3.2535 \(\int \frac {(d+e x)^3}{(a+b x+c x^2)^{3/4}} \, dx\)

Optimal. Leaf size=307 \[ \frac {\sqrt [4]{b^2-4 a c} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) (2 c d-b e) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{4 \sqrt {2} c^{13/4} (b+2 c x)}+\frac {e \sqrt [4]{a+b x+c x^2} \left (-2 c e (8 a e+25 b d)+15 b^2 e^2+6 c e x (2 c d-b e)+56 c^2 d^2\right )}{10 c^3}+\frac {2 e (d+e x)^2 \sqrt [4]{a+b x+c x^2}}{5 c} \]

[Out]

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+1/8*(-4*a*c+b^2)^(1/4)*(-b*e+2*c*d)*(4*c^2*d^2+3*b^2*e^2-4*c*e*(2*a*e+b*d))*(cos(2*ar
ctan(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^(13/4)/(2*c*x+b)*2^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {742, 779, 623, 220} \[ \frac {e \sqrt [4]{a+b x+c x^2} \left (-2 c e (8 a e+25 b d)+15 b^2 e^2+6 c e x (2 c d-b e)+56 c^2 d^2\right )}{10 c^3}+\frac {\sqrt [4]{b^2-4 a c} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) (2 c d-b e) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{4 \sqrt {2} c^{13/4} (b+2 c x)}+\frac {2 e (d+e x)^2 \sqrt [4]{a+b x+c x^2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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))/(10*c^3) + ((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/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4
*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])*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*Sqrt[2]*c^(13/4)*(b + 2*c*x))

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 623

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[(d*Sqrt[(b + 2*c*x)
^2])/(b + 2*c*x), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

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] + Dist[(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] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \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 {2 \int \frac {(d+e x) \left (\frac {1}{4} \left (10 c d^2-e (b d+8 a e)\right )+\frac {9}{4} e (2 c d-b e) x\right )}{\left (a+b x+c x^2\right )^{3/4}} \, dx}{5 c}\\ &=\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 {\left ((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 )\right ) \int \frac {1}{\left (a+b x+c x^2\right )^{3/4}} \, dx}{8 c^3}\\ &=\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 {\left ((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 ) \sqrt {(b+2 c x)^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{2 c^3 (b+2 c x)}\\ &=\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 [4]{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 ) \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 ) F\left (2 \tan ^{-1}\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 )}{4 \sqrt {2} c^{13/4} (b+2 c x)}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 244, normalized size = 0.79 \[ \frac {2 c e \left (-16 a^2 c e^2+a \left (15 b^2 e^2-2 b c e (25 d+11 e x)+4 c^2 \left (15 d^2+5 d e x-3 e^2 x^2\right )\right )+x (b+c x) \left (15 b^2 e^2-2 b c e (25 d+3 e x)+4 c^2 \left (15 d^2+5 d e x+e^2 x^2\right )\right )\right )-5 \sqrt {2} \sqrt {b^2-4 a c} \left (\frac {c (a+x (b+c x))}{4 a c-b^2}\right )^{3/4} (b e-2 c d) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )\right |2\right )}{20 c^4 (a+x (b+c x))^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*e*(-16*a^2*c*e^2 + a*(15*b^2*e^2 - 2*b*c*e*(25*d + 11*e*x) + 4*c^2*(15*d^2 + 5*d*e*x - 3*e^2*x^2)) + x*(b
+ c*x)*(15*b^2*e^2 - 2*b*c*e*(25*d + 3*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))

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fricas [F]  time = 1.19, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{3}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.31, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{3}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{3}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x\right )}^3}{{\left (c\,x^2+b\,x+a\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{3}}{\left (a + b x + c x^{2}\right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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