3.1422 \(\int \frac{(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{16/3}} \, dx\)

Optimal. Leaf size=99 \[ \frac{3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \, _2F_1\left (-\frac{13}{6},-\frac{4}{3};-\frac{7}{6};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{104 c^2 d \sqrt [3]{1-\frac{(b+2 c x)^2}{b^2-4 a c}} (d (b+2 c x))^{13/3}} \]

[Out]

(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/3)*Hypergeometric2F1[-13/6, -4/3, -7/6, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(1
04*c^2*d*(d*(b + 2*c*x))^(13/3)*(1 - (b + 2*c*x)^2/(b^2 - 4*a*c))^(1/3))

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Rubi [A]  time = 0.116381, antiderivative size = 99, normalized size of antiderivative = 1., 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} \[ \frac{3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \, _2F_1\left (-\frac{13}{6},-\frac{4}{3};-\frac{7}{6};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{104 c^2 d \sqrt [3]{1-\frac{(b+2 c x)^2}{b^2-4 a c}} (d (b+2 c x))^{13/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(16/3),x]

[Out]

(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/3)*Hypergeometric2F1[-13/6, -4/3, -7/6, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(1
04*c^2*d*(d*(b + 2*c*x))^(13/3)*(1 - (b + 2*c*x)^2/(b^2 - 4*a*c))^(1/3))

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]

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 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])

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{16/3}} \, 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^{16/3}} \, 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^{16/3}} \, 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{13}{6},-\frac{4}{3};-\frac{7}{6};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{104 c^2 d (d (b+2 c x))^{13/3} \sqrt [3]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}}\\ \end{align*}

Mathematica [A]  time = 0.0898834, size = 112, normalized size = 1.13 \[ \frac{3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} (d (b+2 c x))^{2/3} \, _2F_1\left (-\frac{13}{6},-\frac{4}{3};-\frac{7}{6};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{104\ 2^{2/3} c^2 d^6 (b+2 c x)^5 \sqrt [3]{\frac{c (a+x (b+c x))}{4 a c-b^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(16/3),x]

[Out]

(3*(b^2 - 4*a*c)*(d*(b + 2*c*x))^(2/3)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[-13/6, -4/3, -7/6, (b + 2*c*x
)^2/(b^2 - 4*a*c)])/(104*2^(2/3)*c^2*d^6*(b + 2*c*x)^5*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))

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Maple [F]  time = 1.184, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c{x}^{2}+bx+a \right ) ^{{\frac{4}{3}}} \left ( 2\,cdx+bd \right ) ^{-{\frac{16}{3}}}}\, dx \end{align*}

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)^(16/3),x)

[Out]

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(16/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac{16}{3}}}\,{d x} \end{align*}

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)^(16/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(16/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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}}}{64 \, c^{6} d^{6} x^{6} + 192 \, b c^{5} d^{6} x^{5} + 240 \, b^{2} c^{4} d^{6} x^{4} + 160 \, b^{3} c^{3} d^{6} x^{3} + 60 \, b^{4} c^{2} d^{6} x^{2} + 12 \, b^{5} c d^{6} x + b^{6} d^{6}}, x\right ) \end{align*}

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)^(16/3),x, algorithm="fricas")

[Out]

integral((2*c*d*x + b*d)^(2/3)*(c*x^2 + b*x + a)^(4/3)/(64*c^6*d^6*x^6 + 192*b*c^5*d^6*x^5 + 240*b^2*c^4*d^6*x
^4 + 160*b^3*c^3*d^6*x^3 + 60*b^4*c^2*d^6*x^2 + 12*b^5*c*d^6*x + b^6*d^6), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**(16/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac{16}{3}}}\,{d x} \end{align*}

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)^(16/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(16/3), x)