\(\int (d+e x) (a+b x+c x^2)^{3/4} \, dx\) [730]

Optimal result

Integrand size = 20, antiderivative size = 159 \[ \int (d+e x) \left (a+b x+c x^2\right )^{3/4} \, dx=\frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{10 c^2}+\frac {2 e \left (a+b x+c x^2\right )^{7/4}}{7 c}-\frac {3 \left (b^2-4 a c\right )^{3/2} (2 c d-b e) \sqrt [4]{-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )\right |2\right )}{20 \sqrt {2} c^3 \sqrt [4]{a+b x+c x^2}} \] Output:

1/10*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x+a)^(3/4)/c^2+2/7*e*(c*x^2+b*x+a)^(7 
/4)/c-3/40*(-4*a*c+b^2)^(3/2)*(-b*e+2*c*d)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2)) 
^(1/4)*EllipticE(sin(1/2*arcsin((2*c*x+b)/(-4*a*c+b^2)^(1/2))),2^(1/2))*2^ 
(1/2)/c^3/(c*x^2+b*x+a)^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.19 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.89 \[ \int (d+e x) \left (a+b x+c x^2\right )^{3/4} \, dx=\frac {2 e (a+x (b+c x))^{7/4}}{7 c}+\frac {(2 c d-b e) (b+2 c x) \left (8 c (a+x (b+c x))-3 \sqrt {2} \left (b^2-4 a c\right ) \sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{80 c^3 \sqrt [4]{a+x (b+c x)}} \] Input:

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

Output:

(2*e*(a + x*(b + c*x))^(7/4))/(7*c) + ((2*c*d - b*e)*(b + 2*c*x)*(8*c*(a + 
 x*(b + c*x)) - 3*Sqrt[2]*(b^2 - 4*a*c)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a 
*c))^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])) 
/(80*c^3*(a + x*(b + c*x))^(1/4))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(616\) vs. \(2(159)=318\).

Time = 0.53 (sec) , antiderivative size = 616, normalized size of antiderivative = 3.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1160, 1087, 1094, 834, 761, 1510}

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.

\(\displaystyle \int (d+e x) \left (a+b x+c x^2\right )^{3/4} \, dx\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {(2 c d-b e) \int \left (c x^2+b x+a\right )^{3/4}dx}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{7/4}}{7 c}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt [4]{c x^2+b x+a}}dx}{20 c}\right )}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{7/4}}{7 c}\)

\(\Big \downarrow \) 1094

\(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \int \frac {\sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{5 c (b+2 c x)}\right )}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{7/4}}{7 c}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \left (\frac {\sqrt {b^2-4 a c} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}-\frac {\sqrt {b^2-4 a c} \int \frac {1-\frac {2 \sqrt {c} \sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}\right )}{5 c (b+2 c x)}\right )}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{7/4}}{7 c}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \left (\frac {\left (b^2-4 a c\right )^{3/4} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} c^{3/4} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt {b^2-4 a c} \int \frac {1-\frac {2 \sqrt {c} \sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}\right )}{5 c (b+2 c x)}\right )}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{7/4}}{7 c}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \left (\frac {\left (b^2-4 a c\right )^{3/4} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} c^{3/4} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt {b^2-4 a c} \left (\frac {\sqrt [4]{b^2-4 a c} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt [4]{a+b x+c x^2} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}\right )}{2 \sqrt {c}}\right )}{5 c (b+2 c x)}\right )}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{7/4}}{7 c}\)

Input:

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

Output:

(2*e*(a + b*x + c*x^2)^(7/4))/(7*c) + ((2*c*d - b*e)*(((b + 2*c*x)*(a + b* 
x + c*x^2)^(3/4))/(5*c) - (3*(b^2 - 4*a*c)*Sqrt[(b + 2*c*x)^2]*(-1/2*(Sqrt 
[b^2 - 4*a*c]*(-(((a + b*x + c*x^2)^(1/4)*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]))) + ((b^2 - 4*a*c)^(1/4)*(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)]*EllipticE 
[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 
1/2])/(Sqrt[2]*c^(1/4)*Sqrt[b^2 - 4*a*c + 4*c*(a + b*x + c*x^2)])))/Sqrt[c 
] + ((b^2 - 4*a*c)^(3/4)*(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*ArcTan[ 
(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(4*S 
qrt[2]*c^(3/4)*Sqrt[b^2 - 4*a*c + 4*c*(a + b*x + c*x^2)])))/(5*c*(b + 2*c* 
x))))/(2*c)
 

Defintions of rubi rules used

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 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], 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 1160

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

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (d+e x) \left (a+b x+c x^2\right )^{3/4} \, dx=\frac {-16 \left (c \,x^{2}+b x +a \right )^{\frac {3}{4}} a b e +112 \left (c \,x^{2}+b x +a \right )^{\frac {3}{4}} a c d +12 \left (c \,x^{2}+b x +a \right )^{\frac {3}{4}} b^{2} e x +56 \left (c \,x^{2}+b x +a \right )^{\frac {3}{4}} b c d x +40 \left (c \,x^{2}+b x +a \right )^{\frac {3}{4}} b c e \,x^{2}+84 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}}d x \right ) a b c e -168 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}}d x \right ) a \,c^{2} d -21 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}}d x \right ) b^{3} e +42 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}}d x \right ) b^{2} c d}{140 b c} \] Input:

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

Output:

( - 16*(a + b*x + c*x**2)**(3/4)*a*b*e + 112*(a + b*x + c*x**2)**(3/4)*a*c 
*d + 12*(a + b*x + c*x**2)**(3/4)*b**2*e*x + 56*(a + b*x + c*x**2)**(3/4)* 
b*c*d*x + 40*(a + b*x + c*x**2)**(3/4)*b*c*e*x**2 + 84*int(((a + b*x + c*x 
**2)**(3/4)*x)/(a + b*x + c*x**2),x)*a*b*c*e - 168*int(((a + b*x + c*x**2) 
**(3/4)*x)/(a + b*x + c*x**2),x)*a*c**2*d - 21*int(((a + b*x + c*x**2)**(3 
/4)*x)/(a + b*x + c*x**2),x)*b**3*e + 42*int(((a + b*x + c*x**2)**(3/4)*x) 
/(a + b*x + c*x**2),x)*b**2*c*d)/(140*b*c)