\(\int \frac {x^8 (c+d x^4)}{(a+b x^4)^{7/4}} \, dx\) [140]

Optimal result

Integrand size = 22, antiderivative size = 151 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {a (b c-a d) x}{3 b^3 \left (a+b x^4\right )^{3/4}}+\frac {(6 b c-11 a d) x \sqrt [4]{a+b x^4}}{12 b^3}+\frac {d x^5 \sqrt [4]{a+b x^4}}{6 b^2}+\frac {5 \sqrt {a} (2 b c-3 a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 b^{5/2} \left (a+b x^4\right )^{3/4}} \] Output:

1/3*a*(-a*d+b*c)*x/b^3/(b*x^4+a)^(3/4)+1/12*(-11*a*d+6*b*c)*x*(b*x^4+a)^(1 
/4)/b^3+1/6*d*x^5*(b*x^4+a)^(1/4)/b^2+5/12*a^(1/2)*(-3*a*d+2*b*c)*(1+a/b/x 
^4)^(3/4)*x^3*InverseJacobiAM(1/2*arccot(b^(1/2)*x^2/a^(1/2)),2^(1/2))/b^( 
5/2)/(b*x^4+a)^(3/4)
 

Mathematica [C] (verified)

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

Time = 10.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.67 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {x \left (-15 a^2 d+a b \left (10 c-9 d x^4\right )+2 b^2 x^4 \left (3 c+d x^4\right )+5 a (-2 b c+3 a d) \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{12 b^3 \left (a+b x^4\right )^{3/4}} \] Input:

Integrate[(x^8*(c + d*x^4))/(a + b*x^4)^(7/4),x]
 

Output:

(x*(-15*a^2*d + a*b*(10*c - 9*d*x^4) + 2*b^2*x^4*(3*c + d*x^4) + 5*a*(-2*b 
*c + 3*a*d)*(1 + (b*x^4)/a)^(3/4)*Hypergeometric2F1[1/4, 3/4, 5/4, -((b*x^ 
4)/a)]))/(12*b^3*(a + b*x^4)^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.57 (sec) , antiderivative size = 163, 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, 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.

\(\displaystyle \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {x^9 (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {(2 b c-3 a d) \int \frac {x^8}{\left (b x^4+a\right )^{3/4}}dx}{a b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^9 (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {(2 b c-3 a d) \left (\frac {x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {5 a \int \frac {x^4}{\left (b x^4+a\right )^{3/4}}dx}{6 b}\right )}{a b}\)

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 768

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 229

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

Input:

Int[(x^8*(c + d*x^4))/(a + b*x^4)^(7/4),x]
 

Output:

((b*c - a*d)*x^9)/(3*a*b*(a + b*x^4)^(3/4)) - ((2*b*c - 3*a*d)*((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)))/(a*b)
 

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 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]
 

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]

\[\int \frac {x^{8} \left (d \,x^{4}+c \right )}{\left (b \,x^{4}+a \right )^{\frac {7}{4}}}d x\]

Input:

int(x^8*(d*x^4+c)/(b*x^4+a)^(7/4),x)
 

Output:

int(x^8*(d*x^4+c)/(b*x^4+a)^(7/4),x)
 

Fricas [F]

\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate(x^8*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="fricas")
 

Output:

integral((d*x^12 + c*x^8)*(b*x^4 + a)^(1/4)/(b^2*x^8 + 2*a*b*x^4 + a^2), x 
)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 20.76 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.53 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {c x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {7}{4}} \Gamma \left (\frac {13}{4}\right )} + \frac {d x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {7}{4}} \Gamma \left (\frac {17}{4}\right )} \] Input:

integrate(x**8*(d*x**4+c)/(b*x**4+a)**(7/4),x)
 

Output:

c*x**9*gamma(9/4)*hyper((7/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4* 
a**(7/4)*gamma(13/4)) + d*x**13*gamma(13/4)*hyper((7/4, 13/4), (17/4,), b* 
x**4*exp_polar(I*pi)/a)/(4*a**(7/4)*gamma(17/4))
 

Maxima [F]

\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate(x^8*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)*x^8/(b*x^4 + a)^(7/4), x)
 

Giac [F]

\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate(x^8*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^8/(b*x^4 + a)^(7/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\int \frac {x^8\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{7/4}} \,d x \] Input:

int((x^8*(c + d*x^4))/(a + b*x^4)^(7/4),x)
 

Output:

int((x^8*(c + d*x^4))/(a + b*x^4)^(7/4), x)
 

Reduce [F]

\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\left (\int \frac {x^{12}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) c \] Input:

int(x^8*(d*x^4+c)/(b*x^4+a)^(7/4),x)
 

Output:

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