3.214 \(\int \frac {d+e x}{\sqrt {a+c x^4}} \, dx\)

Optimal. Leaf size=121 \[ \frac {d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}} \]

[Out]

1/2*e*arctanh(x^2*c^(1/2)/(c*x^4+a)^(1/2))/c^(1/2)+1/2*d*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arct
an(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a
)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/c^(1/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1885, 220, 275, 217, 206} \[ \frac {d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/Sqrt[a + c*x^4],x]

[Out]

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

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Mathematica [C]  time = 0.04, size = 79, normalized size = 0.65 \[ \frac {d x \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^4}{a}\right )}{\sqrt {a+c x^4}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e x + d}{\sqrt {c x^{4} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 96, normalized size = 0.79 \[ \frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, d \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {e \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{2 \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(a + c*x^4)^(1/2),x)

[Out]

int((d + e*x)/(a + c*x^4)^(1/2), x)

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sympy [C]  time = 2.32, size = 61, normalized size = 0.50 \[ \frac {e \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {c}} + \frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*asinh(sqrt(c)*x**2/sqrt(a))/(2*sqrt(c)) + d*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(I*pi)/a)
/(4*sqrt(a)*gamma(5/4))

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