3.30.15 \(\int \frac {-b+a x^4}{\sqrt {b+a x^4} (b-c^2 x^2+a x^4)} \, dx\)

Optimal. Leaf size=329 \[ -\frac {\left (\sqrt {2 \sqrt {a} \sqrt {b}+c^2}+c\right ) \sqrt {c \sqrt {2 \sqrt {a} \sqrt {b}+c^2}-\sqrt {a} \sqrt {b}-c^2} \tan ^{-1}\left (\frac {\sqrt {2} x \sqrt {c \sqrt {2 \sqrt {a} \sqrt {b}+c^2}-\sqrt {a} \sqrt {b}-c^2}}{\sqrt {a x^4+b}+\sqrt {a} x^2+\sqrt {b}}\right )}{\sqrt {2} \sqrt {a} \sqrt {b} c}-\frac {\left (\sqrt {2 \sqrt {a} \sqrt {b}+c^2}-c\right ) \sqrt {c \sqrt {2 \sqrt {a} \sqrt {b}+c^2}+\sqrt {a} \sqrt {b}+c^2} \tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt {c \sqrt {2 \sqrt {a} \sqrt {b}+c^2}+\sqrt {a} \sqrt {b}+c^2}}{\sqrt {a x^4+b}+\sqrt {a} x^2+\sqrt {b}}\right )}{\sqrt {2} \sqrt {a} \sqrt {b} c} \]

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Rubi [A]  time = 0.15, antiderivative size = 20, normalized size of antiderivative = 0.06, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2112, 208} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {c x}{\sqrt {a x^4+b}}\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[(c*x)/Sqrt[b + a*x^4]]/c)

Rule 208

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

Rule 2112

Int[((u_)*((A_) + (B_.)*(x_)^4))/Sqrt[v_], x_Symbol] :> With[{a = Coeff[v, x, 0], b = Coeff[v, x, 2], c = Coef
f[v, x, 4], d = Coeff[1/u, x, 0], e = Coeff[1/u, x, 2], f = Coeff[1/u, x, 4]}, Dist[A, Subst[Int[1/(d - (b*d -
a*e)*x^2), x], x, x/Sqrt[v]], x] /; EqQ[a*B + A*c, 0] && EqQ[c*d - a*f, 0]] /; FreeQ[{A, B}, x] && PolyQ[v, x
^2, 2] && PolyQ[1/u, x^2, 2]

Rubi steps

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

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Mathematica [C]  time = 0.95, size = 199, normalized size = 0.60 \begin {gather*} -\frac {i \sqrt {\frac {a x^4}{b}+1} \left (-\Pi \left (\frac {2 i \sqrt {a} \sqrt {b}}{c^2-\sqrt {c^4-4 a b}};\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}} x\right )\right |-1\right )-\Pi \left (\frac {2 i \sqrt {a} \sqrt {b}}{c^2+\sqrt {c^4-4 a b}};\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}} x\right )\right |-1\right )+F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}} x\right )\right |-1\right )\right )}{\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}} \sqrt {a x^4+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[1 + (a*x^4)/b]*(EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[b]]*x], -1] - EllipticPi[((2*I)*Sqrt[a]*S
qrt[b])/(c^2 - Sqrt[-4*a*b + c^4]), I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[b]]*x], -1] - EllipticPi[((2*I)*Sqrt[a]*Sq
rt[b])/(c^2 + Sqrt[-4*a*b + c^4]), I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[b]]*x], -1]))/(Sqrt[(I*Sqrt[a])/Sqrt[b]]*Sq
rt[b + a*x^4])

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IntegrateAlgebraic [A]  time = 1.26, size = 52, normalized size = 0.16 \begin {gather*} -\frac {\log \left (c x+\sqrt {b+a x^4}\right )}{2 c}+\frac {\log \left (c^2 x-c \sqrt {b+a x^4}\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 51, normalized size = 0.16 \begin {gather*} \frac {\log \left (\frac {a x^{4} + c^{2} x^{2} - 2 \, \sqrt {a x^{4} + b} c x + b}{a x^{4} - c^{2} x^{2} + b}\right )}{2 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log((a*x^4 + c^2*x^2 - 2*sqrt(a*x^4 + b)*c*x + b)/(a*x^4 - c^2*x^2 + b))/c

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{4} - b}{{\left (a x^{4} - c^{2} x^{2} + b\right )} \sqrt {a x^{4} + b}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 - b)/((a*x^4 - c^2*x^2 + b)*sqrt(a*x^4 + b)), x)

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maple [A]  time = 0.23, size = 23, normalized size = 0.07

method	result	size
elliptic	\(-\frac {\arctanh \left (\frac {\sqrt {a \,x^{4}+b}}{x c}\right )}{c}\)	\(23\)
default	\(\frac {\sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )}{\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \sqrt {a \,x^{4}+b}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (a \,\textit {\_Z}^{4}-c^{2} \textit {\_Z}^{2}+b \right )}{\sum }\frac {\left (c^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}-2 b \right ) \left (-\frac {\arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{2} a +a \,x^{2}+c^{2}\right )}{\sqrt {c^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {a \,x^{4}+b}}\right )}{\sqrt {c^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}}}+\frac {2 \underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2} a -c^{2}\right ) \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \EllipticPi \left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, \frac {i \left (\underline {\hspace {1.25 ex}}\alpha ^{2} a -c^{2}\right )}{\sqrt {b}\, \sqrt {a}}, \frac {\sqrt {-\frac {i \sqrt {a}}{\sqrt {b}}}}{\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\right )}{\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b \sqrt {a \,x^{4}+b}}\right )}{\underline {\hspace {1.25 ex}}\alpha \left (2 \underline {\hspace {1.25 ex}}\alpha ^{2} a -c^{2}\right )}\right )}{4}\)	\(297\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/c*arctanh((a*x^4+b)^(1/2)/x/c)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{4} - b}{{\left (a x^{4} - c^{2} x^{2} + b\right )} \sqrt {a x^{4} + b}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 - b)/((a*x^4 - c^2*x^2 + b)*sqrt(a*x^4 + b)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {b-a\,x^4}{\sqrt {a\,x^4+b}\,\left (-c^2\,x^2+a\,x^4+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{4} - b}{\sqrt {a x^{4} + b} \left (a x^{4} + b - c^{2} x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x**4 - b)/(sqrt(a*x**4 + b)*(a*x**4 + b - c**2*x**2)), x)

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