\(\int \frac {c+d x^4}{x^8 (a+b x^4)^{9/4}} \, dx\) [154]

Optimal result

Integrand size = 22, antiderivative size = 113 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{9/4}} \, dx=-\frac {c}{7 a x^7 \left (a+b x^4\right )^{5/4}}-\frac {12 b c-7 a d}{35 a^2 x^3 \left (a+b x^4\right )^{5/4}}-\frac {8 (12 b c-7 a d)}{35 a^3 x^3 \sqrt [4]{a+b x^4}}+\frac {32 (12 b c-7 a d) \left (a+b x^4\right )^{3/4}}{105 a^4 x^3} \] Output:

-1/7*c/a/x^7/(b*x^4+a)^(5/4)-1/35*(-7*a*d+12*b*c)/a^2/x^3/(b*x^4+a)^(5/4)- 
8/35*(-7*a*d+12*b*c)/a^3/x^3/(b*x^4+a)^(1/4)+32/105*(-7*a*d+12*b*c)*(b*x^4 
+a)^(3/4)/a^4/x^3
 

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.76 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{9/4}} \, dx=\frac {-15 a^3 c+60 a^2 b c x^4-35 a^3 d x^4+480 a b^2 c x^8-280 a^2 b d x^8+384 b^3 c x^{12}-224 a b^2 d x^{12}}{105 a^4 x^7 \left (a+b x^4\right )^{5/4}} \] Input:

Integrate[(c + d*x^4)/(x^8*(a + b*x^4)^(9/4)),x]
 

Output:

(-15*a^3*c + 60*a^2*b*c*x^4 - 35*a^3*d*x^4 + 480*a*b^2*c*x^8 - 280*a^2*b*d 
*x^8 + 384*b^3*c*x^12 - 224*a*b^2*d*x^12)/(105*a^4*x^7*(a + b*x^4)^(5/4))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {955, 803, 749, 746}

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.

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

\(\Big \downarrow \) 955

\(\displaystyle -\frac {(12 b c-7 a d) \int \frac {1}{x^4 \left (b x^4+a\right )^{9/4}}dx}{7 a}-\frac {c}{7 a x^7 \left (a+b x^4\right )^{5/4}}\)

\(\Big \downarrow \) 803

\(\displaystyle -\frac {(12 b c-7 a d) \left (-\frac {8 b \int \frac {1}{\left (b x^4+a\right )^{9/4}}dx}{3 a}-\frac {1}{3 a x^3 \left (a+b x^4\right )^{5/4}}\right )}{7 a}-\frac {c}{7 a x^7 \left (a+b x^4\right )^{5/4}}\)

\(\Big \downarrow \) 749

\(\displaystyle -\frac {(12 b c-7 a d) \left (-\frac {8 b \left (\frac {4 \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx}{5 a}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{3 a}-\frac {1}{3 a x^3 \left (a+b x^4\right )^{5/4}}\right )}{7 a}-\frac {c}{7 a x^7 \left (a+b x^4\right )^{5/4}}\)

\(\Big \downarrow \) 746

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

Input:

Int[(c + d*x^4)/(x^8*(a + b*x^4)^(9/4)),x]
 

Output:

-1/7*c/(a*x^7*(a + b*x^4)^(5/4)) - ((12*b*c - 7*a*d)*(-1/3*1/(a*x^3*(a + b 
*x^4)^(5/4)) - (8*b*(x/(5*a*(a + b*x^4)^(5/4)) + (4*x)/(5*a^2*(a + b*x^4)^ 
(1/4))))/(3*a)))/(7*a)
 

Defintions of rubi rules used

rule 746

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1) 
/a), x] /; FreeQ[{a, b, n, p}, x] && EqQ[1/n + p + 1, 0]
 

rule 749

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 
 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1))   Int[(a + b*x^ 
n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte 
gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
 

rule 803

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( 
a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 
)))   Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I 
LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
 

rule 955

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

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67

method	result	size
pseudoelliptic	\(\frac {\left (-35 d \,x^{4}-15 c \right ) a^{3}+60 b \,x^{4} \left (-\frac {14 d \,x^{4}}{3}+c \right ) a^{2}+480 b^{2} x^{8} \left (-\frac {7 d \,x^{4}}{15}+c \right ) a +384 b^{3} c \,x^{12}}{105 \left (b \,x^{4}+a \right )^{\frac {5}{4}} x^{7} a^{4}}\)	\(76\)
gosper	\(-\frac {224 a \,b^{2} d \,x^{12}-384 b^{3} c \,x^{12}+280 a^{2} b d \,x^{8}-480 a \,b^{2} c \,x^{8}+35 a^{3} d \,x^{4}-60 a^{2} b c \,x^{4}+15 c \,a^{3}}{105 x^{7} \left (b \,x^{4}+a \right )^{\frac {5}{4}} a^{4}}\)	\(83\)
trager	\(-\frac {224 a \,b^{2} d \,x^{12}-384 b^{3} c \,x^{12}+280 a^{2} b d \,x^{8}-480 a \,b^{2} c \,x^{8}+35 a^{3} d \,x^{4}-60 a^{2} b c \,x^{4}+15 c \,a^{3}}{105 x^{7} \left (b \,x^{4}+a \right )^{\frac {5}{4}} a^{4}}\)	\(83\)
orering	\(-\frac {224 a \,b^{2} d \,x^{12}-384 b^{3} c \,x^{12}+280 a^{2} b d \,x^{8}-480 a \,b^{2} c \,x^{8}+35 a^{3} d \,x^{4}-60 a^{2} b c \,x^{4}+15 c \,a^{3}}{105 x^{7} \left (b \,x^{4}+a \right )^{\frac {5}{4}} a^{4}}\)	\(83\)
risch	\(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (7 a d \,x^{4}-18 b c \,x^{4}+3 a c \right )}{21 a^{4} x^{7}}-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} x \left (9 a b d \,x^{4}-14 b^{2} c \,x^{4}+10 a^{2} d -15 a b c \right ) b}{5 a^{4} \left (x^{8} b^{2}+2 a \,x^{4} b +a^{2}\right )}\)	\(103\)

Input:

int((d*x^4+c)/x^8/(b*x^4+a)^(9/4),x,method=_RETURNVERBOSE)
 

Output:

1/105*((-35*d*x^4-15*c)*a^3+60*b*x^4*(-14/3*d*x^4+c)*a^2+480*b^2*x^8*(-7/1 
5*d*x^4+c)*a+384*b^3*c*x^12)/(b*x^4+a)^(5/4)/x^7/a^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{9/4}} \, dx=\frac {{\left (32 \, {\left (12 \, b^{3} c - 7 \, a b^{2} d\right )} x^{12} + 40 \, {\left (12 \, a b^{2} c - 7 \, a^{2} b d\right )} x^{8} + 5 \, {\left (12 \, a^{2} b c - 7 \, a^{3} d\right )} x^{4} - 15 \, a^{3} c\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{105 \, {\left (a^{4} b^{2} x^{15} + 2 \, a^{5} b x^{11} + a^{6} x^{7}\right )}} \] Input:

integrate((d*x^4+c)/x^8/(b*x^4+a)^(9/4),x, algorithm="fricas")
 

Output:

1/105*(32*(12*b^3*c - 7*a*b^2*d)*x^12 + 40*(12*a*b^2*c - 7*a^2*b*d)*x^8 + 
5*(12*a^2*b*c - 7*a^3*d)*x^4 - 15*a^3*c)*(b*x^4 + a)^(3/4)/(a^4*b^2*x^15 + 
 2*a^5*b*x^11 + a^6*x^7)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 726 vs. \(2 (107) = 214\).

Time = 79.39 (sec) , antiderivative size = 726, normalized size of antiderivative = 6.42 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{9/4}} \, dx =\text {Too large to display} \] Input:

integrate((d*x**4+c)/x**8/(b*x**4+a)**(9/4),x)
 

Output:

c*(-15*a**4*b**(39/4)*(a/(b*x**4) + 1)**(3/4)*gamma(-7/4)/(256*a**7*b**9*x 
**4*gamma(9/4) + 768*a**6*b**10*x**8*gamma(9/4) + 768*a**5*b**11*x**12*gam 
ma(9/4) + 256*a**4*b**12*x**16*gamma(9/4)) + 45*a**3*b**(43/4)*x**4*(a/(b* 
x**4) + 1)**(3/4)*gamma(-7/4)/(256*a**7*b**9*x**4*gamma(9/4) + 768*a**6*b* 
*10*x**8*gamma(9/4) + 768*a**5*b**11*x**12*gamma(9/4) + 256*a**4*b**12*x** 
16*gamma(9/4)) + 540*a**2*b**(47/4)*x**8*(a/(b*x**4) + 1)**(3/4)*gamma(-7/ 
4)/(256*a**7*b**9*x**4*gamma(9/4) + 768*a**6*b**10*x**8*gamma(9/4) + 768*a 
**5*b**11*x**12*gamma(9/4) + 256*a**4*b**12*x**16*gamma(9/4)) + 864*a*b**( 
51/4)*x**12*(a/(b*x**4) + 1)**(3/4)*gamma(-7/4)/(256*a**7*b**9*x**4*gamma( 
9/4) + 768*a**6*b**10*x**8*gamma(9/4) + 768*a**5*b**11*x**12*gamma(9/4) + 
256*a**4*b**12*x**16*gamma(9/4)) + 384*b**(55/4)*x**16*(a/(b*x**4) + 1)**( 
3/4)*gamma(-7/4)/(256*a**7*b**9*x**4*gamma(9/4) + 768*a**6*b**10*x**8*gamm 
a(9/4) + 768*a**5*b**11*x**12*gamma(9/4) + 256*a**4*b**12*x**16*gamma(9/4) 
)) + d*(5*a**2*b**(19/4)*(a/(b*x**4) + 1)**(3/4)*gamma(-3/4)/(64*a**5*b**4 
*gamma(9/4) + 128*a**4*b**5*x**4*gamma(9/4) + 64*a**3*b**6*x**8*gamma(9/4) 
) + 40*a*b**(23/4)*x**4*(a/(b*x**4) + 1)**(3/4)*gamma(-3/4)/(64*a**5*b**4* 
gamma(9/4) + 128*a**4*b**5*x**4*gamma(9/4) + 64*a**3*b**6*x**8*gamma(9/4)) 
 + 32*b**(27/4)*x**8*(a/(b*x**4) + 1)**(3/4)*gamma(-3/4)/(64*a**5*b**4*gam 
ma(9/4) + 128*a**4*b**5*x**4*gamma(9/4) + 64*a**3*b**6*x**8*gamma(9/4)))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.16 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{9/4}} \, dx=\frac {1}{15} \, {\left (\frac {3 \, {\left (b^{2} - \frac {10 \, {\left (b x^{4} + a\right )} b}{x^{4}}\right )} x^{5}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{3}} - \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{a^{3} x^{3}}\right )} d - \frac {1}{35} \, c {\left (\frac {7 \, {\left (b^{3} - \frac {15 \, {\left (b x^{4} + a\right )} b^{2}}{x^{4}}\right )} x^{5}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{4}} - \frac {5 \, {\left (\frac {7 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b}{x^{3}} - \frac {{\left (b x^{4} + a\right )}^{\frac {7}{4}}}{x^{7}}\right )}}{a^{4}}\right )} \] Input:

integrate((d*x^4+c)/x^8/(b*x^4+a)^(9/4),x, algorithm="maxima")
 

Output:

1/15*(3*(b^2 - 10*(b*x^4 + a)*b/x^4)*x^5/((b*x^4 + a)^(5/4)*a^3) - 5*(b*x^ 
4 + a)^(3/4)/(a^3*x^3))*d - 1/35*c*(7*(b^3 - 15*(b*x^4 + a)*b^2/x^4)*x^5/( 
(b*x^4 + a)^(5/4)*a^4) - 5*(7*(b*x^4 + a)^(3/4)*b/x^3 - (b*x^4 + a)^(7/4)/ 
x^7)/a^4)
 

Giac [F]

\[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{9/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {9}{4}} x^{8}} \,d x } \] Input:

integrate((d*x^4+c)/x^8/(b*x^4+a)^(9/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(9/4)*x^8), x)
 

Mupad [B] (verification not implemented)

Time = 4.26 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.30 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{9/4}} \, dx=\frac {x\,\left (\frac {b\,\left (2\,a\,d-3\,b\,c\right )}{2\,a^3}-\frac {a\,\left (\frac {2\,b^3\,c-a\,b^2\,d}{5\,a^4}+\frac {7\,b^2\,\left (2\,a\,d-3\,b\,c\right )}{10\,a^4}\right )}{b}\right )}{{\left (b\,x^4+a\right )}^{5/4}}-\frac {c\,{\left (b\,x^4+a\right )}^{3/4}}{7\,a^3\,x^7}-\frac {{\left (b\,x^4+a\right )}^{3/4}\,\left (7\,a^3\,d-18\,a^2\,b\,c\right )}{21\,a^6\,x^3}+\frac {x\,\left (14\,b^2\,c-9\,a\,b\,d\right )}{5\,a^4\,{\left (b\,x^4+a\right )}^{1/4}} \] Input:

int((c + d*x^4)/(x^8*(a + b*x^4)^(9/4)),x)
 

Output:

(x*((b*(2*a*d - 3*b*c))/(2*a^3) - (a*((2*b^3*c - a*b^2*d)/(5*a^4) + (7*b^2 
*(2*a*d - 3*b*c))/(10*a^4)))/b))/(a + b*x^4)^(5/4) - (c*(a + b*x^4)^(3/4)) 
/(7*a^3*x^7) - ((a + b*x^4)^(3/4)*(7*a^3*d - 18*a^2*b*c))/(21*a^6*x^3) + ( 
x*(14*b^2*c - 9*a*b*d))/(5*a^4*(a + b*x^4)^(1/4))
 

Reduce [F]

\[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{9/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} x^{8}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{16}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} x^{4}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{12}}d x \right ) d \] Input:

int((d*x^4+c)/x^8/(b*x^4+a)^(9/4),x)
 

Output:

int(1/((a + b*x**4)**(1/4)*a**2*x**8 + 2*(a + b*x**4)**(1/4)*a*b*x**12 + ( 
a + b*x**4)**(1/4)*b**2*x**16),x)*c + int(1/((a + b*x**4)**(1/4)*a**2*x**4 
 + 2*(a + b*x**4)**(1/4)*a*b*x**8 + (a + b*x**4)**(1/4)*b**2*x**12),x)*d