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

Optimal. Leaf size=279 \[ -\frac {\log \left (-\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{1-a x}}{\sqrt [4]{b x+1}}+\frac {\sqrt {b} \sqrt {1-a x}}{\sqrt {b x+1}}+\sqrt {a}\right )}{\sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\log \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{1-a x}}{\sqrt [4]{b x+1}}+\frac {\sqrt {b} \sqrt {1-a x}}{\sqrt {b x+1}}+\sqrt {a}\right )}{\sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{1-a x}}{\sqrt [4]{a} \sqrt [4]{b x+1}}\right )}{\sqrt [4]{a} b^{3/4}}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{1-a x}}{\sqrt [4]{a} \sqrt [4]{b x+1}}+1\right )}{\sqrt [4]{a} b^{3/4}} \]

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Rubi [A]  time = 0.30, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {63, 331, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {\log \left (-\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{1-a x}}{\sqrt [4]{b x+1}}+\frac {\sqrt {b} \sqrt {1-a x}}{\sqrt {b x+1}}+\sqrt {a}\right )}{\sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\log \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{1-a x}}{\sqrt [4]{b x+1}}+\frac {\sqrt {b} \sqrt {1-a x}}{\sqrt {b x+1}}+\sqrt {a}\right )}{\sqrt {2} \sqrt [4]{a} b^{3/4}}+\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{1-a x}}{\sqrt [4]{a} \sqrt [4]{b x+1}}\right )}{\sqrt [4]{a} b^{3/4}}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{1-a x}}{\sqrt [4]{a} \sqrt [4]{b x+1}}+1\right )}{\sqrt [4]{a} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*ArcTan[1 - (Sqrt[2]*b^(1/4)*(1 - a*x)^(1/4))/(a^(1/4)*(1 + b*x)^(1/4))])/(a^(1/4)*b^(3/4)) - (Sqrt[2]
*ArcTan[1 + (Sqrt[2]*b^(1/4)*(1 - a*x)^(1/4))/(a^(1/4)*(1 + b*x)^(1/4))])/(a^(1/4)*b^(3/4)) - Log[Sqrt[a] + (S
qrt[b]*Sqrt[1 - a*x])/Sqrt[1 + b*x] - (Sqrt[2]*a^(1/4)*b^(1/4)*(1 - a*x)^(1/4))/(1 + b*x)^(1/4)]/(Sqrt[2]*a^(1
/4)*b^(3/4)) + Log[Sqrt[a] + (Sqrt[b]*Sqrt[1 - a*x])/Sqrt[1 + b*x] + (Sqrt[2]*a^(1/4)*b^(1/4)*(1 - a*x)^(1/4))
/(1 + b*x)^(1/4)]/(Sqrt[2]*a^(1/4)*b^(3/4))

Rule 63

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 204

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

Rule 297

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 331

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 617

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 628

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 1162

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 1165

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

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Mathematica [C]  time = 0.04, size = 65, normalized size = 0.23 \begin {gather*} -\frac {4 (1-a x)^{3/4} \left (\frac {a b x+a}{a+b}\right )^{3/4} \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};\frac {b-a b x}{a+b}\right )}{3 a (b x+1)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(1 - a*x)^(3/4)*((a + a*b*x)/(a + b))^(3/4)*Hypergeometric2F1[3/4, 3/4, 7/4, (b - a*b*x)/(a + b)])/(3*a*(1
+ b*x)^(3/4))

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IntegrateAlgebraic [A]  time = 0.22, size = 176, normalized size = 0.63 \begin {gather*} \frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt [4]{1-a x} \left (\frac {\sqrt [4]{a} \sqrt {b x+1}}{\sqrt {2} \sqrt [4]{b} \sqrt {1-a x}}-\frac {\sqrt [4]{b}}{\sqrt {2} \sqrt [4]{a}}\right )}{\sqrt [4]{b x+1}}\right )}{\sqrt [4]{a} b^{3/4}}+\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{b x+1}}{\sqrt [4]{1-a x} \left (\frac {\sqrt {a} \sqrt {b x+1}}{\sqrt {1-a x}}+\sqrt {b}\right )}\right )}{\sqrt [4]{a} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.11, size = 247, normalized size = 0.89 \begin {gather*} -4 \, \left (-\frac {1}{a b^{3}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (-a x + 1\right )}^{\frac {3}{4}} {\left (b x + 1\right )}^{\frac {1}{4}} a b^{2} \left (-\frac {1}{a b^{3}}\right )^{\frac {3}{4}} - {\left (a^{2} b^{2} x - a b^{2}\right )} \sqrt {\frac {{\left (a b^{2} x - b^{2}\right )} \sqrt {-\frac {1}{a b^{3}}} - \sqrt {-a x + 1} \sqrt {b x + 1}}{a x - 1}} \left (-\frac {1}{a b^{3}}\right )^{\frac {3}{4}}}{a x - 1}\right ) - \left (-\frac {1}{a b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (a b x - b\right )} \left (-\frac {1}{a b^{3}}\right )^{\frac {1}{4}} + {\left (-a x + 1\right )}^{\frac {3}{4}} {\left (b x + 1\right )}^{\frac {1}{4}}}{a x - 1}\right ) + \left (-\frac {1}{a b^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (a b x - b\right )} \left (-\frac {1}{a b^{3}}\right )^{\frac {1}{4}} - {\left (-a x + 1\right )}^{\frac {3}{4}} {\left (b x + 1\right )}^{\frac {1}{4}}}{a x - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (-a x + 1\right )}^{\frac {1}{4}} {\left (b x + 1\right )}^{\frac {3}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (-a x +1\right )^{\frac {1}{4}} \left (b x +1\right )^{\frac {3}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (-a x + 1\right )}^{\frac {1}{4}} {\left (b x + 1\right )}^{\frac {3}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{- a x + 1} \left (b x + 1\right )^{\frac {3}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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