*x)*Hypergeometric2F1[
-1/4, 7/4, 11/4, -((a*x)/b)] + 21*a^2*x^2*(b + a*x)*Hypergeometric2F1[-1/4, 11/4, 15/4, -((a*x)/b)] + 77*b^2*(
(b + a*x)*Hypergeometric2F1[3/4, 3/4, 7/4, -((a*x)/b)] - 2*b*(1 + (a*x)/b)^(1/4)*Hypergeometric2F1[3/4, 1, 7/4
, (2*a*x)/(b + a*x)])))/(231*a^3*(x^3*(b + a*x))^(3/4)*(1 + (a*x)/b)^(1/4))

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IntegrateAlgebraic [A]  time = 0.82, size = 198, normalized size = 1.00 \begin {gather*} \frac {\left (101 b^2+52 a b x+32 a^2 x^2\right ) \sqrt [4]{b x^3+a x^4}}{96 a^3}-\frac {155 b^3 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{64 a^{15/4}}+\frac {2 \sqrt [4]{2} b^3 \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{a^{15/4}}+\frac {155 b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{64 a^{15/4}}-\frac {2 \sqrt [4]{2} b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{a^{15/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2*(b*x^3 + a*x^4)^(1/4))/(-b + a*x),x]

[Out]

((101*b^2 + 52*a*b*x + 32*a^2*x^2)*(b*x^3 + a*x^4)^(1/4))/(96*a^3) - (155*b^3*ArcTan[(a^(1/4)*x)/(b*x^3 + a*x^
4)^(1/4)])/(64*a^(15/4)) + (2*2^(1/4)*b^3*ArcTan[(2^(1/4)*a^(1/4)*x)/(b*x^3 + a*x^4)^(1/4)])/a^(15/4) + (155*b
^3*ArcTanh[(a^(1/4)*x)/(b*x^3 + a*x^4)^(1/4)])/(64*a^(15/4)) - (2*2^(1/4)*b^3*ArcTanh[(2^(1/4)*a^(1/4)*x)/(b*x
^3 + a*x^4)^(1/4)])/a^(15/4)

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fricas [B]  time = 0.50, size = 489, normalized size = 2.47 \begin {gather*} \frac {1536 \cdot 2^{\frac {1}{4}} a^{3} \left (\frac {b^{12}}{a^{15}}\right )^{\frac {1}{4}} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} a^{11} b^{3} \left (\frac {b^{12}}{a^{15}}\right )^{\frac {3}{4}} - 2^{\frac {3}{4}} a^{11} x \sqrt {\frac {\sqrt {2} a^{8} x^{2} \sqrt {\frac {b^{12}}{a^{15}}} + \sqrt {a x^{4} + b x^{3}} b^{6}}{x^{2}}} \left (\frac {b^{12}}{a^{15}}\right )^{\frac {3}{4}}}{2 \, b^{12} x}\right ) - 384 \cdot 2^{\frac {1}{4}} a^{3} \left (\frac {b^{12}}{a^{15}}\right )^{\frac {1}{4}} \log \left (\frac {2^{\frac {1}{4}} a^{4} x \left (\frac {b^{12}}{a^{15}}\right )^{\frac {1}{4}} + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{3}}{x}\right ) + 384 \cdot 2^{\frac {1}{4}} a^{3} \left (\frac {b^{12}}{a^{15}}\right )^{\frac {1}{4}} \log \left (-\frac {2^{\frac {1}{4}} a^{4} x \left (\frac {b^{12}}{a^{15}}\right )^{\frac {1}{4}} - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{3}}{x}\right ) - 1860 \, a^{3} \left (\frac {b^{12}}{a^{15}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} a^{11} b^{3} \left (\frac {b^{12}}{a^{15}}\right )^{\frac {3}{4}} - a^{11} x \sqrt {\frac {a^{8} x^{2} \sqrt {\frac {b^{12}}{a^{15}}} + \sqrt {a x^{4} + b x^{3}} b^{6}}{x^{2}}} \left (\frac {b^{12}}{a^{15}}\right )^{\frac {3}{4}}}{b^{12} x}\right ) + 465 \, a^{3} \left (\frac {b^{12}}{a^{15}}\right )^{\frac {1}{4}} \log \left (\frac {155 \, {\left (a^{4} x \left (\frac {b^{12}}{a^{15}}\right )^{\frac {1}{4}} + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{3}\right )}}{x}\right ) - 465 \, a^{3} \left (\frac {b^{12}}{a^{15}}\right )^{\frac {1}{4}} \log \left (-\frac {155 \, {\left (a^{4} x \left (\frac {b^{12}}{a^{15}}\right )^{\frac {1}{4}} - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{3}\right )}}{x}\right ) + 4 \, {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} {\left (32 \, a^{2} x^{2} + 52 \, a b x + 101 \, b^{2}\right )}}{384 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a*x^4+b*x^3)^(1/4)/(a*x-b),x, algorithm="fricas")

[Out]

1/384*(1536*2^(1/4)*a^3*(b^12/a^15)^(1/4)*arctan(-1/2*(2^(3/4)*(a*x^4 + b*x^3)^(1/4)*a^11*b^3*(b^12/a^15)^(3/4
) - 2^(3/4)*a^11*x*sqrt((sqrt(2)*a^8*x^2*sqrt(b^12/a^15) + sqrt(a*x^4 + b*x^3)*b^6)/x^2)*(b^12/a^15)^(3/4))/(b
^12*x)) - 384*2^(1/4)*a^3*(b^12/a^15)^(1/4)*log((2^(1/4)*a^4*x*(b^12/a^15)^(1/4) + (a*x^4 + b*x^3)^(1/4)*b^3)/
x) + 384*2^(1/4)*a^3*(b^12/a^15)^(1/4)*log(-(2^(1/4)*a^4*x*(b^12/a^15)^(1/4) - (a*x^4 + b*x^3)^(1/4)*b^3)/x) -
1860*a^3*(b^12/a^15)^(1/4)*arctan(-((a*x^4 + b*x^3)^(1/4)*a^11*b^3*(b^12/a^15)^(3/4) - a^11*x*sqrt((a^8*x^2*s
qrt(b^12/a^15) + sqrt(a*x^4 + b*x^3)*b^6)/x^2)*(b^12/a^15)^(3/4))/(b^12*x)) + 465*a^3*(b^12/a^15)^(1/4)*log(15
5*(a^4*x*(b^12/a^15)^(1/4) + (a*x^4 + b*x^3)^(1/4)*b^3)/x) - 465*a^3*(b^12/a^15)^(1/4)*log(-155*(a^4*x*(b^12/a
^15)^(1/4) - (a*x^4 + b*x^3)^(1/4)*b^3)/x) + 4*(a*x^4 + b*x^3)^(1/4)*(32*a^2*x^2 + 52*a*b*x + 101*b^2))/a^3

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giac [B]  time = 2.86, size = 461, normalized size = 2.33 \begin {gather*} -\frac {2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b^{3} \arctan \left (\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \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 )}{a^{4}} - \frac {2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b^{3} \arctan \left (-\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \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 )}{a^{4}} - \frac {2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b^{3} \log \left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{2 \, a^{4}} + \frac {2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b^{3} \log \left (-2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{2 \, a^{4}} + \frac {155 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \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 )}{128 \, a^{4}} + \frac {155 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \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 )}{128 \, a^{4}} + \frac {155 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \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 )}{256 \, a^{4}} + \frac {155 \, \sqrt {2} b^{3} \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 )}{256 \, \left (-a\right )^{\frac {3}{4}} a^{3}} + \frac {{\left (101 \, {\left (a + \frac {b}{x}\right )}^{\frac {9}{4}} b^{3} - 150 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{4}} a b^{3} + 81 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} a^{2} b^{3}\right )} x^{3}}{96 \, a^{3} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a*x^4+b*x^3)^(1/4)/(a*x-b),x, algorithm="giac")

[Out]

-2^(3/4)*(-a)^(1/4)*b^3*arctan(1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) + 2*(a + b/x)^(1/4))/(-a)^(1/4))/a^4 - 2^(3/4)*
(-a)^(1/4)*b^3*arctan(-1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/a^4 - 1/2*2^(3/4)*(-a)
^(1/4)*b^3*log(2^(3/4)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a + b/x))/a^4 + 1/2*2^(3/4)*(-a)^(
1/4)*b^3*log(-2^(3/4)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a + b/x))/a^4 + 155/128*sqrt(2)*(-a
)^(1/4)*b^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x)^(1/4))/(-a)^(1/4))/a^4 + 155/128*sqrt(2)*(-a)
^(1/4)*b^3*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/a^4 + 155/256*sqrt(2)*(-a)
^(1/4)*b^3*log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/a^4 + 155/256*sqrt(2)*b^3*log(-s
qrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/((-a)^(3/4)*a^3) + 1/96*(101*(a + b/x)^(9/4)*b^3
- 150*(a + b/x)^(5/4)*a*b^3 + 81*(a + b/x)^(1/4)*a^2*b^3)*x^3/(a^3*b^3)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (a \,x^{4}+b \,x^{3}\right )^{\frac {1}{4}}}{a x -b}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a*x^4+b*x^3)^(1/4)/(a*x-b),x)

[Out]

int(x^2*(a*x^4+b*x^3)^(1/4)/(a*x-b),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}} x^{2}}{a x - b}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a*x^4+b*x^3)^(1/4)/(a*x-b),x, algorithm="maxima")

[Out]

integrate((a*x^4 + b*x^3)^(1/4)*x^2/(a*x - b), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {x^2\,{\left (a\,x^4+b\,x^3\right )}^{1/4}}{b-a\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2*(a*x^4 + b*x^3)^(1/4))/(b - a*x),x)

[Out]

-int((x^2*(a*x^4 + b*x^3)^(1/4))/(b - a*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \sqrt [4]{x^{3} \left (a x + b\right )}}{a x - b}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a*x**4+b*x**3)**(1/4)/(a*x-b),x)

[Out]

Integral(x**2*(x**3*(a*x + b))**(1/4)/(a*x - b), x)

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