\(\int x^6 (a+b x^4)^{5/4} \, dx\) [518]

Optimal result

Integrand size = 15, antiderivative size = 124 \[ \int x^6 \left (a+b x^4\right )^{5/4} \, dx=\frac {5 a^2 x^3 \sqrt [4]{a+b x^4}}{384 b}+\frac {5}{96} a x^7 \sqrt [4]{a+b x^4}+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}+\frac {5 a^3 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{7/4}}-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{256 b^{7/4}} \] Output:

5/384*a^2*x^3*(b*x^4+a)^(1/4)/b+5/96*a*x^7*(b*x^4+a)^(1/4)+1/12*x^7*(b*x^4 
+a)^(5/4)+5/256*a^3*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(7/4)-5/256*a^3*ar 
ctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(7/4)
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int x^6 \left (a+b x^4\right )^{5/4} \, dx=\frac {2 b^{3/4} x^3 \sqrt [4]{a+b x^4} \left (5 a^2+52 a b x^4+32 b^2 x^8\right )+15 a^3 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-15 a^3 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{768 b^{7/4}} \] Input:

Integrate[x^6*(a + b*x^4)^(5/4),x]
 

Output:

(2*b^(3/4)*x^3*(a + b*x^4)^(1/4)*(5*a^2 + 52*a*b*x^4 + 32*b^2*x^8) + 15*a^ 
3*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] - 15*a^3*ArcTanh[(b^(1/4)*x)/(a + 
b*x^4)^(1/4)])/(768*b^(7/4))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {811, 811, 843, 854, 827, 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.

\(\displaystyle \int x^6 \left (a+b x^4\right )^{5/4} \, dx\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {5}{12} a \int x^6 \sqrt [4]{b x^4+a}dx+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {5}{12} a \left (\frac {1}{8} a \int \frac {x^6}{\left (b x^4+a\right )^{3/4}}dx+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\right )+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {5}{12} a \left (\frac {1}{8} a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \int \frac {x^2}{\left (b x^4+a\right )^{3/4}}dx}{4 b}\right )+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\right )+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {5}{12} a \left (\frac {1}{8} a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^4+a} \left (1-\frac {b x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\right )+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\right )+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {5}{12} a \left (\frac {1}{8} a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}\right )}{4 b}\right )+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\right )+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {5}{12} a \left (\frac {1}{8} a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )}{4 b}\right )+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\right )+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5}{12} a \left (\frac {1}{8} a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )}{4 b}\right )+\frac {1}{8} x^7 \sqrt [4]{a+b x^4}\right )+\frac {1}{12} x^7 \left (a+b x^4\right )^{5/4}\)

Input:

Int[x^6*(a + b*x^4)^(5/4),x]
 

Output:

(x^7*(a + b*x^4)^(5/4))/12 + (5*a*((x^7*(a + b*x^4)^(1/4))/8 + (a*((x^3*(a 
 + b*x^4)^(1/4))/(4*b) - (3*a*(-1/2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/ 
b^(3/4) + ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(3/4))))/(4*b)))/8)) 
/12
 

Defintions of rubi rules used

rule 216

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])
 

rule 219

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])
 

rule 811

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 
))   Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I 
GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m 
, p, x]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]
 

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(\frac {128 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{\frac {11}{4}} x^{11}+208 a \,b^{\frac {7}{4}} x^{7} \left (b \,x^{4}+a \right )^{\frac {1}{4}}+20 a^{2} x^{3} \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{\frac {3}{4}}-15 \ln \left (\frac {b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a^{3}-30 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a^{3}}{1536 b^{\frac {7}{4}}}\)	\(124\)

Input:

int(x^6*(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)
 

Output:

1/1536*(128*(b*x^4+a)^(1/4)*b^(11/4)*x^11+208*a*b^(7/4)*x^7*(b*x^4+a)^(1/4 
)+20*a^2*x^3*(b*x^4+a)^(1/4)*b^(3/4)-15*ln((b^(1/4)*x+(b*x^4+a)^(1/4))/(-b 
^(1/4)*x+(b*x^4+a)^(1/4)))*a^3-30*arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4))*a^3) 
/b^(7/4)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.84 \[ \int x^6 \left (a+b x^4\right )^{5/4} \, dx=-\frac {15 \, \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b \log \left (\frac {5 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} + \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b^{2} x\right )}}{x}\right ) + 15 i \, \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b \log \left (\frac {5 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} + i \, \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b^{2} x\right )}}{x}\right ) - 15 i \, \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b \log \left (\frac {5 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} - i \, \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b^{2} x\right )}}{x}\right ) - 15 \, \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b \log \left (\frac {5 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} - \left (\frac {a^{12}}{b^{7}}\right )^{\frac {1}{4}} b^{2} x\right )}}{x}\right ) - 4 \, {\left (32 \, b^{2} x^{11} + 52 \, a b x^{7} + 5 \, a^{2} x^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{1536 \, b} \] Input:

integrate(x^6*(b*x^4+a)^(5/4),x, algorithm="fricas")
 

Output:

-1/1536*(15*(a^12/b^7)^(1/4)*b*log(5*((b*x^4 + a)^(1/4)*a^3 + (a^12/b^7)^( 
1/4)*b^2*x)/x) + 15*I*(a^12/b^7)^(1/4)*b*log(5*((b*x^4 + a)^(1/4)*a^3 + I* 
(a^12/b^7)^(1/4)*b^2*x)/x) - 15*I*(a^12/b^7)^(1/4)*b*log(5*((b*x^4 + a)^(1 
/4)*a^3 - I*(a^12/b^7)^(1/4)*b^2*x)/x) - 15*(a^12/b^7)^(1/4)*b*log(5*((b*x 
^4 + a)^(1/4)*a^3 - (a^12/b^7)^(1/4)*b^2*x)/x) - 4*(32*b^2*x^11 + 52*a*b*x 
^7 + 5*a^2*x^3)*(b*x^4 + a)^(1/4))/b
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.37 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.31 \[ \int x^6 \left (a+b x^4\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \] Input:

integrate(x**6*(b*x**4+a)**(5/4),x)
 

Output:

a**(5/4)*x**7*gamma(7/4)*hyper((-5/4, 7/4), (11/4,), b*x**4*exp_polar(I*pi 
)/a)/(4*gamma(11/4))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.54 \[ \int x^6 \left (a+b x^4\right )^{5/4} \, dx=\frac {\frac {15 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3} b^{2}}{x} - \frac {42 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{3} b}{x^{5}} - \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{3}}{x^{9}}}{384 \, {\left (b^{4} - \frac {3 \, {\left (b x^{4} + a\right )} b^{3}}{x^{4}} + \frac {3 \, {\left (b x^{4} + a\right )}^{2} b^{2}}{x^{8}} - \frac {{\left (b x^{4} + a\right )}^{3} b}{x^{12}}\right )}} - \frac {5 \, {\left (\frac {2 \, a^{3} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {3}{4}}} - \frac {a^{3} \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {3}{4}}}\right )}}{512 \, b} \] Input:

integrate(x^6*(b*x^4+a)^(5/4),x, algorithm="maxima")
 

Output:

1/384*(15*(b*x^4 + a)^(1/4)*a^3*b^2/x - 42*(b*x^4 + a)^(5/4)*a^3*b/x^5 - 5 
*(b*x^4 + a)^(9/4)*a^3/x^9)/(b^4 - 3*(b*x^4 + a)*b^3/x^4 + 3*(b*x^4 + a)^2 
*b^2/x^8 - (b*x^4 + a)^3*b/x^12) - 5/512*(2*a^3*arctan((b*x^4 + a)^(1/4)/( 
b^(1/4)*x))/b^(3/4) - a^3*log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + 
(b*x^4 + a)^(1/4)/x))/b^(3/4))/b
 

Giac [F]

\[ \int x^6 \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{6} \,d x } \] Input:

integrate(x^6*(b*x^4+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((b*x^4 + a)^(5/4)*x^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^6 \left (a+b x^4\right )^{5/4} \, dx=\int x^6\,{\left (b\,x^4+a\right )}^{5/4} \,d x \] Input:

int(x^6*(a + b*x^4)^(5/4),x)
 

Output:

int(x^6*(a + b*x^4)^(5/4), x)
 

Reduce [F]

\[ \int x^6 \left (a+b x^4\right )^{5/4} \, dx=\frac {5 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} x^{3}+52 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{7}+32 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{11}-15 \left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) a^{3}}{384 b} \] Input:

int(x^6*(b*x^4+a)^(5/4),x)
 

Output:

(5*(a + b*x**4)**(1/4)*a**2*x**3 + 52*(a + b*x**4)**(1/4)*a*b*x**7 + 32*(a 
 + b*x**4)**(1/4)*b**2*x**11 - 15*int(((a + b*x**4)**(1/4)*x**2)/(a + b*x* 
*4),x)*a**3)/(384*b)