\(\int \frac {(a+b x^4)^{3/4}}{(c+d x^4)^2} \, dx\) [207]

Optimal result

Integrand size = 21, antiderivative size = 135 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{\left (c+d x^4\right )^2} \, dx=\frac {x \left (a+b x^4\right )^{3/4}}{4 c \left (c+d x^4\right )}+\frac {3 a \arctan \left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} \sqrt [4]{b c-a d}}+\frac {3 a \text {arctanh}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} \sqrt [4]{b c-a d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {386, 385, 218, 214, 211} \[ \int \frac {\left (a+b x^4\right )^{3/4}}{\left (c+d x^4\right )^2} \, dx=\frac {3 a \arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} \sqrt [4]{b c-a d}}+\frac {3 a \text {arctanh}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} \sqrt [4]{b c-a d}}+\frac {x \left (a+b x^4\right )^{3/4}}{4 c \left (c+d x^4\right )} \]

[In]

Int[(a + b*x^4)^(3/4)/(c + d*x^4)^2,x]

[Out]

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

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 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 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 386

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.46 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.76 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{\left (c+d x^4\right )^2} \, dx=\frac {4 c^{3/4} \sqrt [4]{b c-a d} x \left (a+b x^4\right )^{3/4}+(3+3 i) a \left (c+d x^4\right ) \arctan \left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}-\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )+(3+3 i) a \left (c+d x^4\right ) \text {arctanh}\left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}+\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )}{16 c^{7/4} \sqrt [4]{b c-a d} \left (c+d x^4\right )} \]

[In]

Integrate[(a + b*x^4)^(3/4)/(c + d*x^4)^2,x]

[Out]

(4*c^(3/4)*(b*c - a*d)^(1/4)*x*(a + b*x^4)^(3/4) + (3 + 3*I)*a*(c + d*x^4)*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)] + (3 + 3*I)*a
*(c + d*x^4)*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)])/(16*c^(7/4)*(b*c - a*d)^(1/4)*(c + d*x^4))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(292\) vs. \(2(107)=214\).

Time = 4.43 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.17

method	result	size
pseudoelliptic	\(\frac {\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} x c \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}}}{4}+\frac {3 \sqrt {2}\, a \left (d \,x^{4}+c \right ) \left (2 \arctan \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x -\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}\right )-2 \arctan \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x +\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}\right )-\ln \left (\frac {-\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}\right )\right )}{32}}{c^{2} \left (d \,x^{4}+c \right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}}}\)	\(293\)

[In]

int((b*x^4+a)^(3/4)/(d*x^4+c)^2,x,method=_RETURNVERBOSE)

[Out]

3/16/((a*d-b*c)/c)^(1/4)*(4/3*(b*x^4+a)^(3/4)*x*c*((a*d-b*c)/c)^(1/4)+1/2*2^(1/2)*a*(d*x^4+c)*(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)-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)))))/c^
2/(d*x^4+c)

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^4\right )^{3/4}}{\left (c+d x^4\right )^2} \, dx=\text {Timed out} \]

[In]

integrate((b*x^4+a)^(3/4)/(d*x^4+c)^2,x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {\left (a+b x^4\right )^{3/4}}{\left (c+d x^4\right )^2} \, dx=\int \frac {\left (a + b x^{4}\right )^{\frac {3}{4}}}{\left (c + d x^{4}\right )^{2}}\, dx \]

[In]

integrate((b*x**4+a)**(3/4)/(d*x**4+c)**2,x)

[Out]

Integral((a + b*x**4)**(3/4)/(c + d*x**4)**2, x)

Maxima [F]

\[ \int \frac {\left (a+b x^4\right )^{3/4}}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \]

[In]

integrate((b*x^4+a)^(3/4)/(d*x^4+c)^2,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/4)/(d*x^4 + c)^2, x)

Giac [F]

\[ \int \frac {\left (a+b x^4\right )^{3/4}}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \]

[In]

integrate((b*x^4+a)^(3/4)/(d*x^4+c)^2,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/4)/(d*x^4 + c)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^4\right )^{3/4}}{\left (c+d x^4\right )^2} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{3/4}}{{\left (d\,x^4+c\right )}^2} \,d x \]

[In]

int((a + b*x^4)^(3/4)/(c + d*x^4)^2,x)

[Out]

int((a + b*x^4)^(3/4)/(c + d*x^4)^2, x)