3.3.5 \(\int \frac {(a+b x^4)^{11/4}}{(c+d x^4)^2} \, dx\) [205]

3.3.5.1 Optimal result

Integrand size = 21, antiderivative size = 280 \[ \int \frac {\left (a+b x^4\right )^{11/4}}{\left (c+d x^4\right )^2} \, dx=\frac {b (2 b c-a d) x \left (a+b x^4\right )^{3/4}}{4 c d^2}-\frac {(b c-a d) x \left (a+b x^4\right )^{7/4}}{4 c d \left (c+d x^4\right )}-\frac {b^{7/4} (8 b c-11 a d) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 d^3}+\frac {(b c-a d)^{7/4} (8 b c+3 a d) \arctan \left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} d^3}-\frac {b^{7/4} (8 b c-11 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 d^3}+\frac {(b c-a d)^{7/4} (8 b c+3 a d) \text {arctanh}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} d^3} \]

output

1/4*b*(-a*d+2*b*c)*x*(b*x^4+a)^(3/4)/c/d^2-1/4*(-a*d+b*c)*x*(b*x^4+a)^(7/4 
)/c/d/(d*x^4+c)-1/8*b^(7/4)*(-11*a*d+8*b*c)*arctan(b^(1/4)*x/(b*x^4+a)^(1/ 
4))/d^3+1/8*(-a*d+b*c)^(7/4)*(3*a*d+8*b*c)*arctan((-a*d+b*c)^(1/4)*x/c^(1/ 
4)/(b*x^4+a)^(1/4))/c^(7/4)/d^3-1/8*b^(7/4)*(-11*a*d+8*b*c)*arctanh(b^(1/4 
)*x/(b*x^4+a)^(1/4))/d^3+1/8*(-a*d+b*c)^(7/4)*(3*a*d+8*b*c)*arctanh((-a*d+ 
b*c)^(1/4)*x/c^(1/4)/(b*x^4+a)^(1/4))/c^(7/4)/d^3
 

3.3.5.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.36 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+b x^4\right )^{11/4}}{\left (c+d x^4\right )^2} \, dx=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) \left (\frac {(2-2 i) d x \left (a+b x^4\right )^{3/4} \left (-2 a b c d+a^2 d^2+b^2 c \left (2 c+d x^4\right )\right )}{c \left (c+d x^4\right )}-(1-i) b^{7/4} (8 b c-11 a d) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {(b c-a d)^{7/4} (8 b c+3 a d) \arctan \left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}-\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )}{c^{7/4}}-(1-i) b^{7/4} (8 b c-11 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {(b c-a d)^{7/4} (8 b c+3 a d) \text {arctanh}\left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}+\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )}{c^{7/4}}\right )}{d^3} \]

input

Integrate[(a + b*x^4)^(11/4)/(c + d*x^4)^2,x]
 

output

((1/16 + I/16)*(((2 - 2*I)*d*x*(a + b*x^4)^(3/4)*(-2*a*b*c*d + a^2*d^2 + b 
^2*c*(2*c + d*x^4)))/(c*(c + d*x^4)) - (1 - I)*b^(7/4)*(8*b*c - 11*a*d)*Ar 
cTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] + ((b*c - a*d)^(7/4)*(8*b*c + 3*a*d)*A 
rcTan[(((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)])/c^(7/4) - (1 - I 
)*b^(7/4)*(8*b*c - 11*a*d)*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)] + ((b*c 
- a*d)^(7/4)*(8*b*c + 3*a*d)*ArcTanh[(((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)])/c^(7/4)))/d^3
 

3.3.5.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {930, 1025, 27, 1026, 770, 756, 216, 219, 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.

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

\(\Big \downarrow \) 930

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

\(\Big \downarrow \) 1025

\(\displaystyle \frac {\frac {\int -\frac {4 \left (b^2 c (8 b c-11 a d) x^4+a \left (2 b^2 c^2-2 a b d c-3 a^2 d^2\right )\right )}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{4 d}+\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}-\frac {\int \frac {b^2 c (8 b c-11 a d) x^4+a \left (2 b^2 c^2-2 a b d c-3 a^2 d^2\right )}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 1026

\(\displaystyle \frac {\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}-\frac {\frac {b^2 c (8 b c-11 a d) \int \frac {1}{\sqrt [4]{b x^4+a}}dx}{d}-\frac {(b c-a d)^2 (3 a d+8 b c) \int \frac {1}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{d}}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}-\frac {\frac {b^2 c (8 b c-11 a d) \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{d}-\frac {(b c-a d)^2 (3 a d+8 b c) \int \frac {1}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{d}}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}-\frac {\frac {b^2 c (8 b c-11 a d) \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )}{d}-\frac {(b c-a d)^2 (3 a d+8 b c) \int \frac {1}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{d}}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}-\frac {\frac {b^2 c (8 b c-11 a d) \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{d}-\frac {(b c-a d)^2 (3 a d+8 b c) \int \frac {1}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{d}}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}-\frac {\frac {b^2 c (8 b c-11 a d) \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{d}-\frac {(b c-a d)^2 (3 a d+8 b c) \int \frac {1}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{d}}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 902

\(\displaystyle \frac {\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}-\frac {\frac {b^2 c (8 b c-11 a d) \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{d}-\frac {(b c-a d)^2 (3 a d+8 b c) \int \frac {1}{c-\frac {(b c-a d) x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{d}}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}-\frac {\frac {b^2 c (8 b c-11 a d) \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{d}-\frac {(b c-a d)^2 (3 a d+8 b c) \left (\frac {\int \frac {1}{\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\int \frac {1}{\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}\right )}{d}}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}-\frac {\frac {b^2 c (8 b c-11 a d) \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{d}-\frac {(b c-a d)^2 (3 a d+8 b c) \left (\frac {\int \frac {1}{\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}\right )}{d}}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{d}-\frac {\frac {b^2 c (8 b c-11 a d) \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{d}-\frac {(b c-a d)^2 (3 a d+8 b c) \left (\frac {\arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}+\frac {\text {arctanh}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}\right )}{d}}{d}}{4 c d}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

input

Int[(a + b*x^4)^(11/4)/(c + d*x^4)^2,x]
 

output

-1/4*((b*c - a*d)*x*(a + b*x^4)^(7/4))/(c*d*(c + d*x^4)) + ((b*(2*b*c - a* 
d)*x*(a + b*x^4)^(3/4))/d - ((b^2*c*(8*b*c - 11*a*d)*(ArcTan[(b^(1/4)*x)/( 
a + b*x^4)^(1/4)]/(2*b^(1/4)) + ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2* 
b^(1/4))))/d - ((b*c - a*d)^2*(8*b*c + 3*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))))/d)/d)/(4*c*d)
 

3.3.5.3.1 Defintions of rubi rules used

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 216

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

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

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 930

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

rule 1025

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

rule 1026

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)* 
(x_)^(n_)), x_Symbol] :> Simp[f/d   Int[(a + b*x^n)^p, x], x] + Simp[(d*e - 
 c*f)/d   Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, 
 p, n}, x]
 

3.3.5.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(506\) vs. \(2(232)=464\).

Time = 5.36 (sec) , antiderivative size = 507, normalized size of antiderivative = 1.81

method	result	size
pseudoelliptic	\(-\frac {-8 \left (2 b^{2} c^{2}-2 b \left (-\frac {b \,x^{4}}{2}+a \right ) d c +a^{2} d^{2}\right ) x d \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} c \left (b \,x^{4}+a \right )^{\frac {3}{4}}+\left (16 \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} c^{3} \left (\ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )\right ) b^{\frac {11}{4}}-22 \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} a \,c^{2} d \left (\ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )\right ) b^{\frac {7}{4}}+\sqrt {2}\, \left (3 a d +8 b c \right ) \left (a d -b c \right )^{2} \left (\ln \left (\frac {-\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}\right )-2 \arctan \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x -\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}\right )+2 \arctan \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x +\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}\right )\right )\right ) \left (d \,x^{4}+c \right )}{32 \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} d^{3} c^{2} \left (d \,x^{4}+c \right )}\)	\(507\)

input

int((b*x^4+a)^(11/4)/(d*x^4+c)^2,x,method=_RETURNVERBOSE)
 

output

-1/32*(-8*(2*b^2*c^2-2*b*(-1/2*b*x^4+a)*d*c+a^2*d^2)*x*d*((a*d-b*c)/c)^(1/ 
4)*c*(b*x^4+a)^(3/4)+(16*((a*d-b*c)/c)^(1/4)*c^3*(ln((-b^(1/4)*x-(b*x^4+a) 
^(1/4))/(b^(1/4)*x-(b*x^4+a)^(1/4)))-2*arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4)) 
)*b^(11/4)-22*((a*d-b*c)/c)^(1/4)*a*c^2*d*(ln((-b^(1/4)*x-(b*x^4+a)^(1/4)) 
/(b^(1/4)*x-(b*x^4+a)^(1/4)))-2*arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4)))*b^(7/ 
4)+2^(1/2)*(3*a*d+8*b*c)*(a*d-b*c)^2*(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*a 
rctan((((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)))*(d*x^4+c))/((a*d-b*c)/c)^(1/4)/d^3/c^2/(d*x^4+c)
 

3.3.5.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 10.52 (sec) , antiderivative size = 2764, normalized size of antiderivative = 9.87 \[ \int \frac {\left (a+b x^4\right )^{11/4}}{\left (c+d x^4\right )^2} \, dx=\text {Too large to display} \]

input

integrate((b*x^4+a)^(11/4)/(d*x^4+c)^2,x, algorithm="fricas")
 

output

1/16*((c*d^3*x^4 + c^2*d^2)*((4096*b^11*c^11 - 22528*a*b^10*c^10*d + 46464 
*a^2*b^9*c^9*d^2 - 37664*a^3*b^8*c^8*d^3 - 5071*a^4*b^7*c^7*d^4 + 25641*a^ 
5*b^6*c^6*d^5 - 7931*a^6*b^5*c^5*d^6 - 6259*a^7*b^4*c^4*d^7 + 2739*a^8*b^3 
*c^3*d^8 + 891*a^9*b^2*c^2*d^9 - 297*a^10*b*c*d^10 - 81*a^11*d^11)/(c^7*d^ 
12))^(1/4)*log(-(c^5*d^9*x*((4096*b^11*c^11 - 22528*a*b^10*c^10*d + 46464* 
a^2*b^9*c^9*d^2 - 37664*a^3*b^8*c^8*d^3 - 5071*a^4*b^7*c^7*d^4 + 25641*a^5 
*b^6*c^6*d^5 - 7931*a^6*b^5*c^5*d^6 - 6259*a^7*b^4*c^4*d^7 + 2739*a^8*b^3* 
c^3*d^8 + 891*a^9*b^2*c^2*d^9 - 297*a^10*b*c*d^10 - 81*a^11*d^11)/(c^7*d^1 
2))^(3/4) + (512*b^8*c^8 - 1984*a*b^7*c^7*d + 2456*a^2*b^6*c^6*d^2 - 413*a 
^3*b^5*c^5*d^3 - 1175*a^4*b^4*c^4*d^4 + 478*a^5*b^3*c^3*d^5 + 234*a^6*b^2* 
c^2*d^6 - 81*a^7*b*c*d^7 - 27*a^8*d^8)*(b*x^4 + a)^(1/4))/x) - (c*d^3*x^4 
+ c^2*d^2)*((4096*b^11*c^11 - 22528*a*b^10*c^10*d + 46464*a^2*b^9*c^9*d^2 
- 37664*a^3*b^8*c^8*d^3 - 5071*a^4*b^7*c^7*d^4 + 25641*a^5*b^6*c^6*d^5 - 7 
931*a^6*b^5*c^5*d^6 - 6259*a^7*b^4*c^4*d^7 + 2739*a^8*b^3*c^3*d^8 + 891*a^ 
9*b^2*c^2*d^9 - 297*a^10*b*c*d^10 - 81*a^11*d^11)/(c^7*d^12))^(1/4)*log((c 
^5*d^9*x*((4096*b^11*c^11 - 22528*a*b^10*c^10*d + 46464*a^2*b^9*c^9*d^2 - 
37664*a^3*b^8*c^8*d^3 - 5071*a^4*b^7*c^7*d^4 + 25641*a^5*b^6*c^6*d^5 - 793 
1*a^6*b^5*c^5*d^6 - 6259*a^7*b^4*c^4*d^7 + 2739*a^8*b^3*c^3*d^8 + 891*a^9* 
b^2*c^2*d^9 - 297*a^10*b*c*d^10 - 81*a^11*d^11)/(c^7*d^12))^(3/4) - (512*b 
^8*c^8 - 1984*a*b^7*c^7*d + 2456*a^2*b^6*c^6*d^2 - 413*a^3*b^5*c^5*d^3 ...
 

3.3.5.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^4\right )^{11/4}}{\left (c+d x^4\right )^2} \, dx=\text {Timed out} \]

input

integrate((b*x**4+a)**(11/4)/(d*x**4+c)**2,x)
 

output

Timed out
 

3.3.5.7 Maxima [F]

\[ \int \frac {\left (a+b x^4\right )^{11/4}}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {11}{4}}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \]

input

integrate((b*x^4+a)^(11/4)/(d*x^4+c)^2,x, algorithm="maxima")
 

output

integrate((b*x^4 + a)^(11/4)/(d*x^4 + c)^2, x)
 

3.3.5.8 Giac [F]

\[ \int \frac {\left (a+b x^4\right )^{11/4}}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {11}{4}}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \]

input

integrate((b*x^4+a)^(11/4)/(d*x^4+c)^2,x, algorithm="giac")
 

output

integrate((b*x^4 + a)^(11/4)/(d*x^4 + c)^2, x)
 

3.3.5.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^4\right )^{11/4}}{\left (c+d x^4\right )^2} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{11/4}}{{\left (d\,x^4+c\right )}^2} \,d x \]

input

int((a + b*x^4)^(11/4)/(c + d*x^4)^2,x)
 

output

int((a + b*x^4)^(11/4)/(c + d*x^4)^2, x)