3.748 \(\int \frac{x^{11}}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=253 \[ \frac{a^3}{b^3 \sqrt [3]{a+b x^3} (b c-a d)}-\frac{\left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^3 d^2}-\frac{a \left (a+b x^3\right )^{2/3}}{2 b^3 d}+\frac{\left (a+b x^3\right )^{5/3}}{5 b^3 d}+\frac{c^3 \log \left (c+d x^3\right )}{6 d^{8/3} (b c-a d)^{4/3}}-\frac{c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} (b c-a d)^{4/3}}-\frac{c^3 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{8/3} (b c-a d)^{4/3}} \]

[Out]

a^3/(b^3*(b*c - a*d)*(a + b*x^3)^(1/3)) - (a*(a + b*x^3)^(2/3))/(2*b^3*d) - ((b*c + a*d)*(a + b*x^3)^(2/3))/(2
*b^3*d^2) + (a + b*x^3)^(5/3)/(5*b^3*d) - (c^3*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sq
rt[3]])/(Sqrt[3]*d^(8/3)*(b*c - a*d)^(4/3)) + (c^3*Log[c + d*x^3])/(6*d^(8/3)*(b*c - a*d)^(4/3)) - (c^3*Log[(b
*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(8/3)*(b*c - a*d)^(4/3))

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Rubi [A]  time = 0.340169, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {446, 87, 43, 56, 617, 204, 31} \[ \frac{a^3}{b^3 \sqrt [3]{a+b x^3} (b c-a d)}-\frac{\left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^3 d^2}-\frac{a \left (a+b x^3\right )^{2/3}}{2 b^3 d}+\frac{\left (a+b x^3\right )^{5/3}}{5 b^3 d}+\frac{c^3 \log \left (c+d x^3\right )}{6 d^{8/3} (b c-a d)^{4/3}}-\frac{c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} (b c-a d)^{4/3}}-\frac{c^3 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{8/3} (b c-a d)^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3/(b^3*(b*c - a*d)*(a + b*x^3)^(1/3)) - (a*(a + b*x^3)^(2/3))/(2*b^3*d) - ((b*c + a*d)*(a + b*x^3)^(2/3))/(2
*b^3*d^2) + (a + b*x^3)^(5/3)/(5*b^3*d) - (c^3*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sq
rt[3]])/(Sqrt[3]*d^(8/3)*(b*c - a*d)^(4/3)) + (c^3*Log[c + d*x^3])/(6*d^(8/3)*(b*c - a*d)^(4/3)) - (c^3*Log[(b
*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(8/3)*(b*c - a*d)^(4/3))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(
1/3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && Ne
gQ[(b*c - a*d)/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [C]  time = 0.0878751, size = 147, normalized size = 0.58 \[ \frac{3 a^2 b d^2 \left (2 d x^3-c\right )+18 a^3 d^3-a b^2 d \left (5 c^2+c d x^3+2 d^2 x^6\right )+10 b^3 c^3 \, _2F_1\left (-\frac{1}{3},1;\frac{2}{3};\frac{d \left (b x^3+a\right )}{a d-b c}\right )+b^3 c \left (-10 c^2-5 c d x^3+2 d^2 x^6\right )}{10 b^3 d^3 \sqrt [3]{a+b x^3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(18*a^3*d^3 + 3*a^2*b*d^2*(-c + 2*d*x^3) + b^3*c*(-10*c^2 - 5*c*d*x^3 + 2*d^2*x^6) - a*b^2*d*(5*c^2 + c*d*x^3
+ 2*d^2*x^6) + 10*b^3*c^3*Hypergeometric2F1[-1/3, 1, 2/3, (d*(a + b*x^3))/(-(b*c) + a*d)])/(10*b^3*d^3*(b*c -
a*d)*(a + b*x^3)^(1/3))

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Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{11}}{d{x}^{3}+c} \left ( b{x}^{3}+a \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.60631, size = 2417, normalized size = 9.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[-1/30*(15*sqrt(1/3)*(a*b^4*c^4*d - a^2*b^3*c^3*d^2 + (b^5*c^4*d - a*b^4*c^3*d^2)*x^3)*sqrt(-(b*c*d^2 - a*d^3)
^(1/3)/(b*c - a*d))*log((2*b*d^2*x^3 - b*c*d + 3*a*d^2 - 3*sqrt(1/3)*(2*(b*c*d^2 - a*d^3)^(2/3)*(b*x^3 + a)^(2
/3) + (b*x^3 + a)^(1/3)*(b*c*d - a*d^2) - (b*c*d^2 - a*d^3)^(1/3)*(b*c - a*d))*sqrt(-(b*c*d^2 - a*d^3)^(1/3)/(
b*c - a*d)) - 3*(b*c*d^2 - a*d^3)^(2/3)*(b*x^3 + a)^(1/3))/(d*x^3 + c)) - 5*(b^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2
- a*d^3)^(2/3)*log((b*x^3 + a)^(2/3)*d^2 - (b*c*d^2 - a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(2
/3)) + 10*(b^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2 - a*d^3)^(2/3)*log((b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(1/3))
+ 3*(5*a*b^3*c^3*d^2 - 2*a^2*b^2*c^2*d^3 - 21*a^3*b*c*d^4 + 18*a^4*d^5 - 2*(b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2
*b^2*d^5)*x^6 + (5*b^4*c^3*d^2 - 4*a*b^3*c^2*d^3 - 7*a^2*b^2*c*d^4 + 6*a^3*b*d^5)*x^3)*(b*x^3 + a)^(2/3))/(a*b
^5*c^2*d^4 - 2*a^2*b^4*c*d^5 + a^3*b^3*d^6 + (b^6*c^2*d^4 - 2*a*b^5*c*d^5 + a^2*b^4*d^6)*x^3), 1/30*(30*sqrt(1
/3)*(a*b^4*c^4*d - a^2*b^3*c^3*d^2 + (b^5*c^4*d - a*b^4*c^3*d^2)*x^3)*sqrt((b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d)
)*arctan(sqrt(1/3)*(2*(b*x^3 + a)^(1/3)*d - (b*c*d^2 - a*d^3)^(1/3))*sqrt((b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d))
/d) + 5*(b^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2 - a*d^3)^(2/3)*log((b*x^3 + a)^(2/3)*d^2 - (b*c*d^2 - a*d^3)^(1/3)*
(b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(2/3)) - 10*(b^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2 - a*d^3)^(2/3)*log((b*x
^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(1/3)) - 3*(5*a*b^3*c^3*d^2 - 2*a^2*b^2*c^2*d^3 - 21*a^3*b*c*d^4 + 18*a^4*
d^5 - 2*(b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5)*x^6 + (5*b^4*c^3*d^2 - 4*a*b^3*c^2*d^3 - 7*a^2*b^2*c*d^4 +
6*a^3*b*d^5)*x^3)*(b*x^3 + a)^(2/3))/(a*b^5*c^2*d^4 - 2*a^2*b^4*c*d^5 + a^3*b^3*d^6 + (b^6*c^2*d^4 - 2*a*b^5*
c*d^5 + a^2*b^4*d^6)*x^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23657, size = 502, normalized size = 1.98 \begin{align*} -\frac{{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} c^{3} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} + \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}}\right )}{\sqrt{3} b^{2} c^{2} d^{4} - 2 \, \sqrt{3} a b c d^{5} + \sqrt{3} a^{2} d^{6}} + \frac{{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} c^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} + \left (-\frac{b c - a d}{d}\right )^{\frac{2}{3}}\right )}{6 \,{\left (b^{2} c^{2} d^{4} - 2 \, a b c d^{5} + a^{2} d^{6}\right )}} - \frac{c^{3} \left (-\frac{b c - a d}{d}\right )^{\frac{2}{3}} \log \left ({\left |{\left (b x^{3} + a\right )}^{\frac{1}{3}} - \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} \right |}\right )}{3 \,{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}} + \frac{a^{3}}{{\left (b^{4} c - a b^{3} d\right )}{\left (b x^{3} + a\right )}^{\frac{1}{3}}} - \frac{5 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} b^{13} c d^{3} - 2 \,{\left (b x^{3} + a\right )}^{\frac{5}{3}} b^{12} d^{4} + 10 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}} a b^{12} d^{4}}{10 \, b^{15} d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-(-b*c*d^2 + a*d^3)^(2/3)*c^3*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/
d)^(1/3))/(sqrt(3)*b^2*c^2*d^4 - 2*sqrt(3)*a*b*c*d^5 + sqrt(3)*a^2*d^6) + 1/6*(-b*c*d^2 + a*d^3)^(2/3)*c^3*log
((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/3) + (-(b*c - a*d)/d)^(2/3))/(b^2*c^2*d^4 - 2*a*b*c
*d^5 + a^2*d^6) - 1/3*c^3*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^2*c^2
*d^2 - 2*a*b*c*d^3 + a^2*d^4) + a^3/((b^4*c - a*b^3*d)*(b*x^3 + a)^(1/3)) - 1/10*(5*(b*x^3 + a)^(2/3)*b^13*c*d
^3 - 2*(b*x^3 + a)^(5/3)*b^12*d^4 + 10*(b*x^3 + a)^(2/3)*a*b^12*d^4)/(b^15*d^5)