3.1413 \(\int \frac{(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{17/3}} \, dx\)

Optimal. Leaf size=44 \[ \frac{3 \left (a+b x+c x^2\right )^{7/3}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{14/3}} \]

[Out]

(3*(a + b*x + c*x^2)^(7/3))/(7*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(14/3))

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Rubi [A]  time = 0.0177368, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.036, Rules used = {682} \[ \frac{3 \left (a+b x+c x^2\right )^{7/3}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{14/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(17/3),x]

[Out]

(3*(a + b*x + c*x^2)^(7/3))/(7*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(14/3))

Rule 682

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*
a*c, 0] && EqQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx &=\frac{3 \left (a+b x+c x^2\right )^{7/3}}{7 \left (b^2-4 a c\right ) d (b d+2 c d x)^{14/3}}\\ \end{align*}

Mathematica [A]  time = 0.0500632, size = 50, normalized size = 1.14 \[ \frac{3 (a+x (b+c x))^{7/3} \sqrt [3]{d (b+2 c x)}}{7 d^6 \left (b^2-4 a c\right ) (b+2 c x)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(17/3),x]

[Out]

(3*(d*(b + 2*c*x))^(1/3)*(a + x*(b + c*x))^(7/3))/(7*(b^2 - 4*a*c)*d^6*(b + 2*c*x)^5)

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Maple [A]  time = 0.046, size = 44, normalized size = 1. \begin{align*} -{\frac{6\,cx+3\,b}{28\,ac-7\,{b}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{7}{3}}} \left ( 2\,cdx+bd \right ) ^{-{\frac{17}{3}}}} \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)^(17/3),x)

[Out]

-3/7*(2*c*x+b)*(c*x^2+b*x+a)^(7/3)/(4*a*c-b^2)/(2*c*d*x+b*d)^(17/3)

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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{17}{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)^(17/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(17/3), x)

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Fricas [B]  time = 2.95271, size = 416, normalized size = 9.45 \begin{align*} \frac{3 \,{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}{\left (2 \, c d x + b d\right )}^{\frac{1}{3}}{\left (c x^{2} + b x + a\right )}^{\frac{1}{3}}}{7 \,{\left (32 \,{\left (b^{2} c^{5} - 4 \, a c^{6}\right )} d^{6} x^{5} + 80 \,{\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} d^{6} x^{4} + 80 \,{\left (b^{4} c^{3} - 4 \, a b^{2} c^{4}\right )} d^{6} x^{3} + 40 \,{\left (b^{5} c^{2} - 4 \, a b^{3} c^{3}\right )} d^{6} x^{2} + 10 \,{\left (b^{6} c - 4 \, a b^{4} c^{2}\right )} d^{6} x +{\left (b^{7} - 4 \, a b^{5} c\right )} d^{6}\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)^(17/3),x, algorithm="fricas")

[Out]

3/7*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*(2*c*d*x + b*d)^(1/3)*(c*x^2 + b*x + a)^(1/3)/(3
2*(b^2*c^5 - 4*a*c^6)*d^6*x^5 + 80*(b^3*c^4 - 4*a*b*c^5)*d^6*x^4 + 80*(b^4*c^3 - 4*a*b^2*c^4)*d^6*x^3 + 40*(b^
5*c^2 - 4*a*b^3*c^3)*d^6*x^2 + 10*(b^6*c - 4*a*b^4*c^2)*d^6*x + (b^7 - 4*a*b^5*c)*d^6)

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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)**(17/3),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{17}{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)^(17/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(17/3), x)