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

Optimal. Leaf size=87 \[ \frac {\sqrt [4]{a} c \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{b} \sqrt {a-b x^4}}+\frac {d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}\right )}{2 \sqrt {b}} \]

[Out]

1/2*d*arctan(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)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1885, 224, 221, 275, 217, 203} \[ \frac {\sqrt [4]{a} c \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{b} \sqrt {a-b x^4}}+\frac {d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}\right )}{2 \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/Sqrt[a - b*x^4],x]

[Out]

(d*ArcTan[(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 203

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

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 81, normalized size = 0.93 \[ \frac {c x \sqrt {1-\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {b x^4}{a}\right )}{\sqrt {a-b x^4}}+\frac {d \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}\right )}{2 \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*ArcTan[(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]

________________________________________________________________________________________

fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-b x^{4} + a} {\left (d x + c\right )}}{b x^{4} - a}, x\right ) \]

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

________________________________________________________________________________________

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)

________________________________________________________________________________________

maple [A]  time = 0.17, size = 90, normalized size = 1.03 \[ \frac {\sqrt {-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, c \EllipticF \left (\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}+\frac {d \arctan \left (\frac {\sqrt {b}\, x^{2}}{\sqrt {-b \,x^{4}+a}}\right )}{2 \sqrt {b}} \]

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

________________________________________________________________________________________

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a - b*x^4)^(1/2),x)

[Out]

int((c + d*x)/(a - b*x^4)^(1/2), x)

________________________________________________________________________________________

sympy [A]  time = 2.96, size = 95, normalized size = 1.09 \[ d \left (\begin {cases} - \frac {i \operatorname {acosh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} & \text {for}\: \left |{\frac {b x^{4}}{a}}\right | > 1 \\\frac {\operatorname {asin}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} & \text {otherwise} \end {cases}\right ) + \frac {c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} \]

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

________________________________________________________________________________________