3.1023 \(\int \frac {(a+b x^4)^{3/4}}{x^5} \, dx\)

Optimal. Leaf size=75 \[ -\frac {\left (a+b x^4\right )^{3/4}}{4 x^4}+\frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 \sqrt [4]{a}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 \sqrt [4]{a}} \]

[Out]

-1/4*(b*x^4+a)^(3/4)/x^4+3/8*b*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(1/4)-3/8*b*arctanh((b*x^4+a)^(1/4)/a^(1/4))/
a^(1/4)

________________________________________________________________________________________

Rubi [A]  time = 0.05, 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} \[ -\frac {\left (a+b x^4\right )^{3/4}}{4 x^4}+\frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 \sqrt [4]{a}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 \sqrt [4]{a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(3/4)/x^5,x]

[Out]

-(a + b*x^4)^(3/4)/(4*x^4) + (3*b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(1/4)) - (3*b*ArcTanh[(a + b*x^4)^(1
/4)/a^(1/4)])/(8*a^(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 (a+b x^4\right )^{3/4}}{x^5} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {(a+b x)^{3/4}}{x^2} \, dx,x,x^4\right )\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{4 x^4}+\frac {1}{16} (3 b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt [4]{a+b x}} \, dx,x,x^4\right )\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{4 x^4}+\frac {3}{4} \operatorname {Subst}\left (\int \frac {x^2}{-\frac {a}{b}+\frac {x^4}{b}} \, dx,x,\sqrt [4]{a+b x^4}\right )\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{4 x^4}-\frac {1}{8} (3 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )+\frac {1}{8} (3 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{4 x^4}+\frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 \sqrt [4]{a}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 \sqrt [4]{a}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.03, size = 37, normalized size = 0.49 \[ \frac {b \left (a+b x^4\right )^{7/4} \, _2F_1\left (\frac {7}{4},2;\frac {11}{4};\frac {b x^4}{a}+1\right )}{7 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(3/4)/x^5,x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.66, size = 185, normalized size = 2.47 \[ -\frac {12 \, \left (\frac {b^{4}}{a}\right )^{\frac {1}{4}} x^{4} \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (\frac {b^{4}}{a}\right )^{\frac {1}{4}} b^{3} - \sqrt {\sqrt {b x^{4} + a} b^{6} + \sqrt {\frac {b^{4}}{a}} a b^{4}} \left (\frac {b^{4}}{a}\right )^{\frac {1}{4}}}{b^{4}}\right ) + 3 \, \left (\frac {b^{4}}{a}\right )^{\frac {1}{4}} x^{4} \log \left (27 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3} + 27 \, \left (\frac {b^{4}}{a}\right )^{\frac {3}{4}} a\right ) - 3 \, \left (\frac {b^{4}}{a}\right )^{\frac {1}{4}} x^{4} \log \left (27 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3} - 27 \, \left (\frac {b^{4}}{a}\right )^{\frac {3}{4}} a\right ) + 4 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{16 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^5,x, algorithm="fricas")

[Out]

-1/16*(12*(b^4/a)^(1/4)*x^4*arctan(-((b*x^4 + a)^(1/4)*(b^4/a)^(1/4)*b^3 - sqrt(sqrt(b*x^4 + a)*b^6 + sqrt(b^4
/a)*a*b^4)*(b^4/a)^(1/4))/b^4) + 3*(b^4/a)^(1/4)*x^4*log(27*(b*x^4 + a)^(1/4)*b^3 + 27*(b^4/a)^(3/4)*a) - 3*(b
^4/a)^(1/4)*x^4*log(27*(b*x^4 + a)^(1/4)*b^3 - 27*(b^4/a)^(3/4)*a) + 4*(b*x^4 + a)^(3/4))/x^4

________________________________________________________________________________________

giac [B]  time = 0.21, size = 209, normalized size = 2.79 \[ \frac {\frac {6 \, \sqrt {2} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {1}{4}}} + \frac {6 \, \sqrt {2} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {1}{4}}} + \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a} + \frac {3 \, \sqrt {2} b^{2} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {1}{4}}} - \frac {8 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b}{x^{4}}}{32 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^5,x, algorithm="giac")

[Out]

1/32*(6*sqrt(2)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(1/4) + 6*s
qrt(2)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(1/4) + 3*sqrt(2)*(
-a)^(3/4)*b^2*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a + 3*sqrt(2)*b^2*log(-sq
rt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/(-a)^(1/4) - 8*(b*x^4 + a)^(3/4)*b/x^4)/b

________________________________________________________________________________________

maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(3/4)/x^5,x)

[Out]

int((b*x^4+a)^(3/4)/x^5,x)

________________________________________________________________________________________

maxima [A]  time = 3.01, size = 74, normalized size = 0.99 \[ \frac {3}{16} \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )} - \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^5,x, algorithm="maxima")

[Out]

3/16*b*(2*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(1/4) + log(((b*x^4 + a)^(1/4) - a^(1/4))/((b*x^4 + a)^(1/4) + a
^(1/4)))/a^(1/4)) - 1/4*(b*x^4 + a)^(3/4)/x^4

________________________________________________________________________________________

mupad [B]  time = 1.26, size = 55, normalized size = 0.73 \[ \frac {3\,b\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{1/4}}-\frac {{\left (b\,x^4+a\right )}^{3/4}}{4\,x^4}-\frac {3\,b\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(3/4)/x^5,x)

[Out]

(3*b*atan((a + b*x^4)^(1/4)/a^(1/4)))/(8*a^(1/4)) - (a + b*x^4)^(3/4)/(4*x^4) - (3*b*atanh((a + b*x^4)^(1/4)/a
^(1/4)))/(8*a^(1/4))

________________________________________________________________________________________

sympy [C]  time = 3.05, size = 39, normalized size = 0.52 \[ - \frac {b^{\frac {3}{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 x \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(3/4)/x**5,x)

[Out]

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

________________________________________________________________________________________