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

Optimal. Leaf size=135 \[ \frac {3 a \tan ^{-1}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} \sqrt [4]{b c-a d}}+\frac {3 a \tanh ^{-1}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} \sqrt [4]{b c-a d}}+\frac {x \left (a+b x^4\right )^{3/4}}{4 c \left (c+d x^4\right )} \]

[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)

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Rubi [A]  time = 0.08, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {378, 377, 212, 208, 205} \[ \frac {3 a \tan ^{-1}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} \sqrt [4]{b c-a d}}+\frac {3 a \tanh ^{-1}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{8 c^{7/4} \sqrt [4]{b c-a d}}+\frac {x \left (a+b x^4\right )^{3/4}}{4 c \left (c+d x^4\right )} \]

Antiderivative was successfully verified.

[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 205

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

Rule 208

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 377

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 378

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*} \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) \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) \operatorname {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) \operatorname {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) \operatorname {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*}

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Mathematica [C]  time = 0.06, size = 78, normalized size = 0.58 \[ \frac {x \left (a+b x^4\right )^{3/4} \, _2F_1\left (-\frac {3}{4},\frac {1}{4};\frac {5}{4};\frac {(a d-b c) x^4}{a \left (d x^4+c\right )}\right )}{c^2 \left (\frac {b x^4}{a}+1\right )^{3/4} \sqrt [4]{\frac {d x^4}{c}+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(a + b*x^4)^(3/4)*Hypergeometric2F1[-3/4, 1/4, 5/4, ((-(b*c) + a*d)*x^4)/(a*(c + d*x^4))])/(c^2*(1 + (b*x^4
)/a)^(3/4)*(1 + (d*x^4)/c)^(1/4))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{{\left (d x^{4} + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{\left (d \,x^{4}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{{\left (d x^{4} + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^4+a\right )}^{3/4}}{{\left (d\,x^4+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{4}\right )^{\frac {3}{4}}}{\left (c + d x^{4}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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