3.2.3 \(\int \frac {1}{(a+b x^3)^{7/3} (c+d x^3)^2} \, dx\) [103]

Optimal. Leaf size=324 \[ \frac {b (3 b c+4 a d) x}{12 a c (b c-a d)^2 \left (a+b x^3\right )^{4/3}}+\frac {b \left (9 b^2 c^2-33 a b c d-4 a^2 d^2\right ) x}{12 a^2 c (b c-a d)^3 \sqrt [3]{a+b x^3}}-\frac {d x}{3 c (b c-a d) \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}+\frac {d^2 (9 b c-2 a d) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} (b c-a d)^{10/3}}+\frac {d^2 (9 b c-2 a d) \log \left (c+d x^3\right )}{18 c^{5/3} (b c-a d)^{10/3}}-\frac {d^2 (9 b c-2 a d) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{6 c^{5/3} (b c-a d)^{10/3}} \]

[Out]

1/12*b*(4*a*d+3*b*c)*x/a/c/(-a*d+b*c)^2/(b*x^3+a)^(4/3)+1/12*b*(-4*a^2*d^2-33*a*b*c*d+9*b^2*c^2)*x/a^2/c/(-a*d
+b*c)^3/(b*x^3+a)^(1/3)-1/3*d*x/c/(-a*d+b*c)/(b*x^3+a)^(4/3)/(d*x^3+c)+1/18*d^2*(-2*a*d+9*b*c)*ln(d*x^3+c)/c^(
5/3)/(-a*d+b*c)^(10/3)-1/6*d^2*(-2*a*d+9*b*c)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(5/3)/(-a*d+b*c
)^(10/3)+1/9*d^2*(-2*a*d+9*b*c)*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3^(1/2))/c^(5/3)/(
-a*d+b*c)^(10/3)*3^(1/2)

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Rubi [A]
time = 0.25, antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {425, 541, 12, 384} \begin {gather*} \frac {b x \left (-4 a^2 d^2-33 a b c d+9 b^2 c^2\right )}{12 a^2 c \sqrt [3]{a+b x^3} (b c-a d)^3}+\frac {d^2 (9 b c-2 a d) \text {ArcTan}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} (b c-a d)^{10/3}}+\frac {d^2 (9 b c-2 a d) \log \left (c+d x^3\right )}{18 c^{5/3} (b c-a d)^{10/3}}-\frac {d^2 (9 b c-2 a d) \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{6 c^{5/3} (b c-a d)^{10/3}}-\frac {d x}{3 c \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) (b c-a d)}+\frac {b x (4 a d+3 b c)}{12 a c \left (a+b x^3\right )^{4/3} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^3)^(7/3)*(c + d*x^3)^2),x]

[Out]

(b*(3*b*c + 4*a*d)*x)/(12*a*c*(b*c - a*d)^2*(a + b*x^3)^(4/3)) + (b*(9*b^2*c^2 - 33*a*b*c*d - 4*a^2*d^2)*x)/(1
2*a^2*c*(b*c - a*d)^3*(a + b*x^3)^(1/3)) - (d*x)/(3*c*(b*c - a*d)*(a + b*x^3)^(4/3)*(c + d*x^3)) + (d^2*(9*b*c
- 2*a*d)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]])/(3*Sqrt[3]*c^(5/3)*(b*c -
a*d)^(10/3)) + (d^2*(9*b*c - 2*a*d)*Log[c + d*x^3])/(18*c^(5/3)*(b*c - a*d)^(10/3)) - (d^2*(9*b*c - 2*a*d)*Lo
g[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(6*c^(5/3)*(b*c - a*d)^(10/3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 541

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

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \, dx &=\frac {\sqrt [3]{1+\frac {b x^3}{a}} \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{7/3} \left (c+d x^3\right )^2} \, dx}{a^2 \sqrt [3]{a+b x^3}}\\ &=\frac {26130 c^5 (b c-a d)^2 x^6 \left (a+b x^3\right )^2+89505 c^4 d (b c-a d)^2 x^9 \left (a+b x^3\right )^2+84240 c^3 d^2 (b c-a d)^2 x^{12} \left (a+b x^3\right )^2+26325 c^2 d^3 (b c-a d)^2 x^{15} \left (a+b x^3\right )^2+748020 c^6 (b c-a d) x^3 \left (a+b x^3\right )^3+2113020 c^5 d (b c-a d) x^6 \left (a+b x^3\right )^3+1916460 c^4 d^2 (b c-a d) x^9 \left (a+b x^3\right )^3+589680 c^3 d^3 (b c-a d) x^{12} \left (a+b x^3\right )^3-2002000 c^7 \left (a+b x^3\right )^4-5460000 c^6 d x^3 \left (a+b x^3\right )^4-4914000 c^5 d^2 x^6 \left (a+b x^3\right )^4-1506960 c^4 d^3 x^9 \left (a+b x^3\right )^4-1248520 c^6 (b c-a d) x^3 \left (a+b x^3\right )^3 \, _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 )-3478020 c^5 d (b c-a d) x^6 \left (a+b x^3\right )^3 \, _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 )-3144960 c^4 d^2 (b c-a d) x^9 \left (a+b x^3\right )^3 \, _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 )-966420 c^3 d^3 (b c-a d) x^{12} \left (a+b x^3\right )^3 \, _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 )+2002000 c^7 \left (a+b x^3\right )^4 \, _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 )+5460000 c^6 d x^3 \left (a+b x^3\right )^4 \, _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 )+4914000 c^5 d^2 x^6 \left (a+b x^3\right )^4 \, _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 )+1506960 c^4 d^3 x^9 \left (a+b x^3\right )^4 \, _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 )+7938 c^3 (b c-a d)^4 x^{12} \, _3F_2\left (2,2,\frac {10}{3};1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+22680 c^2 d (b c-a d)^4 x^{15} \, _3F_2\left (2,2,\frac {10}{3};1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+21546 c d^2 (b c-a d)^4 x^{18} \, _3F_2\left (2,2,\frac {10}{3};1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+6804 d^3 (b c-a d)^4 x^{21} \, _3F_2\left (2,2,\frac {10}{3};1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+1134 c^3 (b c-a d)^4 x^{12} \, _4F_3\left (2,2,2,\frac {10}{3};1,1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+3402 c^2 d (b c-a d)^4 x^{15} \, _4F_3\left (2,2,2,\frac {10}{3};1,1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+3402 c d^2 (b c-a d)^4 x^{18} \, _4F_3\left (2,2,2,\frac {10}{3};1,1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+1134 d^3 (b c-a d)^4 x^{21} \, _4F_3\left (2,2,2,\frac {10}{3};1,1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{21840 c^5 (b c-a d)^3 x^8 \left (a+b x^3\right )^{10/3} \left (c+d x^3\right )}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 6.31, size = 443, normalized size = 1.37 \begin {gather*} \frac {\frac {3 c^{2/3} x \left (4 a^4 d^3+8 a^3 b d^3 x^3-9 b^4 c^2 x^3 \left (c+d x^3\right )+4 a^2 b^2 d \left (9 c^2+9 c d x^3+d^2 x^6\right )+3 a b^3 c \left (-4 c^2+7 c d x^3+11 d^2 x^6\right )\right )}{a^2 (-b c+a d)^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}+\frac {2 i \left (3 i+\sqrt {3}\right ) d^2 (-9 b c+2 a d) \tanh ^{-1}\left (\frac {i+\frac {\left (-i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d} x}}{\sqrt {3}}\right )}{(b c-a d)^{10/3}}+\frac {2 \left (1+i \sqrt {3}\right ) d^2 (9 b c-2 a d) \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{10/3}}+\frac {\left (1+i \sqrt {3}\right ) d^2 (-9 b c+2 a d) \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{10/3}}}{36 c^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x^3)^(7/3)*(c + d*x^3)^2),x]

[Out]

((3*c^(2/3)*x*(4*a^4*d^3 + 8*a^3*b*d^3*x^3 - 9*b^4*c^2*x^3*(c + d*x^3) + 4*a^2*b^2*d*(9*c^2 + 9*c*d*x^3 + d^2*
x^6) + 3*a*b^3*c*(-4*c^2 + 7*c*d*x^3 + 11*d^2*x^6)))/(a^2*(-(b*c) + a*d)^3*(a + b*x^3)^(4/3)*(c + d*x^3)) + ((
2*I)*(3*I + Sqrt[3])*d^2*(-9*b*c + 2*a*d)*ArcTanh[(I + ((-I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3))/((b*c - a*d)
^(1/3)*x))/Sqrt[3]])/(b*c - a*d)^(10/3) + (2*(1 + I*Sqrt[3])*d^2*(9*b*c - 2*a*d)*Log[2*(b*c - a*d)^(1/3)*x + (
1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(b*c - a*d)^(10/3) + ((1 + I*Sqrt[3])*d^2*(-9*b*c + 2*a*d)*Log[2*(b
*c - a*d)^(2/3)*x^2 + (-1 - I*Sqrt[3])*c^(1/3)*(b*c - a*d)^(1/3)*x*(a + b*x^3)^(1/3) + I*(I + Sqrt[3])*c^(2/3)
*(a + b*x^3)^(2/3)])/(b*c - a*d)^(10/3))/(36*c^(5/3))

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {7}{3}} \left (d \,x^{3}+c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^3+a)^(7/3)/(d*x^3+c)^2,x)

[Out]

int(1/(b*x^3+a)^(7/3)/(d*x^3+c)^2,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(1/(b*x^3+a)^(7/3)/(d*x^3+c)^2,x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(7/3)*(d*x^3 + c)^2), 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(1/(b*x^3+a)^(7/3)/(d*x^3+c)^2,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 {1}{\left (a + b x^{3}\right )^{\frac {7}{3}} \left (c + d x^{3}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**3+a)**(7/3)/(d*x**3+c)**2,x)

[Out]

Integral(1/((a + b*x**3)**(7/3)*(c + d*x**3)**2), 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(1/(b*x^3+a)^(7/3)/(d*x^3+c)^2,x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(7/3)*(d*x^3 + c)^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (b\,x^3+a\right )}^{7/3}\,{\left (d\,x^3+c\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^3)^(7/3)*(c + d*x^3)^2),x)

[Out]

int(1/((a + b*x^3)^(7/3)*(c + d*x^3)^2), x)

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