3.1414 \(\int \frac{(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{23/3}} \, dx\)

Optimal. Leaf size=89 \[ \frac{9 \left (a+b x+c x^2\right )^{7/3}}{70 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{14/3}}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{10 d \left (b^2-4 a c\right ) (b d+2 c d x)^{20/3}} \]

[Out]

(3*(a + b*x + c*x^2)^(7/3))/(10*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(20/3)) + (9*(a + b*x + c*x^2)^(7/3))/(70*(b^2
- 4*a*c)^2*d^3*(b*d + 2*c*d*x)^(14/3))

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Rubi [A]  time = 0.0395993, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {693, 682} \[ \frac{9 \left (a+b x+c x^2\right )^{7/3}}{70 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{14/3}}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{10 d \left (b^2-4 a c\right ) (b d+2 c d x)^{20/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(23/3),x]

[Out]

(3*(a + b*x + c*x^2)^(7/3))/(10*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(20/3)) + (9*(a + b*x + c*x^2)^(7/3))/(70*(b^2
- 4*a*c)^2*d^3*(b*d + 2*c*d*x)^(14/3))

Rule 693

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

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)^{23/3}} \, dx &=\frac{3 \left (a+b x+c x^2\right )^{7/3}}{10 \left (b^2-4 a c\right ) d (b d+2 c d x)^{20/3}}+\frac{3 \int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx}{10 \left (b^2-4 a c\right ) d^2}\\ &=\frac{3 \left (a+b x+c x^2\right )^{7/3}}{10 \left (b^2-4 a c\right ) d (b d+2 c d x)^{20/3}}+\frac{9 \left (a+b x+c x^2\right )^{7/3}}{70 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{14/3}}\\ \end{align*}

Mathematica [A]  time = 0.0776626, size = 74, normalized size = 0.83 \[ \frac{3 (a+x (b+c x))^{7/3} \left (2 c \left (3 c x^2-7 a\right )+5 b^2+6 b c x\right ) \sqrt [3]{d (b+2 c x)}}{35 d^8 \left (b^2-4 a c\right )^2 (b+2 c x)^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(23/3),x]

[Out]

(3*(d*(b + 2*c*x))^(1/3)*(a + x*(b + c*x))^(7/3)*(5*b^2 + 6*b*c*x + 2*c*(-7*a + 3*c*x^2)))/(35*(b^2 - 4*a*c)^2
*d^8*(b + 2*c*x)^7)

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Maple [A]  time = 0.044, size = 76, normalized size = 0.9 \begin{align*} -{\frac{ \left ( 6\,cx+3\,b \right ) \left ( -6\,{c}^{2}{x}^{2}-6\,bcx+14\,ac-5\,{b}^{2} \right ) }{560\,{a}^{2}{c}^{2}-280\,ac{b}^{2}+35\,{b}^{4}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{7}{3}}} \left ( 2\,cdx+bd \right ) ^{-{\frac{23}{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)^(23/3),x)

[Out]

-3/35*(2*c*x+b)*(c*x^2+b*x+a)^(7/3)*(-6*c^2*x^2-6*b*c*x+14*a*c-5*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)/(2*c*d*x+b*d)
^(23/3)

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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{23}{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)^(23/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(23/3), x)

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Fricas [B]  time = 3.58114, size = 868, normalized size = 9.75 \begin{align*} \frac{3 \,{\left (6 \, c^{4} x^{6} + 18 \, b c^{3} x^{5} +{\left (23 \, b^{2} c^{2} - 2 \, a c^{3}\right )} x^{4} + 5 \, a^{2} b^{2} - 14 \, a^{3} c + 4 \,{\left (4 \, b^{3} c - a b c^{2}\right )} x^{3} +{\left (5 \, b^{4} + 8 \, a b^{2} c - 22 \, a^{2} c^{2}\right )} x^{2} + 2 \,{\left (5 \, a b^{3} - 11 \, a^{2} b c\right )} x\right )}{\left (2 \, c d x + b d\right )}^{\frac{1}{3}}{\left (c x^{2} + b x + a\right )}^{\frac{1}{3}}}{35 \,{\left (128 \,{\left (b^{4} c^{7} - 8 \, a b^{2} c^{8} + 16 \, a^{2} c^{9}\right )} d^{8} x^{7} + 448 \,{\left (b^{5} c^{6} - 8 \, a b^{3} c^{7} + 16 \, a^{2} b c^{8}\right )} d^{8} x^{6} + 672 \,{\left (b^{6} c^{5} - 8 \, a b^{4} c^{6} + 16 \, a^{2} b^{2} c^{7}\right )} d^{8} x^{5} + 560 \,{\left (b^{7} c^{4} - 8 \, a b^{5} c^{5} + 16 \, a^{2} b^{3} c^{6}\right )} d^{8} x^{4} + 280 \,{\left (b^{8} c^{3} - 8 \, a b^{6} c^{4} + 16 \, a^{2} b^{4} c^{5}\right )} d^{8} x^{3} + 84 \,{\left (b^{9} c^{2} - 8 \, a b^{7} c^{3} + 16 \, a^{2} b^{5} c^{4}\right )} d^{8} x^{2} + 14 \,{\left (b^{10} c - 8 \, a b^{8} c^{2} + 16 \, a^{2} b^{6} c^{3}\right )} d^{8} x +{\left (b^{11} - 8 \, a b^{9} c + 16 \, a^{2} b^{7} c^{2}\right )} d^{8}\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)^(23/3),x, algorithm="fricas")

[Out]

3/35*(6*c^4*x^6 + 18*b*c^3*x^5 + (23*b^2*c^2 - 2*a*c^3)*x^4 + 5*a^2*b^2 - 14*a^3*c + 4*(4*b^3*c - a*b*c^2)*x^3
+ (5*b^4 + 8*a*b^2*c - 22*a^2*c^2)*x^2 + 2*(5*a*b^3 - 11*a^2*b*c)*x)*(2*c*d*x + b*d)^(1/3)*(c*x^2 + b*x + a)^
(1/3)/(128*(b^4*c^7 - 8*a*b^2*c^8 + 16*a^2*c^9)*d^8*x^7 + 448*(b^5*c^6 - 8*a*b^3*c^7 + 16*a^2*b*c^8)*d^8*x^6 +
672*(b^6*c^5 - 8*a*b^4*c^6 + 16*a^2*b^2*c^7)*d^8*x^5 + 560*(b^7*c^4 - 8*a*b^5*c^5 + 16*a^2*b^3*c^6)*d^8*x^4 +
280*(b^8*c^3 - 8*a*b^6*c^4 + 16*a^2*b^4*c^5)*d^8*x^3 + 84*(b^9*c^2 - 8*a*b^7*c^3 + 16*a^2*b^5*c^4)*d^8*x^2 +
14*(b^10*c - 8*a*b^8*c^2 + 16*a^2*b^6*c^3)*d^8*x + (b^11 - 8*a*b^9*c + 16*a^2*b^7*c^2)*d^8)

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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)**(23/3),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{23}{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)^(23/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(23/3), x)