3.2.15 \(\int \frac {1}{(a+b x^3)^{4/3} (c+d x^3)^3} \, dx\) [115]

Optimal. Leaf size=377 \[ -\frac {d x}{6 c (b c-a d) \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2}+\frac {b (6 b c+a d) x}{6 a c (b c-a d)^2 \sqrt [3]{a+b x^3} \left (c+d x^3\right )}+\frac {d \left (18 b^2 c^2+15 a b c d-5 a^2 d^2\right ) x \left (a+b x^3\right )^{2/3}}{18 a c^2 (b c-a d)^3 \left (c+d x^3\right )}-\frac {d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} c^{8/3} (b c-a d)^{10/3}}-\frac {d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \log \left (c+d x^3\right )}{54 c^{8/3} (b c-a d)^{10/3}}+\frac {d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} (b c-a d)^{10/3}} \]

[Out]

-1/6*d*x/c/(-a*d+b*c)/(b*x^3+a)^(1/3)/(d*x^3+c)^2+1/6*b*(a*d+6*b*c)*x/a/c/(-a*d+b*c)^2/(b*x^3+a)^(1/3)/(d*x^3+
c)+1/18*d*(-5*a^2*d^2+15*a*b*c*d+18*b^2*c^2)*x*(b*x^3+a)^(2/3)/a/c^2/(-a*d+b*c)^3/(d*x^3+c)-1/54*d*(5*a^2*d^2-
18*a*b*c*d+27*b^2*c^2)*ln(d*x^3+c)/c^(8/3)/(-a*d+b*c)^(10/3)+1/18*d*(5*a^2*d^2-18*a*b*c*d+27*b^2*c^2)*ln((-a*d
+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(8/3)/(-a*d+b*c)^(10/3)-1/27*d*(5*a^2*d^2-18*a*b*c*d+27*b^2*c^2)*arct
an(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3^(1/2))/c^(8/3)/(-a*d+b*c)^(10/3)*3^(1/2)

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Rubi [A]
time = 0.25, antiderivative size = 377, 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 {d \left (5 a^2 d^2-18 a b c d+27 b^2 c^2\right ) \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 )}{9 \sqrt {3} c^{8/3} (b c-a d)^{10/3}}+\frac {d x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2+15 a b c d+18 b^2 c^2\right )}{18 a c^2 \left (c+d x^3\right ) (b c-a d)^3}-\frac {d \left (5 a^2 d^2-18 a b c d+27 b^2 c^2\right ) \log \left (c+d x^3\right )}{54 c^{8/3} (b c-a d)^{10/3}}+\frac {d \left (5 a^2 d^2-18 a b c d+27 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)^{10/3}}+\frac {b x (a d+6 b c)}{6 a c \sqrt [3]{a+b x^3} \left (c+d x^3\right ) (b c-a d)^2}-\frac {d x}{6 c \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^3)^(4/3)*(c + d*x^3)^3),x]

[Out]

-1/6*(d*x)/(c*(b*c - a*d)*(a + b*x^3)^(1/3)*(c + d*x^3)^2) + (b*(6*b*c + a*d)*x)/(6*a*c*(b*c - a*d)^2*(a + b*x
^3)^(1/3)*(c + d*x^3)) + (d*(18*b^2*c^2 + 15*a*b*c*d - 5*a^2*d^2)*x*(a + b*x^3)^(2/3))/(18*a*c^2*(b*c - a*d)^3
*(c + d*x^3)) - (d*(27*b^2*c^2 - 18*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)^(10/3)) - (d*(27*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*Log[c
+ d*x^3])/(54*c^(8/3)*(b*c - a*d)^(10/3)) + (d*(27*b^2*c^2 - 18*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)^(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 )^{4/3} \left (c+d x^3\right )^3} \, dx &=\frac {\sqrt [3]{1+\frac {b x^3}{a}} \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{4/3} \left (c+d x^3\right )^3} \, dx}{a \sqrt [3]{a+b x^3}}\\ &=-\frac {65 c^2 \left (a+b x^3\right )^2 \left (14000 a^2 c^5+21896 a b c^5 x^3+48104 a^2 c^4 d x^3+8391 b^2 c^5 x^6+70802 a b c^4 d x^6+60807 a^2 c^3 d^2 x^6+24417 b^2 c^4 d x^9+81534 a b c^3 d^2 x^9+33657 a^2 c^2 d^3 x^9+23409 b^2 c^3 d^2 x^{12}+38652 a b c^2 d^3 x^{12}+7155 a^2 c d^4 x^{12}+7425 b^2 c^2 d^3 x^{15}+5940 a b c d^4 x^{15}+243 a^2 d^5 x^{15}-28 \left (c+d x^3\right )^2 \left (27 b^2 c^2 x^6 \left (7 c+6 d x^3\right )+9 a b c x^3 \left (73 c^2+104 c d x^3+33 d^2 x^6\right )+a^2 \left (500 c^3+843 c^2 d x^3+375 c d^2 x^6+27 d^3 x^9\right )\right ) \, _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 )-486 (b c-a d)^4 x^{12} \left (c+d x^3\right )^3 \, _4F_3\left (2,2,2,\frac {7}{3};1,1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{16380 c^5 (b c-a d)^3 x^8 \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 7.99, size = 471, normalized size = 1.25 \begin {gather*} \frac {-\frac {6 c^{2/3} x \left (18 b^3 c^2 \left (c+d x^3\right )^2+3 a b^2 c d^2 x^3 \left (6 c+5 d x^3\right )-a^3 d^3 \left (8 c+5 d x^3\right )+a^2 b d^2 \left (18 c^2+7 c d x^3-5 d^2 x^6\right )\right )}{a (-b c+a d)^3 \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2}+\frac {2 i \left (3 i+\sqrt {3}\right ) d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \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 i \left (-i+\sqrt {3}\right ) d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \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 \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \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}}}{108 c^{8/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x^3)^(4/3)*(c + d*x^3)^3),x]

[Out]

((-6*c^(2/3)*x*(18*b^3*c^2*(c + d*x^3)^2 + 3*a*b^2*c*d^2*x^3*(6*c + 5*d*x^3) - a^3*d^3*(8*c + 5*d*x^3) + a^2*b
*d^2*(18*c^2 + 7*c*d*x^3 - 5*d^2*x^6)))/(a*(-(b*c) + a*d)^3*(a + b*x^3)^(1/3)*(c + d*x^3)^2) + ((2*I)*(3*I + S
qrt[3])*d*(27*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*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*I)*(-I + Sqrt[3])*d*(27*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*
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*(27*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*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))/(108*c^(8/3))

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^3+a)^(4/3)/(d*x^3+c)^3,x)

[Out]

int(1/(b*x^3+a)^(4/3)/(d*x^3+c)^3,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)^(4/3)/(d*x^3+c)^3,x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)^3), 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)^(4/3)/(d*x^3+c)^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [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)**(4/3)/(d*x**3+c)**3,x)

[Out]

Timed out

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

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)^3), 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 )}^{4/3}\,{\left (d\,x^3+c\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^3)^(4/3)*(c + d*x^3)^3),x)

[Out]

int(1/((a + b*x^3)^(4/3)*(c + d*x^3)^3), x)

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