Integrand size = 17, antiderivative size = 154 \[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a+b (c+d x)^4}}\right )}{2 \sqrt {b} d^2}-\frac {c \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt {a+b (c+d x)^4}} \]
1/2*arctanh((d*x+c)^2*b^(1/2)/(a+b*(d*x+c)^4)^(1/2))/d^2/b^(1/2)-1/2*c*(co s(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*x+c)^2*b^(1/2))*((a+b*(d*x+c)^4)/(a^(1/2)+(d*x+c)^2*b^(1/2))^2) ^(1/2)/a^(1/4)/b^(1/4)/d^2/(a+b*(d*x+c)^4)^(1/2)
Result contains complex when optimal does not.
Time = 10.56 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.53 \[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\frac {\sqrt [4]{-1} \left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)\right )^2 \sqrt {\frac {(1-i) \left ((-1)^{3/4} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \sqrt {-\frac {(1+i) \left ((-1)^{3/4} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \left (\left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} c\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right ),-1\right )-2 \sqrt [4]{-1} \sqrt [4]{a} \operatorname {EllipticPi}\left (-i,\arcsin \left (\sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right ),-1\right )\right )}{\sqrt [4]{a} \sqrt {b} d^2 \sqrt {a+b (c+d x)^4}} \]
((-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 = 0.30 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {896, 25, 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.
\(\displaystyle \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx\) |
\(\Big \downarrow \) 896 |
\(\displaystyle \frac {\int \frac {d x}{\sqrt {b (c+d x)^4+a}}d(c+d x)}{d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int -\frac {d x}{\sqrt {b (c+d x)^4+a}}d(c+d x)}{d^2}\) |
\(\Big \downarrow \) 2424 |
\(\displaystyle -\frac {\int \left (\frac {c}{\sqrt {b (c+d x)^4+a}}-\frac {c+d x}{\sqrt {b (c+d x)^4+a}}\right )d(c+d x)}{d^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a+b (c+d x)^4}}\right )}{2 \sqrt {b}}-\frac {c \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt {a+b (c+d x)^4}}}{d^2}\) |
(ArcTanh[(Sqrt[b]*(c + d*x)^2)/Sqrt[a + b*(c + d*x)^4]]/(2*Sqrt[b]) - (c*( Sqrt[a] + Sqrt[b]*(c + d*x)^2)*Sqrt[(a + b*(c + d*x)^4)/(Sqrt[a] + Sqrt[b] *(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)*Sqrt[a + b*(c + d*x)^4]))/d^2
3.30.18.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1) Subst[Int[Si mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
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)]
Result contains complex when optimal does not.
Time = 2.08 (sec) , antiderivative size = 1528, normalized size of antiderivative = 9.92
method | result | size |
default | \(\text {Expression too large to display}\) | \(1528\) |
elliptic | \(\text {Expression too large to display}\) | \(1528\) |
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))+((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*...
\[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\int { \frac {x}{\sqrt {{\left (d x + c\right )}^{4} b + a}} \,d x } \]
\[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\int \frac {x}{\sqrt {a + b c^{4} + 4 b c^{3} d x + 6 b c^{2} d^{2} x^{2} + 4 b c d^{3} x^{3} + b d^{4} x^{4}}}\, dx \]
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)
\[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\int { \frac {x}{\sqrt {{\left (d x + c\right )}^{4} b + a}} \,d x } \]
\[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\int { \frac {x}{\sqrt {{\left (d x + c\right )}^{4} b + a}} \,d x } \]
Timed out. \[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\int \frac {x}{\sqrt {a+b\,{\left (c+d\,x\right )}^4}} \,d x \]