Integrand size = 49, antiderivative size = 111 \[ \int \frac {1}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\frac {\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d \sqrt {a+b (c+d x)^4}} \] Output:
1/2*(a^(1/2)+b^(1/2)*(d*x+c)^2)*((a+b*(d*x+c)^4)/(a^(1/2)+b^(1/2)*(d*x+c)^ 2)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*(d*x+c)/a^(1/4)),1/2*2^(1/2)) /a^(1/4)/b^(1/4)/d/(a+b*(d*x+c)^4)^(1/2)
Result contains complex when optimal does not.
Time = 20.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=-\frac {i \sqrt {\frac {a+b (c+d x)^4}{a}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} (c+d x)\right ),-1\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} d \sqrt {a+b (c+d x)^4}} \] Input:
Integrate[1/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]
Output:
((-I)*Sqrt[(a + b*(c + d*x)^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqr t[a]]*(c + d*x)], -1])/(Sqrt[(I*Sqrt[b])/Sqrt[a]]*d*Sqrt[a + b*(c + d*x)^4 ])
Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.21, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {2458, 761}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx\) |
\(\Big \downarrow \) 2458 |
\(\displaystyle \int \frac {1}{\sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}}d\left (\frac {c}{d}+x\right )\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\left (\sqrt {a}+\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2\right ) \sqrt {\frac {a+b d^4 \left (\frac {c}{d}+x\right )^4}{\left (\sqrt {a}+\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} d \left (\frac {c}{d}+x\right )}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d \sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}}\) |
Input:
Int[1/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]
Output:
((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)*d*(c/d + x))/a^ (1/4)], 1/2])/(2*a^(1/4)*b^(1/4)*d*Sqrt[a + b*d^4*(c/d + x)^4])
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]
Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
Result contains complex when optimal does not.
Time = 1.43 (sec) , antiderivative size = 1036, normalized size of antiderivative = 9.33
method | result | size |
default | \(\text {Expression too large to display}\) | \(1036\) |
elliptic | \(\text {Expression too large to display}\) | \(1036\) |
Input:
int(1/(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)
Output:
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)*(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)/(b*d^4*(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)*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)*(-(-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-(-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))
\[ \int \frac {1}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int { \frac {1}{\sqrt {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}} \,d x } \] Input:
integrate(1/(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")
Output:
integral(1/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)
\[ \int \frac {1}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int \frac {1}{\sqrt {a + b c^{4} + 4 b c^{3} d x + 6 b c^{2} d^{2} x^{2} + 4 b c d^{3} x^{3} + b d^{4} x^{4}}}\, dx \] Input:
integrate(1/(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)
Output:
Integral(1/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)
\[ \int \frac {1}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int { \frac {1}{\sqrt {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}} \,d x } \] Input:
integrate(1/(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")
Output:
integrate(1/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)
\[ \int \frac {1}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int { \frac {1}{\sqrt {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}} \,d x } \] Input:
integrate(1/(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")
Output:
integrate(1/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)
Timed out. \[ \int \frac {1}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int \frac {1}{\sqrt {b\,c^4+4\,b\,c^3\,d\,x+6\,b\,c^2\,d^2\,x^2+4\,b\,c\,d^3\,x^3+b\,d^4\,x^4+a}} \,d x \] Input:
int(1/(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)
Output:
int(1/(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)
\[ \int \frac {1}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int \frac {\sqrt {b \,d^{4} x^{4}+4 b c \,d^{3} x^{3}+6 b \,c^{2} d^{2} x^{2}+4 b \,c^{3} d x +b \,c^{4}+a}}{b \,d^{4} x^{4}+4 b c \,d^{3} x^{3}+6 b \,c^{2} d^{2} x^{2}+4 b \,c^{3} d x +b \,c^{4}+a}d x \] Input:
int(1/(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)
Output:
int(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)/(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)