\(\int \frac {x^3}{\sqrt {a+b (c+d x)^4}} \, dx\) [123]

Optimal result

Integrand size = 19, antiderivative size = 361 \[ \int \frac {x^3}{\sqrt {a+b (c+d x)^4}} \, dx=\frac {\sqrt {a+b (c+d x)^4}}{2 b d^4}-\frac {3 c (c+d x) \sqrt {a+b (c+d x)^4}}{\sqrt {b} d^4 \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )}+\frac {3 c^2 \text {arctanh}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a+b (c+d x)^4}}\right )}{2 \sqrt {b} d^4}+\frac {3 \sqrt [4]{a} c \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} d^4 \sqrt {a+b (c+d x)^4}}-\frac {c \left (3 \sqrt {a}+\sqrt {b} c^2\right ) \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} b^{3/4} d^4 \sqrt {a+b (c+d x)^4}} \] Output:

1/2*(a+b*(d*x+c)^4)^(1/2)/b/d^4-3*c*(d*x+c)*(a+b*(d*x+c)^4)^(1/2)/b^(1/2)/ 
d^4/(a^(1/2)+b^(1/2)*(d*x+c)^2)+3/2*c^2*arctanh(b^(1/2)*(d*x+c)^2/(a+b*(d* 
x+c)^4)^(1/2))/b^(1/2)/d^4+3*a^(1/4)*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)*EllipticE(sin(2*arctan(b^(1 
/4)*(d*x+c)/a^(1/4))),1/2*2^(1/2))/b^(3/4)/d^4/(a+b*(d*x+c)^4)^(1/2)-1/2*c 
*(3*a^(1/2)+b^(1/2)*c^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^(3/4)/d^4/(a+b*(d*x+c)^4)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 15.65 (sec) , antiderivative size = 1258, normalized size of antiderivative = 3.48 \[ \int \frac {x^3}{\sqrt {a+b (c+d x)^4}} \, dx =\text {Too large to display} \] Input:

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

Output:

(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 + ( 
-6*(-1)^(3/4)*a*b^(1/4)*c - (6*I)*a^(3/4)*Sqrt[b]*c*(c + d*x) - 6*(-1)^(1/ 
4)*Sqrt[a]*b^(3/4)*c*(c + d*x)^2 - 6*a^(1/4)*b*c*(c + d*x)^3 - (1 + I)*Sqr 
t[2]*b^(3/4)*c*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[-(((-1)^(3/4)*a^(1/4) + I*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)*c - b^(1/4)*c*(c + d*x))^2*EllipticF[ArcSin[S 
qrt[-(((-1)^(3/4)*a^(1/4) + I*b^(1/4)*(c + d*x))/((-1)^(1/4)*a^(1/4) - b^( 
1/4)*(c + d*x)))]], -1] - 6*(-1)^(1/4)*Sqrt[b]*c^2*((-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]) + (3*I)*b^(1/4)*c*((-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/...
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 349, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, 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.

\(\displaystyle \int \frac {x^3}{\sqrt {a+b (c+d x)^4}} \, dx\)

\(\Big \downarrow \) 896

\(\displaystyle \frac {\int \frac {d^3 x^3}{\sqrt {b (c+d x)^4+a}}d(c+d x)}{d^4}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int -\frac {d^3 x^3}{\sqrt {b (c+d x)^4+a}}d(c+d x)}{d^4}\)

\(\Big \downarrow \) 2424

\(\displaystyle -\frac {\int \left (\frac {(c+d x) \left (-3 c^2-(c+d x)^2\right )}{\sqrt {b (c+d x)^4+a}}+\frac {c^3+3 (c+d x)^2 c}{\sqrt {b (c+d x)^4+a}}\right )d(c+d x)}{d^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {\sqrt [4]{a} c \left (\frac {\sqrt {b} c^2}{\sqrt {a}}+3\right ) \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b (c+d x)^4}}+\frac {3 \sqrt [4]{a} c \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right ) \sqrt {\frac {a+b (c+d x)^4}{\left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b (c+d x)^4}}+\frac {3 c^2 \text {arctanh}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a+b (c+d x)^4}}\right )}{2 \sqrt {b}}-\frac {3 c (c+d x) \sqrt {a+b (c+d x)^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} (c+d x)^2\right )}+\frac {\sqrt {a+b (c+d x)^4}}{2 b}}{d^4}\)

Input:

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

Output:

(Sqrt[a + b*(c + d*x)^4]/(2*b) - (3*c*(c + d*x)*Sqrt[a + b*(c + d*x)^4])/( 
Sqrt[b]*(Sqrt[a] + Sqrt[b]*(c + d*x)^2)) + (3*c^2*ArcTanh[(Sqrt[b]*(c + d* 
x)^2)/Sqrt[a + b*(c + d*x)^4]])/(2*Sqrt[b]) + (3*a^(1/4)*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]*EllipticE[2*ArcTan[(b^(1/4)*(c + d*x))/a^(1/4)], 1/2])/(b^(3/4)*Sqrt[a 
+ b*(c + d*x)^4]) - (a^(1/4)*c*(3 + (Sqrt[b]*c^2)/Sqrt[a])*(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*b^(3/4)*Sqrt[ 
a + b*(c + d*x)^4]))/d^4
 

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 = 18.93 (sec) , antiderivative size = 4787, normalized size of antiderivative = 13.26

method	result	size
risch	\(\text {Expression too large to display}\)	\(4787\)
default	\(\text {Expression too large to display}\)	\(4788\)
elliptic	\(\text {Expression too large to display}\)	\(4788\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(x^3/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]

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

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

Output:

Integral(x**3/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]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\sqrt {a+b (c+d x)^4}} \, dx=\int \frac {x^3}{\sqrt {a+b\,{\left (c+d\,x\right )}^4}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

(3*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)*c 
**2 - 6*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**3*d + 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)*x**3)/(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*d**4 - 6*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)*x)/(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**2*d**2)/(2*b*d**4)