3.26.18 \(\int (d+e x) (a+b x+c x^2)^{3/4} \, dx\) [2518]

Optimal. Leaf size=510 \[ \frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{10 c^2}+\frac {2 e \left (a+b x+c x^2\right )^{7/4}}{7 c}-\frac {3 \sqrt {b^2-4 a c} (2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{20 c^{5/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}+\frac {3 \left (b^2-4 a c\right )^{7/4} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{20 \sqrt {2} c^{11/4} (b+2 c x)}-\frac {3 \left (b^2-4 a c\right )^{7/4} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{40 \sqrt {2} c^{11/4} (b+2 c x)} \]

[Out]

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-3/20*(-b*e+2*c*d)*(2*c*x+b)*(c
*x^2+b*x+a)^(1/4)*(-4*a*c+b^2)^(1/2)/c^(5/2)/(1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))+3/40*(-4*a*c
+b^2)^(7/4)*(-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)))*EllipticE(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^(11/4)/(2*c*x+b)*2^(1/2)-3/80
*(-4*a*c+b^2)^(7/4)*(-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^(11/4)/(2*c*x+b)*2^(1
/2)

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Rubi [A]
time = 0.31, antiderivative size = 510, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {654, 626, 637, 311, 226, 1210} \begin {gather*} -\frac {3 \left (b^2-4 a c\right )^{7/4} \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 )}{40 \sqrt {2} c^{11/4} (b+2 c x)}+\frac {3 \left (b^2-4 a c\right )^{7/4} \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) E\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 )}{20 \sqrt {2} c^{11/4} (b+2 c x)}-\frac {3 \sqrt {b^2-4 a c} (b+2 c x) \sqrt [4]{a+b x+c x^2} (2 c d-b e)}{20 c^{5/2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{10 c^2}+\frac {2 e \left (a+b x+c x^2\right )^{7/4}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*c*d - b*e)*(b + 2*c*x)*(a + b*x + c*x^2)^(3/4))/(10*c^2) + (2*e*(a + b*x + c*x^2)^(7/4))/(7*c) - (3*Sqrt[b
^2 - 4*a*c]*(2*c*d - b*e)*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(20*c^(5/2)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^
2])/Sqrt[b^2 - 4*a*c])) + (3*(b^2 - 4*a*c)^(7/4)*(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])*Ell
ipticE[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(20*Sqrt[2]*c^(11/4)*(b
+ 2*c*x)) - (3*(b^2 - 4*a*c)^(7/4)*(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*ArcTa
n[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(40*Sqrt[2]*c^(11/4)*(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 311

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

Rule 1210

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
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int (d+e x) \left (a+b x+c x^2\right )^{3/4} \, dx &=\frac {2 e \left (a+b x+c x^2\right )^{7/4}}{7 c}+\frac {(2 c d-b e) \int \left (a+b x+c x^2\right )^{3/4} \, dx}{2 c}\\ &=\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 {\left (3 \left (b^2-4 a c\right ) (2 c d-b e)\right ) \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx}{40 c^2}\\ &=\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 {\left (3 \left (b^2-4 a c\right ) (2 c d-b e) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{10 c^2 (b+2 c x)}\\ &=\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 {\left (3 \left (b^2-4 a c\right )^{3/2} (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 )}{20 c^{5/2} (b+2 c x)}+\frac {\left (3 \left (b^2-4 a c\right )^{3/2} (2 c d-b e) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {1-\frac {2 \sqrt {c} x^2}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{20 c^{5/2} (b+2 c x)}\\ &=\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 \sqrt {b^2-4 a c} (2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{20 c^{5/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}+\frac {3 \left (b^2-4 a c\right )^{7/4} (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 ) E\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 )}{20 \sqrt {2} c^{11/4} (b+2 c x)}-\frac {3 \left (b^2-4 a c\right )^{7/4} (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 )}{40 \sqrt {2} c^{11/4} (b+2 c x)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.17, size = 141, normalized size = 0.28 \begin {gather*} \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}} \, _2F_1\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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

integrate((c*x^2 + b*x + a)^(3/4)*(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 )}^{3/4} \,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)^(3/4),x)

[Out]

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

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