\(\int x^{10} (a+b x^4)^{3/4} \, dx\) [495]

Optimal result

Integrand size = 15, antiderivative size = 150 \[ \int x^{10} \left (a+b x^4\right )^{3/4} \, dx=\frac {3 a^3 x^3}{80 b^2 \sqrt [4]{a+b x^4}}-\frac {a^2 x^3 \left (a+b x^4\right )^{3/4}}{40 b^2}+\frac {3 a x^7 \left (a+b x^4\right )^{3/4}}{140 b}+\frac {1}{14} x^{11} \left (a+b x^4\right )^{3/4}+\frac {3 a^{7/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{80 b^{5/2} \sqrt [4]{a+b x^4}} \] Output:

3/80*a^3*x^3/b^2/(b*x^4+a)^(1/4)-1/40*a^2*x^3*(b*x^4+a)^(3/4)/b^2+3/140*a* 
x^7*(b*x^4+a)^(3/4)/b+1/14*x^11*(b*x^4+a)^(3/4)+3/80*a^(7/2)*(1+a/b/x^4)^( 
1/4)*x*EllipticE(sin(1/2*arccot(b^(1/2)*x^2/a^(1/2))),2^(1/2))/b^(5/2)/(b* 
x^4+a)^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.63 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.64 \[ \int x^{10} \left (a+b x^4\right )^{3/4} \, dx=\frac {x^3 \left (a+b x^4\right )^{3/4} \left (\left (1+\frac {b x^4}{a}\right )^{3/4} \left (-7 a^2+3 a b x^4+10 b^2 x^8\right )+7 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{140 b^2 \left (1+\frac {b x^4}{a}\right )^{3/4}} \] Input:

Integrate[x^10*(a + b*x^4)^(3/4),x]
 

Output:

(x^3*(a + b*x^4)^(3/4)*((1 + (b*x^4)/a)^(3/4)*(-7*a^2 + 3*a*b*x^4 + 10*b^2 
*x^8) + 7*a^2*Hypergeometric2F1[-3/4, 3/4, 7/4, -((b*x^4)/a)]))/(140*b^2*( 
1 + (b*x^4)/a)^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.51 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {811, 843, 843, 839, 813, 858, 807, 212}

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.

\(\displaystyle \int x^{10} \left (a+b x^4\right )^{3/4} \, dx\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {3}{14} a \int \frac {x^{10}}{\sqrt [4]{b x^4+a}}dx+\frac {1}{14} x^{11} \left (a+b x^4\right )^{3/4}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {3}{14} a \left (\frac {x^7 \left (a+b x^4\right )^{3/4}}{10 b}-\frac {7 a \int \frac {x^6}{\sqrt [4]{b x^4+a}}dx}{10 b}\right )+\frac {1}{14} x^{11} \left (a+b x^4\right )^{3/4}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {3}{14} a \left (\frac {x^7 \left (a+b x^4\right )^{3/4}}{10 b}-\frac {7 a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \int \frac {x^2}{\sqrt [4]{b x^4+a}}dx}{2 b}\right )}{10 b}\right )+\frac {1}{14} x^{11} \left (a+b x^4\right )^{3/4}\)

\(\Big \downarrow \) 839

\(\displaystyle \frac {3}{14} a \left (\frac {x^7 \left (a+b x^4\right )^{3/4}}{10 b}-\frac {7 a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \left (\frac {x^3}{2 \sqrt [4]{a+b x^4}}-\frac {1}{2} a \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx\right )}{2 b}\right )}{10 b}\right )+\frac {1}{14} x^{11} \left (a+b x^4\right )^{3/4}\)

\(\Big \downarrow \) 813

\(\displaystyle \frac {3}{14} a \left (\frac {x^7 \left (a+b x^4\right )^{3/4}}{10 b}-\frac {7 a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \left (\frac {x^3}{2 \sqrt [4]{a+b x^4}}-\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{2 b \sqrt [4]{a+b x^4}}\right )}{2 b}\right )}{10 b}\right )+\frac {1}{14} x^{11} \left (a+b x^4\right )^{3/4}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {3}{14} a \left (\frac {x^7 \left (a+b x^4\right )^{3/4}}{10 b}-\frac {7 a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \left (\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{2 b \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )}{2 b}\right )}{10 b}\right )+\frac {1}{14} x^{11} \left (a+b x^4\right )^{3/4}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {3}{14} a \left (\frac {x^7 \left (a+b x^4\right )^{3/4}}{10 b}-\frac {7 a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \left (\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{4 b \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )}{2 b}\right )}{10 b}\right )+\frac {1}{14} x^{11} \left (a+b x^4\right )^{3/4}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {3}{14} a \left (\frac {x^7 \left (a+b x^4\right )^{3/4}}{10 b}-\frac {7 a \left (\frac {x^3 \left (a+b x^4\right )^{3/4}}{6 b}-\frac {a \left (\frac {\sqrt {a} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{2 \sqrt {b} \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )}{2 b}\right )}{10 b}\right )+\frac {1}{14} x^{11} \left (a+b x^4\right )^{3/4}\)

Input:

Int[x^10*(a + b*x^4)^(3/4),x]
 

Output:

(x^11*(a + b*x^4)^(3/4))/14 + (3*a*((x^7*(a + b*x^4)^(3/4))/(10*b) - (7*a* 
((x^3*(a + b*x^4)^(3/4))/(6*b) - (a*(x^3/(2*(a + b*x^4)^(1/4)) + (Sqrt[a]* 
(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(2* 
Sqrt[b]*(a + b*x^4)^(1/4))))/(2*b)))/(10*b)))/14
 

Defintions of rubi rules used

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

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 813

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) 
)^(1/4)/(b*(a + b*x^4)^(1/4)))   Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] 
/; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 839

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 
)^(1/4)), x] - Simp[a/2   Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} 
, x] && PosQ[b/a]
 

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 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

\[\int x^{10} \left (b \,x^{4}+a \right )^{\frac {3}{4}}d x\]

Input:

int(x^10*(b*x^4+a)^(3/4),x)
 

Output:

int(x^10*(b*x^4+a)^(3/4),x)
 

Fricas [F]

\[ \int x^{10} \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{10} \,d x } \] Input:

integrate(x^10*(b*x^4+a)^(3/4),x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(3/4)*x^10, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.26 \[ \int x^{10} \left (a+b x^4\right )^{3/4} \, dx=\frac {a^{\frac {3}{4}} x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {15}{4}\right )} \] Input:

integrate(x**10*(b*x**4+a)**(3/4),x)
 

Output:

a**(3/4)*x**11*gamma(11/4)*hyper((-3/4, 11/4), (15/4,), b*x**4*exp_polar(I 
*pi)/a)/(4*gamma(15/4))
 

Maxima [F]

\[ \int x^{10} \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{10} \,d x } \] Input:

integrate(x^10*(b*x^4+a)^(3/4),x, algorithm="maxima")
 

Output:

integrate((b*x^4 + a)^(3/4)*x^10, x)
 

Giac [F]

\[ \int x^{10} \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{10} \,d x } \] Input:

integrate(x^10*(b*x^4+a)^(3/4),x, algorithm="giac")
 

Output:

integrate((b*x^4 + a)^(3/4)*x^10, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^{10} \left (a+b x^4\right )^{3/4} \, dx=\int x^{10}\,{\left (b\,x^4+a\right )}^{3/4} \,d x \] Input:

int(x^10*(a + b*x^4)^(3/4),x)
 

Output:

int(x^10*(a + b*x^4)^(3/4), x)
 

Reduce [F]

\[ \int x^{10} \left (a+b x^4\right )^{3/4} \, dx=\frac {-7 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{2} x^{3}+6 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a b \,x^{7}+20 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{2} x^{11}+21 \left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) a^{3}}{280 b^{2}} \] Input:

int(x^10*(b*x^4+a)^(3/4),x)
 

Output:

( - 7*(a + b*x**4)**(3/4)*a**2*x**3 + 6*(a + b*x**4)**(3/4)*a*b*x**7 + 20* 
(a + b*x**4)**(3/4)*b**2*x**11 + 21*int(((a + b*x**4)**(3/4)*x**2)/(a + b* 
x**4),x)*a**3)/(280*b**2)