3.30.35 \(\int \frac {1}{(b+a x) \sqrt [4]{b^2 x+a^2 x^3}} \, dx\)

Optimal. Leaf size=345 \[ \frac {\tan ^{-1}\left (\frac {2 \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{a^2 x^3+b^2 x}}{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{a^2 x^3+b^2 x}-2^{3/4} a x+2^{3/4} b}\right )}{2\ 2^{3/4} a^{3/4} b^{3/4}}-\frac {\tan ^{-1}\left (\frac {2 \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{a^2 x^3+b^2 x}}{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{a^2 x^3+b^2 x}+2^{3/4} a x-2^{3/4} b}\right )}{2\ 2^{3/4} a^{3/4} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {-2 \sqrt [4]{2} \sqrt {a} \sqrt {b} \sqrt {a^2 x^3+b^2 x}-\frac {a^2 x^2}{\sqrt [4]{2}}+2^{3/4} a b x-\frac {b^2}{\sqrt [4]{2}}}{2 a^{5/4} \sqrt [4]{b} x \sqrt [4]{a^2 x^3+b^2 x}-2 \sqrt [4]{a} b^{5/4} \sqrt [4]{a^2 x^3+b^2 x}}\right )}{2\ 2^{3/4} a^{3/4} b^{3/4}} \]

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Rubi [C]  time = 0.30, antiderivative size = 150, normalized size of antiderivative = 0.43, number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2056, 959, 466, 511, 510} \begin {gather*} \frac {4 x \sqrt [4]{\frac {a^2 x^2}{b^2}+1} F_1\left (\frac {3}{8};1,\frac {1}{4};\frac {11}{8};\frac {a^2 x^2}{b^2},-\frac {a^2 x^2}{b^2}\right )}{3 b \sqrt [4]{a^2 x^3+b^2 x}}-\frac {4 a x^2 \sqrt [4]{\frac {a^2 x^2}{b^2}+1} F_1\left (\frac {7}{8};1,\frac {1}{4};\frac {15}{8};\frac {a^2 x^2}{b^2},-\frac {a^2 x^2}{b^2}\right )}{7 b^2 \sqrt [4]{a^2 x^3+b^2 x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[1/((b + a*x)*(b^2*x + a^2*x^3)^(1/4)),x]

[Out]

(4*x*(1 + (a^2*x^2)/b^2)^(1/4)*AppellF1[3/8, 1, 1/4, 11/8, (a^2*x^2)/b^2, -((a^2*x^2)/b^2)])/(3*b*(b^2*x + a^2
*x^3)^(1/4)) - (4*a*x^2*(1 + (a^2*x^2)/b^2)^(1/4)*AppellF1[7/8, 1, 1/4, 15/8, (a^2*x^2)/b^2, -((a^2*x^2)/b^2)]
)/(7*b^2*(b^2*x + a^2*x^3)^(1/4))

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 959

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[(d*(g*x)^n)/x^n, In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[(e*(g*x)^n)/x^n, Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*p]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

\begin {align*} \int \frac {1}{(b+a x) \sqrt [4]{b^2 x+a^2 x^3}} \, dx &=\frac {\left (\sqrt [4]{x} \sqrt [4]{b^2+a^2 x^2}\right ) \int \frac {1}{\sqrt [4]{x} (b+a x) \sqrt [4]{b^2+a^2 x^2}} \, dx}{\sqrt [4]{b^2 x+a^2 x^3}}\\ &=-\frac {\left (a \sqrt [4]{x} \sqrt [4]{b^2+a^2 x^2}\right ) \int \frac {x^{3/4}}{\left (b^2-a^2 x^2\right ) \sqrt [4]{b^2+a^2 x^2}} \, dx}{\sqrt [4]{b^2 x+a^2 x^3}}+\frac {\left (b \sqrt [4]{x} \sqrt [4]{b^2+a^2 x^2}\right ) \int \frac {1}{\sqrt [4]{x} \left (b^2-a^2 x^2\right ) \sqrt [4]{b^2+a^2 x^2}} \, dx}{\sqrt [4]{b^2 x+a^2 x^3}}\\ &=-\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{b^2+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (b^2-a^2 x^8\right ) \sqrt [4]{b^2+a^2 x^8}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b^2 x+a^2 x^3}}+\frac {\left (4 b \sqrt [4]{x} \sqrt [4]{b^2+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (b^2-a^2 x^8\right ) \sqrt [4]{b^2+a^2 x^8}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b^2 x+a^2 x^3}}\\ &=-\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{1+\frac {a^2 x^2}{b^2}}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (b^2-a^2 x^8\right ) \sqrt [4]{1+\frac {a^2 x^8}{b^2}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b^2 x+a^2 x^3}}+\frac {\left (4 b \sqrt [4]{x} \sqrt [4]{1+\frac {a^2 x^2}{b^2}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (b^2-a^2 x^8\right ) \sqrt [4]{1+\frac {a^2 x^8}{b^2}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b^2 x+a^2 x^3}}\\ &=\frac {4 x \sqrt [4]{1+\frac {a^2 x^2}{b^2}} F_1\left (\frac {3}{8};1,\frac {1}{4};\frac {11}{8};\frac {a^2 x^2}{b^2},-\frac {a^2 x^2}{b^2}\right )}{3 b \sqrt [4]{b^2 x+a^2 x^3}}-\frac {4 a x^2 \sqrt [4]{1+\frac {a^2 x^2}{b^2}} F_1\left (\frac {7}{8};1,\frac {1}{4};\frac {15}{8};\frac {a^2 x^2}{b^2},-\frac {a^2 x^2}{b^2}\right )}{7 b^2 \sqrt [4]{b^2 x+a^2 x^3}}\\ \end {align*}

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Mathematica [F]  time = 0.83, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(b+a x) \sqrt [4]{b^2 x+a^2 x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[1/((b + a*x)*(b^2*x + a^2*x^3)^(1/4)),x]

[Out]

Integrate[1/((b + a*x)*(b^2*x + a^2*x^3)^(1/4)), x]

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IntegrateAlgebraic [A]  time = 1.24, size = 345, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {2 \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{b^2 x+a^2 x^3}}{2^{3/4} b-2^{3/4} a x+2 \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{b^2 x+a^2 x^3}}\right )}{2\ 2^{3/4} a^{3/4} b^{3/4}}-\frac {\tan ^{-1}\left (\frac {2 \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{b^2 x+a^2 x^3}}{-2^{3/4} b+2^{3/4} a x+2 \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{b^2 x+a^2 x^3}}\right )}{2\ 2^{3/4} a^{3/4} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {-\frac {b^2}{\sqrt [4]{2}}+2^{3/4} a b x-\frac {a^2 x^2}{\sqrt [4]{2}}-2 \sqrt [4]{2} \sqrt {a} \sqrt {b} \sqrt {b^2 x+a^2 x^3}}{-2 \sqrt [4]{a} b^{5/4} \sqrt [4]{b^2 x+a^2 x^3}+2 a^{5/4} \sqrt [4]{b} x \sqrt [4]{b^2 x+a^2 x^3}}\right )}{2\ 2^{3/4} a^{3/4} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((b + a*x)*(b^2*x + a^2*x^3)^(1/4)),x]

[Out]

ArcTan[(2*a^(1/4)*b^(1/4)*(b^2*x + a^2*x^3)^(1/4))/(2^(3/4)*b - 2^(3/4)*a*x + 2*a^(1/4)*b^(1/4)*(b^2*x + a^2*x
^3)^(1/4))]/(2*2^(3/4)*a^(3/4)*b^(3/4)) - ArcTan[(2*a^(1/4)*b^(1/4)*(b^2*x + a^2*x^3)^(1/4))/(-(2^(3/4)*b) + 2
^(3/4)*a*x + 2*a^(1/4)*b^(1/4)*(b^2*x + a^2*x^3)^(1/4))]/(2*2^(3/4)*a^(3/4)*b^(3/4)) - ArcTanh[(-(b^2/2^(1/4))
+ 2^(3/4)*a*b*x - (a^2*x^2)/2^(1/4) - 2*2^(1/4)*Sqrt[a]*Sqrt[b]*Sqrt[b^2*x + a^2*x^3])/(-2*a^(1/4)*b^(5/4)*(b
^2*x + a^2*x^3)^(1/4) + 2*a^(5/4)*b^(1/4)*x*(b^2*x + a^2*x^3)^(1/4))]/(2*2^(3/4)*a^(3/4)*b^(3/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+b)/(a^2*x^3+b^2*x)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (a^{2} x^{3} + b^{2} x\right )}^{\frac {1}{4}} {\left (a x + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+b)/(a^2*x^3+b^2*x)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((a^2*x^3 + b^2*x)^(1/4)*(a*x + b)), x)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a x +b \right ) \left (a^{2} x^{3}+b^{2} x \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+b)/(a^2*x^3+b^2*x)^(1/4),x)

[Out]

int(1/(a*x+b)/(a^2*x^3+b^2*x)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (a^{2} x^{3} + b^{2} x\right )}^{\frac {1}{4}} {\left (a x + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+b)/(a^2*x^3+b^2*x)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((a^2*x^3 + b^2*x)^(1/4)*(a*x + b)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b^2*x + a^2*x^3)^(1/4)*(b + a*x)),x)

[Out]

int(1/((b^2*x + a^2*x^3)^(1/4)*(b + a*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{x \left (a^{2} x^{2} + b^{2}\right )} \left (a x + b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+b)/(a**2*x**3+b**2*x)**(1/4),x)

[Out]

Integral(1/((x*(a**2*x**2 + b**2))**(1/4)*(a*x + b)), x)

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