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

Optimal. Leaf size=123 \[ \frac {77 b^4 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{1024 a^{15/4}}-\frac {77 b^4 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{1024 a^{15/4}}+\frac {\left (384 a^3 x^3+32 a^2 b x^2-44 a b^2 x+77 b^3\right ) \sqrt [4]{a x^4+b x^3}}{1536 a^3} \]

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Rubi [A]  time = 0.35, antiderivative size = 224, normalized size of antiderivative = 1.82, number of steps used = 10, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {2021, 2024, 2032, 63, 331, 298, 203, 206} \begin {gather*} \frac {77 b^4 x^{9/4} (a x+b)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{1024 a^{15/4} \left (a x^4+b x^3\right )^{3/4}}-\frac {77 b^4 x^{9/4} (a x+b)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{1024 a^{15/4} \left (a x^4+b x^3\right )^{3/4}}+\frac {77 b^3 \sqrt [4]{a x^4+b x^3}}{1536 a^3}-\frac {11 b^2 x \sqrt [4]{a x^4+b x^3}}{384 a^2}+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}+\frac {b x^2 \sqrt [4]{a x^4+b x^3}}{48 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(b*x^3 + a*x^4)^(1/4),x]

[Out]

(77*b^3*(b*x^3 + a*x^4)^(1/4))/(1536*a^3) - (11*b^2*x*(b*x^3 + a*x^4)^(1/4))/(384*a^2) + (b*x^2*(b*x^3 + a*x^4
)^(1/4))/(48*a) + (x^3*(b*x^3 + a*x^4)^(1/4))/4 + (77*b^4*x^(9/4)*(b + a*x)^(3/4)*ArcTan[(a^(1/4)*x^(1/4))/(b
+ a*x)^(1/4)])/(1024*a^(15/4)*(b*x^3 + a*x^4)^(3/4)) - (77*b^4*x^(9/4)*(b + a*x)^(3/4)*ArcTanh[(a^(1/4)*x^(1/4
))/(b + a*x)^(1/4)])/(1024*a^(15/4)*(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 2021

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

Rubi steps

\begin {align*} \int x^2 \sqrt [4]{b x^3+a x^4} \, dx &=\frac {1}{4} x^3 \sqrt [4]{b x^3+a x^4}+\frac {1}{16} b \int \frac {x^5}{\left (b x^3+a x^4\right )^{3/4}} \, dx\\ &=\frac {b x^2 \sqrt [4]{b x^3+a x^4}}{48 a}+\frac {1}{4} x^3 \sqrt [4]{b x^3+a x^4}-\frac {\left (11 b^2\right ) \int \frac {x^4}{\left (b x^3+a x^4\right )^{3/4}} \, dx}{192 a}\\ &=-\frac {11 b^2 x \sqrt [4]{b x^3+a x^4}}{384 a^2}+\frac {b x^2 \sqrt [4]{b x^3+a x^4}}{48 a}+\frac {1}{4} x^3 \sqrt [4]{b x^3+a x^4}+\frac {\left (77 b^3\right ) \int \frac {x^3}{\left (b x^3+a x^4\right )^{3/4}} \, dx}{1536 a^2}\\ &=\frac {77 b^3 \sqrt [4]{b x^3+a x^4}}{1536 a^3}-\frac {11 b^2 x \sqrt [4]{b x^3+a x^4}}{384 a^2}+\frac {b x^2 \sqrt [4]{b x^3+a x^4}}{48 a}+\frac {1}{4} x^3 \sqrt [4]{b x^3+a x^4}-\frac {\left (77 b^4\right ) \int \frac {x^2}{\left (b x^3+a x^4\right )^{3/4}} \, dx}{2048 a^3}\\ &=\frac {77 b^3 \sqrt [4]{b x^3+a x^4}}{1536 a^3}-\frac {11 b^2 x \sqrt [4]{b x^3+a x^4}}{384 a^2}+\frac {b x^2 \sqrt [4]{b x^3+a x^4}}{48 a}+\frac {1}{4} x^3 \sqrt [4]{b x^3+a x^4}-\frac {\left (77 b^4 x^{9/4} (b+a x)^{3/4}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{2048 a^3 \left (b x^3+a x^4\right )^{3/4}}\\ &=\frac {77 b^3 \sqrt [4]{b x^3+a x^4}}{1536 a^3}-\frac {11 b^2 x \sqrt [4]{b x^3+a x^4}}{384 a^2}+\frac {b x^2 \sqrt [4]{b x^3+a x^4}}{48 a}+\frac {1}{4} x^3 \sqrt [4]{b x^3+a x^4}-\frac {\left (77 b^4 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 )}{512 a^3 \left (b x^3+a x^4\right )^{3/4}}\\ &=\frac {77 b^3 \sqrt [4]{b x^3+a x^4}}{1536 a^3}-\frac {11 b^2 x \sqrt [4]{b x^3+a x^4}}{384 a^2}+\frac {b x^2 \sqrt [4]{b x^3+a x^4}}{48 a}+\frac {1}{4} x^3 \sqrt [4]{b x^3+a x^4}-\frac {\left (77 b^4 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 )}{512 a^3 \left (b x^3+a x^4\right )^{3/4}}\\ &=\frac {77 b^3 \sqrt [4]{b x^3+a x^4}}{1536 a^3}-\frac {11 b^2 x \sqrt [4]{b x^3+a x^4}}{384 a^2}+\frac {b x^2 \sqrt [4]{b x^3+a x^4}}{48 a}+\frac {1}{4} x^3 \sqrt [4]{b x^3+a x^4}-\frac {\left (77 b^4 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 )}{1024 a^{7/2} \left (b x^3+a x^4\right )^{3/4}}+\frac {\left (77 b^4 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 )}{1024 a^{7/2} \left (b x^3+a x^4\right )^{3/4}}\\ &=\frac {77 b^3 \sqrt [4]{b x^3+a x^4}}{1536 a^3}-\frac {11 b^2 x \sqrt [4]{b x^3+a x^4}}{384 a^2}+\frac {b x^2 \sqrt [4]{b x^3+a x^4}}{48 a}+\frac {1}{4} x^3 \sqrt [4]{b x^3+a x^4}+\frac {77 b^4 x^{9/4} (b+a x)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{1024 a^{15/4} \left (b x^3+a x^4\right )^{3/4}}-\frac {77 b^4 x^{9/4} (b+a x)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{1024 a^{15/4} \left (b x^3+a x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 49, normalized size = 0.40 \begin {gather*} \frac {4 x^3 \sqrt [4]{x^3 (a x+b)} \, _2F_1\left (-\frac {1}{4},\frac {15}{4};\frac {19}{4};-\frac {a x}{b}\right )}{15 \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),x]

[Out]

(4*x^3*(x^3*(b + a*x))^(1/4)*Hypergeometric2F1[-1/4, 15/4, 19/4, -((a*x)/b)])/(15*(1 + (a*x)/b)^(1/4))

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IntegrateAlgebraic [A]  time = 0.52, size = 123, normalized size = 1.00 \begin {gather*} \frac {\left (77 b^3-44 a b^2 x+32 a^2 b x^2+384 a^3 x^3\right ) \sqrt [4]{b x^3+a x^4}}{1536 a^3}+\frac {77 b^4 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{1024 a^{15/4}}-\frac {77 b^4 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{1024 a^{15/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((77*b^3 - 44*a*b^2*x + 32*a^2*b*x^2 + 384*a^3*x^3)*(b*x^3 + a*x^4)^(1/4))/(1536*a^3) + (77*b^4*ArcTan[(a^(1/4
)*x)/(b*x^3 + a*x^4)^(1/4)])/(1024*a^(15/4)) - (77*b^4*ArcTanh[(a^(1/4)*x)/(b*x^3 + a*x^4)^(1/4)])/(1024*a^(15
/4))

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fricas [B]  time = 0.46, size = 264, normalized size = 2.15 \begin {gather*} \frac {924 \, a^{3} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} a^{11} b^{4} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {3}{4}} - a^{11} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {3}{4}} x \sqrt {\frac {a^{8} \sqrt {\frac {b^{16}}{a^{15}}} x^{2} + \sqrt {a x^{4} + b x^{3}} b^{8}}{x^{2}}}}{b^{16} x}\right ) - 231 \, a^{3} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} \log \left (\frac {77 \, {\left (a^{4} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{4}\right )}}{x}\right ) + 231 \, a^{3} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} \log \left (-\frac {77 \, {\left (a^{4} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{4}\right )}}{x}\right ) + 4 \, {\left (384 \, a^{3} x^{3} + 32 \, a^{2} b x^{2} - 44 \, a b^{2} x + 77 \, b^{3}\right )} {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{6144 \, 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),x, algorithm="fricas")

[Out]

1/6144*(924*a^3*(b^16/a^15)^(1/4)*arctan(-((a*x^4 + b*x^3)^(1/4)*a^11*b^4*(b^16/a^15)^(3/4) - a^11*(b^16/a^15)
^(3/4)*x*sqrt((a^8*sqrt(b^16/a^15)*x^2 + sqrt(a*x^4 + b*x^3)*b^8)/x^2))/(b^16*x)) - 231*a^3*(b^16/a^15)^(1/4)*
log(77*(a^4*(b^16/a^15)^(1/4)*x + (a*x^4 + b*x^3)^(1/4)*b^4)/x) + 231*a^3*(b^16/a^15)^(1/4)*log(-77*(a^4*(b^16
/a^15)^(1/4)*x - (a*x^4 + b*x^3)^(1/4)*b^4)/x) + 4*(384*a^3*x^3 + 32*a^2*b*x^2 - 44*a*b^2*x + 77*b^3)*(a*x^4 +
b*x^3)^(1/4))/a^3

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giac [B]  time = 0.27, size = 278, normalized size = 2.26 \begin {gather*} \frac {\frac {462 \, \sqrt {2} b^{5} \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^{3}} + \frac {462 \, \sqrt {2} b^{5} \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^{3}} + \frac {231 \, \sqrt {2} b^{5} \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^{3}} + \frac {231 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{5} \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^{4}} + \frac {8 \, {\left (77 \, {\left (a + \frac {b}{x}\right )}^{\frac {13}{4}} b^{5} - 275 \, {\left (a + \frac {b}{x}\right )}^{\frac {9}{4}} a b^{5} + 351 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{4}} a^{2} b^{5} + 231 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} a^{3} b^{5}\right )} x^{4}}{a^{3} b^{4}}}{12288 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12288*(462*sqrt(2)*b^5*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a
^3) + 462*sqrt(2)*b^5*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a^3
) + 231*sqrt(2)*b^5*log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/((-a)^(3/4)*a^3) + 231*
sqrt(2)*(-a)^(1/4)*b^5*log(-sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/a^4 + 8*(77*(a + b/
x)^(13/4)*b^5 - 275*(a + b/x)^(9/4)*a*b^5 + 351*(a + b/x)^(5/4)*a^2*b^5 + 231*(a + b/x)^(1/4)*a^3*b^5)*x^4/(a^
3*b^4))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} x^{2}\,{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),x, algorithm="maxima")

[Out]

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

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

[Out]

int(x^2*(a*x^4 + b*x^3)^(1/4), x)

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

[Out]

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

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