3.26.36 \(\int \frac {(d+e x)^2}{(a+b x+c x^2)^{3/4}} \, dx\) [2536]

Optimal. Leaf size=262 \[ \frac {5 e (2 c d-b e) \sqrt [4]{a+b x+c x^2}}{3 c^2}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}+\frac {\sqrt [4]{b^2-4 a c} \left (12 c^2 d^2+5 b^2 e^2-4 c e (3 b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{6 \sqrt {2} c^{9/4} (b+2 c x)} \]

[Out]

5/3*e*(-b*e+2*c*d)*(c*x^2+b*x+a)^(1/4)/c^2+2/3*e*(e*x+d)*(c*x^2+b*x+a)^(1/4)/c+1/12*(-4*a*c+b^2)^(1/4)*(12*c^2
*d^2+5*b^2*e^2-4*c*e*(2*a*e+3*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^(9/4)/(2*c*x+b)*2^
(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 262, 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 = {756, 654, 637, 226} \begin {gather*} \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 ) \left (-4 c e (2 a e+3 b d)+5 b^2 e^2+12 c^2 d^2\right ) F\left (2 \text {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^{9/4} (b+2 c x)}+\frac {5 e \sqrt [4]{a+b x+c x^2} (2 c d-b e)}{3 c^2}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*e*(2*c*d - b*e)*(a + b*x + c*x^2)^(1/4))/(3*c^2) + (2*e*(d + e*x)*(a + b*x + c*x^2)^(1/4))/(3*c) + ((b^2 -
4*a*c)^(1/4)*(12*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(3*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])*Elli
pticF[2*ArcTan[(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^(9/4)*(b + 2
*c*x))

Rule 226

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 637

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 654

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] + Dist[(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[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 756

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]

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{3/4}} \, dx &=\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}+\frac {2 \int \frac {\frac {1}{4} \left (6 c d^2-e (b d+4 a e)\right )+\frac {5}{4} e (2 c d-b e) x}{\left (a+b x+c x^2\right )^{3/4}} \, dx}{3 c}\\ &=\frac {5 e (2 c d-b e) \sqrt [4]{a+b x+c x^2}}{3 c^2}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}+\frac {\left (-\frac {5}{4} b e (2 c d-b e)+\frac {1}{2} c \left (6 c d^2-e (b d+4 a e)\right )\right ) \int \frac {1}{\left (a+b x+c x^2\right )^{3/4}} \, dx}{3 c^2}\\ &=\frac {5 e (2 c d-b e) \sqrt [4]{a+b x+c x^2}}{3 c^2}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}+\frac {\left (4 \left (-\frac {5}{4} b e (2 c d-b e)+\frac {1}{2} c \left (6 c d^2-e (b d+4 a e)\right )\right ) \sqrt {(b+2 c x)^2}\right ) \text {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 )}{3 c^2 (b+2 c x)}\\ &=\frac {5 e (2 c d-b e) \sqrt [4]{a+b x+c x^2}}{3 c^2}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}+\frac {\sqrt [4]{b^2-4 a c} \left (12 c^2 d^2+5 b^2 e^2-4 c e (3 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 )}{6 \sqrt {2} c^{9/4} (b+2 c x)}\\ \end {align*}

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Mathematica [A]
time = 10.20, size = 150, normalized size = 0.57 \begin {gather*} \frac {2 c e (a+x (b+c x)) (-5 b e+2 c (6 d+e x))+\sqrt {2} \sqrt {b^2-4 a c} \left (12 c^2 d^2+5 b^2 e^2-4 c e (3 b d+2 a e)\right ) \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )\right |2\right )}{6 c^3 (a+x (b+c x))^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*e*(a + x*(b + c*x))*(-5*b*e + 2*c*(6*d + e*x)) + Sqrt[2]*Sqrt[b^2 - 4*a*c]*(12*c^2*d^2 + 5*b^2*e^2 - 4*c*
e*(3*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])/(6*c^3*(a + x*(b + c*x))^(3/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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