3.114 \(\int \frac{1}{\sqrt [3]{a+b x^3} (c+d x^3)^3} \, dx\)

Optimal. Leaf size=307 \[ \frac{\left (5 a^2 d^2-12 a b c d+9 b^2 c^2\right ) \log \left (c+d x^3\right )}{54 c^{8/3} (b c-a d)^{7/3}}-\frac{\left (5 a^2 d^2-12 a b c d+9 b^2 c^2\right ) \log \left (\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} (b c-a d)^{7/3}}+\frac{\left (5 a^2 d^2-12 a b c d+9 b^2 c^2\right ) \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} c^{8/3} (b c-a d)^{7/3}}-\frac{d x \left (a+b x^3\right )^{2/3} (9 b c-5 a d)}{18 c^2 \left (c+d x^3\right ) (b c-a d)^2}-\frac{d x \left (a+b x^3\right )^{2/3}}{6 c \left (c+d x^3\right )^2 (b c-a d)} \]

[Out]

-(d*x*(a + b*x^3)^(2/3))/(6*c*(b*c - a*d)*(c + d*x^3)^2) - (d*(9*b*c - 5*a*d)*x*(a + b*x^3)^(2/3))/(18*c^2*(b*
c - a*d)^2*(c + d*x^3)) + ((9*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(
a + b*x^3)^(1/3)))/Sqrt[3]])/(9*Sqrt[3]*c^(8/3)*(b*c - a*d)^(7/3)) + ((9*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*Log
[c + d*x^3])/(54*c^(8/3)*(b*c - a*d)^(7/3)) - ((9*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*Log[((b*c - a*d)^(1/3)*x)/
c^(1/3) - (a + b*x^3)^(1/3)])/(18*c^(8/3)*(b*c - a*d)^(7/3))

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Rubi [C]  time = 0.310109, antiderivative size = 167, normalized size of antiderivative = 0.54, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {430, 429} \[ -\frac{x \left (c d \left (-a^2 d \left (8 c+5 d x^3\right )+a b \left (12 c^2+c d x^3-5 d^2 x^6\right )+3 b^2 c x^3 \left (4 c+3 d x^3\right )\right )-2 \left (c+d x^3\right )^2 \left (5 a^2 d^2-12 a b c d+9 b^2 c^2\right ) \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )\right )}{18 c^3 \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^3)^(1/3)*(c + d*x^3)^3),x]

[Out]

-(x*(c*d*(3*b^2*c*x^3*(4*c + 3*d*x^3) - a^2*d*(8*c + 5*d*x^3) + a*b*(12*c^2 + c*d*x^3 - 5*d^2*x^6)) - 2*(9*b^2
*c^2 - 12*a*b*c*d + 5*a^2*d^2)*(c + d*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))]
))/(18*c^3*(b*c - a*d)^2*(a + b*x^3)^(1/3)*(c + d*x^3)^2)

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )^3} \, dx &=\frac{\sqrt [3]{1+\frac{b x^3}{a}} \int \frac{1}{\sqrt [3]{1+\frac{b x^3}{a}} \left (c+d x^3\right )^3} \, dx}{\sqrt [3]{a+b x^3}}\\ &=-\frac{x \left (c d \left (3 b^2 c x^3 \left (4 c+3 d x^3\right )-a^2 d \left (8 c+5 d x^3\right )+a b \left (12 c^2+c d x^3-5 d^2 x^6\right )\right )-2 \left (9 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \left (c+d x^3\right )^2 \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (a+b x^3\right )}\right )\right )}{18 c^3 (b c-a d)^2 \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2}\\ \end{align*}

Mathematica [C]  time = 0.225834, size = 168, normalized size = 0.55 \[ \frac{x \left (2 \left (c+d x^3\right )^2 \left (5 a^2 d^2-12 a b c d+9 b^2 c^2\right ) \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-c d \left (-a^2 d \left (8 c+5 d x^3\right )+a b \left (12 c^2+c d x^3-5 d^2 x^6\right )+3 b^2 c x^3 \left (4 c+3 d x^3\right )\right )\right )}{18 c^3 \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x^3)^(1/3)*(c + d*x^3)^3),x]

[Out]

(x*(-(c*d*(3*b^2*c*x^3*(4*c + 3*d*x^3) - a^2*d*(8*c + 5*d*x^3) + a*b*(12*c^2 + c*d*x^3 - 5*d^2*x^6))) + 2*(9*b
^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*(c + d*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3)
)]))/(18*c^3*(b*c - a*d)^2*(a + b*x^3)^(1/3)*(c + d*x^3)^2)

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Maple [F]  time = 0.441, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d{x}^{3}+c \right ) ^{3}}{\frac{1}{\sqrt [3]{b{x}^{3}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^3+a)^(1/3)/(d*x^3+c)^3,x)

[Out]

int(1/(b*x^3+a)^(1/3)/(d*x^3+c)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x^{3} + c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a)^(1/3)/(d*x^3+c)^3,x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(1/3)*(d*x^3 + c)^3), x)

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Fricas [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(1/(b*x^3+a)^(1/3)/(d*x^3+c)^3,x, algorithm="fricas")

[Out]

Timed out

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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(1/(b*x**3+a)**(1/3)/(d*x**3+c)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x^{3} + c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a)^(1/3)/(d*x^3+c)^3,x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(1/3)*(d*x^3 + c)^3), x)