3.212 \(\int \frac {c+d x}{\sqrt {-a+b x^4}} \, dx\)

Optimal. Leaf size=89 \[ \frac {\sqrt [4]{a} c \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{b} \sqrt {b x^4-a}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {b x^4-a}}\right )}{2 \sqrt {b}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1885, 224, 221, 275, 217, 206} \[ \frac {\sqrt [4]{a} c \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{b} \sqrt {b x^4-a}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {b x^4-a}}\right )}{2 \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/Sqrt[-a + b*x^4],x]

[Out]

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

Rule 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 1885

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[P
q, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b
, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rubi steps

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

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Mathematica [C]  time = 0.05, size = 83, normalized size = 0.93 \[ \frac {c x \sqrt {1-\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {b x^4}{a}\right )}{\sqrt {b x^4-a}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {b x^4-a}}\right )}{2 \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[-a + b*x^4]])/(2*Sqrt[b]) + (c*x*Sqrt[1 - (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2,
5/4, (b*x^4)/a])/Sqrt[-a + b*x^4]

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d x + c}{\sqrt {b x^{4} - a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x + c)/sqrt(b*x^4 - a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x + c}{\sqrt {b x^{4} - a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)/sqrt(b*x^4 - a), x)

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maple [A]  time = 0.17, size = 95, normalized size = 1.07 \[ \frac {\sqrt {\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, c \EllipticF \left (\sqrt {-\frac {\sqrt {b}}{\sqrt {a}}}\, x , i\right )}{\sqrt {-\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}-a}}+\frac {d \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}-a}\right )}{2 \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*d*ln(b^(1/2)*x^2+(b*x^4-a)^(1/2))/b^(1/2)+c/(-1/a^(1/2)*b^(1/2))^(1/2)*(1/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(-1
/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(b*x^4-a)^(1/2)*EllipticF(x*(-1/a^(1/2)*b^(1/2))^(1/2),I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x + c}{\sqrt {b x^{4} - a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)/sqrt(b*x^4 - a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {c+d\,x}{\sqrt {b\,x^4-a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(b*x^4 - a)^(1/2),x)

[Out]

int((c + d*x)/(b*x^4 - a)^(1/2), x)

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sympy [A]  time = 2.89, size = 90, normalized size = 1.01 \[ d \left (\begin {cases} \frac {\operatorname {acosh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} & \text {for}\: \left |{\frac {b x^{4}}{a}}\right | > 1 \\- \frac {i \operatorname {asin}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} & \text {otherwise} \end {cases}\right ) - \frac {i c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4}}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*Piecewise((acosh(sqrt(b)*x**2/sqrt(a))/(2*sqrt(b)), Abs(b*x**4/a) > 1), (-I*asin(sqrt(b)*x**2/sqrt(a))/(2*sq
rt(b)), True)) - I*c*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), b*x**4/a)/(4*sqrt(a)*gamma(5/4))

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