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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {424, 542, 544, 246, 218, 212, 209, 385, 214, 211} \[ \int \frac {\left (a+b x^4\right )^{11/4}}{\left (c+d x^4\right )^2} \, dx=-\frac {b^{7/4} \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ) (8 b c-11 a d)}{8 d^3}+\frac {(b c-a d)^{7/4} (3 a d+8 b c) \arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} d^3}-\frac {b^{7/4} \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ) (8 b c-11 a d)}{8 d^3}+\frac {(b c-a d)^{7/4} (3 a d+8 b c) \text {arctanh}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} d^3}+\frac {b x \left (a+b x^4\right )^{3/4} (2 b c-a d)}{4 c d^2}-\frac {x \left (a+b x^4\right )^{7/4} (b c-a d)}{4 c d \left (c+d x^4\right )} \]

[In]

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

[Out]

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

Rule 209

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

Rule 211

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

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 218

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[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 424

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] - Dist[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 542

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] + Dist[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 544

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[f/d,
Int[(a + b*x^n)^p, x], x] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {(b c-a d) x \left (a+b x^4\right )^{7/4}}{4 c d \left (c+d x^4\right )}+\frac {\int \frac {\left (a+b x^4\right )^{3/4} \left (a (b c+3 a d)+4 b (2 b c-a d) x^4\right )}{c+d x^4} \, dx}{4 c d} \\ & = \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 {\int \frac {-4 a \left (2 b^2 c^2-2 a b c d-3 a^2 d^2\right )-4 b^2 c (8 b c-11 a d) x^4}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )} \, dx}{16 c d^2} \\ & = \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 {\left (b^2 (8 b c-11 a d)\right ) \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx}{4 d^3}+\frac {\left ((b c-a d)^2 (8 b c+3 a d)\right ) \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )} \, dx}{4 c d^3} \\ & = \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 {\left (b^2 (8 b c-11 a d)\right ) \text {Subst}\left (\int \frac {1}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{4 d^3}+\frac {\left ((b c-a d)^2 (8 b c+3 a d)\right ) \text {Subst}\left (\int \frac {1}{c-(b c-a d) x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{4 c d^3} \\ & = \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 {\left (b^2 (8 b c-11 a d)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 d^3}-\frac {\left (b^2 (8 b c-11 a d)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 d^3}+\frac {\left ((b c-a d)^2 (8 b c+3 a d)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c}-\sqrt {b c-a d} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 c^{3/2} d^3}+\frac {\left ((b c-a d)^2 (8 b c+3 a d)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c}+\sqrt {b c-a d} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 c^{3/2} d^3} \\ & = \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) \tan ^{-1}\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) \tan ^{-1}\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) \tanh ^{-1}\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) \tanh ^{-1}\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} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

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

[In]

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

[Out]

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

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

[In]

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

[Out]

-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*ar
ctan(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*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)))*(d*x^4+c))/((a*d-b*
c)/c)^(1/4)/d^3/c^2/(d*x^4+c)

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

[In]

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

[Out]

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^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 - 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 - 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 - 50
71*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))^(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) - (-I*c*d^3*x^4 - I*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((I*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^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 - 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) - (I*c*d^3*x^4 + I*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((-I*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^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 - 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^4 - 22528*a*b^10*c^3*d + 46464*a^2*b^9*c^2*d^2 - 42592*a^3*
b^8*c*d^3 + 14641*a^4*b^7*d^4)/d^12)^(1/4)*log(-(d^9*x*((4096*b^11*c^4 - 22528*a*b^10*c^3*d + 46464*a^2*b^9*c^
2*d^2 - 42592*a^3*b^8*c*d^3 + 14641*a^4*b^7*d^4)/d^12)^(3/4) + (512*b^8*c^3 - 2112*a*b^7*c^2*d + 2904*a^2*b^6*
c*d^2 - 1331*a^3*b^5*d^3)*(b*x^4 + a)^(1/4))/x) + (c*d^3*x^4 + c^2*d^2)*((4096*b^11*c^4 - 22528*a*b^10*c^3*d +
46464*a^2*b^9*c^2*d^2 - 42592*a^3*b^8*c*d^3 + 14641*a^4*b^7*d^4)/d^12)^(1/4)*log((d^9*x*((4096*b^11*c^4 - 225
28*a*b^10*c^3*d + 46464*a^2*b^9*c^2*d^2 - 42592*a^3*b^8*c*d^3 + 14641*a^4*b^7*d^4)/d^12)^(3/4) - (512*b^8*c^3
- 2112*a*b^7*c^2*d + 2904*a^2*b^6*c*d^2 - 1331*a^3*b^5*d^3)*(b*x^4 + a)^(1/4))/x) - (I*c*d^3*x^4 + I*c^2*d^2)*
((4096*b^11*c^4 - 22528*a*b^10*c^3*d + 46464*a^2*b^9*c^2*d^2 - 42592*a^3*b^8*c*d^3 + 14641*a^4*b^7*d^4)/d^12)^
(1/4)*log((I*d^9*x*((4096*b^11*c^4 - 22528*a*b^10*c^3*d + 46464*a^2*b^9*c^2*d^2 - 42592*a^3*b^8*c*d^3 + 14641*
a^4*b^7*d^4)/d^12)^(3/4) - (512*b^8*c^3 - 2112*a*b^7*c^2*d + 2904*a^2*b^6*c*d^2 - 1331*a^3*b^5*d^3)*(b*x^4 + a
)^(1/4))/x) - (-I*c*d^3*x^4 - I*c^2*d^2)*((4096*b^11*c^4 - 22528*a*b^10*c^3*d + 46464*a^2*b^9*c^2*d^2 - 42592*
a^3*b^8*c*d^3 + 14641*a^4*b^7*d^4)/d^12)^(1/4)*log((-I*d^9*x*((4096*b^11*c^4 - 22528*a*b^10*c^3*d + 46464*a^2*
b^9*c^2*d^2 - 42592*a^3*b^8*c*d^3 + 14641*a^4*b^7*d^4)/d^12)^(3/4) - (512*b^8*c^3 - 2112*a*b^7*c^2*d + 2904*a^
2*b^6*c*d^2 - 1331*a^3*b^5*d^3)*(b*x^4 + a)^(1/4))/x) + 4*(b^2*c*d*x^5 + (2*b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x)*
(b*x^4 + a)^(3/4))/(c*d^3*x^4 + c^2*d^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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