### 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[(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)

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Rubi [A]  time = 0.282689, antiderivative size = 278, normalized size of antiderivative = 1., 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 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 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 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 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]

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 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 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 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]

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*}

Mathematica [C]  time = 0.102812, 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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Maple [F]  time = 0.567, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{bx+a}{\frac{1}{\sqrt [4]{d{x}^{2}+c}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (d x^{2} + c\right )}^{\frac{1}{4}}{\left (b x + a\right )}}\,{d x} \end{align*}

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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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x\right ) \sqrt [4]{c + d x^{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (d x^{2} + c\right )}^{\frac{1}{4}}{\left (b x + a\right )}}\,{d x} \end{align*}

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)