3.15.22 \(\int \frac {(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{16/3}} \, dx\) [1422]

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

[Out]

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

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Rubi [A]
time = 0.08, 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 = {708, 372, 371} \begin {gather*} \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}} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 708

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)^{16/3}} \, dx &=\frac {\text {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 ) \text {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*}

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Mathematica [A]
time = 10.09, size = 112, normalized size = 1.13 \begin {gather*} \frac {3 \left (b^2-4 a c\right ) (d (b+2 c x))^{2/3} \sqrt [3]{a+x (b+c x)} \, _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))}{-b^2+4 a c}}} \end {gather*}

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 = 0.40, size = 0, normalized size = 0.00 \[\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {16}{3}}}\, dx\]

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)^(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 = 0.00, 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((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 = 0.00, 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((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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 4496 deep

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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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{16/3}} \,d x \end {gather*}

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)^(16/3),x)

[Out]

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

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