3.11.71 \(\int x^{12} (a+b x^4)^{5/4} \, dx\) [1071]

3.11.71.1 Optimal result

Integrand size = 15, antiderivative size = 171 \[ \int x^{12} \left (a+b x^4\right )^{5/4} \, dx=\frac {5 a^4 x \sqrt [4]{a+b x^4}}{672 b^3}-\frac {a^3 x^5 \sqrt [4]{a+b x^4}}{336 b^2}+\frac {a^2 x^9 \sqrt [4]{a+b x^4}}{504 b}+\frac {5}{252} a x^{13} \sqrt [4]{a+b x^4}+\frac {1}{18} x^{13} \left (a+b x^4\right )^{5/4}+\frac {5 a^{9/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{672 b^{5/2} \left (a+b x^4\right )^{3/4}} \]

output

5/672*a^4*x*(b*x^4+a)^(1/4)/b^3-1/336*a^3*x^5*(b*x^4+a)^(1/4)/b^2+1/504*a^ 
2*x^9*(b*x^4+a)^(1/4)/b+5/252*a*x^13*(b*x^4+a)^(1/4)+1/18*x^13*(b*x^4+a)^( 
5/4)+5/672*a^(9/2)*(1+a/b/x^4)^(3/4)*x^3*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/ 
2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arcco 
t(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(5/2)/(b*x^4+a)^(3/4)
 

3.11.71.2 Mathematica [C] (verified)

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

Time = 9.47 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.52 \[ \int x^{12} \left (a+b x^4\right )^{5/4} \, dx=\frac {x \sqrt [4]{a+b x^4} \left (\left (a+b x^4\right )^2 \left (9 a^2-18 a b x^4+28 b^2 x^8\right )-\frac {9 a^4 \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt [4]{1+\frac {b x^4}{a}}}\right )}{504 b^3} \]

input

Integrate[x^12*(a + b*x^4)^(5/4),x]
 

output

(x*(a + b*x^4)^(1/4)*((a + b*x^4)^2*(9*a^2 - 18*a*b*x^4 + 28*b^2*x^8) - (9 
*a^4*Hypergeometric2F1[-5/4, 1/4, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^(1/4 
)))/(504*b^3)
 

3.11.71.3 Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {811, 811, 843, 843, 843, 768, 858, 807, 229}

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.

\(\displaystyle \int x^{12} \left (a+b x^4\right )^{5/4} \, dx\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {5}{18} a \int x^{12} \sqrt [4]{b x^4+a}dx+\frac {1}{18} x^{13} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {5}{18} a \left (\frac {1}{14} a \int \frac {x^{12}}{\left (b x^4+a\right )^{3/4}}dx+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}\right )+\frac {1}{18} x^{13} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {5}{18} a \left (\frac {1}{14} a \left (\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}-\frac {9 a \int \frac {x^8}{\left (b x^4+a\right )^{3/4}}dx}{10 b}\right )+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}\right )+\frac {1}{18} x^{13} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {5}{18} a \left (\frac {1}{14} a \left (\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}-\frac {9 a \left (\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {5 a \int \frac {x^4}{\left (b x^4+a\right )^{3/4}}dx}{6 b}\right )}{10 b}\right )+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}\right )+\frac {1}{18} x^{13} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {5}{18} a \left (\frac {1}{14} a \left (\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}-\frac {9 a \left (\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {5 a \left (\frac {x \sqrt [4]{a+b x^4}}{2 b}-\frac {a \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx}{2 b}\right )}{6 b}\right )}{10 b}\right )+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}\right )+\frac {1}{18} x^{13} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {5}{18} a \left (\frac {1}{14} a \left (\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}-\frac {9 a \left (\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {5 a \left (\frac {x \sqrt [4]{a+b x^4}}{2 b}-\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{2 b \left (a+b x^4\right )^{3/4}}\right )}{6 b}\right )}{10 b}\right )+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}\right )+\frac {1}{18} x^{13} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {5}{18} a \left (\frac {1}{14} a \left (\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}-\frac {9 a \left (\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {5 a \left (\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{2 b \left (a+b x^4\right )^{3/4}}+\frac {x \sqrt [4]{a+b x^4}}{2 b}\right )}{6 b}\right )}{10 b}\right )+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}\right )+\frac {1}{18} x^{13} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {5}{18} a \left (\frac {1}{14} a \left (\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}-\frac {9 a \left (\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {5 a \left (\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{4 b \left (a+b x^4\right )^{3/4}}+\frac {x \sqrt [4]{a+b x^4}}{2 b}\right )}{6 b}\right )}{10 b}\right )+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}\right )+\frac {1}{18} x^{13} \left (a+b x^4\right )^{5/4}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {5}{18} a \left (\frac {1}{14} a \left (\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}-\frac {9 a \left (\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {5 a \left (\frac {\sqrt {a} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{2 \sqrt {b} \left (a+b x^4\right )^{3/4}}+\frac {x \sqrt [4]{a+b x^4}}{2 b}\right )}{6 b}\right )}{10 b}\right )+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}\right )+\frac {1}{18} x^{13} \left (a+b x^4\right )^{5/4}\)

input

Int[x^12*(a + b*x^4)^(5/4),x]
 

output

(x^13*(a + b*x^4)^(5/4))/18 + (5*a*((x^13*(a + b*x^4)^(1/4))/14 + (a*((x^9 
*(a + b*x^4)^(1/4))/(10*b) - (9*a*((x^5*(a + b*x^4)^(1/4))/(6*b) - (5*a*(( 
x*(a + b*x^4)^(1/4))/(2*b) + (Sqrt[a]*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ 
ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(2*Sqrt[b]*(a + b*x^4)^(3/4))))/(6*b) 
))/(10*b)))/14))/18
 

3.11.71.3.1 Defintions of rubi rules used

rule 229

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

rule 768

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

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 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]
 

3.11.71.4 Maple [F]

\[\int x^{12} \left (b \,x^{4}+a \right )^{\frac {5}{4}}d x\]

input

int(x^12*(b*x^4+a)^(5/4),x)
 

output

int(x^12*(b*x^4+a)^(5/4),x)
 

3.11.71.5 Fricas [F]

\[ \int x^{12} \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{12} \,d x } \]

input

integrate(x^12*(b*x^4+a)^(5/4),x, algorithm="fricas")
 

output

integral((b*x^16 + a*x^12)*(b*x^4 + a)^(1/4), x)
 

3.11.71.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.23 \[ \int x^{12} \left (a+b x^4\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {17}{4}\right )} \]

input

integrate(x**12*(b*x**4+a)**(5/4),x)
 

output

a**(5/4)*x**13*gamma(13/4)*hyper((-5/4, 13/4), (17/4,), b*x**4*exp_polar(I 
*pi)/a)/(4*gamma(17/4))
 

3.11.71.7 Maxima [F]

\[ \int x^{12} \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{12} \,d x } \]

input

integrate(x^12*(b*x^4+a)^(5/4),x, algorithm="maxima")
 

output

integrate((b*x^4 + a)^(5/4)*x^12, x)
 

3.11.71.8 Giac [F]

\[ \int x^{12} \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{12} \,d x } \]

input

integrate(x^12*(b*x^4+a)^(5/4),x, algorithm="giac")
 

output

integrate((b*x^4 + a)^(5/4)*x^12, x)
 

3.11.71.9 Mupad [F(-1)]

Timed out. \[ \int x^{12} \left (a+b x^4\right )^{5/4} \, dx=\int x^{12}\,{\left (b\,x^4+a\right )}^{5/4} \,d x \]

input

int(x^12*(a + b*x^4)^(5/4),x)
 

output

int(x^12*(a + b*x^4)^(5/4), x)