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

Optimal result

Integrand size = 21, antiderivative size = 377 \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^3} \, dx=-\frac {d x}{6 c (b c-a d) \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2}+\frac {b (6 b c+a d) x}{6 a c (b c-a d)^2 \sqrt [3]{a+b x^3} \left (c+d x^3\right )}+\frac {d \left (18 b^2 c^2+15 a b c d-5 a^2 d^2\right ) x \left (a+b x^3\right )^{2/3}}{18 a c^2 (b c-a d)^3 \left (c+d x^3\right )}-\frac {d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} c^{8/3} (b c-a d)^{10/3}}-\frac {d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \log \left (c+d x^3\right )}{54 c^{8/3} (b c-a d)^{10/3}}+\frac {d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} (b c-a d)^{10/3}} \]

[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)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {425, 541, 12, 384} \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^3} \, dx=-\frac {d \left (5 a^2 d^2-18 a b c d+27 b^2 c^2\right ) \arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} c^{8/3} (b c-a d)^{10/3}}+\frac {d x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2+15 a b c d+18 b^2 c^2\right )}{18 a c^2 \left (c+d x^3\right ) (b c-a d)^3}-\frac {d \left (5 a^2 d^2-18 a b c d+27 b^2 c^2\right ) \log \left (c+d x^3\right )}{54 c^{8/3} (b c-a d)^{10/3}}+\frac {d \left (5 a^2 d^2-18 a b c d+27 b^2 c^2\right ) \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} (b c-a d)^{10/3}}+\frac {b x (a d+6 b c)}{6 a c \sqrt [3]{a+b x^3} \left (c+d x^3\right ) (b c-a d)^2}-\frac {d x}{6 c \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)} \]

[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*} \text {integral}& = -\frac {d x}{6 c (b c-a d) \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2}+\frac {\int \frac {6 b c-5 a d-6 b d x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2} \, dx}{6 c (b c-a d)} \\ & = -\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 {\int \frac {a d (12 b c-5 a d)-3 b d (6 b c+a d) x^3}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )^2} \, dx}{6 a c (b c-a d)^2} \\ & = -\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 {\int \frac {2 a d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right )}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{18 a c^2 (b c-a d)^3} \\ & = -\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 {\left (d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{9 c^2 (b c-a d)^3} \\ & = -\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}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 13.68 (sec) , antiderivative size = 428, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^3} \, dx=-\frac {65 c^2 \left (a+b x^3\right )^2 \left (-14000 a^2 c^5-21896 a b c^5 x^3-48104 a^2 c^4 d x^3-8391 b^2 c^5 x^6-70802 a b c^4 d x^6-60807 a^2 c^3 d^2 x^6-24417 b^2 c^4 d x^9-81534 a b c^3 d^2 x^9-33657 a^2 c^2 d^3 x^9-23409 b^2 c^3 d^2 x^{12}-38652 a b c^2 d^3 x^{12}-7155 a^2 c d^4 x^{12}-7425 b^2 c^2 d^3 x^{15}-5940 a b c d^4 x^{15}-243 a^2 d^5 x^{15}+28 \left (c+d x^3\right )^2 \left (27 b^2 c^2 x^6 \left (7 c+6 d x^3\right )+9 a b c x^3 \left (73 c^2+104 c d x^3+33 d^2 x^6\right )+a^2 \left (500 c^3+843 c^2 d x^3+375 c d^2 x^6+27 d^3 x^9\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},1,\frac {4}{3},\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )\right )+486 (b c-a d)^4 x^{12} \left (c+d x^3\right )^3 \, _4F_3\left (2,2,2,\frac {7}{3};1,1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{16380 c^5 (-b c+a d)^3 x^8 \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \]

[In]

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

[Out]

-1/16380*(65*c^2*(a + b*x^3)^2*(-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*(c + d*x^3)^2*(27*b^2*c^2*x^6*(7*c + 6*d*x^3) + 9*a*b*c*x^3*(73*c^2 + 104*
c*d*x^3 + 33*d^2*x^6) + a^2*(500*c^3 + 843*c^2*d*x^3 + 375*c*d^2*x^6 + 27*d^3*x^9))*Hypergeometric2F1[1/3, 1,
4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))]) + 486*(b*c - a*d)^4*x^12*(c + d*x^3)^3*HypergeometricPFQ[{2, 2, 2, 7/
3}, {1, 1, 16/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))])/(c^5*(-(b*c) + a*d)^3*x^8*(a + b*x^3)^(7/3)*(c + d*x^3)^
2)

Maple [A] (verified)

Time = 4.46 (sec) , antiderivative size = 426, normalized size of antiderivative = 1.13

method	result	size
pseudoelliptic	\(-\frac {-2 \left (5 a^{2} d^{2}-18 a b c d +27 b^{2} c^{2}\right ) d a \left (d \,x^{3}+c \right )^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-3 x c \left (5 a^{2} b \,d^{4} x^{6}-15 a \,b^{2} c \,d^{3} x^{6}-18 b^{3} c^{2} d^{2} x^{6}+5 a^{3} d^{4} x^{3}-7 a^{2} b c \,d^{3} x^{3}-18 a \,b^{2} c^{2} d^{2} x^{3}-36 b^{3} c^{3} d \,x^{3}+8 a^{3} c \,d^{3}-18 a^{2} b \,c^{2} d^{2}-18 c^{4} b^{3}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}+\left (5 a^{2} d^{2}-18 a b c d +27 b^{2} c^{2}\right ) d a \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )\right ) \left (d \,x^{3}+c \right )^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{54 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (d \,x^{3}+c \right )^{2} c^{3} \left (a d -b c \right )^{3} a}\)	\(426\)

[In]

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

[Out]

-1/54*(-2*(5*a^2*d^2-18*a*b*c*d+27*b^2*c^2)*d*a*(d*x^3+c)^2*(b*x^3+a)^(1/3)*ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a
)^(1/3))/x)-3*x*c*(5*a^2*b*d^4*x^6-15*a*b^2*c*d^3*x^6-18*b^3*c^2*d^2*x^6+5*a^3*d^4*x^3-7*a^2*b*c*d^3*x^3-18*a*
b^2*c^2*d^2*x^3-36*b^3*c^3*d*x^3+8*a^3*c*d^3-18*a^2*b*c^2*d^2-18*b^3*c^4)*((a*d-b*c)/c)^(1/3)+(5*a^2*d^2-18*a*
b*c*d+27*b^2*c^2)*d*a*(-2*arctan(1/3*3^(1/2)*(((a*d-b*c)/c)^(1/3)*x-2*(b*x^3+a)^(1/3))/((a*d-b*c)/c)^(1/3)/x)*
3^(1/2)+ln((((a*d-b*c)/c)^(2/3)*x^2-((a*d-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2))*(d*x^3+c)^2*(
b*x^3+a)^(1/3))/((a*d-b*c)/c)^(1/3)/(b*x^3+a)^(1/3)/(d*x^3+c)^2/c^3/(a*d-b*c)^3/a

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^3} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{4/3}\,{\left (d\,x^3+c\right )}^3} \,d x \]

[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)