3.1420 \(\int \frac {(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{4/3}} \, dx\)

Optimal. Leaf size=99 \[ \frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \, _2F_1\left (-\frac {4}{3},-\frac {1}{6};\frac {5}{6};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8 c^2 d \sqrt [3]{1-\frac {(b+2 c x)^2}{b^2-4 a c}} \sqrt [3]{d (b+2 c x)}} \]

[Out]

3/8*(-4*a*c+b^2)*(c*x^2+b*x+a)^(1/3)*hypergeom([-4/3, -1/6],[5/6],(2*c*x+b)^2/(-4*a*c+b^2))/c^2/d/(d*(2*c*x+b)
)^(1/3)/(1-(2*c*x+b)^2/(-4*a*c+b^2))^(1/3)

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Rubi [A]  time = 0.12, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {694, 365, 364} \[ \frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \, _2F_1\left (-\frac {4}{3},-\frac {1}{6};\frac {5}{6};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8 c^2 d \sqrt [3]{1-\frac {(b+2 c x)^2}{b^2-4 a c}} \sqrt [3]{d (b+2 c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/3)*Hypergeometric2F1[-4/3, -1/6, 5/6, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(8*c
^2*d*(d*(b + 2*c*x))^(1/3)*(1 - (b + 2*c*x)^2/(b^2 - 4*a*c))^(1/3))

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

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

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)^{4/3}} \, 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^{4/3}} \, dx,x,b d+2 c d x\right )}{2 c d}\\ &=\frac {\left (\left (a-\frac {b^2}{4 c}\right ) \sqrt [3]{a+b x+c x^2}\right ) \operatorname {Subst}\left (\int \frac {\left (1+\frac {x^2}{4 \left (a-\frac {b^2}{4 c}\right ) c d^2}\right )^{4/3}}{x^{4/3}} \, dx,x,b d+2 c d x\right )}{\sqrt [3]{2} c d \sqrt [3]{4+\frac {(b d+2 c d x)^2}{\left (a-\frac {b^2}{4 c}\right ) c d^2}}}\\ &=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \, _2F_1\left (-\frac {4}{3},-\frac {1}{6};\frac {5}{6};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8 c^2 d \sqrt [3]{d (b+2 c x)} \sqrt [3]{1-\frac {(b+2 c x)^2}{b^2-4 a c}}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 104, normalized size = 1.05 \[ \frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \, _2F_1\left (-\frac {4}{3},-\frac {1}{6};\frac {5}{6};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8\ 2^{2/3} c^2 d \sqrt [3]{\frac {c (a+x (b+c x))}{4 a c-b^2}} \sqrt [3]{d (b+2 c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b^2 - 4*a*c)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[-4/3, -1/6, 5/6, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(8*2
^(2/3)*c^2*d*(d*(b + 2*c*x))^(1/3)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))

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fricas [F]  time = 1.21, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (2 \, c d x + b d\right )}^{\frac {2}{3}} {\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{4 \, c^{2} d^{2} x^{2} + 4 \, b c d^{2} x + b^{2} d^{2}}, x\right ) \]

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)^(4/3),x, algorithm="fricas")

[Out]

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

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

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)^(4/3),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {4}{3}}}\, dx \]

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)^(4/3),x)

[Out]

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

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

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)^(4/3),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{4/3}} \,d x \]

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)^(4/3),x)

[Out]

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

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

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)**(4/3),x)

[Out]

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

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