\(\int \frac {1}{(a+b x^3)^{4/3} (c+d x^3)^3} \, dx\) [167]

Optimal result

Integrand size = 21, antiderivative size = 366 \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^3} \, dx=\frac {b \left (18 b^2 c^2+15 a b c d-5 a^2 d^2\right ) x}{18 a c^2 (b c-a d)^3 \sqrt [3]{a+b x^3}}-\frac {d x}{6 c (b c-a d) \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2}-\frac {d (12 b c-5 a d) x}{18 c^2 (b c-a d)^2 \sqrt [3]{a+b x^3} \left (c+d x^3\right )}-\frac {d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} c^{8/3} (b c-a d)^{10/3}}-\frac {d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \log \left (c+d x^3\right )}{54 c^{8/3} (b c-a d)^{10/3}}+\frac {d \left (27 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} (b c-a d)^{10/3}} \] Output:

1/18*b*(-5*a^2*d^2+15*a*b*c*d+18*b^2*c^2)*x/a/c^2/(-a*d+b*c)^3/(b*x^3+a)^( 
1/3)-1/6*d*x/c/(-a*d+b*c)/(b*x^3+a)^(1/3)/(d*x^3+c)^2-1/18*d*(-5*a*d+12*b* 
c)*x/c^2/(-a*d+b*c)^2/(b*x^3+a)^(1/3)/(d*x^3+c)-1/27*d*(5*a^2*d^2-18*a*b*c 
*d+27*b^2*c^2)*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3)) 
*3^(1/2))*3^(1/2)/c^(8/3)/(-a*d+b*c)^(10/3)-1/54*d*(5*a^2*d^2-18*a*b*c*d+2 
7*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)
 

Mathematica [C] (warning: unable to verify)

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

Time = 14.17 (sec) , antiderivative size = 428, normalized size of antiderivative = 1.17 \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^3} \, dx=-\frac {65 c^2 \left (a+b x^3\right )^2 \left (-14000 a^2 c^5-21896 a b c^5 x^3-48104 a^2 c^4 d x^3-8391 b^2 c^5 x^6-70802 a b c^4 d x^6-60807 a^2 c^3 d^2 x^6-24417 b^2 c^4 d x^9-81534 a b c^3 d^2 x^9-33657 a^2 c^2 d^3 x^9-23409 b^2 c^3 d^2 x^{12}-38652 a b c^2 d^3 x^{12}-7155 a^2 c d^4 x^{12}-7425 b^2 c^2 d^3 x^{15}-5940 a b c d^4 x^{15}-243 a^2 d^5 x^{15}+28 \left (c+d x^3\right )^2 \left (27 b^2 c^2 x^6 \left (7 c+6 d x^3\right )+9 a b c x^3 \left (73 c^2+104 c d x^3+33 d^2 x^6\right )+a^2 \left (500 c^3+843 c^2 d x^3+375 c d^2 x^6+27 d^3 x^9\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},1,\frac {4}{3},\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )\right )+486 (b c-a d)^4 x^{12} \left (c+d x^3\right )^3 \, _4F_3\left (2,2,2,\frac {7}{3};1,1,\frac {16}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{16380 c^5 (-b c+a d)^3 x^8 \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \] Input:

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

Output:

-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)
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {931, 1024, 27, 1024, 27, 901}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 931

\(\displaystyle \frac {\int \frac {-6 b d x^3+6 b c-5 a d}{\left (b x^3+a\right )^{4/3} \left (d x^3+c\right )^2}dx}{6 c (b c-a d)}-\frac {d x}{6 c \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 1024

\(\displaystyle \frac {\frac {b x (a d+6 b c)}{a \sqrt [3]{a+b x^3} \left (c+d x^3\right ) (b c-a d)}-\frac {\int \frac {d \left (a (12 b c-5 a d)-3 b (6 b c+a d) x^3\right )}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )^2}dx}{a (b c-a d)}}{6 c (b c-a d)}-\frac {d x}{6 c \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {b x (a d+6 b c)}{a \sqrt [3]{a+b x^3} \left (c+d x^3\right ) (b c-a d)}-\frac {d \int \frac {a (12 b c-5 a d)-3 b (6 b c+a d) x^3}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )^2}dx}{a (b c-a d)}}{6 c (b c-a d)}-\frac {d x}{6 c \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 1024

\(\displaystyle \frac {\frac {b x (a d+6 b c)}{a \sqrt [3]{a+b x^3} \left (c+d x^3\right ) (b c-a d)}-\frac {d \left (\frac {\int \frac {2 a \left (27 b^2 c^2-18 a b d c+5 a^2 d^2\right )}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{3 c (b c-a d)}-\frac {x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2+15 a b c d+18 b^2 c^2\right )}{3 c \left (c+d x^3\right ) (b c-a d)}\right )}{a (b c-a d)}}{6 c (b c-a d)}-\frac {d x}{6 c \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {b x (a d+6 b c)}{a \sqrt [3]{a+b x^3} \left (c+d x^3\right ) (b c-a d)}-\frac {d \left (\frac {2 a \left (5 a^2 d^2-18 a b c d+27 b^2 c^2\right ) \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{3 c (b c-a d)}-\frac {x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2+15 a b c d+18 b^2 c^2\right )}{3 c \left (c+d x^3\right ) (b c-a d)}\right )}{a (b c-a d)}}{6 c (b c-a d)}-\frac {d x}{6 c \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 901

\(\displaystyle \frac {\frac {b x (a d+6 b c)}{a \sqrt [3]{a+b x^3} \left (c+d x^3\right ) (b c-a d)}-\frac {d \left (\frac {2 a \left (5 a^2 d^2-18 a b c d+27 b^2 c^2\right ) \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{3 c (b c-a d)}-\frac {x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2+15 a b c d+18 b^2 c^2\right )}{3 c \left (c+d x^3\right ) (b c-a d)}\right )}{a (b c-a d)}}{6 c (b c-a d)}-\frac {d x}{6 c \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)}\)

Input:

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

Output:

-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)/(a*(b*c - a*d)*(a + b*x^3)^(1/3)*(c + d*x^3)) - (d*(-1/3*((18*b^2* 
c^2 + 15*a*b*c*d - 5*a^2*d^2)*x*(a + b*x^3)^(2/3))/(c*(b*c - a*d)*(c + d*x 
^3)) + (2*a*(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]]/(Sqrt[3]*c^(2/3)*(b*c - 
 a*d)^(1/3)) + Log[c + d*x^3]/(6*c^(2/3)*(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^(2/3)*(b*c - a*d)^(1/3)))) 
/(3*c*(b*c - a*d))))/(a*(b*c - a*d)))/(6*c*(b*c - a*d))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 901

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit 
h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S 
qrt[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 931

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] + Simp[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]) && IntBinomialQ[a, b, 
 c, d, n, p, q, x]
 

rule 1024

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] + Simp[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] /; Fr 
eeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(\frac {-\frac {5 a d \left (a^{2} d^{2}-\frac {18}{5} a b c d +\frac {27}{5} b^{2} c^{2}\right ) \left (d \,x^{3}+c \right )^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{54}+\frac {5 \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) a d \left (a^{2} d^{2}-\frac {18}{5} a b c d +\frac {27}{5} b^{2} c^{2}\right ) \left (d \,x^{3}+c \right )^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \sqrt {3}}{27}+\frac {5 a d \left (a^{2} d^{2}-\frac {18}{5} a b c d +\frac {27}{5} b^{2} c^{2}\right ) \left (d \,x^{3}+c \right )^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )}{27}+\frac {4 c \left (\frac {5 a^{2} x^{3} \left (b \,x^{3}+a \right ) d^{4}}{8}+a \left (-\frac {15 b \,x^{3}}{8}+a \right ) c \left (b \,x^{3}+a \right ) d^{3}-\frac {9 b \,c^{2} \left (b^{2} x^{6}+a b \,x^{3}+a^{2}\right ) d^{2}}{4}-\frac {9 b^{3} c^{3} d \,x^{3}}{2}-\frac {9 c^{4} b^{3}}{4}\right ) x \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{9}}{\left (d \,x^{3}+c \right )^{2} c^{3} \left (a d -b c \right )^{3} \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} a}\)	\(416\)

Input:

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

Output:

4/9/((a*d-b*c)/c)^(1/3)/(b*x^3+a)^(1/3)*(-5/24*a*d*(a^2*d^2-18/5*a*b*c*d+2 
7/5*b^2*c^2)*(d*x^3+c)^2*(b*x^3+a)^(1/3)*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)+5/12*arctan(1/3*3^( 
1/2)*(-2/((a*d-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)+x)/x)*a*d*(a^2*d^2-18/5*a*b*c 
*d+27/5*b^2*c^2)*(d*x^3+c)^2*(b*x^3+a)^(1/3)*3^(1/2)+5/12*a*d*(a^2*d^2-18/ 
5*a*b*c*d+27/5*b^2*c^2)*(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)+c*(5/8*a^2*x^3*(b*x^3+a)*d^4+a*(-15/8*b*x^3+a)*c*( 
b*x^3+a)*d^3-9/4*b*c^2*(b^2*x^6+a*b*x^3+a^2)*d^2-9/2*b^3*c^3*d*x^3-9/4*c^4 
*b^3)*x*((a*d-b*c)/c)^(1/3))/(d*x^3+c)^2/c^3/(a*d-b*c)^3/a
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^3} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,c^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,c^{2} d \,x^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a c \,d^{2} x^{6}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,d^{3} x^{9}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,c^{3} x^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,c^{2} d \,x^{6}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b c \,d^{2} x^{9}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,d^{3} x^{12}}d x \] Input:

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

Output:

int(1/((a + b*x**3)**(1/3)*a*c**3 + 3*(a + b*x**3)**(1/3)*a*c**2*d*x**3 + 
3*(a + b*x**3)**(1/3)*a*c*d**2*x**6 + (a + b*x**3)**(1/3)*a*d**3*x**9 + (a 
 + b*x**3)**(1/3)*b*c**3*x**3 + 3*(a + b*x**3)**(1/3)*b*c**2*d*x**6 + 3*(a 
 + b*x**3)**(1/3)*b*c*d**2*x**9 + (a + b*x**3)**(1/3)*b*d**3*x**12),x)