\(\int \frac {x^{10} (c+d x^4)}{(a+b x^4)^{13/4}} \, dx\) [172]

Optimal result

Integrand size = 22, antiderivative size = 152 \[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=-\frac {(b c-a d) x^7}{9 b^2 \left (a+b x^4\right )^{9/4}}-\frac {(7 b c-16 a d) x^3}{45 b^3 \left (a+b x^4\right )^{5/4}}+\frac {d x^3}{2 b^3 \sqrt [4]{a+b x^4}}-\frac {7 (2 b c-11 a d) \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{30 \sqrt {a} b^{7/2} \sqrt [4]{a+b x^4}} \] Output:

-1/9*(-a*d+b*c)*x^7/b^2/(b*x^4+a)^(9/4)-1/45*(-16*a*d+7*b*c)*x^3/b^3/(b*x^ 
4+a)^(5/4)+1/2*d*x^3/b^3/(b*x^4+a)^(1/4)-7/30*(-11*a*d+2*b*c)*(1+a/b/x^4)^ 
(1/4)*x*EllipticE(sin(1/2*arccot(b^(1/2)*x^2/a^(1/2))),2^(1/2))/a^(1/2)/b^ 
(7/2)/(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 = 10.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.77 \[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\frac {x^3 \left (a \left (77 a^2 d+12 b^2 x^4 \left (-c+d x^4\right )+a b \left (-14 c+66 d x^4\right )\right )+7 (2 b c-11 a d) \left (a+b x^4\right )^2 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {13}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{24 a b^3 \left (a+b x^4\right )^{9/4}} \] Input:

Integrate[(x^10*(c + d*x^4))/(a + b*x^4)^(13/4),x]
 

Output:

(x^3*(a*(77*a^2*d + 12*b^2*x^4*(-c + d*x^4) + a*b*(-14*c + 66*d*x^4)) + 7* 
(2*b*c - 11*a*d)*(a + b*x^4)^2*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[3/4 
, 13/4, 7/4, -((b*x^4)/a)]))/(24*a*b^3*(a + b*x^4)^(9/4))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {957, 817, 815, 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.

\(\displaystyle \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {x^{11} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(2 b c-11 a d) \int \frac {x^{10}}{\left (b x^4+a\right )^{9/4}}dx}{9 a b}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {x^{11} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(2 b c-11 a d) \left (\frac {7 \int \frac {x^6}{\left (b x^4+a\right )^{5/4}}dx}{5 b}-\frac {x^7}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 815

\(\displaystyle \frac {x^{11} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(2 b c-11 a d) \left (\frac {7 \left (\frac {x^3}{2 b \sqrt [4]{a+b x^4}}-\frac {3 a \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx}{2 b}\right )}{5 b}-\frac {x^7}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 813

\(\displaystyle \frac {x^{11} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(2 b c-11 a d) \left (\frac {7 \left (\frac {x^3}{2 b \sqrt [4]{a+b x^4}}-\frac {3 a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{2 b^2 \sqrt [4]{a+b x^4}}\right )}{5 b}-\frac {x^7}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {x^{11} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(2 b c-11 a d) \left (\frac {7 \left (\frac {3 a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{2 b^2 \sqrt [4]{a+b x^4}}+\frac {x^3}{2 b \sqrt [4]{a+b x^4}}\right )}{5 b}-\frac {x^7}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {x^{11} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(2 b c-11 a d) \left (\frac {7 \left (\frac {3 a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{4 b^2 \sqrt [4]{a+b x^4}}+\frac {x^3}{2 b \sqrt [4]{a+b x^4}}\right )}{5 b}-\frac {x^7}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {x^{11} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(2 b c-11 a d) \left (\frac {7 \left (\frac {3 \sqrt {a} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{2 b^{3/2} \sqrt [4]{a+b x^4}}+\frac {x^3}{2 b \sqrt [4]{a+b x^4}}\right )}{5 b}-\frac {x^7}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

Input:

Int[(x^10*(c + d*x^4))/(a + b*x^4)^(13/4),x]
 

Output:

((b*c - a*d)*x^11)/(9*a*b*(a + b*x^4)^(9/4)) - ((2*b*c - 11*a*d)*(-1/5*x^7 
/(b*(a + b*x^4)^(5/4)) + (7*(x^3/(2*b*(a + b*x^4)^(1/4)) + (3*Sqrt[a]*(1 + 
 a/(b*x^4))^(1/4)*x*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(2*b^(3 
/2)*(a + b*x^4)^(1/4))))/(5*b)))/(9*a*b)
 

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 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 815

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x^(m - 3)/(b*( 
m - 4)*(a + b*x^4)^(1/4)), x] - Simp[a*((m - 3)/(b*(m - 4)))   Int[x^(m - 4 
)/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a] && IGtQ[(m - 2 
)/4, 0]
 

rule 817

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*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 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]
 

rule 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

Maple [F]

\[\int \frac {x^{10} \left (d \,x^{4}+c \right )}{\left (b \,x^{4}+a \right )^{\frac {13}{4}}}d x\]

Input:

int(x^10*(d*x^4+c)/(b*x^4+a)^(13/4),x)
 

Output:

int(x^10*(d*x^4+c)/(b*x^4+a)^(13/4),x)
 

Fricas [F]

\[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{10}}{{\left (b x^{4} + a\right )}^{\frac {13}{4}}} \,d x } \] Input:

integrate(x^10*(d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="fricas")
 

Output:

integral((d*x^14 + c*x^10)*(b*x^4 + a)^(3/4)/(b^4*x^16 + 4*a*b^3*x^12 + 6* 
a^2*b^2*x^8 + 4*a^3*b*x^4 + a^4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 148.98 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.53 \[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\frac {c x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {11}{4}, \frac {13}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {13}{4}} \Gamma \left (\frac {15}{4}\right )} + \frac {d x^{15} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {13}{4}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {13}{4}} \Gamma \left (\frac {19}{4}\right )} \] Input:

integrate(x**10*(d*x**4+c)/(b*x**4+a)**(13/4),x)
 

Output:

c*x**11*gamma(11/4)*hyper((11/4, 13/4), (15/4,), b*x**4*exp_polar(I*pi)/a) 
/(4*a**(13/4)*gamma(15/4)) + d*x**15*gamma(15/4)*hyper((13/4, 15/4), (19/4 
,), b*x**4*exp_polar(I*pi)/a)/(4*a**(13/4)*gamma(19/4))
 

Maxima [F]

\[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{10}}{{\left (b x^{4} + a\right )}^{\frac {13}{4}}} \,d x } \] Input:

integrate(x^10*(d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)*x^10/(b*x^4 + a)^(13/4), x)
 

Giac [F]

\[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{10}}{{\left (b x^{4} + a\right )}^{\frac {13}{4}}} \,d x } \] Input:

integrate(x^10*(d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^10/(b*x^4 + a)^(13/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\int \frac {x^{10}\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{13/4}} \,d x \] Input:

int((x^10*(c + d*x^4))/(a + b*x^4)^(13/4),x)
 

Output:

int((x^10*(c + d*x^4))/(a + b*x^4)^(13/4), x)
 

Reduce [F]

\[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\left (\int \frac {x^{14}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{12}}d x \right ) d +\left (\int \frac {x^{10}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{12}}d x \right ) c \] Input:

int(x^10*(d*x^4+c)/(b*x^4+a)^(13/4),x)
 

Output:

int(x**14/((a + b*x**4)**(1/4)*a**3 + 3*(a + b*x**4)**(1/4)*a**2*b*x**4 + 
3*(a + b*x**4)**(1/4)*a*b**2*x**8 + (a + b*x**4)**(1/4)*b**3*x**12),x)*d + 
 int(x**10/((a + b*x**4)**(1/4)*a**3 + 3*(a + b*x**4)**(1/4)*a**2*b*x**4 + 
 3*(a + b*x**4)**(1/4)*a*b**2*x**8 + (a + b*x**4)**(1/4)*b**3*x**12),x)*c