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

Optimal. Leaf size=278 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\sqrt {b} \sqrt [4]{a^2 d+b^2 c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\sqrt {b} \sqrt [4]{a^2 d+b^2 c}}-\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (-\frac {b \sqrt {c}}{\sqrt {d a^2+b^2 c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{b x \sqrt {a^2 d+b^2 c}}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (\frac {b \sqrt {c}}{\sqrt {d a^2+b^2 c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{b x \sqrt {a^2 d+b^2 c}} \]

[Out]

arctan((d*x^2+c)^(1/4)*b^(1/2)/(a^2*d+b^2*c)^(1/4))/(a^2*d+b^2*c)^(1/4)/b^(1/2)-arctanh((d*x^2+c)^(1/4)*b^(1/2
)/(a^2*d+b^2*c)^(1/4))/(a^2*d+b^2*c)^(1/4)/b^(1/2)-a*c^(1/4)*EllipticPi((d*x^2+c)^(1/4)/c^(1/4),-b*c^(1/2)/(a^
2*d+b^2*c)^(1/2),I)*(-d*x^2/c)^(1/2)/b/x/(a^2*d+b^2*c)^(1/2)+a*c^(1/4)*EllipticPi((d*x^2+c)^(1/4)/c^(1/4),b*c^
(1/2)/(a^2*d+b^2*c)^(1/2),I)*(-d*x^2/c)^(1/2)/b/x/(a^2*d+b^2*c)^(1/2)

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Rubi [A]  time = 0.28, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {746, 399, 490, 1218, 444, 63, 298, 205, 208} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\sqrt {b} \sqrt [4]{a^2 d+b^2 c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\sqrt {b} \sqrt [4]{a^2 d+b^2 c}}-\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (-\frac {b \sqrt {c}}{\sqrt {d a^2+b^2 c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{b x \sqrt {a^2 d+b^2 c}}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (\frac {b \sqrt {c}}{\sqrt {d a^2+b^2 c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{b x \sqrt {a^2 d+b^2 c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[b]*(c + d*x^2)^(1/4))/(b^2*c + a^2*d)^(1/4)]/(Sqrt[b]*(b^2*c + a^2*d)^(1/4)) - ArcTanh[(Sqrt[b]*(
c + d*x^2)^(1/4))/(b^2*c + a^2*d)^(1/4)]/(Sqrt[b]*(b^2*c + a^2*d)^(1/4)) - (a*c^(1/4)*Sqrt[-((d*x^2)/c)]*Ellip
ticPi[-((b*Sqrt[c])/Sqrt[b^2*c + a^2*d]), ArcSin[(c + d*x^2)^(1/4)/c^(1/4)], -1])/(b*Sqrt[b^2*c + a^2*d]*x) +
(a*c^(1/4)*Sqrt[-((d*x^2)/c)]*EllipticPi[(b*Sqrt[c])/Sqrt[b^2*c + a^2*d], ArcSin[(c + d*x^2)^(1/4)/c^(1/4)], -
1])/(b*Sqrt[b^2*c + a^2*d]*x)

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 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 298

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

Rule 399

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/x, Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 746

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(1/4)), x_Symbol] :> Dist[d, Int[1/((d^2 - e^2*x^2)*(a + c*x^
2)^(1/4)), x], x] - Dist[e, Int[x/((d^2 - e^2*x^2)*(a + c*x^2)^(1/4)), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ
[c*d^2 + a*e^2, 0]

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rubi steps

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

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Mathematica [C]  time = 0.11, size = 126, normalized size = 0.45 \[ -\frac {2 \sqrt [4]{\frac {b \left (x-\sqrt {-\frac {c}{d}}\right )}{a+b x}} \sqrt [4]{\frac {b \left (\sqrt {-\frac {c}{d}}+x\right )}{a+b x}} F_1\left (\frac {1}{2};\frac {1}{4},\frac {1}{4};\frac {3}{2};\frac {a-b \sqrt {-\frac {c}{d}}}{a+b x},\frac {a+b \sqrt {-\frac {c}{d}}}{a+b x}\right )}{b \sqrt [4]{c+d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*((b*(-Sqrt[-(c/d)] + x))/(a + b*x))^(1/4)*((b*(Sqrt[-(c/d)] + x))/(a + b*x))^(1/4)*AppellF1[1/2, 1/4, 1/4,
3/2, (a - b*Sqrt[-(c/d)])/(a + b*x), (a + b*Sqrt[-(c/d)])/(a + b*x)])/(b*(c + d*x^2)^(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(1/(b*x+a)/(d*x^2+c)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.80, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x +a \right ) \left (d \,x^{2}+c \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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