\(\int \frac {(d+e x)^3}{(a+b x+c x^2)^{3/4}} \, dx\) [747]

Optimal result

Integrand size = 22, antiderivative size = 229 \[ \int \frac {(d+e x)^3}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\frac {2 e (d+e x)^2 \sqrt [4]{a+b x+c x^2}}{5 c}+\frac {e \left (56 c^2 d^2+15 b^2 e^2-2 c e (25 b d+8 a e)+6 c e (2 c d-b e) x\right ) \sqrt [4]{a+b x+c x^2}}{10 c^3}+\frac {\sqrt {-b^2+4 a c} (2 c d-b e) \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \left (-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right ),2\right )}{2 \sqrt {2} c^4 \left (a+b x+c x^2\right )^{3/4}} \] Output:

2/5*e*(e*x+d)^2*(c*x^2+b*x+a)^(1/4)/c+1/10*e*(56*c^2*d^2+15*b^2*e^2-2*c*e* 
(8*a*e+25*b*d)+6*c*e*(-b*e+2*c*d)*x)*(c*x^2+b*x+a)^(1/4)/c^3+1/4*(4*a*c-b^ 
2)^(1/2)*(-b*e+2*c*d)*(4*c^2*d^2+3*b^2*e^2-4*c*e*(2*a*e+b*d))*(-c*(c*x^2+b 
*x+a)/(-4*a*c+b^2))^(3/4)*InverseJacobiAM(1/2*arctan((2*c*x+b)/(4*a*c-b^2) 
^(1/2)),2^(1/2))*2^(1/2)/c^4/(c*x^2+b*x+a)^(3/4)
 

Mathematica [A] (warning: unable to verify)

Time = 10.29 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^3}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\frac {-2 c e (a+x (b+c x)) \left (-15 b^2 e^2+2 c e (25 b d+8 a e+3 b e x)-4 c^2 \left (15 d^2+5 d e x+e^2 x^2\right )\right )-5 \sqrt {2} \sqrt {b^2-4 a c} (-2 c d+b e) \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )}{20 c^4 (a+x (b+c x))^{3/4}} \] Input:

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

Output:

(-2*c*e*(a + x*(b + c*x))*(-15*b^2*e^2 + 2*c*e*(25*b*d + 8*a*e + 3*b*e*x) 
- 4*c^2*(15*d^2 + 5*d*e*x + e^2*x^2)) - 5*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(-2*c* 
d + b*e)*(4*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(b*d + 2*a*e))*((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])/(20*c^4*(a + x*(b + c*x))^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.45 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.58, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1166, 27, 1225, 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.

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

\(\Big \downarrow \) 1166

\(\displaystyle \frac {2 \int \frac {(d+e x) \left (10 c d^2-e (b d+8 a e)+9 e (2 c d-b e) x\right )}{4 \left (c x^2+b x+a\right )^{3/4}}dx}{5 c}+\frac {2 e (d+e x)^2 \sqrt [4]{a+b x+c x^2}}{5 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(d+e x) \left (10 c d^2-e (b d+8 a e)+9 e (2 c d-b e) x\right )}{\left (c x^2+b x+a\right )^{3/4}}dx}{10 c}+\frac {2 e (d+e x)^2 \sqrt [4]{a+b x+c x^2}}{5 c}\)

\(\Big \downarrow \) 1225

\(\displaystyle \frac {\frac {5 (2 c d-b e) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \int \frac {1}{\left (c x^2+b x+a\right )^{3/4}}dx}{4 c^2}+\frac {e \sqrt [4]{a+b x+c x^2} \left (-2 c e (8 a e+25 b d)+15 b^2 e^2+6 c e x (2 c d-b e)+56 c^2 d^2\right )}{c^2}}{10 c}+\frac {2 e (d+e x)^2 \sqrt [4]{a+b x+c x^2}}{5 c}\)

\(\Big \downarrow \) 1094

\(\displaystyle \frac {\frac {5 \sqrt {(b+2 c x)^2} (2 c d-b e) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{c^2 (b+2 c x)}+\frac {e \sqrt [4]{a+b x+c x^2} \left (-2 c e (8 a e+25 b d)+15 b^2 e^2+6 c e x (2 c d-b e)+56 c^2 d^2\right )}{c^2}}{10 c}+\frac {2 e (d+e x)^2 \sqrt [4]{a+b x+c x^2}}{5 c}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {\frac {5 \sqrt [4]{b^2-4 a c} \sqrt {(b+2 c x)^2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} (2 c d-b e) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{2 \sqrt {2} c^{9/4} (b+2 c x) \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}+\frac {e \sqrt [4]{a+b x+c x^2} \left (-2 c e (8 a e+25 b d)+15 b^2 e^2+6 c e x (2 c d-b e)+56 c^2 d^2\right )}{c^2}}{10 c}+\frac {2 e (d+e x)^2 \sqrt [4]{a+b x+c x^2}}{5 c}\)

Input:

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

Output:

(2*e*(d + e*x)^2*(a + b*x + c*x^2)^(1/4))/(5*c) + ((e*(56*c^2*d^2 + 15*b^2 
*e^2 - 2*c*e*(25*b*d + 8*a*e) + 6*c*e*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^( 
1/4))/c^2 + (5*(b^2 - 4*a*c)^(1/4)*(2*c*d - b*e)*(4*c^2*d^2 + 3*b^2*e^2 - 
4*c*e*(b*d + 2*a*e))*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)]*Ell 
ipticF[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1 
/4)], 1/2])/(2*Sqrt[2]*c^(9/4)*(b + 2*c*x)*Sqrt[b^2 - 4*a*c + 4*c*(a + b*x 
 + c*x^2)]))/(10*c)
 

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 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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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