\(\int x^{10} (a+b x^4)^{5/4} \, dx\) [517]

Optimal result

Integrand size = 15, antiderivative size = 148 \[ \int x^{10} \left (a+b x^4\right )^{5/4} \, dx=-\frac {35 a^3 x^3 \sqrt [4]{a+b x^4}}{6144 b^2}+\frac {5 a^2 x^7 \sqrt [4]{a+b x^4}}{1536 b}+\frac {5}{192} a x^{11} \sqrt [4]{a+b x^4}+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}-\frac {35 a^4 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{11/4}}+\frac {35 a^4 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{11/4}} \] Output:

-35/6144*a^3*x^3*(b*x^4+a)^(1/4)/b^2+5/1536*a^2*x^7*(b*x^4+a)^(1/4)/b+5/19 
2*a*x^11*(b*x^4+a)^(1/4)+1/16*x^11*(b*x^4+a)^(5/4)-35/4096*a^4*arctan(b^(1 
/4)*x/(b*x^4+a)^(1/4))/b^(11/4)+35/4096*a^4*arctanh(b^(1/4)*x/(b*x^4+a)^(1 
/4))/b^(11/4)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.75 \[ \int x^{10} \left (a+b x^4\right )^{5/4} \, dx=\frac {2 b^{3/4} x^3 \sqrt [4]{a+b x^4} \left (-35 a^3+20 a^2 b x^4+544 a b^2 x^8+384 b^3 x^{12}\right )-105 a^4 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+105 a^4 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{12288 b^{11/4}} \] Input:

Integrate[x^10*(a + b*x^4)^(5/4),x]
 

Output:

(2*b^(3/4)*x^3*(a + b*x^4)^(1/4)*(-35*a^3 + 20*a^2*b*x^4 + 544*a*b^2*x^8 + 
 384*b^3*x^12) - 105*a^4*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] + 105*a^4*A 
rcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(12288*b^(11/4))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {811, 811, 843, 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.

\(\displaystyle \int x^{10} \left (a+b x^4\right )^{5/4} \, dx\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {5}{16} a \int x^{10} \sqrt [4]{b x^4+a}dx+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {5}{16} a \left (\frac {1}{12} a \int \frac {x^{10}}{\left (b x^4+a\right )^{3/4}}dx+\frac {1}{12} x^{11} \sqrt [4]{a+b x^4}\right )+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {5}{16} a \left (\frac {1}{12} a \left (\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \int \frac {x^6}{\left (b x^4+a\right )^{3/4}}dx}{8 b}\right )+\frac {1}{12} x^{11} \sqrt [4]{a+b x^4}\right )+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {5}{16} a \left (\frac {1}{12} a \left (\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \int \frac {x^2}{\left (b x^4+a\right )^{3/4}}dx}{4 b}\right )}{8 b}\right )+\frac {1}{12} x^{11} \sqrt [4]{a+b x^4}\right )+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {5}{16} a \left (\frac {1}{12} a \left (\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^4+a} \left (1-\frac {b x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\right )}{8 b}\right )+\frac {1}{12} x^{11} \sqrt [4]{a+b x^4}\right )+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {5}{16} a \left (\frac {1}{12} a \left (\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}\right )}{4 b}\right )}{8 b}\right )+\frac {1}{12} x^{11} \sqrt [4]{a+b x^4}\right )+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {5}{16} a \left (\frac {1}{12} a \left (\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )}{4 b}\right )}{8 b}\right )+\frac {1}{12} x^{11} \sqrt [4]{a+b x^4}\right )+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5}{16} a \left (\frac {1}{12} a \left (\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )}{4 b}\right )}{8 b}\right )+\frac {1}{12} x^{11} \sqrt [4]{a+b x^4}\right )+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}\)

Input:

Int[x^10*(a + b*x^4)^(5/4),x]
 

Output:

(x^11*(a + b*x^4)^(5/4))/16 + (5*a*((x^11*(a + b*x^4)^(1/4))/12 + (a*((x^7 
*(a + b*x^4)^(1/4))/(8*b) - (7*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*b)))/12))/16
 

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 = 0.74 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(\frac {\frac {35 \ln \left (\frac {b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a^{4}}{8192}+\frac {35 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a^{4}}{4096}-\frac {35 x^{3} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-\frac {384 b^{\frac {15}{4}} x^{12}}{35}-\frac {544 a \,b^{\frac {11}{4}} x^{8}}{35}-\frac {4 a^{2} b^{\frac {7}{4}} x^{4}}{7}+a^{3} b^{\frac {3}{4}}\right )}{6144}}{b^{\frac {11}{4}}}\)	\(118\)

Input:

int(x^10*(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)
 

Output:

35/4096*(1/2*ln((b^(1/4)*x+(b*x^4+a)^(1/4))/(-b^(1/4)*x+(b*x^4+a)^(1/4)))* 
a^4+arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4))*a^4-2/3*x^3*(b*x^4+a)^(1/4)*(-384/ 
35*b^(15/4)*x^12-544/35*a*b^(11/4)*x^8-4/7*a^2*b^(7/4)*x^4+a^3*b^(3/4)))/b 
^(11/4)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.67 \[ \int x^{10} \left (a+b x^4\right )^{5/4} \, dx=\frac {105 \, \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{2} \log \left (\frac {35 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{4} + \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{3} x\right )}}{x}\right ) + 105 i \, \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{2} \log \left (\frac {35 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{4} + i \, \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{3} x\right )}}{x}\right ) - 105 i \, \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{2} \log \left (\frac {35 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{4} - i \, \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{3} x\right )}}{x}\right ) - 105 \, \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{2} \log \left (\frac {35 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{4} - \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{3} x\right )}}{x}\right ) + 4 \, {\left (384 \, b^{3} x^{15} + 544 \, a b^{2} x^{11} + 20 \, a^{2} b x^{7} - 35 \, a^{3} x^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{24576 \, b^{2}} \] Input:

integrate(x^10*(b*x^4+a)^(5/4),x, algorithm="fricas")
 

Output:

1/24576*(105*(a^16/b^11)^(1/4)*b^2*log(35*((b*x^4 + a)^(1/4)*a^4 + (a^16/b 
^11)^(1/4)*b^3*x)/x) + 105*I*(a^16/b^11)^(1/4)*b^2*log(35*((b*x^4 + a)^(1/ 
4)*a^4 + I*(a^16/b^11)^(1/4)*b^3*x)/x) - 105*I*(a^16/b^11)^(1/4)*b^2*log(3 
5*((b*x^4 + a)^(1/4)*a^4 - I*(a^16/b^11)^(1/4)*b^3*x)/x) - 105*(a^16/b^11) 
^(1/4)*b^2*log(35*((b*x^4 + a)^(1/4)*a^4 - (a^16/b^11)^(1/4)*b^3*x)/x) + 4 
*(384*b^3*x^15 + 544*a*b^2*x^11 + 20*a^2*b*x^7 - 35*a^3*x^3)*(b*x^4 + a)^( 
1/4))/b^2
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 10.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.26 \[ \int x^{10} \left (a+b x^4\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {15}{4}\right )} \] Input:

integrate(x**10*(b*x**4+a)**(5/4),x)
 

Output:

a**(5/4)*x**11*gamma(11/4)*hyper((-5/4, 11/4), (15/4,), b*x**4*exp_polar(I 
*pi)/a)/(4*gamma(15/4))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.55 \[ \int x^{10} \left (a+b x^4\right )^{5/4} \, dx=-\frac {\frac {105 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{4} b^{3}}{x} - \frac {399 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{4} b^{2}}{x^{5}} - \frac {125 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{4} b}{x^{9}} + \frac {35 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{4}}{x^{13}}}{6144 \, {\left (b^{6} - \frac {4 \, {\left (b x^{4} + a\right )} b^{5}}{x^{4}} + \frac {6 \, {\left (b x^{4} + a\right )}^{2} b^{4}}{x^{8}} - \frac {4 \, {\left (b x^{4} + a\right )}^{3} b^{3}}{x^{12}} + \frac {{\left (b x^{4} + a\right )}^{4} b^{2}}{x^{16}}\right )}} + \frac {35 \, {\left (\frac {2 \, a^{4} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {3}{4}}} - \frac {a^{4} \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {3}{4}}}\right )}}{8192 \, b^{2}} \] Input:

integrate(x^10*(b*x^4+a)^(5/4),x, algorithm="maxima")
 

Output:

-1/6144*(105*(b*x^4 + a)^(1/4)*a^4*b^3/x - 399*(b*x^4 + a)^(5/4)*a^4*b^2/x 
^5 - 125*(b*x^4 + a)^(9/4)*a^4*b/x^9 + 35*(b*x^4 + a)^(13/4)*a^4/x^13)/(b^ 
6 - 4*(b*x^4 + a)*b^5/x^4 + 6*(b*x^4 + a)^2*b^4/x^8 - 4*(b*x^4 + a)^3*b^3/ 
x^12 + (b*x^4 + a)^4*b^2/x^16) + 35/8192*(2*a^4*arctan((b*x^4 + a)^(1/4)/( 
b^(1/4)*x))/b^(3/4) - a^4*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^2
 

Giac [F]

\[ \int x^{10} \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{10} \,d x } \] Input:

integrate(x^10*(b*x^4+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((b*x^4 + a)^(5/4)*x^10, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^{10} \left (a+b x^4\right )^{5/4} \, dx=\int x^{10}\,{\left (b\,x^4+a\right )}^{5/4} \,d x \] Input:

int(x^10*(a + b*x^4)^(5/4),x)
 

Output:

int(x^10*(a + b*x^4)^(5/4), x)
 

Reduce [F]

\[ \int x^{10} \left (a+b x^4\right )^{5/4} \, dx=\frac {-35 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} x^{3}+20 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{7}+544 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{11}+384 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{15}+105 \left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) a^{4}}{6144 b^{2}} \] Input:

int(x^10*(b*x^4+a)^(5/4),x)
 

Output:

( - 35*(a + b*x**4)**(1/4)*a**3*x**3 + 20*(a + b*x**4)**(1/4)*a**2*b*x**7 
+ 544*(a + b*x**4)**(1/4)*a*b**2*x**11 + 384*(a + b*x**4)**(1/4)*b**3*x**1 
5 + 105*int(((a + b*x**4)**(1/4)*x**2)/(a + b*x**4),x)*a**4)/(6144*b**2)