\(\int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx\) [125]

Optimal result

Integrand size = 17, antiderivative size = 154 \[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a+b (c+d x)^4}}\right )}{2 \sqrt {b} d^2}-\frac {c \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d^2 \sqrt {a+b (c+d x)^4}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 10.38 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.53 \[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\frac {\sqrt [4]{-1} \left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)\right )^2 \sqrt {\frac {(1-i) \left ((-1)^{3/4} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \sqrt {-\frac {(1+i) \left ((-1)^{3/4} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}} \left (\left (\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} c\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right ),-1\right )-2 \sqrt [4]{-1} \sqrt [4]{a} \operatorname {EllipticPi}\left (-i,\arcsin \left (\sqrt {-\frac {i \left (\sqrt [4]{-1} \sqrt [4]{a}+\sqrt [4]{b} (c+d x)\right )}{\sqrt [4]{-1} \sqrt [4]{a}-\sqrt [4]{b} (c+d x)}}\right ),-1\right )\right )}{\sqrt [4]{a} \sqrt {b} d^2 \sqrt {a+b (c+d x)^4}} \] Input:

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

Output:

((-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])
 

Rubi [A] (verified)

Time = 0.47 (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

\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a+b (c+d x)^4}}\right )}{2 \sqrt {b}}-\frac {c \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt {a+b (c+d x)^4}}}{d^2}\)

Input:

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

Output:

(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
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 896

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2424

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)]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.63 (sec) , antiderivative size = 1528, normalized size of antiderivative = 9.92

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

Input:

int(x/(a+b*(d*x+c)^4)^(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)/(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)*(-(-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...
 

Fricas [F]

\[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\int { \frac {x}{\sqrt {{\left (d x + c\right )}^{4} b + a}} \,d x } \] Input:

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

Output:

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)
 

Sympy [F]

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

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

Output:

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)
 

Maxima [F]

\[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\int { \frac {x}{\sqrt {{\left (d x + c\right )}^{4} b + a}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\int { \frac {x}{\sqrt {{\left (d x + c\right )}^{4} b + a}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\int \frac {x}{\sqrt {a+b\,{\left (c+d\,x\right )}^4}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x}{\sqrt {a+b (c+d x)^4}} \, dx=\frac {\sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,d^{4} x^{4}+4 b c \,d^{3} x^{3}+6 b \,c^{2} d^{2} x^{2}+4 b \,c^{3} d x +b \,c^{4}+a}-b \,c^{2}-2 b c d x -b \,d^{2} x^{2}\right )-2 \left (\int \frac {\sqrt {b \,d^{4} x^{4}+4 b c \,d^{3} x^{3}+6 b \,c^{2} d^{2} x^{2}+4 b \,c^{3} d x +b \,c^{4}+a}}{b \,d^{4} x^{4}+4 b c \,d^{3} x^{3}+6 b \,c^{2} d^{2} x^{2}+4 b \,c^{3} d x +b \,c^{4}+a}d x \right ) b c d}{2 b \,d^{2}} \] Input:

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

Output:

(sqrt(b)*log( - sqrt(b)*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) - b*c**2 - 2*b*c*d*x - b*d**2*x**2) - 2 
*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)*b*c*d)/(2*b*d**2)