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

Optimal. Leaf size=82 \[ \frac {7 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{2 a^{3/4}}-\frac {7 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^3}}\right )}{2 a^{3/4}}+\sqrt [4]{a x^4+b x^3} \]

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Rubi [A]  time = 0.36, antiderivative size = 136, normalized size of antiderivative = 1.66, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2056, 80, 63, 331, 298, 203, 206} \begin {gather*} \frac {7 b \sqrt [4]{a x^4+b x^3} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4} x^{3/4} \sqrt [4]{a x+b}}-\frac {7 b \sqrt [4]{a x^4+b x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4} x^{3/4} \sqrt [4]{a x+b}}+\sqrt [4]{a x^4+b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

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

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 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

\begin {align*} \int \frac {(-b+a x) \sqrt [4]{b x^3+a x^4}}{x (b+a x)} \, dx &=\frac {\sqrt [4]{b x^3+a x^4} \int \frac {-b+a x}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{x^{3/4} \sqrt [4]{b+a x}}\\ &=\sqrt [4]{b x^3+a x^4}-\frac {\left (7 b \sqrt [4]{b x^3+a x^4}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}} \, dx}{4 x^{3/4} \sqrt [4]{b+a x}}\\ &=\sqrt [4]{b x^3+a x^4}-\frac {\left (7 b \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 )}{x^{3/4} \sqrt [4]{b+a x}}\\ &=\sqrt [4]{b x^3+a x^4}-\frac {\left (7 b \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 )}{x^{3/4} \sqrt [4]{b+a x}}\\ &=\sqrt [4]{b x^3+a x^4}-\frac {\left (7 b \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 )}{2 \sqrt {a} x^{3/4} \sqrt [4]{b+a x}}+\frac {\left (7 b \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 )}{2 \sqrt {a} x^{3/4} \sqrt [4]{b+a x}}\\ &=\sqrt [4]{b x^3+a x^4}+\frac {7 b \sqrt [4]{b x^3+a x^4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{2 a^{3/4} x^{3/4} \sqrt [4]{b+a x}}-\frac {7 b \sqrt [4]{b x^3+a x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{2 a^{3/4} x^{3/4} \sqrt [4]{b+a x}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 60, normalized size = 0.73 \begin {gather*} \frac {x^3 \left (3 (a x+b)-7 b \left (\frac {a x}{b}+1\right )^{3/4} \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};-\frac {a x}{b}\right )\right )}{3 \left (x^3 (a x+b)\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(3*(b + a*x) - 7*b*(1 + (a*x)/b)^(3/4)*Hypergeometric2F1[3/4, 3/4, 7/4, -((a*x)/b)]))/(3*(x^3*(b + a*x))^
(3/4))

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IntegrateAlgebraic [A]  time = 0.41, size = 82, normalized size = 1.00 \begin {gather*} \sqrt [4]{b x^3+a x^4}+\frac {7 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{2 a^{3/4}}-\frac {7 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{2 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.49, size = 207, normalized size = 2.52 \begin {gather*} 7 \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {a^{2} \left (\frac {b^{4}}{a^{3}}\right )^{\frac {3}{4}} x \sqrt {\frac {a^{2} \sqrt {\frac {b^{4}}{a^{3}}} x^{2} + \sqrt {a x^{4} + b x^{3}} b^{2}}{x^{2}}} - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} a^{2} b \left (\frac {b^{4}}{a^{3}}\right )^{\frac {3}{4}}}{b^{4} x}\right ) - \frac {7}{4} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (\frac {7 \, {\left (a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b\right )}}{x}\right ) + \frac {7}{4} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b\right )}}{x}\right ) + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7*(b^4/a^3)^(1/4)*arctan((a^2*(b^4/a^3)^(3/4)*x*sqrt((a^2*sqrt(b^4/a^3)*x^2 + sqrt(a*x^4 + b*x^3)*b^2)/x^2) -
(a*x^4 + b*x^3)^(1/4)*a^2*b*(b^4/a^3)^(3/4))/(b^4*x)) - 7/4*(b^4/a^3)^(1/4)*log(7*(a*(b^4/a^3)^(1/4)*x + (a*x^
4 + b*x^3)^(1/4)*b)/x) + 7/4*(b^4/a^3)^(1/4)*log(-7*(a*(b^4/a^3)^(1/4)*x - (a*x^4 + b*x^3)^(1/4)*b)/x) + (a*x^
4 + b*x^3)^(1/4)

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giac [B]  time = 0.38, size = 207, normalized size = 2.52 \begin {gather*} \frac {\frac {14 \, \sqrt {2} b^{2} \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}}} + \frac {14 \, \sqrt {2} b^{2} \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}}} + \frac {7 \, \sqrt {2} b^{2} \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}}} + \frac {7 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \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} + 8 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} b x}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(14*sqrt(2)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 14*sq
rt(2)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 7*sqrt(2)*b^2*
log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/(-a)^(3/4) + 7*sqrt(2)*(-a)^(1/4)*b^2*log(-
sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/a + 8*(a + b/x)^(1/4)*b*x)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 (a x - b\right )}{x \left (a x + b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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