\(\int \frac {1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx\) [1730]

Optimal result

Integrand size = 20, antiderivative size = 193 \[ \int \frac {1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx=\frac {\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{a}-\frac {\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{a}-\frac {\log \left (1+\frac {\sqrt {1-a x}}{\sqrt {1+a x}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{\sqrt {2} a}+\frac {\log \left (1+\frac {\sqrt {1-a x}}{\sqrt {1+a x}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{\sqrt {2} a} \]

[Out]

-1/2*ln(1-(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4)+(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a*2^(1/2)+1/2*ln(1+(-a*x+1)^(1/4)
*2^(1/2)/(a*x+1)^(1/4)+(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a*2^(1/2)-arctan(-1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4))
*2^(1/2)/a-arctan(1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4))*2^(1/2)/a

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {65, 338, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx=\frac {\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{a}-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{a}-\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2} a}+\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2} a} \]

[In]

Int[1/((1 - a*x)^(1/4)*(1 + a*x)^(3/4)),x]

[Out]

(Sqrt[2]*ArcTan[1 - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/a - (Sqrt[2]*ArcTan[1 + (Sqrt[2]*(1 - a*x)^(1/
4))/(1 + a*x)^(1/4)])/a - Log[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/(Sq
rt[2]*a) + Log[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/(Sqrt[2]*a)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 210

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

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.51 \[ \int \frac {1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx=\frac {\sqrt {2} \left (\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a^2 x^2}}{\sqrt {1-a x}-\sqrt {1+a x}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-a^2 x^2}}{\sqrt {1-a x}+\sqrt {1+a x}}\right )\right )}{a} \]

[In]

Integrate[1/((1 - a*x)^(1/4)*(1 + a*x)^(3/4)),x]

[Out]

(Sqrt[2]*(ArcTan[(Sqrt[2]*(1 - a^2*x^2)^(1/4))/(Sqrt[1 - a*x] - Sqrt[1 + a*x])] + ArcTanh[(Sqrt[2]*(1 - a^2*x^
2)^(1/4))/(Sqrt[1 - a*x] + Sqrt[1 + a*x])]))/a

Maple [F]

\[\int \frac {1}{\left (-a x +1\right )^{\frac {1}{4}} \left (a x +1\right )^{\frac {3}{4}}}d x\]

[In]

int(1/(-a*x+1)^(1/4)/(a*x+1)^(3/4),x)

[Out]

int(1/(-a*x+1)^(1/4)/(a*x+1)^(3/4),x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx=-\left (-\frac {1}{a^{4}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (a^{2} x - a\right )} \left (-\frac {1}{a^{4}}\right )^{\frac {1}{4}} + {\left (a x + 1\right )}^{\frac {1}{4}} {\left (-a x + 1\right )}^{\frac {3}{4}}}{a x - 1}\right ) + \left (-\frac {1}{a^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (a^{2} x - a\right )} \left (-\frac {1}{a^{4}}\right )^{\frac {1}{4}} - {\left (a x + 1\right )}^{\frac {1}{4}} {\left (-a x + 1\right )}^{\frac {3}{4}}}{a x - 1}\right ) - i \, \left (-\frac {1}{a^{4}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (i \, a^{2} x - i \, a\right )} \left (-\frac {1}{a^{4}}\right )^{\frac {1}{4}} + {\left (a x + 1\right )}^{\frac {1}{4}} {\left (-a x + 1\right )}^{\frac {3}{4}}}{a x - 1}\right ) + i \, \left (-\frac {1}{a^{4}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (-i \, a^{2} x + i \, a\right )} \left (-\frac {1}{a^{4}}\right )^{\frac {1}{4}} + {\left (a x + 1\right )}^{\frac {1}{4}} {\left (-a x + 1\right )}^{\frac {3}{4}}}{a x - 1}\right ) \]

[In]

integrate(1/(-a*x+1)^(1/4)/(a*x+1)^(3/4),x, algorithm="fricas")

[Out]

-(-1/a^4)^(1/4)*log(((a^2*x - a)*(-1/a^4)^(1/4) + (a*x + 1)^(1/4)*(-a*x + 1)^(3/4))/(a*x - 1)) + (-1/a^4)^(1/4
)*log(-((a^2*x - a)*(-1/a^4)^(1/4) - (a*x + 1)^(1/4)*(-a*x + 1)^(3/4))/(a*x - 1)) - I*(-1/a^4)^(1/4)*log(((I*a
^2*x - I*a)*(-1/a^4)^(1/4) + (a*x + 1)^(1/4)*(-a*x + 1)^(3/4))/(a*x - 1)) + I*(-1/a^4)^(1/4)*log(((-I*a^2*x +
I*a)*(-1/a^4)^(1/4) + (a*x + 1)^(1/4)*(-a*x + 1)^(3/4))/(a*x - 1))

Sympy [F]

\[ \int \frac {1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx=\int \frac {1}{\sqrt [4]{- a x + 1} \left (a x + 1\right )^{\frac {3}{4}}}\, dx \]

[In]

integrate(1/(-a*x+1)**(1/4)/(a*x+1)**(3/4),x)

[Out]

Integral(1/((-a*x + 1)**(1/4)*(a*x + 1)**(3/4)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx=\int { \frac {1}{{\left (a x + 1\right )}^{\frac {3}{4}} {\left (-a x + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate(1/(-a*x+1)^(1/4)/(a*x+1)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((a*x + 1)^(3/4)*(-a*x + 1)^(1/4)), x)

Giac [F]

\[ \int \frac {1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx=\int { \frac {1}{{\left (a x + 1\right )}^{\frac {3}{4}} {\left (-a x + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate(1/(-a*x+1)^(1/4)/(a*x+1)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((a*x + 1)^(3/4)*(-a*x + 1)^(1/4)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx=\int \frac {1}{{\left (1-a\,x\right )}^{1/4}\,{\left (a\,x+1\right )}^{3/4}} \,d x \]

[In]

int(1/((1 - a*x)^(1/4)*(a*x + 1)^(3/4)),x)

[Out]

int(1/((1 - a*x)^(1/4)*(a*x + 1)^(3/4)), x)