3.216 \(\int \frac {1}{(d+e x) \sqrt {a+c x^4}} \, dx\)

Optimal. Leaf size=405 \[ \frac {e \tan ^{-1}\left (\frac {x \sqrt {-a e^4-c d^4}}{d e \sqrt {a+c x^4}}\right )}{2 \sqrt {-a e^4-c d^4}}+\frac {\sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {e \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{2 \sqrt {a e^4+c d^4}} \]

[Out]

1/2*e*arctan(x*(-a*e^4-c*d^4)^(1/2)/d/e/(c*x^4+a)^(1/2))/(-a*e^4-c*d^4)^(1/2)-1/2*e*arctanh((c*d^2*x^2+a*e^2)/
(a*e^4+c*d^4)^(1/2)/(c*x^4+a)^(1/2))/(a*e^4+c*d^4)^(1/2)+1/2*c^(1/4)*d*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1
/2)/cos(2*arctan(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)/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+a)^(1/2)-1/4*(cos(2*ar
ctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticPi(sin(2*arctan(c^(1/4)*x/a^(1/4)))
,1/4*(e^2*a^(1/2)+d^2*c^(1/2))^2/d^2/e^2/a^(1/2)/c^(1/2),1/2*2^(1/2))*(-e^2*a^(1/2)+d^2*c^(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)/d/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 405, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1725, 1217, 220, 1707, 1248, 725, 206} \[ \frac {e \tan ^{-1}\left (\frac {x \sqrt {-a e^4-c d^4}}{d e \sqrt {a+c x^4}}\right )}{2 \sqrt {-a e^4-c d^4}}-\frac {e \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{2 \sqrt {a e^4+c d^4}}+\frac {\sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*Sqrt[a + c*x^4]),x]

[Out]

(e*ArcTan[(Sqrt[-(c*d^4) - a*e^4]*x)/(d*e*Sqrt[a + c*x^4])])/(2*Sqrt[-(c*d^4) - a*e^4]) - (e*ArcTanh[(a*e^2 +
c*d^2*x^2)/(Sqrt[c*d^4 + a*e^4]*Sqrt[a + c*x^4])])/(2*Sqrt[c*d^4 + a*e^4]) + (c^(1/4)*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)*(Sqrt[
c]*d^2 + Sqrt[a]*e^2)*Sqrt[a + c*x^4]) - ((Sqrt[c]*d^2 - Sqrt[a]*e^2)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)
/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticPi[(Sqrt[c]*d^2 + Sqrt[a]*e^2)^2/(4*Sqrt[a]*Sqrt[c]*d^2*e^2), 2*ArcTan[(c^
(1/4)*x)/a^(1/4)], 1/2])/(4*a^(1/4)*c^(1/4)*d*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*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 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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 1217

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(c*d + a*e*q
)/(c*d^2 - a*e^2), Int[1/Sqrt[a + c*x^4], x], x] - Dist[(a*e*(e + d*q))/(c*d^2 - a*e^2), Int[(1 + q*x^2)/((d +
e*x^2)*Sqrt[a + c*x^4]), x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0]
&& PosQ[c/a]

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1707

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, -Simp[((B*d - A*e)*ArcTan[(Rt[(c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + c*x^4]])/(2*d*e*Rt[(c*d)/e + (a*e)/d, 2]),
x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + c*x^4))/(a*(A + B*x^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2
/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2])/(4*d*e*A*q*Sqrt[a + c*x^4]), x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rule 1725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[d, Int[1/((d^2 - e^2*x^2)*Sqrt[a + c*
x^4]), x], x] - Dist[e, Int[x/((d^2 - e^2*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x]

Rubi steps

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

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Mathematica [C]  time = 0.26, size = 200, normalized size = 0.49 \[ \frac {\sqrt {\frac {c x^4}{a}+1} \left (\sqrt [4]{c} d \log \left (\frac {e^2 x^2-d^2}{a e^2 \left (\sqrt {\frac {c x^4}{a}+1} \sqrt {\frac {c d^4}{a e^4}+1}+1\right )+c d^2 x^2}\right )-2 \sqrt [4]{-1} \sqrt [4]{a} e \sqrt {\frac {c d^4}{a e^4}+1} \Pi \left (\frac {i \sqrt {a} e^2}{\sqrt {c} d^2};\left .\sin ^{-1}\left (\frac {(-1)^{3/4} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )\right )}{2 \sqrt [4]{c} d e \sqrt {a+c x^4} \sqrt {\frac {c d^4}{a e^4}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*Sqrt[a + c*x^4]),x]

[Out]

(Sqrt[1 + (c*x^4)/a]*(-2*(-1)^(1/4)*a^(1/4)*Sqrt[1 + (c*d^4)/(a*e^4)]*e*EllipticPi[(I*Sqrt[a]*e^2)/(Sqrt[c]*d^
2), ArcSin[((-1)^(3/4)*c^(1/4)*x)/a^(1/4)], -1] + c^(1/4)*d*Log[(-d^2 + e^2*x^2)/(c*d^2*x^2 + a*e^2*(1 + Sqrt[
1 + (c*d^4)/(a*e^4)]*Sqrt[1 + (c*x^4)/a]))]))/(2*c^(1/4)*d*Sqrt[1 + (c*d^4)/(a*e^4)]*e*Sqrt[a + c*x^4])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{4} + a} {\left (e x + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 169, normalized size = 0.42 \[ \frac {\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, e \EllipticPi \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, d}-\frac {\arctanh \left (\frac {\frac {2 c \,d^{2} x^{2}}{e^{2}}+2 a}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}}}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*(-1/2/(a+c*d^4/e^4)^(1/2)*arctanh(1/2*(2*c*d^2/e^2*x^2+2*a)/(a+c*d^4/e^4)^(1/2)/(c*x^4+a)^(1/2))+1/(I/a^(1
/2)*c^(1/2))^(1/2)/d*e*(-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)*Ellipt
icPi((I/a^(1/2)*c^(1/2))^(1/2)*x,-I*a^(1/2)/c^(1/2)/d^2*e^2,(-I/a^(1/2)*c^(1/2))^(1/2)/(I/a^(1/2)*c^(1/2))^(1/
2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{4} + a} {\left (e x + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + c x^{4}} \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(a + c*x**4)*(d + e*x)), x)

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