3.15.12 \(\int \frac {(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{11/3}} \, dx\) [1412]

Optimal. Leaf size=320 \[ -\frac {3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac {9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}}}{\sqrt {3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}-\frac {3 \log \left ((d (b+2 c x))^{2/3}-2^{2/3} \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}} \]

[Out]

-3/16*(d*(2*c*x+b))^(4/3)*(c*x^2+b*x+a)^(1/3)/c^2/(-4*a*c+b^2)/d^5+9/16*(d*(2*c*x+b))^(4/3)*(c*x^2+b*x+a)^(4/3
)/c/(-4*a*c+b^2)^2/d^5+3/4*(c*x^2+b*x+a)^(7/3)/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(8/3)-9/4*(c*x^2+b*x+a)^(7/3)/(-4*
a*c+b^2)^2/d^3/(2*c*d*x+b*d)^(2/3)-3/64*ln((d*(2*c*x+b))^(2/3)-2^(2/3)*c^(1/3)*d^(2/3)*(c*x^2+b*x+a)^(1/3))*2^
(1/3)/c^(7/3)/d^(11/3)-1/32*arctan(1/3*(1+2^(1/3)*(d*(2*c*x+b))^(2/3)/c^(1/3)/d^(2/3)/(c*x^2+b*x+a)^(1/3))*3^(
1/2))*3^(1/2)*2^(1/3)/c^(7/3)/d^(11/3)

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Rubi [A]
time = 0.45, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {707, 708, 285, 335, 281, 337} \begin {gather*} -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\frac {\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}}+1}{\sqrt {3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}-\frac {3 \sqrt [3]{a+b x+c x^2} (d (b+2 c x))^{4/3}}{16 c^2 d^5 \left (b^2-4 a c\right )}+\frac {9 \left (a+b x+c x^2\right )^{4/3} (d (b+2 c x))^{4/3}}{16 c d^5 \left (b^2-4 a c\right )^2}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{2/3}}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac {3 \log \left ((d (b+2 c x))^{2/3}-2^{2/3} \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(11/3),x]

[Out]

(-3*(d*(b + 2*c*x))^(4/3)*(a + b*x + c*x^2)^(1/3))/(16*c^2*(b^2 - 4*a*c)*d^5) + (9*(d*(b + 2*c*x))^(4/3)*(a +
b*x + c*x^2)^(4/3))/(16*c*(b^2 - 4*a*c)^2*d^5) + (3*(a + b*x + c*x^2)^(7/3))/(4*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x
)^(8/3)) - (9*(a + b*x + c*x^2)^(7/3))/(4*(b^2 - 4*a*c)^2*d^3*(b*d + 2*c*d*x)^(2/3)) - (Sqrt[3]*ArcTan[(1 + (2
^(1/3)*(d*(b + 2*c*x))^(2/3))/(c^(1/3)*d^(2/3)*(a + b*x + c*x^2)^(1/3)))/Sqrt[3]])/(16*2^(2/3)*c^(7/3)*d^(11/3
)) - (3*Log[(d*(b + 2*c*x))^(2/3) - 2^(2/3)*c^(1/3)*d^(2/3)*(a + b*x + c*x^2)^(1/3)])/(32*2^(2/3)*c^(7/3)*d^(1
1/3))

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 285

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 337

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Simp[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(
1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]

Rule 707

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 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)^{11/3}} \, dx &=\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {3 \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{5/3}} \, dx}{4 \left (b^2-4 a c\right ) d^2}\\ &=\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac {9 \int \sqrt [3]{b d+2 c d x} \left (a+b x+c x^2\right )^{4/3} \, dx}{2 \left (b^2-4 a c\right )^2 d^4}\\ &=\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac {9 \text {Subst}\left (\int \sqrt [3]{x} \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^{4/3} \, dx,x,b d+2 c d x\right )}{4 c \left (b^2-4 a c\right )^2 d^5}\\ &=\frac {9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}-\frac {3 \text {Subst}\left (\int \sqrt [3]{x} \sqrt [3]{a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}} \, dx,x,b d+2 c d x\right )}{8 c^2 \left (b^2-4 a c\right ) d^5}\\ &=-\frac {3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac {9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac {\text {Subst}\left (\int \frac {\sqrt [3]{x}}{\left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^{2/3}} \, dx,x,b d+2 c d x\right )}{32 c^3 d^5}\\ &=-\frac {3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac {9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac {3 \text {Subst}\left (\int \frac {x^3}{\left (a-\frac {b^2}{4 c}+\frac {x^6}{4 c d^2}\right )^{2/3}} \, dx,x,\sqrt [3]{d (b+2 c x)}\right )}{32 c^3 d^5}\\ &=-\frac {3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac {9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac {3 \text {Subst}\left (\int \frac {x}{\left (a-\frac {b^2}{4 c}+\frac {x^3}{4 c d^2}\right )^{2/3}} \, dx,x,(d (b+2 c x))^{2/3}\right )}{64 c^3 d^5}\\ &=-\frac {3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac {9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac {3 \text {Subst}\left (\int \frac {x}{1-\frac {x^3}{4 c d^2}} \, dx,x,\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{64 c^3 d^5}\\ &=-\frac {3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac {9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}+\frac {\text {Subst}\left (\int \frac {1}{1-\frac {x}{2^{2/3} \sqrt [3]{c} d^{2/3}}} \, dx,x,\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{32 \sqrt [3]{2} c^{8/3} d^{13/3}}-\frac {\text {Subst}\left (\int \frac {1-\frac {x}{2^{2/3} \sqrt [3]{c} d^{2/3}}}{1+\frac {x}{2^{2/3} \sqrt [3]{c} d^{2/3}}+\frac {x^2}{2 \sqrt [3]{2} c^{2/3} d^{4/3}}} \, dx,x,\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{32 \sqrt [3]{2} c^{8/3} d^{13/3}}\\ &=-\frac {3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac {9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}-\frac {\log \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}-\frac {3 \text {Subst}\left (\int \frac {1}{1+\frac {x}{2^{2/3} \sqrt [3]{c} d^{2/3}}+\frac {x^2}{2 \sqrt [3]{2} c^{2/3} d^{4/3}}} \, dx,x,\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{64 \sqrt [3]{2} c^{8/3} d^{13/3}}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2^{2/3} \sqrt [3]{c} d^{2/3}}+\frac {x}{\sqrt [3]{2} c^{2/3} d^{4/3}}}{1+\frac {x}{2^{2/3} \sqrt [3]{c} d^{2/3}}+\frac {x^2}{2 \sqrt [3]{2} c^{2/3} d^{4/3}}} \, dx,x,\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}\\ &=-\frac {3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac {9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}-\frac {\log \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}+\frac {\log \left (4 c^{2/3} d^{4/3}+\frac {2^{2/3} (d (b+2 c x))^{4/3}}{(a+x (b+c x))^{2/3}}+\frac {2 \sqrt [3]{2} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}+\frac {3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{c} d^{2/3} \sqrt [3]{a+x (b+c x)}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}\\ &=-\frac {3 (d (b+2 c x))^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac {9 (d (b+2 c x))^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{c} d^{2/3} \sqrt [3]{a+x (b+c x)}}}{\sqrt {3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}-\frac {\log \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}+\frac {\log \left (4 c^{2/3} d^{4/3}+\frac {2^{2/3} (d (b+2 c x))^{4/3}}{(a+x (b+c x))^{2/3}}+\frac {2 \sqrt [3]{2} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 10.08, size = 104, normalized size = 0.32 \begin {gather*} \frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \, _2F_1\left (-\frac {4}{3},-\frac {4}{3};-\frac {1}{3};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{64\ 2^{2/3} c^2 d (d (b+2 c x))^{8/3} \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)^(11/3),x]

[Out]

(3*(b^2 - 4*a*c)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[-4/3, -4/3, -1/3, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(64
*2^(2/3)*c^2*d*(d*(b + 2*c*x))^(8/3)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))

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Maple [F]
time = 0.35, size = 0, normalized size = 0.00 \[\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {11}{3}}}\, dx\]

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)^(11/3),x)

[Out]

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(11/3),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)^(11/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(11/3), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)^(11/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x + c x^{2}\right )^{\frac {4}{3}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {11}{3}}}\, dx \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)**(11/3),x)

[Out]

Integral((a + b*x + c*x**2)**(4/3)/(d*(b + 2*c*x))**(11/3), x)

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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)^(11/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(11/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{11/3}} \,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)^(11/3),x)

[Out]

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(11/3), x)

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