3.10.86 \(\int \frac {(b+a x^3)^{3/4}}{x^4} \, dx\)

Optimal. Leaf size=75 \[ -\frac {\left (a x^3+b\right )^{3/4}}{3 x^3}+\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}} \]

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Rubi [A]  time = 0.07, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {266, 47, 63, 298, 203, 206} \begin {gather*} -\frac {\left (a x^3+b\right )^{3/4}}{3 x^3}+\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b + a*x^3)^(3/4)/x^4,x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.49 \begin {gather*} \frac {4 a \left (a x^3+b\right )^{7/4} \, _2F_1\left (\frac {7}{4},2;\frac {11}{4};\frac {a x^3}{b}+1\right )}{21 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x^3)^(3/4)/x^4,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.10, size = 75, normalized size = 1.00 \begin {gather*} -\frac {\left (b+a x^3\right )^{3/4}}{3 x^3}+\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.48, size = 182, normalized size = 2.43 \begin {gather*} -\frac {12 \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{3} \arctan \left (-\frac {{\left (a x^{3} + b\right )}^{\frac {1}{4}} \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} a^{3} - \sqrt {\sqrt {a x^{3} + b} a^{6} + \sqrt {\frac {a^{4}}{b}} a^{4} b} \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}}}{a^{4}}\right ) + 3 \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{3} \log \left ({\left (a x^{3} + b\right )}^{\frac {1}{4}} a^{3} + \left (\frac {a^{4}}{b}\right )^{\frac {3}{4}} b\right ) - 3 \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{3} \log \left ({\left (a x^{3} + b\right )}^{\frac {1}{4}} a^{3} - \left (\frac {a^{4}}{b}\right )^{\frac {3}{4}} b\right ) + 4 \, {\left (a x^{3} + b\right )}^{\frac {3}{4}}}{12 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*(a^4/b)^(1/4)*x^3*arctan(-((a*x^3 + b)^(1/4)*(a^4/b)^(1/4)*a^3 - sqrt(sqrt(a*x^3 + b)*a^6 + sqrt(a^4
/b)*a^4*b)*(a^4/b)^(1/4))/a^4) + 3*(a^4/b)^(1/4)*x^3*log((a*x^3 + b)^(1/4)*a^3 + (a^4/b)^(3/4)*b) - 3*(a^4/b)^
(1/4)*x^3*log((a*x^3 + b)^(1/4)*a^3 - (a^4/b)^(3/4)*b) + 4*(a*x^3 + b)^(3/4))/x^3

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giac [B]  time = 0.61, size = 209, normalized size = 2.79 \begin {gather*} \frac {\frac {6 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{\left (-b\right )^{\frac {1}{4}}} + \frac {6 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{\left (-b\right )^{\frac {1}{4}}} + \frac {3 \, \sqrt {2} a^{2} \left (-b\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (a x^{3} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{3} + b} + \sqrt {-b}\right )}{b} + \frac {3 \, \sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{3} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{3} + b} + \sqrt {-b}\right )}{\left (-b\right )^{\frac {1}{4}}} - \frac {8 \, {\left (a x^{3} + b\right )}^{\frac {3}{4}} a}{x^{3}}}{24 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(a*x^3 + b)^(1/4))/(-b)^(1/4))/(-b)^(1/4) + 6*s
qrt(2)*a^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(a*x^3 + b)^(1/4))/(-b)^(1/4))/(-b)^(1/4) + 3*sqrt(2)*a
^2*(-b)^(3/4)*log(sqrt(2)*(a*x^3 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^3 + b) + sqrt(-b))/b + 3*sqrt(2)*a^2*log(-sq
rt(2)*(a*x^3 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^3 + b) + sqrt(-b))/(-b)^(1/4) - 8*(a*x^3 + b)^(3/4)*a/x^3)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 74, normalized size = 0.99 \begin {gather*} \frac {1}{4} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{3} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (a x^{3} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{3} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}}\right )} - \frac {{\left (a x^{3} + b\right )}^{\frac {3}{4}}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.97, size = 55, normalized size = 0.73 \begin {gather*} \frac {a\,\mathrm {atan}\left (\frac {{\left (a\,x^3+b\right )}^{1/4}}{b^{1/4}}\right )}{2\,b^{1/4}}-\frac {{\left (a\,x^3+b\right )}^{3/4}}{3\,x^3}-\frac {a\,\mathrm {atanh}\left (\frac {{\left (a\,x^3+b\right )}^{1/4}}{b^{1/4}}\right )}{2\,b^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b + a*x^3)^(3/4)/x^4,x)

[Out]

(a*atan((b + a*x^3)^(1/4)/b^(1/4)))/(2*b^(1/4)) - (b + a*x^3)^(3/4)/(3*x^3) - (a*atanh((b + a*x^3)^(1/4)/b^(1/
4)))/(2*b^(1/4))

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sympy [C]  time = 1.23, size = 42, normalized size = 0.56 \begin {gather*} - \frac {a^{\frac {3}{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 x^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(3/4)*gamma(1/4)*hyper((-3/4, 1/4), (5/4,), b*exp_polar(I*pi)/(a*x**3))/(3*x**(3/4)*gamma(5/4))

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