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

Optimal. Leaf size=127 \[ \frac{c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt{-a-b x^4}}+\frac{d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{-a-b x^4}}\right )}{2 \sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.0721643, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1885, 220, 275, 217, 203} \[ \frac{c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt{-a-b x^4}}+\frac{d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{-a-b x^4}}\right )}{2 \sqrt{b}} \]

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

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)]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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 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 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])

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

Mathematica [C]  time = 0.0451096, size = 85, normalized size = 0.67 \[ \frac{c x \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^4}{a}\right )}{\sqrt{-a-b x^4}}+\frac{d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{-a-b x^4}}\right )}{2 \sqrt{b}} \]

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]

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Maple [C]  time = 0.018, size = 101, normalized size = 0.8 \begin{align*}{\frac{d}{2}\arctan \left ({{x}^{2}\sqrt{b}{\frac{1}{\sqrt{-b{x}^{4}-a}}}} \right ){\frac{1}{\sqrt{b}}}}+{c\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{-i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{-i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{-b{x}^{4}-a}}}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{\sqrt{-b x^{4} - a}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-b x^{4} - a}{\left (d x + c\right )}}{b x^{4} + a}, x\right ) \end{align*}

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)

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Sympy [C]  time = 1.96542, size = 66, normalized size = 0.52 \begin{align*} - \frac{i d \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2 \sqrt{b}} - \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} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*d*asinh(sqrt(b)*x**2/sqrt(a))/(2*sqrt(b)) - I*c*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), b*x**4*exp_polar(I*p
i)/a)/(4*sqrt(a)*gamma(5/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{\sqrt{-b x^{4} - a}}\,{d x} \end{align*}

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)