3.26.12 \(\int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx\) [2512]

Optimal. Leaf size=241 \[ \frac {(2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{6 c^2}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}-\frac {\left (b^2-4 a c\right )^{5/4} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{12 \sqrt {2} c^{9/4} (b+2 c x)} \]

[Out]

1/6*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)/c^2+2/5*e*(c*x^2+b*x+a)^(5/4)/c-1/24*(-4*a*c+b^2)^(5/4)*(-b*e+2
*c*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 = 0.12, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {654, 626, 637, 226} \begin {gather*} -\frac {\left (b^2-4 a c\right )^{5/4} \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) 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 )}{12 \sqrt {2} c^{9/4} (b+2 c x)}+\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2} (2 c d-b e)}{6 c^2}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*c*d - b*e)*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(6*c^2) + (2*e*(a + b*x + c*x^2)^(5/4))/(5*c) - ((b^2 - 4*
a*c)^(5/4)*(2*c*d - b*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])/(12*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 626

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] - Dist[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]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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]

Rubi steps

\begin {align*} \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx &=\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}+\frac {(2 c d-b e) \int \sqrt [4]{a+b x+c x^2} \, dx}{2 c}\\ &=\frac {(2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{6 c^2}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}-\frac {\left (\left (b^2-4 a c\right ) (2 c d-b e)\right ) \int \frac {1}{\left (a+b x+c x^2\right )^{3/4}} \, dx}{24 c^2}\\ &=\frac {(2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{6 c^2}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}-\frac {\left (\left (b^2-4 a c\right ) (2 c d-b e) \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 )}{6 c^2 (b+2 c x)}\\ &=\frac {(2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{6 c^2}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}-\frac {\left (b^2-4 a c\right )^{5/4} (2 c d-b e) \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 )}{12 \sqrt {2} c^{9/4} (b+2 c x)}\\ \end {align*}

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Mathematica [A]
time = 9.73, size = 140, normalized size = 0.58 \begin {gather*} \frac {24 c^2 e (a+x (b+c x))^2+5 (2 c d-b e) \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} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )\right |2\right )\right )}{60 c^3 (a+x (b+c x))^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(24*c^2*e*(a + x*(b + c*x))^2 + 5*(2*c*d - b*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]))/(60*c^
3*(a + x*(b + c*x))^(3/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (e x +d \right ) \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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