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

Optimal. Leaf size=173 \[ \frac{b^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 d}+\frac{b^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 d}-\frac{(b c-a d)^{3/4} \tan ^{-1}\left (\frac{x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} d}-\frac{(b c-a d)^{3/4} \tanh ^{-1}\left (\frac{x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.103124, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {408, 240, 212, 206, 203, 377, 208, 205} \[ \frac{b^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 d}+\frac{b^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 d}-\frac{(b c-a d)^{3/4} \tan ^{-1}\left (\frac{x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} d}-\frac{(b c-a d)^{3/4} \tanh ^{-1}\left (\frac{x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(3/4)/(c + d*x^4),x]

[Out]

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

Rule 408

Int[((a_) + (b_.)*(x_)^4)^(p_)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[b/d, Int[(a + b*x^4)^(p - 1), x], x] -
Dist[(b*c - a*d)/d, Int[(a + b*x^4)^(p - 1)/(c + d*x^4), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0
] && (EqQ[p, 3/4] || EqQ[p, 5/4])

Rule 240

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 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 206

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

Rule 203

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

Rubi steps

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

Mathematica [C]  time = 0.158452, size = 161, normalized size = 0.93 \[ \frac{5 a c x \left (a+b x^4\right )^{3/4} F_1\left (\frac{1}{4};-\frac{3}{4},1;\frac{5}{4};-\frac{b x^4}{a},-\frac{d x^4}{c}\right )}{\left (c+d x^4\right ) \left (x^4 \left (3 b c F_1\left (\frac{5}{4};\frac{1}{4},1;\frac{9}{4};-\frac{b x^4}{a},-\frac{d x^4}{c}\right )-4 a d F_1\left (\frac{5}{4};-\frac{3}{4},2;\frac{9}{4};-\frac{b x^4}{a},-\frac{d x^4}{c}\right )\right )+5 a c F_1\left (\frac{1}{4};-\frac{3}{4},1;\frac{5}{4};-\frac{b x^4}{a},-\frac{d x^4}{c}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(5*a*c*x*(a + b*x^4)^(3/4)*AppellF1[1/4, -3/4, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)])/((c + d*x^4)*(5*a*c*Appell
F1[1/4, -3/4, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] + x^4*(-4*a*d*AppellF1[5/4, -3/4, 2, 9/4, -((b*x^4)/a), -((d
*x^4)/c)] + 3*b*c*AppellF1[5/4, 1/4, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])))

________________________________________________________________________________________

Maple [F]  time = 0.415, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{d{x}^{4}+c} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{d x^{4} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.1963, size = 1794, normalized size = 10.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)/(c^3*d^4))^(1/4)*arctan(-(c*d*x*sqrt(((b^3*c^4*d^2 - 3*a*
b^2*c^3*d^3 + 3*a^2*b*c^2*d^4 - a^3*c*d^5)*x^2*sqrt((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)/(c^3*d
^4)) + (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*sqrt(b*x^4 + a))/x^2)*((b^3*c^3
- 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)/(c^3*d^4))^(1/4) - (b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*(b*x^4
+ a)^(1/4)*((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)/(c^3*d^4))^(1/4))/((b^3*c^3 - 3*a*b^2*c^2*d +
3*a^2*b*c*d^2 - a^3*d^3)*x)) + (b^3/d^4)^(1/4)*arctan(-((b*x^4 + a)^(1/4)*b^2*d*(b^3/d^4)^(1/4) - d*x*(b^3/d^4
)^(1/4)*sqrt((b^3*d^2*x^2*sqrt(b^3/d^4) + sqrt(b*x^4 + a)*b^4)/x^2))/(b^3*x)) - 1/4*((b^3*c^3 - 3*a*b^2*c^2*d
+ 3*a^2*b*c*d^2 - a^3*d^3)/(c^3*d^4))^(1/4)*log((c^2*d^3*x*((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3
)/(c^3*d^4))^(3/4) + (b*x^4 + a)^(1/4)*(b^2*c^2 - 2*a*b*c*d + a^2*d^2))/x) + 1/4*((b^3*c^3 - 3*a*b^2*c^2*d + 3
*a^2*b*c*d^2 - a^3*d^3)/(c^3*d^4))^(1/4)*log(-(c^2*d^3*x*((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)/
(c^3*d^4))^(3/4) - (b*x^4 + a)^(1/4)*(b^2*c^2 - 2*a*b*c*d + a^2*d^2))/x) + 1/4*(b^3/d^4)^(1/4)*log((d^3*x*(b^3
/d^4)^(3/4) + (b*x^4 + a)^(1/4)*b^2)/x) - 1/4*(b^3/d^4)^(1/4)*log(-(d^3*x*(b^3/d^4)^(3/4) - (b*x^4 + a)^(1/4)*
b^2)/x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{4}\right )^{\frac{3}{4}}}{c + d x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{d x^{4} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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