3.670 \(\int \frac{\sqrt [3]{a+b x^3}}{x^8 (c+d x^3)} \, dx\)

Optimal. Leaf size=258 \[ \frac{\sqrt [3]{a+b x^3} \left (-28 a^2 d^2+7 a b c d+3 b^2 c^2\right )}{28 a^2 c^3 x}+\frac{d^2 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 c^{10/3}}-\frac{d^2 \sqrt [3]{b c-a d} \log \left (\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{10/3}}-\frac{d^2 \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} c^{10/3}}-\frac{\sqrt [3]{a+b x^3} (b c-7 a d)}{28 a c^2 x^4}-\frac{\sqrt [3]{a+b x^3}}{7 c x^7} \]

[Out]

-(a + b*x^3)^(1/3)/(7*c*x^7) - ((b*c - 7*a*d)*(a + b*x^3)^(1/3))/(28*a*c^2*x^4) + ((3*b^2*c^2 + 7*a*b*c*d - 28
*a^2*d^2)*(a + b*x^3)^(1/3))/(28*a^2*c^3*x) - (d^2*(b*c - a*d)^(1/3)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1
/3)*(a + b*x^3)^(1/3)))/Sqrt[3]])/(Sqrt[3]*c^(10/3)) + (d^2*(b*c - a*d)^(1/3)*Log[c + d*x^3])/(6*c^(10/3)) - (
d^2*(b*c - a*d)^(1/3)*Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(2*c^(10/3))

________________________________________________________________________________________

Rubi [C]  time = 0.939877, antiderivative size = 451, normalized size of antiderivative = 1.75, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {511, 510} \[ -\frac{-9 x^3 \left (c+d x^3\right )^2 (b c-a d) \, _3F_2\left (\frac{2}{3},2,2;1,\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-2 x^3 \left (2 c^2-3 c d x^3+9 d^2 x^6\right ) (b c-a d) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-12 b c^2 d x^6 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+15 b c^3 x^3 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-15 a c^2 d x^3 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+27 a d^3 x^9 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-27 b c d^2 x^9 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+12 a c d^2 x^6 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-12 a c^2 d x^3+8 a c^3+36 a c d^2 x^6-12 b c^2 d x^6+8 b c^3 x^3+36 b c d^2 x^9}{56 c^4 x^7 \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*x^3)^(1/3)/(x^8*(c + d*x^3)),x]

[Out]

-(8*a*c^3 + 8*b*c^3*x^3 - 12*a*c^2*d*x^3 - 12*b*c^2*d*x^6 + 36*a*c*d^2*x^6 + 36*b*c*d^2*x^9 - 2*(b*c - a*d)*x^
3*(2*c^2 - 3*c*d*x^3 + 9*d^2*x^6)*Hypergeometric2F1[2/3, 1, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 15*b*c^3
*x^3*Hypergeometric2F1[2/3, 2, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 15*a*c^2*d*x^3*Hypergeometric2F1[2/3,
2, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 12*b*c^2*d*x^6*Hypergeometric2F1[2/3, 2, 5/3, ((b*c - a*d)*x^3)/
(c*(a + b*x^3))] + 12*a*c*d^2*x^6*Hypergeometric2F1[2/3, 2, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 27*b*c*d
^2*x^9*Hypergeometric2F1[2/3, 2, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 27*a*d^3*x^9*Hypergeometric2F1[2/3,
2, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 9*(b*c - a*d)*x^3*(c + d*x^3)^2*HypergeometricPFQ[{2/3, 2, 2}, {
1, 5/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))])/(56*c^4*x^7*(a + b*x^3)^(2/3))

Rule 511

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

Rule 510

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

Rubi steps

\begin{align*} \int \frac{\sqrt [3]{a+b x^3}}{x^8 \left (c+d x^3\right )} \, dx &=\frac{\sqrt [3]{a+b x^3} \int \frac{\sqrt [3]{1+\frac{b x^3}{a}}}{x^8 \left (c+d x^3\right )} \, dx}{\sqrt [3]{1+\frac{b x^3}{a}}}\\ &=-\frac{8 a c^3+8 b c^3 x^3-12 a c^2 d x^3-12 b c^2 d x^6+36 a c d^2 x^6+36 b c d^2 x^9-2 (b c-a d) x^3 \left (2 c^2-3 c d x^3+9 d^2 x^6\right ) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+15 b c^3 x^3 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-15 a c^2 d x^3 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-12 b c^2 d x^6 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+12 a c d^2 x^6 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-27 b c d^2 x^9 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+27 a d^3 x^9 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-9 (b c-a d) x^3 \left (c+d x^3\right )^2 \, _3F_2\left (\frac{2}{3},2,2;1,\frac{5}{3};\frac{(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{56 c^4 x^7 \left (a+b x^3\right )^{2/3}}\\ \end{align*}

Mathematica [C]  time = 1.59537, size = 451, normalized size = 1.75 \[ -\frac{-9 x^3 \left (c+d x^3\right )^2 (b c-a d) \text{HypergeometricPFQ}\left (\left \{\frac{2}{3},2,2\right \},\left \{1,\frac{5}{3}\right \},\frac{x^3 (b c-a d)}{c \left (a+b x^3\right )}\right )-2 x^3 \left (2 c^2-3 c d x^3+9 d^2 x^6\right ) (b c-a d) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-12 b c^2 d x^6 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+15 b c^3 x^3 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-15 a c^2 d x^3 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+27 a d^3 x^9 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-27 b c d^2 x^9 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+12 a c d^2 x^6 \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-12 a c^2 d x^3+8 a c^3+36 a c d^2 x^6-12 b c^2 d x^6+8 b c^3 x^3+36 b c d^2 x^9}{56 c^4 x^7 \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(1/3)/(x^8*(c + d*x^3)),x]

[Out]

-(8*a*c^3 + 8*b*c^3*x^3 - 12*a*c^2*d*x^3 - 12*b*c^2*d*x^6 + 36*a*c*d^2*x^6 + 36*b*c*d^2*x^9 - 2*(b*c - a*d)*x^
3*(2*c^2 - 3*c*d*x^3 + 9*d^2*x^6)*Hypergeometric2F1[2/3, 1, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 15*b*c^3
*x^3*Hypergeometric2F1[2/3, 2, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 15*a*c^2*d*x^3*Hypergeometric2F1[2/3,
2, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 12*b*c^2*d*x^6*Hypergeometric2F1[2/3, 2, 5/3, ((b*c - a*d)*x^3)/
(c*(a + b*x^3))] + 12*a*c*d^2*x^6*Hypergeometric2F1[2/3, 2, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 27*b*c*d
^2*x^9*Hypergeometric2F1[2/3, 2, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 27*a*d^3*x^9*Hypergeometric2F1[2/3,
2, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 9*(b*c - a*d)*x^3*(c + d*x^3)^2*HypergeometricPFQ[{2/3, 2, 2}, {
1, 5/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))])/(56*c^4*x^7*(a + b*x^3)^(2/3))

________________________________________________________________________________________

Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{8} \left ( d{x}^{3}+c \right ) }\sqrt [3]{b{x}^{3}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(1/3)/x^8/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(1/3)/x^8/(d*x^3+c),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{{\left (d x^{3} + c\right )} x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)/x^8/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*x^8), x)

________________________________________________________________________________________

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((b*x^3+a)^(1/3)/x^8/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt [3]{a + b x^{3}}}{x^{8} \left (c + d x^{3}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/3)/x**8/(d*x**3+c),x)

[Out]

Integral((a + b*x**3)**(1/3)/(x**8*(c + d*x**3)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{{\left (d x^{3} + c\right )} x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)/x^8/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*x^8), x)