3.1418 \(\int \frac {(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{14/3}} \, dx\)

Optimal. Leaf size=591 \[ \frac {3^{3/4} \left (-4 a c+b^2-(b+2 c x)^2\right ) \sqrt [3]{d (b+2 c x)} \left (2 \sqrt [3]{c} d^{2/3}-\frac {\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right ) \sqrt {\frac {\frac {2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}+\frac {(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}+2 \sqrt [3]{2} c^{2/3} d^{4/3}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac {2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1-\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{c x^2+b x+a}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{c x^2+b x+a}}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{440 c^{10/3} d^{17/3} \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{2/3} \sqrt {-\frac {(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )}{\sqrt [3]{a+b x+c x^2} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}}}-\frac {3 \sqrt [3]{a+b x+c x^2}}{55 c^2 d^3 (d (b+2 c x))^{5/3}}-\frac {3 \left (a+b x+c x^2\right )^{4/3}}{22 c d (d (b+2 c x))^{11/3}} \]

[Out]

-3/55*(c*x^2+b*x+a)^(1/3)/c^2/d^3/(d*(2*c*x+b))^(5/3)-3/22*(c*x^2+b*x+a)^(4/3)/c/d/(d*(2*c*x+b))^(11/3)+1/440*
3^(3/4)*(d*(2*c*x+b))^(1/3)*(b^2-4*a*c-(2*c*x+b)^2)*(2*c^(1/3)*d^(2/3)-2^(1/3)*(d*(2*c*x+b))^(2/3)/(c*x^2+b*x+
a)^(1/3))*((2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1-3^(1/2))/(c*x^2+b*x+a)^(1/3))^2/(2^(2/3)*c^(1/3)*d^
(2/3)-(d*(2*c*x+b))^(2/3)*(1+3^(1/2))/(c*x^2+b*x+a)^(1/3))^2)^(1/2)/(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/
3)*(1-3^(1/2))/(c*x^2+b*x+a)^(1/3))*(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1+3^(1/2))/(c*x^2+b*x+a)^(1/
3))*EllipticF((1-(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1-3^(1/2))/(c*x^2+b*x+a)^(1/3))^2/(2^(2/3)*c^(1
/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1+3^(1/2))/(c*x^2+b*x+a)^(1/3))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*((2*2^(1/3)
*c^(2/3)*d^(4/3)+(d*(2*c*x+b))^(4/3)/(c*x^2+b*x+a)^(2/3)+2^(2/3)*c^(1/3)*d^(2/3)*(d*(2*c*x+b))^(2/3)/(c*x^2+b*
x+a)^(1/3))/(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1+3^(1/2))/(c*x^2+b*x+a)^(1/3))^2)^(1/2)/c^(10/3)/(-
4*a*c+b^2)/d^(17/3)/(c*x^2+b*x+a)^(2/3)/(-(d*(2*c*x+b))^(2/3)*(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)/(c*
x^2+b*x+a)^(1/3))/(c*x^2+b*x+a)^(1/3)/(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1+3^(1/2))/(c*x^2+b*x+a)^(
1/3))^2)^(1/2)

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Rubi [A]  time = 2.00, antiderivative size = 591, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {694, 277, 329, 241, 225} \[ \frac {3^{3/4} \left (-4 a c+b^2-(b+2 c x)^2\right ) \sqrt [3]{d (b+2 c x)} \left (2 \sqrt [3]{c} d^{2/3}-\frac {\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right ) \sqrt {\frac {\frac {2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}+\frac {(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}+2 \sqrt [3]{2} c^{2/3} d^{4/3}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac {2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1-\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{c x^2+b x+a}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{c x^2+b x+a}}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{440 c^{10/3} d^{17/3} \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{2/3} \sqrt {-\frac {(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )}{\sqrt [3]{a+b x+c x^2} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}}}-\frac {3 \sqrt [3]{a+b x+c x^2}}{55 c^2 d^3 (d (b+2 c x))^{5/3}}-\frac {3 \left (a+b x+c x^2\right )^{4/3}}{22 c d (d (b+2 c x))^{11/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(14/3),x]

[Out]

(-3*(a + b*x + c*x^2)^(1/3))/(55*c^2*d^3*(d*(b + 2*c*x))^(5/3)) - (3*(a + b*x + c*x^2)^(4/3))/(22*c*d*(d*(b +
2*c*x))^(11/3)) + (3^(3/4)*(d*(b + 2*c*x))^(1/3)*(b^2 - 4*a*c - (b + 2*c*x)^2)*(2*c^(1/3)*d^(2/3) - (2^(1/3)*(
d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))*Sqrt[(2*2^(1/3)*c^(2/3)*d^(4/3) + (d*(b + 2*c*x))^(4/3)/(a + b*
x + c*x^2)^(2/3) + (2^(2/3)*c^(1/3)*d^(2/3)*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))/(2^(2/3)*c^(1/3)*d
^(2/3) - ((1 + Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))^2]*EllipticF[ArcCos[(2^(2/3)*c^(1/3)*d
^(2/3) - ((1 - Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))/(2^(2/3)*c^(1/3)*d^(2/3) - ((1 + Sqrt[
3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))], (2 + Sqrt[3])/4])/(440*c^(10/3)*(b^2 - 4*a*c)*d^(17/3)*(
a + b*x + c*x^2)^(2/3)*Sqrt[-(((d*(b + 2*c*x))^(2/3)*(2^(2/3)*c^(1/3)*d^(2/3) - (d*(b + 2*c*x))^(2/3)/(a + b*x
+ c*x^2)^(1/3)))/((a + b*x + c*x^2)^(1/3)*(2^(2/3)*c^(1/3)*d^(2/3) - ((1 + Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a
+ b*x + c*x^2)^(1/3))^2))])

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rule 241

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a/(a + b*x^n))^(p + 1/n)*(a + b*x^n)^(p + 1/n), Subst[In
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p,
0] && NeQ[p, -2^(-1)] && LtQ[Denominator[p + 1/n], Denominator[p]]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{14/3}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^{4/3}}{x^{14/3}} \, dx,x,b d+2 c d x\right )}{2 c d}\\ &=-\frac {3 \left (a+b x+c x^2\right )^{4/3}}{22 c d (d (b+2 c x))^{11/3}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt [3]{a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}}}{x^{8/3}} \, dx,x,b d+2 c d x\right )}{11 c^2 d^3}\\ &=-\frac {3 \sqrt [3]{a+b x+c x^2}}{55 c^2 d^3 (d (b+2 c x))^{5/3}}-\frac {3 \left (a+b x+c x^2\right )^{4/3}}{22 c d (d (b+2 c x))^{11/3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x^{2/3} \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^{2/3}} \, dx,x,b d+2 c d x\right )}{110 c^3 d^5}\\ &=-\frac {3 \sqrt [3]{a+b x+c x^2}}{55 c^2 d^3 (d (b+2 c x))^{5/3}}-\frac {3 \left (a+b x+c x^2\right )^{4/3}}{22 c d (d (b+2 c x))^{11/3}}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{\left (a-\frac {b^2}{4 c}+\frac {x^6}{4 c d^2}\right )^{2/3}} \, dx,x,\sqrt [3]{d (b+2 c x)}\right )}{110 c^3 d^5}\\ &=-\frac {3 \sqrt [3]{a+b x+c x^2}}{55 c^2 d^3 (d (b+2 c x))^{5/3}}-\frac {3 \left (a+b x+c x^2\right )^{4/3}}{22 c d (d (b+2 c x))^{11/3}}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^6}{4 c d^2}}} \, dx,x,\frac {\sqrt [3]{d (b+2 c x)}}{\sqrt [6]{a+x (b+c x)}}\right )}{110 c^3 d^5 \sqrt {\frac {a-\frac {b^2}{4 c}}{a+x (b+c x)}} \sqrt {a+x (b+c x)}}\\ &=-\frac {3 \sqrt [3]{a+b x+c x^2}}{55 c^2 d^3 (d (b+2 c x))^{5/3}}-\frac {3 \left (a+b x+c x^2\right )^{4/3}}{22 c d (d (b+2 c x))^{11/3}}+\frac {3^{3/4} \sqrt [3]{d (b+2 c x)} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right ) \sqrt {\frac {2 \sqrt [3]{2} c^{2/3} d^{4/3}+\frac {(d (b+2 c x))^{4/3}}{(a+x (b+c x))^{2/3}}+\frac {2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )^2}} F\left (\cos ^{-1}\left (\frac {2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1-\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{55\ 2^{2/3} c^{10/3} d^{17/3} \sqrt {\frac {4 a-\frac {b^2}{c}}{a+x (b+c x)}} (a+x (b+c x))^{2/3} \sqrt {4-\frac {(b+2 c x)^2}{c (a+x (b+c x))}} \sqrt {-\frac {(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{\sqrt [3]{a+x (b+c x)} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )^2}}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 112, normalized size = 0.19 \[ \frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \sqrt [3]{d (b+2 c x)} \, _2F_1\left (-\frac {11}{6},-\frac {4}{3};-\frac {5}{6};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{88\ 2^{2/3} c^2 d^5 (b+2 c x)^4 \sqrt [3]{\frac {c (a+x (b+c x))}{4 a c-b^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(14/3),x]

[Out]

(3*(b^2 - 4*a*c)*(d*(b + 2*c*x))^(1/3)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[-11/6, -4/3, -5/6, (b + 2*c*x
)^2/(b^2 - 4*a*c)])/(88*2^(2/3)*c^2*d^5*(b + 2*c*x)^4*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))

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fricas [F]  time = 1.35, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (2 \, c d x + b d\right )}^{\frac {1}{3}} {\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{32 \, c^{5} d^{5} x^{5} + 80 \, b c^{4} d^{5} x^{4} + 80 \, b^{2} c^{3} d^{5} x^{3} + 40 \, b^{3} c^{2} d^{5} x^{2} + 10 \, b^{4} c d^{5} x + b^{5} d^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x, algorithm="fricas")

[Out]

integral((2*c*d*x + b*d)^(1/3)*(c*x^2 + b*x + a)^(4/3)/(32*c^5*d^5*x^5 + 80*b*c^4*d^5*x^4 + 80*b^2*c^3*d^5*x^3
+ 40*b^3*c^2*d^5*x^2 + 10*b^4*c*d^5*x + b^5*d^5), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {14}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(14/3), x)

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maple [F]  time = 1.50, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {14}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x)

[Out]

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {14}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(14/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{14/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(14/3),x)

[Out]

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(14/3), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(14/3),x)

[Out]

Timed out

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