Integrand size = 11, antiderivative size = 96 \[ \int \left (a+b x^4\right )^{7/4} \, dx=\frac {7}{32} a x \left (a+b x^4\right )^{3/4}+\frac {1}{8} x \left (a+b x^4\right )^{7/4}+\frac {21 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt [4]{b}}+\frac {21 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt [4]{b}} \] Output:
7/32*a*x*(b*x^4+a)^(3/4)+1/8*x*(b*x^4+a)^(7/4)+21/64*a^2*arctan(b^(1/4)*x/ (b*x^4+a)^(1/4))/b^(1/4)+21/64*a^2*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(1 /4)
Time = 0.39 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.93 \[ \int \left (a+b x^4\right )^{7/4} \, dx=\frac {1}{32} x \left (a+b x^4\right )^{3/4} \left (11 a+4 b x^4\right )+\frac {21 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt [4]{b}}+\frac {21 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt [4]{b}} \] Input:
Integrate[(a + b*x^4)^(7/4),x]
Output:
(x*(a + b*x^4)^(3/4)*(11*a + 4*b*x^4))/32 + (21*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(1/4)) + (21*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1 /4)])/(64*b^(1/4))
Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {748, 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.
\(\displaystyle \int \left (a+b x^4\right )^{7/4} \, dx\) |
\(\Big \downarrow \) 748 |
\(\displaystyle \frac {7}{8} a \int \left (b x^4+a\right )^{3/4}dx+\frac {1}{8} x \left (a+b x^4\right )^{7/4}\) |
\(\Big \downarrow \) 748 |
\(\displaystyle \frac {7}{8} a \left (\frac {3}{4} a \int \frac {1}{\sqrt [4]{b x^4+a}}dx+\frac {1}{4} x \left (a+b x^4\right )^{3/4}\right )+\frac {1}{8} x \left (a+b x^4\right )^{7/4}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {7}{8} a \left (\frac {3}{4} a \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{4} x \left (a+b x^4\right )^{3/4}\right )+\frac {1}{8} x \left (a+b x^4\right )^{7/4}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {7}{8} a \left (\frac {3}{4} a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )+\frac {1}{4} x \left (a+b x^4\right )^{3/4}\right )+\frac {1}{8} x \left (a+b x^4\right )^{7/4}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {7}{8} a \left (\frac {3}{4} a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )+\frac {1}{4} x \left (a+b x^4\right )^{3/4}\right )+\frac {1}{8} x \left (a+b x^4\right )^{7/4}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {7}{8} a \left (\frac {3}{4} a \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )+\frac {1}{4} x \left (a+b x^4\right )^{3/4}\right )+\frac {1}{8} x \left (a+b x^4\right )^{7/4}\) |
Input:
Int[(a + b*x^4)^(7/4),x]
Output:
(x*(a + b*x^4)^(7/4))/8 + (7*a*((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))/8
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 = 1.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.06
method | result | size |
pseudoelliptic | \(\frac {16 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{\frac {5}{4}} x^{5}+44 a x \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{\frac {1}{4}}-42 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a^{2}+21 \ln \left (\frac {b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a^{2}}{128 b^{\frac {1}{4}}}\) | \(102\) |
Input:
int((b*x^4+a)^(7/4),x,method=_RETURNVERBOSE)
Output:
1/128*(16*(b*x^4+a)^(3/4)*b^(5/4)*x^5+44*a*x*(b*x^4+a)^(3/4)*b^(1/4)-42*ar ctan(1/b^(1/4)/x*(b*x^4+a)^(1/4))*a^2+21*ln((b^(1/4)*x+(b*x^4+a)^(1/4))/(- b^(1/4)*x+(b*x^4+a)^(1/4)))*a^2)/b^(1/4)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.06 \[ \int \left (a+b x^4\right )^{7/4} \, dx=\frac {1}{32} \, {\left (4 \, b x^{5} + 11 \, a x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}} + \frac {21}{128} \, \left (\frac {a^{8}}{b}\right )^{\frac {1}{4}} \log \left (\frac {9261 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6} + \left (\frac {a^{8}}{b}\right )^{\frac {3}{4}} b x\right )}}{x}\right ) - \frac {21}{128} i \, \left (\frac {a^{8}}{b}\right )^{\frac {1}{4}} \log \left (\frac {9261 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6} + i \, \left (\frac {a^{8}}{b}\right )^{\frac {3}{4}} b x\right )}}{x}\right ) + \frac {21}{128} i \, \left (\frac {a^{8}}{b}\right )^{\frac {1}{4}} \log \left (\frac {9261 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6} - i \, \left (\frac {a^{8}}{b}\right )^{\frac {3}{4}} b x\right )}}{x}\right ) - \frac {21}{128} \, \left (\frac {a^{8}}{b}\right )^{\frac {1}{4}} \log \left (\frac {9261 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6} - \left (\frac {a^{8}}{b}\right )^{\frac {3}{4}} b x\right )}}{x}\right ) \] Input:
integrate((b*x^4+a)^(7/4),x, algorithm="fricas")
Output:
1/32*(4*b*x^5 + 11*a*x)*(b*x^4 + a)^(3/4) + 21/128*(a^8/b)^(1/4)*log(9261* ((b*x^4 + a)^(1/4)*a^6 + (a^8/b)^(3/4)*b*x)/x) - 21/128*I*(a^8/b)^(1/4)*lo g(9261*((b*x^4 + a)^(1/4)*a^6 + I*(a^8/b)^(3/4)*b*x)/x) + 21/128*I*(a^8/b) ^(1/4)*log(9261*((b*x^4 + a)^(1/4)*a^6 - I*(a^8/b)^(3/4)*b*x)/x) - 21/128* (a^8/b)^(1/4)*log(9261*((b*x^4 + a)^(1/4)*a^6 - (a^8/b)^(3/4)*b*x)/x)
Result contains complex when optimal does not.
Time = 1.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.39 \[ \int \left (a+b x^4\right )^{7/4} \, dx=\frac {a^{\frac {7}{4}} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**4+a)**(7/4),x)
Output:
a**(7/4)*x*gamma(1/4)*hyper((-7/4, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a) /(4*gamma(5/4))
Time = 0.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.49 \[ \int \left (a+b x^4\right )^{7/4} \, dx=-\frac {21}{128} \, a^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )} - \frac {\frac {7 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2} b}{x^{3}} - \frac {11 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2}}{x^{7}}}{32 \, {\left (b^{2} - \frac {2 \, {\left (b x^{4} + a\right )} b}{x^{4}} + \frac {{\left (b x^{4} + a\right )}^{2}}{x^{8}}\right )}} \] Input:
integrate((b*x^4+a)^(7/4),x, algorithm="maxima")
Output:
-21/128*a^2*(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)) - 1/32 *(7*(b*x^4 + a)^(3/4)*a^2*b/x^3 - 11*(b*x^4 + a)^(7/4)*a^2/x^7)/(b^2 - 2*( b*x^4 + a)*b/x^4 + (b*x^4 + a)^2/x^8)
\[ \int \left (a+b x^4\right )^{7/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {7}{4}} \,d x } \] Input:
integrate((b*x^4+a)^(7/4),x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(7/4), x)
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.39 \[ \int \left (a+b x^4\right )^{7/4} \, dx=\frac {x\,{\left (b\,x^4+a\right )}^{7/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (\frac {b\,x^4}{a}+1\right )}^{7/4}} \] Input:
int((a + b*x^4)^(7/4),x)
Output:
(x*(a + b*x^4)^(7/4)*hypergeom([-7/4, 1/4], 5/4, -(b*x^4)/a))/((b*x^4)/a + 1)^(7/4)
\[ \int \left (a+b x^4\right )^{7/4} \, dx=\frac {11 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a x}{32}+\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{5}}{8}+\frac {21 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) a^{2}}{32} \] Input:
int((b*x^4+a)^(7/4),x)
Output:
(11*(a + b*x**4)**(3/4)*a*x + 4*(a + b*x**4)**(3/4)*b*x**5 + 21*int((a + b *x**4)**(3/4)/(a + b*x**4),x)*a**2)/32