Integrand size = 51, antiderivative size = 184 \[ \int \frac {x}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}}\right )}{2 \sqrt {b} d^2}-\frac {c \left (\sqrt {a}+\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2\right ) \sqrt {\frac {a+b d^4 \left (\frac {c}{d}+x\right )^4}{\left (\sqrt {a}+\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}} \]
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)
Result contains complex when optimal does not.
Time = 10.36 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.11 \[ \int \frac {x}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\frac {\sqrt [4]{-1} \left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)\right )^2 \sqrt {\frac {(1-i) \left ((-1)^{3/4} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \sqrt {-\frac {(1+i) \left ((-1)^{3/4} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \left (\left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} c\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right ),-1\right )-2 \sqrt [4]{-1} \sqrt [4]{a} \operatorname {EllipticPi}\left (-i,\arcsin \left (\sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right ),-1\right )\right )}{\sqrt [4]{a} \sqrt {b} d^2 \sqrt {a+b (c+d x)^4}} \]
((-1)^(1/4)*((-1)^(1/4)*a^(1/4) - b^(1/4)*(c + d*x))^2*Sqrt[((1 - I)*((-1) ^(3/4)*a^(1/4) - b^(1/4)*(c + d*x)))/((-1)^(1/4)*a^(1/4) - b^(1/4)*(c + d* x))]*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))]*Sqrt[((-1 - I)*((-1)^(3/4)*a^(1/4) + b^(1/4)*(c + d*x)))/((-1)^(1/4)*a^(1/4) - b^(1/4)*(c + d*x))]*(((-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)*EllipticPi[-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[a + b*(c + d*x)^4])
Time = 0.38 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2459, 2424, 2009}
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.
| \(\displaystyle \int \frac {x}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx\) |
| \(\Big \downarrow \) 2459 |
| \(\displaystyle \int \frac {x}{\sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}}d\left (\frac {c}{d}+x\right )\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \int \left (\frac {\frac {c}{d}+x}{\sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}}-\frac {c}{d \sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}}\right )d\left (\frac {c}{d}+x\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}}\right )}{2 \sqrt {b} d^2}-\frac {c \left (\sqrt {a}+\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2\right ) \sqrt {\frac {a+b d^4 \left (\frac {c}{d}+x\right )^4}{\left (\sqrt {a}+\sqrt {b} d^2 \left (\frac {c}{d}+x\right )^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} d \left (\frac {c}{d}+x\right )}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt {a+b d^4 \left (\frac {c}{d}+x\right )^4}}\) |
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)*d*(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])
3.11.2.3.1 Defintions of rubi rules used
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[Pq, 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)]
Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
Result contains complex when optimal does not.
Time = 4.99 (sec) , antiderivative size = 1528, normalized size of antiderivative = 8.30
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1528\) |
| elliptic | \(\text {Expression too large to display}\) | \(1528\) |
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)
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)*((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)*(-(-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))+(-(I/b*(-a*b^3)^(1/4)-c)/d+(1/b*(-a*b^3)^(1/4)-c)/d)*EllipticP i((((-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...
\[ \int \frac {x}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int { \frac {x}{\sqrt {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}} \,d x } \]
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")
\[ \int \frac {x}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int \frac {x}{\sqrt {a + b c^{4} + 4 b c^{3} d x + 6 b c^{2} d^{2} x^{2} + 4 b c d^{3} x^{3} + b d^{4} x^{4}}}\, dx \]
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)
\[ \int \frac {x}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int { \frac {x}{\sqrt {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}} \,d x } \]
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")
\[ \int \frac {x}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int { \frac {x}{\sqrt {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}} \,d x } \]
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")
Timed out. \[ \int \frac {x}{\sqrt {a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx=\int \frac {x}{\sqrt {b\,c^4+4\,b\,c^3\,d\,x+6\,b\,c^2\,d^2\,x^2+4\,b\,c\,d^3\,x^3+b\,d^4\,x^4+a}} \,d x \]