### 3.1415 $$\int \frac{(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{29/3}} \, dx$$

Optimal. Leaf size=133 $\frac{27 \left (a+b x+c x^2\right )^{7/3}}{455 d^5 \left (b^2-4 a c\right )^3 (b d+2 c d x)^{14/3}}+\frac{9 \left (a+b x+c x^2\right )^{7/3}}{65 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{20/3}}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{13 d \left (b^2-4 a c\right ) (b d+2 c d x)^{26/3}}$

[Out]

(3*(a + b*x + c*x^2)^(7/3))/(13*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(26/3)) + (9*(a + b*x + c*x^2)^(7/3))/(65*(b^2
- 4*a*c)^2*d^3*(b*d + 2*c*d*x)^(20/3)) + (27*(a + b*x + c*x^2)^(7/3))/(455*(b^2 - 4*a*c)^3*d^5*(b*d + 2*c*d*x
)^(14/3))

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Rubi [A]  time = 0.0658108, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.071, Rules used = {693, 682} $\frac{27 \left (a+b x+c x^2\right )^{7/3}}{455 d^5 \left (b^2-4 a c\right )^3 (b d+2 c d x)^{14/3}}+\frac{9 \left (a+b x+c x^2\right )^{7/3}}{65 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{20/3}}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{13 d \left (b^2-4 a c\right ) (b d+2 c d x)^{26/3}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(29/3),x]

[Out]

(3*(a + b*x + c*x^2)^(7/3))/(13*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(26/3)) + (9*(a + b*x + c*x^2)^(7/3))/(65*(b^2
- 4*a*c)^2*d^3*(b*d + 2*c*d*x)^(20/3)) + (27*(a + b*x + c*x^2)^(7/3))/(455*(b^2 - 4*a*c)^3*d^5*(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)^{29/3}} \, dx &=\frac{3 \left (a+b x+c x^2\right )^{7/3}}{13 \left (b^2-4 a c\right ) d (b d+2 c d x)^{26/3}}+\frac{6 \int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx}{13 \left (b^2-4 a c\right ) d^2}\\ &=\frac{3 \left (a+b x+c x^2\right )^{7/3}}{13 \left (b^2-4 a c\right ) d (b d+2 c d x)^{26/3}}+\frac{9 \left (a+b x+c x^2\right )^{7/3}}{65 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{20/3}}+\frac{9 \int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{17/3}} \, dx}{65 \left (b^2-4 a c\right )^2 d^4}\\ &=\frac{3 \left (a+b x+c x^2\right )^{7/3}}{13 \left (b^2-4 a c\right ) d (b d+2 c d x)^{26/3}}+\frac{9 \left (a+b x+c x^2\right )^{7/3}}{65 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{20/3}}+\frac{27 \left (a+b x+c x^2\right )^{7/3}}{455 \left (b^2-4 a c\right )^3 d^5 (b d+2 c d x)^{14/3}}\\ \end{align*}

Mathematica [A]  time = 0.104086, size = 122, normalized size = 0.92 $\frac{3 (a+x (b+c x))^{7/3} \left (16 c^2 \left (35 a^2-21 a c x^2+9 c^2 x^4\right )+4 b^2 c \left (75 c x^2-91 a\right )+48 b c^2 x \left (6 c x^2-7 a\right )+156 b^3 c x+65 b^4\right )}{455 d^9 \left (b^2-4 a c\right )^3 (b+2 c x)^8 (d (b+2 c x))^{2/3}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(29/3),x]

[Out]

(3*(a + x*(b + c*x))^(7/3)*(65*b^4 + 156*b^3*c*x + 48*b*c^2*x*(-7*a + 6*c*x^2) + 4*b^2*c*(-91*a + 75*c*x^2) +
16*c^2*(35*a^2 - 21*a*c*x^2 + 9*c^2*x^4)))/(455*(b^2 - 4*a*c)^3*d^9*(b + 2*c*x)^8*(d*(b + 2*c*x))^(2/3))

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Maple [A]  time = 0.048, size = 139, normalized size = 1.1 \begin{align*} -{\frac{ \left ( 6\,cx+3\,b \right ) \left ( 144\,{c}^{4}{x}^{4}+288\,b{c}^{3}{x}^{3}-336\,{x}^{2}a{c}^{3}+300\,{x}^{2}{b}^{2}{c}^{2}-336\,ba{c}^{2}x+156\,{b}^{3}cx+560\,{a}^{2}{c}^{2}-364\,ac{b}^{2}+65\,{b}^{4} \right ) }{29120\,{a}^{3}{c}^{3}-21840\,{a}^{2}{b}^{2}{c}^{2}+5460\,a{b}^{4}c-455\,{b}^{6}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{7}{3}}} \left ( 2\,cdx+bd \right ) ^{-{\frac{29}{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)^(29/3),x)

[Out]

-3/455*(2*c*x+b)*(c*x^2+b*x+a)^(7/3)*(144*c^4*x^4+288*b*c^3*x^3-336*a*c^3*x^2+300*b^2*c^2*x^2-336*a*b*c^2*x+15
6*b^3*c*x+560*a^2*c^2-364*a*b^2*c+65*b^4)/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)/(2*c*d*x+b*d)^(29/3)

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Maxima [B]  time = 2.55788, size = 1057, normalized size = 7.95 \begin{align*} \text{result too large to display} \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)^(29/3),x, algorithm="maxima")

[Out]

3/455*(144*c^6*d^(1/3)*x^8 + 576*b*c^5*d^(1/3)*x^7 + 12*(85*b^2*c^4*d^(1/3) - 4*a*c^5*d^(1/3))*x^6 + 65*a^2*b^
4*d^(1/3) - 364*a^3*b^2*c*d^(1/3) + 560*a^4*c^2*d^(1/3) + 36*(29*b^3*c^3*d^(1/3) - 4*a*b*c^4*d^(1/3))*x^5 + (6
77*b^4*c^2*d^(1/3) - 196*a*b^2*c^3*d^(1/3) + 32*a^2*c^4*d^(1/3))*x^4 + 2*(143*b^5*c*d^(1/3) - 76*a*b^3*c^2*d^(
1/3) + 32*a^2*b*c^3*d^(1/3))*x^3 + (65*b^6*d^(1/3) + 78*a*b^4*c*d^(1/3) - 540*a^2*b^2*c^2*d^(1/3) + 784*a^3*c^
3*d^(1/3))*x^2 + 2*(65*a*b^5*d^(1/3) - 286*a^2*b^3*c*d^(1/3) + 392*a^3*b*c^2*d^(1/3))*x)*(c*x^2 + b*x + a)^(1/
3)/((b^14*d^10 - 12*a*b^12*c*d^10 + 48*a^2*b^10*c^2*d^10 - 64*a^3*b^8*c^3*d^10 + 256*(b^6*c^8*d^10 - 12*a*b^4*
c^9*d^10 + 48*a^2*b^2*c^10*d^10 - 64*a^3*c^11*d^10)*x^8 + 1024*(b^7*c^7*d^10 - 12*a*b^5*c^8*d^10 + 48*a^2*b^3*
c^9*d^10 - 64*a^3*b*c^10*d^10)*x^7 + 1792*(b^8*c^6*d^10 - 12*a*b^6*c^7*d^10 + 48*a^2*b^4*c^8*d^10 - 64*a^3*b^2
*c^9*d^10)*x^6 + 1792*(b^9*c^5*d^10 - 12*a*b^7*c^6*d^10 + 48*a^2*b^5*c^7*d^10 - 64*a^3*b^3*c^8*d^10)*x^5 + 112
0*(b^10*c^4*d^10 - 12*a*b^8*c^5*d^10 + 48*a^2*b^6*c^6*d^10 - 64*a^3*b^4*c^7*d^10)*x^4 + 448*(b^11*c^3*d^10 - 1
2*a*b^9*c^4*d^10 + 48*a^2*b^7*c^5*d^10 - 64*a^3*b^5*c^6*d^10)*x^3 + 112*(b^12*c^2*d^10 - 12*a*b^10*c^3*d^10 +
48*a^2*b^8*c^4*d^10 - 64*a^3*b^6*c^5*d^10)*x^2 + 16*(b^13*c*d^10 - 12*a*b^11*c^2*d^10 + 48*a^2*b^9*c^3*d^10 -
64*a^3*b^7*c^4*d^10)*x)*(2*c*x + b)^(2/3))

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Fricas [B]  time = 4.3618, size = 1536, normalized size = 11.55 \begin{align*} \frac{3 \,{\left (144 \, c^{6} x^{8} + 576 \, b c^{5} x^{7} + 12 \,{\left (85 \, b^{2} c^{4} - 4 \, a c^{5}\right )} x^{6} + 65 \, a^{2} b^{4} - 364 \, a^{3} b^{2} c + 560 \, a^{4} c^{2} + 36 \,{\left (29 \, b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{5} +{\left (677 \, b^{4} c^{2} - 196 \, a b^{2} c^{3} + 32 \, a^{2} c^{4}\right )} x^{4} + 2 \,{\left (143 \, b^{5} c - 76 \, a b^{3} c^{2} + 32 \, a^{2} b c^{3}\right )} x^{3} +{\left (65 \, b^{6} + 78 \, a b^{4} c - 540 \, a^{2} b^{2} c^{2} + 784 \, a^{3} c^{3}\right )} x^{2} + 2 \,{\left (65 \, a b^{5} - 286 \, a^{2} b^{3} c + 392 \, a^{3} b c^{2}\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}}}{455 \,{\left (512 \,{\left (b^{6} c^{9} - 12 \, a b^{4} c^{10} + 48 \, a^{2} b^{2} c^{11} - 64 \, a^{3} c^{12}\right )} d^{10} x^{9} + 2304 \,{\left (b^{7} c^{8} - 12 \, a b^{5} c^{9} + 48 \, a^{2} b^{3} c^{10} - 64 \, a^{3} b c^{11}\right )} d^{10} x^{8} + 4608 \,{\left (b^{8} c^{7} - 12 \, a b^{6} c^{8} + 48 \, a^{2} b^{4} c^{9} - 64 \, a^{3} b^{2} c^{10}\right )} d^{10} x^{7} + 5376 \,{\left (b^{9} c^{6} - 12 \, a b^{7} c^{7} + 48 \, a^{2} b^{5} c^{8} - 64 \, a^{3} b^{3} c^{9}\right )} d^{10} x^{6} + 4032 \,{\left (b^{10} c^{5} - 12 \, a b^{8} c^{6} + 48 \, a^{2} b^{6} c^{7} - 64 \, a^{3} b^{4} c^{8}\right )} d^{10} x^{5} + 2016 \,{\left (b^{11} c^{4} - 12 \, a b^{9} c^{5} + 48 \, a^{2} b^{7} c^{6} - 64 \, a^{3} b^{5} c^{7}\right )} d^{10} x^{4} + 672 \,{\left (b^{12} c^{3} - 12 \, a b^{10} c^{4} + 48 \, a^{2} b^{8} c^{5} - 64 \, a^{3} b^{6} c^{6}\right )} d^{10} x^{3} + 144 \,{\left (b^{13} c^{2} - 12 \, a b^{11} c^{3} + 48 \, a^{2} b^{9} c^{4} - 64 \, a^{3} b^{7} c^{5}\right )} d^{10} x^{2} + 18 \,{\left (b^{14} c - 12 \, a b^{12} c^{2} + 48 \, a^{2} b^{10} c^{3} - 64 \, a^{3} b^{8} c^{4}\right )} d^{10} x +{\left (b^{15} - 12 \, a b^{13} c + 48 \, a^{2} b^{11} c^{2} - 64 \, a^{3} b^{9} c^{3}\right )} d^{10}\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)^(29/3),x, algorithm="fricas")

[Out]

3/455*(144*c^6*x^8 + 576*b*c^5*x^7 + 12*(85*b^2*c^4 - 4*a*c^5)*x^6 + 65*a^2*b^4 - 364*a^3*b^2*c + 560*a^4*c^2
+ 36*(29*b^3*c^3 - 4*a*b*c^4)*x^5 + (677*b^4*c^2 - 196*a*b^2*c^3 + 32*a^2*c^4)*x^4 + 2*(143*b^5*c - 76*a*b^3*c
^2 + 32*a^2*b*c^3)*x^3 + (65*b^6 + 78*a*b^4*c - 540*a^2*b^2*c^2 + 784*a^3*c^3)*x^2 + 2*(65*a*b^5 - 286*a^2*b^3
*c + 392*a^3*b*c^2)*x)*(2*c*d*x + b*d)^(1/3)*(c*x^2 + b*x + a)^(1/3)/(512*(b^6*c^9 - 12*a*b^4*c^10 + 48*a^2*b^
2*c^11 - 64*a^3*c^12)*d^10*x^9 + 2304*(b^7*c^8 - 12*a*b^5*c^9 + 48*a^2*b^3*c^10 - 64*a^3*b*c^11)*d^10*x^8 + 46
08*(b^8*c^7 - 12*a*b^6*c^8 + 48*a^2*b^4*c^9 - 64*a^3*b^2*c^10)*d^10*x^7 + 5376*(b^9*c^6 - 12*a*b^7*c^7 + 48*a^
2*b^5*c^8 - 64*a^3*b^3*c^9)*d^10*x^6 + 4032*(b^10*c^5 - 12*a*b^8*c^6 + 48*a^2*b^6*c^7 - 64*a^3*b^4*c^8)*d^10*x
^5 + 2016*(b^11*c^4 - 12*a*b^9*c^5 + 48*a^2*b^7*c^6 - 64*a^3*b^5*c^7)*d^10*x^4 + 672*(b^12*c^3 - 12*a*b^10*c^4
+ 48*a^2*b^8*c^5 - 64*a^3*b^6*c^6)*d^10*x^3 + 144*(b^13*c^2 - 12*a*b^11*c^3 + 48*a^2*b^9*c^4 - 64*a^3*b^7*c^5
)*d^10*x^2 + 18*(b^14*c - 12*a*b^12*c^2 + 48*a^2*b^10*c^3 - 64*a^3*b^8*c^4)*d^10*x + (b^15 - 12*a*b^13*c + 48*
a^2*b^11*c^2 - 64*a^3*b^9*c^3)*d^10)

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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)**(29/3),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{29}{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)^(29/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(29/3), x)