Integrand size = 11, antiderivative size = 75 \[ \int \left (a+b x^4\right )^{3/4} \, dx=\frac {1}{4} x \left (a+b x^4\right )^{3/4}+\frac {3 a \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}}+\frac {3 a \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}} \] Output:
1/4*x*(b*x^4+a)^(3/4)+3/8*a*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(1/4)+3/8* a*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(1/4)
Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^4\right )^{3/4} \, dx=\frac {1}{4} x \left (a+b x^4\right )^{3/4}+\frac {3 a \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}}+\frac {3 a \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 \sqrt [4]{b}} \] Input:
Integrate[(a + b*x^4)^(3/4),x]
Output:
(x*(a + b*x^4)^(3/4))/4 + (3*a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b ^(1/4)) + (3*a*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(1/4))
Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {748, 770, 756, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (a+b x^4\right )^{3/4} \, dx\) |
\(\Big \downarrow \) 748 |
\(\displaystyle \frac {3}{4} a \int \frac {1}{\sqrt [4]{b x^4+a}}dx+\frac {1}{4} x \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {3}{4} a \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{4} x \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {3}{4} a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )+\frac {1}{4} x \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {3}{4} a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )+\frac {1}{4} x \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3}{4} a \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )+\frac {1}{4} x \left (a+b x^4\right )^{3/4}\) |
Input:
Int[(a + b*x^4)^(3/4),x]
Output:
(x*(a + b*x^4)^(3/4))/4 + (3*a*(ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b ^(1/4)) + ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(1/4))))/4
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Simp[a*n*(p/(n*p + 1)) Int[(a + b*x^n)^(p - 1), x], x] /; Fre eQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || LtQ[Denominat or[p + 1/n], Denominator[p]])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p + 1 /n]
Time = 0.01 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07
method | result | size |
pseudoelliptic | \(\frac {4 \left (b \,x^{4}+a \right )^{\frac {3}{4}} x \,b^{\frac {1}{4}}-6 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a +3 \ln \left (\frac {b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a}{16 b^{\frac {1}{4}}}\) | \(80\) |
Input:
int((b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)
Output:
1/16*(4*(b*x^4+a)^(3/4)*x*b^(1/4)-6*arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4))*a+ 3*ln((b^(1/4)*x+(b*x^4+a)^(1/4))/(-b^(1/4)*x+(b*x^4+a)^(1/4)))*a)/b^(1/4)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.51 \[ \int \left (a+b x^4\right )^{3/4} \, dx=\frac {1}{4} \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} x + \frac {3}{16} \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} \log \left (\frac {27 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} + \left (\frac {a^{4}}{b}\right )^{\frac {3}{4}} b x\right )}}{x}\right ) - \frac {3}{16} i \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} \log \left (\frac {27 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} + i \, \left (\frac {a^{4}}{b}\right )^{\frac {3}{4}} b x\right )}}{x}\right ) + \frac {3}{16} i \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} \log \left (\frac {27 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} - i \, \left (\frac {a^{4}}{b}\right )^{\frac {3}{4}} b x\right )}}{x}\right ) - \frac {3}{16} \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} \log \left (\frac {27 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} - \left (\frac {a^{4}}{b}\right )^{\frac {3}{4}} b x\right )}}{x}\right ) \] Input:
integrate((b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
1/4*(b*x^4 + a)^(3/4)*x + 3/16*(a^4/b)^(1/4)*log(27*((b*x^4 + a)^(1/4)*a^3 + (a^4/b)^(3/4)*b*x)/x) - 3/16*I*(a^4/b)^(1/4)*log(27*((b*x^4 + a)^(1/4)* a^3 + I*(a^4/b)^(3/4)*b*x)/x) + 3/16*I*(a^4/b)^(1/4)*log(27*((b*x^4 + a)^( 1/4)*a^3 - I*(a^4/b)^(3/4)*b*x)/x) - 3/16*(a^4/b)^(1/4)*log(27*((b*x^4 + a )^(1/4)*a^3 - (a^4/b)^(3/4)*b*x)/x)
Result contains complex when optimal does not.
Time = 0.65 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.49 \[ \int \left (a+b x^4\right )^{3/4} \, dx=\frac {a^{\frac {3}{4}} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**4+a)**(3/4),x)
Output:
a**(3/4)*x*gamma(1/4)*hyper((-3/4, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a) /(4*gamma(5/4))
Time = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.36 \[ \int \left (a+b x^4\right )^{3/4} \, dx=-\frac {3}{16} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )} - \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} a}{4 \, {\left (b - \frac {b x^{4} + a}{x^{4}}\right )} x^{3}} \] Input:
integrate((b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
-3/16*a*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) + log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(1/4)) - 1/4*(b*x ^4 + a)^(3/4)*a/((b - (b*x^4 + a)/x^4)*x^3)
\[ \int \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} \,d x } \] Input:
integrate((b*x^4+a)^(3/4),x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(3/4), x)
Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.49 \[ \int \left (a+b x^4\right )^{3/4} \, dx=\frac {x\,{\left (b\,x^4+a\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (\frac {b\,x^4}{a}+1\right )}^{3/4}} \] Input:
int((a + b*x^4)^(3/4),x)
Output:
(x*(a + b*x^4)^(3/4)*hypergeom([-3/4, 1/4], 5/4, -(b*x^4)/a))/((b*x^4)/a + 1)^(3/4)
\[ \int \left (a+b x^4\right )^{3/4} \, dx=\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} x}{4}+\frac {3 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) a}{4} \] Input:
int((b*x^4+a)^(3/4),x)
Output:
((a + b*x**4)**(3/4)*x + 3*int((a + b*x**4)**(3/4)/(a + b*x**4),x)*a)/4