3.11.2 \(\int \frac {x}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx\) [1002]

Optimal. Leaf size=184 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}}\right )}{2 \sqrt {b} d^2}-\frac {c \left (\sqrt {a}+\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2\right ) \sqrt {\frac {a+b d^4 \left (\frac {c}{d}+x\right )^4}{\left (\sqrt {a}+\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}} \]

[Out]

1/2*arctanh(d^2*(c/d+x)^2*b^(1/2)/(a+b*d^4*(c/d+x)^4)^(1/2))/d^2/b^(1/2)-1/2*c*(cos(2*arctan(b^(1/4)*(d*x+c)/a
^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(d*x+c)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*(d*x+c)/a^(1/4))),1/2*
2^(1/2))*(a^(1/2)+d^2*(c/d+x)^2*b^(1/2))*((a+b*d^4*(c/d+x)^4)/(a^(1/2)+d^2*(c/d+x)^2*b^(1/2))^2)^(1/2)/a^(1/4)
/b^(1/4)/d^2/(a+b*d^4*(c/d+x)^4)^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1694, 1899, 226, 281, 223, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}}\right )}{2 \sqrt {b} d^2}-\frac {c \left (\sqrt {a}+\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2\right ) \sqrt {\frac {a+b d^4 \left (\frac {c}{d}+x\right )^4}{\left (\sqrt {a}+\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b*c^4 + 4*b*c^3*d*x + 6*b*c^2*d^2*x^2 + 4*b*c*d^3*x^3 + b*d^4*x^4],x]

[Out]

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

Rule 212

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

Rule 223

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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 281

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 1694

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -d/(4*e) + x)*(a + d^4/(256*e^3)
- b*(d/(8*e)) + (c - 3*(d^2/(8*e)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0]
&& NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1899

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 {x}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx &=\text {Subst}\left (\int \frac {-\frac {c}{d}+x}{\sqrt {a+b d^4 x^4}} \, dx,x,\frac {c}{d}+x\right )\\ &=\text {Subst}\left (\int \left (-\frac {c}{d \sqrt {a+b d^4 x^4}}+\frac {x}{\sqrt {a+b d^4 x^4}}\right ) \, dx,x,\frac {c}{d}+x\right )\\ &=-\frac {c \text {Subst}\left (\int \frac {1}{\sqrt {a+b d^4 x^4}} \, dx,x,\frac {c}{d}+x\right )}{d}+\text {Subst}\left (\int \frac {x}{\sqrt {a+b d^4 x^4}} \, dx,x,\frac {c}{d}+x\right )\\ &=-\frac {c \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt {a+b (c+d x)^4}}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {a+b d^4 x^2}} \, dx,x,\left (\frac {c}{d}+x\right )^2\right )\\ &=-\frac {c \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt {a+b (c+d x)^4}}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-b d^4 x^2} \, dx,x,\frac {\left (\frac {c}{d}+x\right )^2}{\sqrt {a+b (c+d x)^4}}\right )\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a+b (c+d x)^4}}\right )}{2 \sqrt {b} d^2}-\frac {c \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt {a+b (c+d x)^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.40, size = 330, normalized size = 1.79 \begin {gather*} \frac {\sqrt [4]{-1} \sqrt {2} \sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \left (i \sqrt {a}+\sqrt {b} (c+d x)^2\right ) \left (\left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} c\right ) F\left (\left .\sin ^{-1}\left (\sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right )\right |-1\right )-2 \sqrt [4]{-1} \sqrt [4]{a} \Pi \left (-i;\left .\sin ^{-1}\left (\sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right )\right |-1\right )\right )}{\sqrt [4]{a} \sqrt {b} d^2 \sqrt {\frac {i \sqrt {a}+\sqrt {b} (c+d x)^2}{\left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)\right )^2}} \sqrt {a+b (c+d x)^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a + b*c^4 + 4*b*c^3*d*x + 6*b*c^2*d^2*x^2 + 4*b*c*d^3*x^3 + b*d^4*x^4],x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.56, size = 1528, normalized size = 8.30

method	result	size
default	\(\text {Expression too large to display}\)	\(1528\)
elliptic	\(\text {Expression too large to display}\)	\(1528\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*d^4*x^4+4*b*c*d^3*x^3+6*b*c^2*d^2*x^2+4*b*c^3*d*x+b*c^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*((1/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d)*(((-I/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)*(
x-(1/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))
^(1/2)*(x-(I/b*(-a*b^3)^(1/4)-c)/d)^2*(((I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)*(x-(-1/b*(-a*b^3)^(
1/4)-c)/d)/((-1/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)*(((I/b*(-a
*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)*(x-(-I/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b
^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)/((-I/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)/((I/b
*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(d^4*b*(x-(1/b*(-a*b^3)^(1/4)-c)/d)*(x-(I/b*(-a*b^3)^(1/4)-c)/d
)*(x-(-1/b*(-a*b^3)^(1/4)-c)/d)*(x-(-I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)*((I/b*(-a*b^3)^(1/4)-c)/d*EllipticF((((-I
/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)*(x-(1/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*
(-a*b^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))^(1/2),(((I/b*(-a*b^3)^(1/4)-c)/d-(-1/b*(-a*b^3)^(1/4)-c)/d)
*((1/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d)/((1/b*(-a*b^3)^(1/4)-c)/d-(-1/b*(-a*b^3)^(1/4)-c)/d)/((I
/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d))^(1/2))+((1/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)*
EllipticPi((((-I/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)*(x-(1/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^
(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))^(1/2),((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b
^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d),(((I/b*(-a*b^3)^(1/4)-c)/d-(-1/b*(-a*b^3)
^(1/4)-c)/d)*((1/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d)/((1/b*(-a*b^3)^(1/4)-c)/d-(-1/b*(-a*b^3)^(1/
4)-c)/d)/((I/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*d^4*x^4+4*b*c*d^3*x^3+6*b*c^2*d^2*x^2+4*b*c^3*d*x+b*c^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(x/sqrt(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*d^4*x^4+4*b*c*d^3*x^3+6*b*c^2*d^2*x^2+4*b*c^3*d*x+b*c^4+a)^(1/2),x, algorithm="fricas")

[Out]

integral(x/sqrt(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {a + b c^{4} + 4 b c^{3} d x + 6 b c^{2} d^{2} x^{2} + 4 b c d^{3} x^{3} + b d^{4} x^{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*d**4*x**4+4*b*c*d**3*x**3+6*b*c**2*d**2*x**2+4*b*c**3*d*x+b*c**4+a)**(1/2),x)

[Out]

Integral(x/sqrt(a + b*c**4 + 4*b*c**3*d*x + 6*b*c**2*d**2*x**2 + 4*b*c*d**3*x**3 + b*d**4*x**4), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*d^4*x^4+4*b*c*d^3*x^3+6*b*c^2*d^2*x^2+4*b*c^3*d*x+b*c^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate(x/sqrt(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{\sqrt {b\,c^4+4\,b\,c^3\,d\,x+6\,b\,c^2\,d^2\,x^2+4\,b\,c\,d^3\,x^3+b\,d^4\,x^4+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*c^4 + b*d^4*x^4 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + 4*b*c*d^3*x^3)^(1/2),x)

[Out]

int(x/(a + b*c^4 + b*d^4*x^4 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + 4*b*c*d^3*x^3)^(1/2), x)

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