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

Optimal. Leaf size=112 \[ -\frac {7 b^3 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{64 a^{11/4}}+\frac {7 b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{64 a^{11/4}}+\frac {\left (32 a^2 x^2+4 a b x-7 b^2\right ) \sqrt [4]{a x^4+b x^3}}{96 a^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 196, normalized size of antiderivative = 1.75, number of steps used = 9, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {2021, 2024, 2032, 63, 331, 298, 203, 206} \begin {gather*} -\frac {7 b^3 x^{9/4} (a x+b)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{64 a^{11/4} \left (a x^4+b x^3\right )^{3/4}}+\frac {7 b^3 x^{9/4} (a x+b)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{64 a^{11/4} \left (a x^4+b x^3\right )^{3/4}}-\frac {7 b^2 \sqrt [4]{a x^4+b x^3}}{96 a^2}+\frac {b x \sqrt [4]{a x^4+b x^3}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{a x^4+b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(b*x^3 + a*x^4)^(1/4),x]

[Out]

(-7*b^2*(b*x^3 + a*x^4)^(1/4))/(96*a^2) + (b*x*(b*x^3 + a*x^4)^(1/4))/(24*a) + (x^2*(b*x^3 + a*x^4)^(1/4))/3 -
(7*b^3*x^(9/4)*(b + a*x)^(3/4)*ArcTan[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(64*a^(11/4)*(b*x^3 + a*x^4)^(3/4))
+ (7*b^3*x^(9/4)*(b + a*x)^(3/4)*ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)])/(64*a^(11/4)*(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 \sqrt [4]{b x^3+a x^4} \, dx &=\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {1}{12} b \int \frac {x^4}{\left (b x^3+a x^4\right )^{3/4}} \, dx\\ &=\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}-\frac {\left (7 b^2\right ) \int \frac {x^3}{\left (b x^3+a x^4\right )^{3/4}} \, dx}{96 a}\\ &=-\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {\left (7 b^3\right ) \int \frac {x^2}{\left (b x^3+a x^4\right )^{3/4}} \, dx}{128 a^2}\\ &=-\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {\left (7 b^3 x^{9/4} (b+a x)^{3/4}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{128 a^2 \left (b x^3+a x^4\right )^{3/4}}\\ &=-\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {\left (7 b^3 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 )}{32 a^2 \left (b x^3+a x^4\right )^{3/4}}\\ &=-\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {\left (7 b^3 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 )}{32 a^2 \left (b x^3+a x^4\right )^{3/4}}\\ &=-\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}+\frac {\left (7 b^3 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 )}{64 a^{5/2} \left (b x^3+a x^4\right )^{3/4}}-\frac {\left (7 b^3 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 )}{64 a^{5/2} \left (b x^3+a x^4\right )^{3/4}}\\ &=-\frac {7 b^2 \sqrt [4]{b x^3+a x^4}}{96 a^2}+\frac {b x \sqrt [4]{b x^3+a x^4}}{24 a}+\frac {1}{3} x^2 \sqrt [4]{b x^3+a x^4}-\frac {7 b^3 x^{9/4} (b+a x)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{11/4} \left (b x^3+a x^4\right )^{3/4}}+\frac {7 b^3 x^{9/4} (b+a x)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{64 a^{11/4} \left (b x^3+a x^4\right )^{3/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 49, normalized size = 0.44 \begin {gather*} \frac {4 x^2 \sqrt [4]{x^3 (a x+b)} \, _2F_1\left (-\frac {1}{4},\frac {11}{4};\frac {15}{4};-\frac {a x}{b}\right )}{11 \sqrt [4]{\frac {a x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(b*x^3 + a*x^4)^(1/4),x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.49, size = 112, normalized size = 1.00 \begin {gather*} \frac {\left (-7 b^2+4 a b x+32 a^2 x^2\right ) \sqrt [4]{b x^3+a x^4}}{96 a^2}-\frac {7 b^3 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{64 a^{11/4}}+\frac {7 b^3 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{64 a^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-7*b^2 + 4*a*b*x + 32*a^2*x^2)*(b*x^3 + a*x^4)^(1/4))/(96*a^2) - (7*b^3*ArcTan[(a^(1/4)*x)/(b*x^3 + a*x^4)^(
1/4)])/(64*a^(11/4)) + (7*b^3*ArcTanh[(a^(1/4)*x)/(b*x^3 + a*x^4)^(1/4)])/(64*a^(11/4))

________________________________________________________________________________________

fricas [B]  time = 0.46, size = 253, normalized size = 2.26 \begin {gather*} -\frac {84 \, a^{2} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} a^{8} b^{3} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {3}{4}} - a^{8} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {3}{4}} x \sqrt {\frac {a^{6} \sqrt {\frac {b^{12}}{a^{11}}} x^{2} + \sqrt {a x^{4} + b x^{3}} b^{6}}{x^{2}}}}{b^{12} x}\right ) - 21 \, a^{2} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} \log \left (\frac {7 \, {\left (a^{3} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{3}\right )}}{x}\right ) + 21 \, a^{2} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (a^{3} \left (\frac {b^{12}}{a^{11}}\right )^{\frac {1}{4}} x - {\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} + 4 \, a b x - 7 \, b^{2}\right )}}{384 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(84*a^2*(b^12/a^11)^(1/4)*arctan(-((a*x^4 + b*x^3)^(1/4)*a^8*b^3*(b^12/a^11)^(3/4) - a^8*(b^12/a^11)^(3
/4)*x*sqrt((a^6*sqrt(b^12/a^11)*x^2 + sqrt(a*x^4 + b*x^3)*b^6)/x^2))/(b^12*x)) - 21*a^2*(b^12/a^11)^(1/4)*log(
7*(a^3*(b^12/a^11)^(1/4)*x + (a*x^4 + b*x^3)^(1/4)*b^3)/x) + 21*a^2*(b^12/a^11)^(1/4)*log(-7*(a^3*(b^12/a^11)^
(1/4)*x - (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 + 4*a*b*x - 7*b^2))/a^2

________________________________________________________________________________________

giac [B]  time = 0.38, size = 261, normalized size = 2.33 \begin {gather*} \frac {\frac {42 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{4} \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 )}{a^{3}} + \frac {42 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{4} \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 )}{a^{3}} + \frac {21 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{4} \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^{3}} + \frac {21 \, \sqrt {2} b^{4} \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^{2}} - \frac {8 \, {\left (7 \, {\left (a + \frac {b}{x}\right )}^{\frac {9}{4}} b^{4} - 18 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{4}} a b^{4} - 21 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} a^{2} b^{4}\right )} x^{3}}{a^{2} b^{3}}}{768 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(42*sqrt(2)*(-a)^(1/4)*b^4*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x)^(1/4))/(-a)^(1/4))/a^3 +
42*sqrt(2)*(-a)^(1/4)*b^4*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/a^3 + 21*s
qrt(2)*(-a)^(1/4)*b^4*log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/a^3 + 21*sqrt(2)*b^4*
log(-sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/((-a)^(3/4)*a^2) - 8*(7*(a + b/x)^(9/4)*b^
4 - 18*(a + b/x)^(5/4)*a*b^4 - 21*(a + b/x)^(1/4)*a^2*b^4)*x^3/(a^2*b^3))/b

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 + b*x^3)^(1/4)*x, x)

________________________________________________________________________________________

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

[Out]

int(x*(a*x^4 + b*x^3)^(1/4), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \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*(a*x**4+b*x**3)**(1/4),x)

[Out]

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

________________________________________________________________________________________