3.750 \(\int \frac{x^5}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=174 \[ \frac{c \log \left (c+d x^3\right )}{6 d^{2/3} (b c-a d)^{4/3}}-\frac{c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} (b c-a d)^{4/3}}-\frac{c \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{2/3} (b c-a d)^{4/3}}+\frac{a}{b \sqrt [3]{a+b x^3} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.171119, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 78, 56, 617, 204, 31} \[ \frac{c \log \left (c+d x^3\right )}{6 d^{2/3} (b c-a d)^{4/3}}-\frac{c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} (b c-a d)^{4/3}}-\frac{c \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{2/3} (b c-a d)^{4/3}}+\frac{a}{b \sqrt [3]{a+b x^3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

a/(b*(b*c - a*d)*(a + b*x^3)^(1/3)) - (c*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]]
)/(Sqrt[3]*d^(2/3)*(b*c - a*d)^(4/3)) + (c*Log[c + d*x^3])/(6*d^(2/3)*(b*c - a*d)^(4/3)) - (c*Log[(b*c - a*d)^
(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(2/3)*(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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

Mathematica [C]  time = 0.0307372, size = 77, normalized size = 0.44 \[ \frac{b c \left (a+b x^3\right ) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{d \left (b x^3+a\right )}{a d-b c}\right )+2 a (b c-a d)}{2 b \sqrt [3]{a+b x^3} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

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

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5}}{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^5/(b*x^3+a)^(4/3)/(d*x^3+c),x)

[Out]

int(x^5/(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^5/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.48355, size = 1890, normalized size = 10.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[-1/6*(3*sqrt(1/3)*(a*b^2*c^2*d - a^2*b*c*d^2 + (b^3*c^2*d - a*b^2*c*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)) - (b^2*c*x^3 + a*b*c)*(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)) + 2*(b^2*c*
x^3 + a*b*c)*(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)) - 6*(a*b*c*d^2 - a^2*d
^3)*(b*x^3 + a)^(2/3))/(a*b^3*c^2*d^2 - 2*a^2*b^2*c*d^3 + a^3*b*d^4 + (b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d
^4)*x^3), 1/6*(6*sqrt(1/3)*(a*b^2*c^2*d - a^2*b*c*d^2 + (b^3*c^2*d - a*b^2*c*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) + (b^2*c*x^3 + a*b*c)*(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)) - 2*(b^2*c*x^3 + a*b*c)*(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)) + 6*(a*b*c*d^2 - a^2*d^3)*(b*x^3 + a)^(2/3))/(a*b^3*c^2*d^2 - 2*a
^2*b^2*c*d^3 + a^3*b*d^4 + (b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\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**5/(b*x**3+a)**(4/3)/(d*x**3+c),x)

[Out]

Integral(x**5/((a + b*x**3)**(4/3)*(c + d*x**3)), x)

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Giac [B]  time = 1.19788, size = 406, normalized size = 2.33 \begin{align*} -\frac{\frac{6 \,{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} b c \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^{2} - 2 \, \sqrt{3} a b c d^{3} + \sqrt{3} a^{2} d^{4}} - \frac{{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} b c \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 )}{b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}} + \frac{2 \, b c \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 )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} - \frac{6 \, a}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (b c - a d\right )}}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

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