3.3.10 \(\int \frac {1}{(a+b x^4)^{9/4} (c+d x^4)^2} \, dx\) [210]

3.3.10.1 Optimal result

Integrand size = 21, antiderivative size = 266 \[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )^2} \, dx=\frac {b (4 b c+5 a d) x}{20 a c (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (16 b^2 c^2-56 a b c d-5 a^2 d^2\right ) x}{20 a^2 c (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d x}{4 c (b c-a d) \left (a+b x^4\right )^{5/4} \left (c+d x^4\right )}+\frac {3 d^2 (4 b c-a d) \arctan \left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} (b c-a d)^{13/4}}+\frac {3 d^2 (4 b c-a d) \text {arctanh}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} (b c-a d)^{13/4}} \]

output

1/20*b*(5*a*d+4*b*c)*x/a/c/(-a*d+b*c)^2/(b*x^4+a)^(5/4)+1/20*b*(-5*a^2*d^2 
-56*a*b*c*d+16*b^2*c^2)*x/a^2/c/(-a*d+b*c)^3/(b*x^4+a)^(1/4)-1/4*d*x/c/(-a 
*d+b*c)/(b*x^4+a)^(5/4)/(d*x^4+c)+3/8*d^2*(-a*d+4*b*c)*arctan((-a*d+b*c)^( 
1/4)*x/c^(1/4)/(b*x^4+a)^(1/4))/c^(7/4)/(-a*d+b*c)^(13/4)+3/8*d^2*(-a*d+4* 
b*c)*arctanh((-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x^4+a)^(1/4))/c^(7/4)/(-a*d+b*c 
)^(13/4)
 

3.3.10.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.69 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.34 \[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )^2} \, dx=\frac {\left (\frac {1}{80}+\frac {i}{80}\right ) \left (\frac {(2-2 i) c^{3/4} x \left (5 a^4 d^3+10 a^3 b d^3 x^4-16 b^4 c^2 x^4 \left (c+d x^4\right )+5 a^2 b^2 d \left (12 c^2+12 c d x^4+d^2 x^8\right )+4 a b^3 c \left (-5 c^2+9 c d x^4+14 d^2 x^8\right )\right )}{a^2 (-b c+a d)^3 \left (a+b x^4\right )^{5/4} \left (c+d x^4\right )}+\frac {15 d^2 (4 b c-a d) \arctan \left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}-\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )}{(b c-a d)^{13/4}}+\frac {15 d^2 (4 b c-a d) \text {arctanh}\left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}+\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )}{(b c-a d)^{13/4}}\right )}{c^{7/4}} \]

input

Integrate[1/((a + b*x^4)^(9/4)*(c + d*x^4)^2),x]
 

output

((1/80 + I/80)*(((2 - 2*I)*c^(3/4)*x*(5*a^4*d^3 + 10*a^3*b*d^3*x^4 - 16*b^ 
4*c^2*x^4*(c + d*x^4) + 5*a^2*b^2*d*(12*c^2 + 12*c*d*x^4 + d^2*x^8) + 4*a* 
b^3*c*(-5*c^2 + 9*c*d*x^4 + 14*d^2*x^8)))/(a^2*(-(b*c) + a*d)^3*(a + b*x^4 
)^(5/4)*(c + d*x^4)) + (15*d^2*(4*b*c - a*d)*ArcTan[(((1 - I)*(b*c - a*d)^ 
(1/4)*x^2)/(c^(1/4)*(a + b*x^4)^(1/4)) - ((1 + I)*c^(1/4)*(a + b*x^4)^(1/4 
))/(b*c - a*d)^(1/4))/(2*x)])/(b*c - a*d)^(13/4) + (15*d^2*(4*b*c - a*d)*A 
rcTanh[(((1 - I)*(b*c - a*d)^(1/4)*x^2)/(c^(1/4)*(a + b*x^4)^(1/4)) + ((1 
+ I)*c^(1/4)*(a + b*x^4)^(1/4))/(b*c - a*d)^(1/4))/(2*x)])/(b*c - a*d)^(13 
/4)))/c^(7/4)
 

3.3.10.3 Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {931, 1024, 25, 1024, 27, 902, 756, 218, 221}

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.

\(\displaystyle \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )^2} \, dx\)

\(\Big \downarrow \) 931

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

\(\Big \downarrow \) 1024

\(\displaystyle \frac {\frac {b x (5 a d+4 b c)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}-\frac {\int -\frac {4 b d (4 b c+5 a d) x^4+16 b^2 c^2+15 a^2 d^2-40 a b c d}{\left (b x^4+a\right )^{5/4} \left (d x^4+c\right )}dx}{5 a (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^4\right )^{5/4} \left (c+d x^4\right ) (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {4 b d (4 b c+5 a d) x^4+16 b^2 c^2+15 a^2 d^2-40 a b c d}{\left (b x^4+a\right )^{5/4} \left (d x^4+c\right )}dx}{5 a (b c-a d)}+\frac {b x (5 a d+4 b c)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^4\right )^{5/4} \left (c+d x^4\right ) (b c-a d)}\)

\(\Big \downarrow \) 1024

\(\displaystyle \frac {\frac {\frac {b x \left (-5 a^2 d^2-56 a b c d+16 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}-\frac {\int -\frac {15 a^2 d^2 (4 b c-a d)}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{a (b c-a d)}}{5 a (b c-a d)}+\frac {b x (5 a d+4 b c)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^4\right )^{5/4} \left (c+d x^4\right ) (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {15 a d^2 (4 b c-a d) \int \frac {1}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{b c-a d}+\frac {b x \left (-5 a^2 d^2-56 a b c d+16 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}}{5 a (b c-a d)}+\frac {b x (5 a d+4 b c)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^4\right )^{5/4} \left (c+d x^4\right ) (b c-a d)}\)

\(\Big \downarrow \) 902

\(\displaystyle \frac {\frac {\frac {15 a d^2 (4 b c-a d) \int \frac {1}{c-\frac {(b c-a d) x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{b c-a d}+\frac {b x \left (-5 a^2 d^2-56 a b c d+16 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}}{5 a (b c-a d)}+\frac {b x (5 a d+4 b c)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^4\right )^{5/4} \left (c+d x^4\right ) (b c-a d)}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {\frac {\frac {15 a d^2 (4 b c-a d) \left (\frac {\int \frac {1}{\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\int \frac {1}{\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}\right )}{b c-a d}+\frac {b x \left (-5 a^2 d^2-56 a b c d+16 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}}{5 a (b c-a d)}+\frac {b x (5 a d+4 b c)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^4\right )^{5/4} \left (c+d x^4\right ) (b c-a d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {15 a d^2 (4 b c-a d) \left (\frac {\int \frac {1}{\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}\right )}{b c-a d}+\frac {b x \left (-5 a^2 d^2-56 a b c d+16 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}}{5 a (b c-a d)}+\frac {b x (5 a d+4 b c)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^4\right )^{5/4} \left (c+d x^4\right ) (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {b x \left (-5 a^2 d^2-56 a b c d+16 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}+\frac {15 a d^2 (4 b c-a d) \left (\frac {\arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}+\frac {\text {arctanh}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}\right )}{b c-a d}}{5 a (b c-a d)}+\frac {b x (5 a d+4 b c)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \left (a+b x^4\right )^{5/4} \left (c+d x^4\right ) (b c-a d)}\)

input

Int[1/((a + b*x^4)^(9/4)*(c + d*x^4)^2),x]
 

output

-1/4*(d*x)/(c*(b*c - a*d)*(a + b*x^4)^(5/4)*(c + d*x^4)) + ((b*(4*b*c + 5* 
a*d)*x)/(5*a*(b*c - a*d)*(a + b*x^4)^(5/4)) + ((b*(16*b^2*c^2 - 56*a*b*c*d 
 - 5*a^2*d^2)*x)/(a*(b*c - a*d)*(a + b*x^4)^(1/4)) + (15*a*d^2*(4*b*c - a* 
d)*(ArcTan[((b*c - a*d)^(1/4)*x)/(c^(1/4)*(a + b*x^4)^(1/4))]/(2*c^(3/4)*( 
b*c - a*d)^(1/4)) + ArcTanh[((b*c - a*d)^(1/4)*x)/(c^(1/4)*(a + b*x^4)^(1/ 
4))]/(2*c^(3/4)*(b*c - a*d)^(1/4))))/(b*c - a*d))/(5*a*(b*c - a*d)))/(4*c* 
(b*c - a*d))
 

3.3.10.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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

rule 221

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

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 902

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Su 
bst[Int[1/(c - (b*c - a*d)*x^n), x], x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b 
, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
 

rule 931

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[(-b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - 
 a*d))), x] + Simp[1/(a*n*(p + 1)*(b*c - a*d))   Int[(a + b*x^n)^(p + 1)*(c 
 + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, 
 x], x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, 
-1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, 
 c, d, n, p, q, x]
 

rule 1024

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f 
_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c 
+ d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*( 
p + 1))   Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b 
*c - a*d)*(p + 1) + d*(b*e - a*f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
 

3.3.10.4 Maple [A] (verified)

Time = 4.66 (sec) , antiderivative size = 445, normalized size of antiderivative = 1.67

method	result	size
pseudoelliptic	\(-\frac {3 \left (\left (b \,x^{4}+a \right )^{\frac {5}{4}} a^{2} d^{2} \left (d \,x^{4}+c \right ) \left (a d -4 b c \right ) \left (\ln \left (\frac {-\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}\right )-2 \arctan \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x -\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}\right )+2 \arctan \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x +\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}\right )\right ) \sqrt {2}-\frac {8 \left (5 a^{2} b^{2} d^{3} x^{8}+56 a \,b^{3} c \,d^{2} x^{8}-16 b^{4} c^{2} d \,x^{8}+10 a^{3} b \,d^{3} x^{4}+60 a^{2} b^{2} c \,d^{2} x^{4}+36 a \,b^{3} c^{2} d \,x^{4}-16 b^{4} c^{3} x^{4}+5 a^{4} d^{3}+60 a^{2} b^{2} c^{2} d -20 a \,b^{3} c^{3}\right ) x c \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}}}{15}\right )}{32 \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {5}{4}} c^{2} \left (d \,x^{4}+c \right ) \left (a d -b c \right )^{3} a^{2}}\)	\(445\)

input

int(1/(b*x^4+a)^(9/4)/(d*x^4+c)^2,x,method=_RETURNVERBOSE)
 

output

-3/32/((a*d-b*c)/c)^(1/4)/(b*x^4+a)^(5/4)*((b*x^4+a)^(5/4)*a^2*d^2*(d*x^4+ 
c)*(a*d-4*b*c)*(ln((-((a*d-b*c)/c)^(1/4)*(b*x^4+a)^(1/4)*2^(1/2)*x+((a*d-b 
*c)/c)^(1/2)*x^2+(b*x^4+a)^(1/2))/(((a*d-b*c)/c)^(1/4)*(b*x^4+a)^(1/4)*2^( 
1/2)*x+((a*d-b*c)/c)^(1/2)*x^2+(b*x^4+a)^(1/2)))-2*arctan((((a*d-b*c)/c)^( 
1/4)*x-2^(1/2)*(b*x^4+a)^(1/4))/((a*d-b*c)/c)^(1/4)/x)+2*arctan((((a*d-b*c 
)/c)^(1/4)*x+2^(1/2)*(b*x^4+a)^(1/4))/((a*d-b*c)/c)^(1/4)/x))*2^(1/2)-8/15 
*(5*a^2*b^2*d^3*x^8+56*a*b^3*c*d^2*x^8-16*b^4*c^2*d*x^8+10*a^3*b*d^3*x^4+6 
0*a^2*b^2*c*d^2*x^4+36*a*b^3*c^2*d*x^4-16*b^4*c^3*x^4+5*a^4*d^3+60*a^2*b^2 
*c^2*d-20*a*b^3*c^3)*x*c*((a*d-b*c)/c)^(1/4))/c^2/(d*x^4+c)/(a*d-b*c)^3/a^ 
2
 

3.3.10.5 Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )^2} \, dx=\text {Timed out} \]

input

integrate(1/(b*x^4+a)^(9/4)/(d*x^4+c)^2,x, algorithm="fricas")
 

output

Timed out
 

3.3.10.6 Sympy [F]

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

input

integrate(1/(b*x**4+a)**(9/4)/(d*x**4+c)**2,x)
 

output

Integral(1/((a + b*x**4)**(9/4)*(c + d*x**4)**2), x)
 

3.3.10.7 Maxima [F]

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

input

integrate(1/(b*x^4+a)^(9/4)/(d*x^4+c)^2,x, algorithm="maxima")
 

output

integrate(1/((b*x^4 + a)^(9/4)*(d*x^4 + c)^2), x)
 

3.3.10.8 Giac [F]

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

input

integrate(1/(b*x^4+a)^(9/4)/(d*x^4+c)^2,x, algorithm="giac")
 

output

integrate(1/((b*x^4 + a)^(9/4)*(d*x^4 + c)^2), x)
 

3.3.10.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )^2} \, dx=\int \frac {1}{{\left (b\,x^4+a\right )}^{9/4}\,{\left (d\,x^4+c\right )}^2} \,d x \]

input

int(1/((a + b*x^4)^(9/4)*(c + d*x^4)^2),x)
 

output

int(1/((a + b*x^4)^(9/4)*(c + d*x^4)^2), x)