3.26.10 \(\int (d+e x)^3 \sqrt [4]{a+b x+c x^2} \, dx\) [2510]

3.26.10.1 Optimal result

Integrand size = 22, antiderivative size = 374 \[ \int (d+e x)^3 \sqrt [4]{a+b x+c x^2} \, dx=\frac {(2 c d-b e) \left (28 c^2 d^2+13 b^2 e^2-4 c e (7 b d+6 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{168 c^4}+\frac {2 e (d+e x)^2 \left (a+b x+c x^2\right )^{5/4}}{9 c}+\frac {e \left (616 c^2 d^2+117 b^2 e^2-2 c e (243 b d+56 a e)+130 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{5/4}}{630 c^3}-\frac {\left (b^2-4 a c\right )^{5/4} (2 c d-b e) \left (28 c^2 d^2+13 b^2 e^2-4 c e (7 b d+6 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{336 \sqrt {2} c^{17/4} (b+2 c x)} \]

output

1/168*(-b*e+2*c*d)*(28*c^2*d^2+13*b^2*e^2-4*c*e*(6*a*e+7*b*d))*(2*c*x+b)*( 
c*x^2+b*x+a)^(1/4)/c^4+2/9*e*(e*x+d)^2*(c*x^2+b*x+a)^(5/4)/c+1/630*e*(616* 
c^2*d^2+117*b^2*e^2-2*c*e*(56*a*e+243*b*d)+130*c*e*(-b*e+2*c*d)*x)*(c*x^2+ 
b*x+a)^(5/4)/c^3-1/672*(-4*a*c+b^2)^(5/4)*(-b*e+2*c*d)*(28*c^2*d^2+13*b^2* 
e^2-4*c*e*(6*a*e+7*b*d))*(cos(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2) 
/(-4*a*c+b^2)^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^ 
(1/2)/(-4*a*c+b^2)^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1 
/4)*2^(1/2)/(-4*a*c+b^2)^(1/4))),1/2*2^(1/2))*(1+2*c^(1/2)*(c*x^2+b*x+a)^( 
1/2)/(-4*a*c+b^2)^(1/2))*((2*c*x+b)^2/(-4*a*c+b^2)/(1+2*c^(1/2)*(c*x^2+b*x 
+a)^(1/2)/(-4*a*c+b^2)^(1/2))^2)^(1/2)/c^(17/4)/(2*c*x+b)*2^(1/2)
 

3.26.10.2 Mathematica [A] (verified)

Time = 9.30 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.63 \[ \int (d+e x)^3 \sqrt [4]{a+b x+c x^2} \, dx=\frac {1120 c^4 e (d+e x)^2 (a+x (b+c x))^2+8 c^2 e (a+x (b+c x))^2 \left (117 b^2 e^2+4 c^2 d (154 d+65 e x)-2 c e (243 b d+56 a e+65 b e x)\right )+15 (2 c d-b e) \left (28 c^2 d^2+13 b^2 e^2-4 c e (7 b d+6 a e)\right ) \left (2 c (b+2 c x) (a+x (b+c x))-\sqrt {2} \left (b^2-4 a c\right )^{3/2} \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )\right )}{5040 c^5 (a+x (b+c x))^{3/4}} \]

input

Integrate[(d + e*x)^3*(a + b*x + c*x^2)^(1/4),x]
 

output

(1120*c^4*e*(d + e*x)^2*(a + x*(b + c*x))^2 + 8*c^2*e*(a + x*(b + c*x))^2* 
(117*b^2*e^2 + 4*c^2*d*(154*d + 65*e*x) - 2*c*e*(243*b*d + 56*a*e + 65*b*e 
*x)) + 15*(2*c*d - b*e)*(28*c^2*d^2 + 13*b^2*e^2 - 4*c*e*(7*b*d + 6*a*e))* 
(2*c*(b + 2*c*x)*(a + x*(b + c*x)) - Sqrt[2]*(b^2 - 4*a*c)^(3/2)*((c*(a + 
x*(b + c*x)))/(-b^2 + 4*a*c))^(3/4)*EllipticF[ArcSin[(b + 2*c*x)/Sqrt[b^2 
- 4*a*c]]/2, 2]))/(5040*c^5*(a + x*(b + c*x))^(3/4))
 

3.26.10.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1166, 27, 1225, 1087, 1094, 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.

\(\displaystyle \int (d+e x)^3 \sqrt [4]{a+b x+c x^2} \, dx\)

\(\Big \downarrow \) 1166

\(\displaystyle \frac {2 \int \frac {1}{4} (d+e x) \left (18 c d^2-5 b e d-8 a e^2+13 e (2 c d-b e) x\right ) \sqrt [4]{c x^2+b x+a}dx}{9 c}+\frac {2 e (d+e x)^2 \left (a+b x+c x^2\right )^{5/4}}{9 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (d+e x) \left (18 c d^2-5 b e d-8 a e^2+13 e (2 c d-b e) x\right ) \sqrt [4]{c x^2+b x+a}dx}{18 c}+\frac {2 e (d+e x)^2 \left (a+b x+c x^2\right )^{5/4}}{9 c}\)

\(\Big \downarrow \) 1225

\(\displaystyle \frac {\frac {9 (2 c d-b e) \left (-4 c e (6 a e+7 b d)+13 b^2 e^2+28 c^2 d^2\right ) \int \sqrt [4]{c x^2+b x+a}dx}{28 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/4} \left (-2 c e (56 a e+243 b d)+117 b^2 e^2+130 c e x (2 c d-b e)+616 c^2 d^2\right )}{35 c^2}}{18 c}+\frac {2 e (d+e x)^2 \left (a+b x+c x^2\right )^{5/4}}{9 c}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {\frac {9 (2 c d-b e) \left (-4 c e (6 a e+7 b d)+13 b^2 e^2+28 c^2 d^2\right ) \left (\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\left (c x^2+b x+a\right )^{3/4}}dx}{12 c}\right )}{28 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/4} \left (-2 c e (56 a e+243 b d)+117 b^2 e^2+130 c e x (2 c d-b e)+616 c^2 d^2\right )}{35 c^2}}{18 c}+\frac {2 e (d+e x)^2 \left (a+b x+c x^2\right )^{5/4}}{9 c}\)

\(\Big \downarrow \) 1094

\(\displaystyle \frac {\frac {9 (2 c d-b e) \left (-4 c e (6 a e+7 b d)+13 b^2 e^2+28 c^2 d^2\right ) \left (\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{3 c (b+2 c x)}\right )}{28 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/4} \left (-2 c e (56 a e+243 b d)+117 b^2 e^2+130 c e x (2 c d-b e)+616 c^2 d^2\right )}{35 c^2}}{18 c}+\frac {2 e (d+e x)^2 \left (a+b x+c x^2\right )^{5/4}}{9 c}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {\frac {9 (2 c d-b e) \left (-4 c e (6 a e+7 b d)+13 b^2 e^2+28 c^2 d^2\right ) \left (\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {(b+2 c x)^2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} c^{5/4} (b+2 c x) \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}\right )}{28 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/4} \left (-2 c e (56 a e+243 b d)+117 b^2 e^2+130 c e x (2 c d-b e)+616 c^2 d^2\right )}{35 c^2}}{18 c}+\frac {2 e (d+e x)^2 \left (a+b x+c x^2\right )^{5/4}}{9 c}\)

input

Int[(d + e*x)^3*(a + b*x + c*x^2)^(1/4),x]
 

output

(2*e*(d + e*x)^2*(a + b*x + c*x^2)^(5/4))/(9*c) + ((e*(616*c^2*d^2 + 117*b 
^2*e^2 - 2*c*e*(243*b*d + 56*a*e) + 130*c*e*(2*c*d - b*e)*x)*(a + b*x + c* 
x^2)^(5/4))/(35*c^2) + (9*(2*c*d - b*e)*(28*c^2*d^2 + 13*b^2*e^2 - 4*c*e*( 
7*b*d + 6*a*e))*(((b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(3*c) - ((b^2 - 4*a 
*c)^(5/4)*Sqrt[(b + 2*c*x)^2]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[ 
b^2 - 4*a*c])*Sqrt[(b^2 - 4*a*c + 4*c*(a + b*x + c*x^2))/((b^2 - 4*a*c)*(1 
 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*EllipticF[2*Ar 
cTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2]) 
/(6*Sqrt[2]*c^(5/4)*(b + 2*c*x)*Sqrt[b^2 - 4*a*c + 4*c*(a + b*x + c*x^2)]) 
))/(28*c^2))/(18*c)
 

3.26.10.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 761

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]
 

rule 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1094

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[4*(Sqrt[(b 
+ 2*c*x)^2]/(b + 2*c*x))   Subst[Int[x^(4*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4 
*c*x^4], x], x, (a + b*x + c*x^2)^(1/4)], x] /; FreeQ[{a, b, c}, x] && Inte 
gerQ[4*p]
 

rule 1166

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[1/(c*(m + 2*p + 1))   Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m 
+ 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* 
(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration 
alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat 
icQ[a, b, c, d, e, m, p, x]
 

rule 1225

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( 
x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 
 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), 
 x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p 
+ 3))/(2*c^2*(2*p + 3))   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, p}, x] &&  !LeQ[p, -1]
 

3.26.10.4 Maple [F]

\[\int \left (e x +d \right )^{3} \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}d x\]

input

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

output

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

3.26.10.5 Fricas [F]

\[ \int (d+e x)^3 \sqrt [4]{a+b x+c x^2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{3} \,d x } \]

input

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

output

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

3.26.10.6 Sympy [F]

\[ \int (d+e x)^3 \sqrt [4]{a+b x+c x^2} \, dx=\int \left (d + e x\right )^{3} \sqrt [4]{a + b x + c x^{2}}\, dx \]

input

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

output

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

3.26.10.7 Maxima [F]

\[ \int (d+e x)^3 \sqrt [4]{a+b x+c x^2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{3} \,d x } \]

input

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

output

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

3.26.10.8 Giac [F]

\[ \int (d+e x)^3 \sqrt [4]{a+b x+c x^2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{3} \,d x } \]

input

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

output

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

3.26.10.9 Mupad [F(-1)]

Timed out. \[ \int (d+e x)^3 \sqrt [4]{a+b x+c x^2} \, dx=\int {\left (d+e\,x\right )}^3\,{\left (c\,x^2+b\,x+a\right )}^{1/4} \,d x \]

input

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

output

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