\(\int \frac {\sqrt {x} (A+B x^3)}{(a+b x^3)^3} \, dx\) [151]

Optimal result

Integrand size = 22, antiderivative size = 104 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac {(3 A b+a B) x^{3/2}}{12 a^2 b \left (a+b x^3\right )}+\frac {(3 A b+a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{5/2} b^{3/2}} \] Output:

1/6*(A*b-B*a)*x^(3/2)/a/b/(b*x^3+a)^2+1/12*(3*A*b+B*a)*x^(3/2)/a^2/b/(b*x^ 
3+a)+1/12*(3*A*b+B*a)*arctan(b^(1/2)*x^(3/2)/a^(1/2))/a^(5/2)/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=-\frac {x^{3/2} \left (-5 a A b+a^2 B-3 A b^2 x^3-a b B x^3\right )}{12 a^2 b \left (a+b x^3\right )^2}+\frac {(3 A b+a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{5/2} b^{3/2}} \] Input:

Integrate[(Sqrt[x]*(A + B*x^3))/(a + b*x^3)^3,x]
 

Output:

-1/12*(x^(3/2)*(-5*a*A*b + a^2*B - 3*A*b^2*x^3 - a*b*B*x^3))/(a^2*b*(a + b 
*x^3)^2) + ((3*A*b + a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(12*a^(5/2)*b 
^(3/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {957, 819, 851, 807, 218}

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.

\(\displaystyle \int \frac {\sqrt {x} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {(a B+3 A b) \int \frac {\sqrt {x}}{\left (b x^3+a\right )^2}dx}{4 a b}+\frac {x^{3/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}\)

\(\Big \downarrow \) 819

\(\displaystyle \frac {(a B+3 A b) \left (\frac {\int \frac {\sqrt {x}}{b x^3+a}dx}{2 a}+\frac {x^{3/2}}{3 a \left (a+b x^3\right )}\right )}{4 a b}+\frac {x^{3/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {(a B+3 A b) \left (\frac {\int \frac {x}{b x^3+a}d\sqrt {x}}{a}+\frac {x^{3/2}}{3 a \left (a+b x^3\right )}\right )}{4 a b}+\frac {x^{3/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(a B+3 A b) \left (\frac {\int \frac {1}{a+b x}dx^{3/2}}{3 a}+\frac {x^{3/2}}{3 a \left (a+b x^3\right )}\right )}{4 a b}+\frac {x^{3/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(a B+3 A b) \left (\frac {\arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{3/2} \sqrt {b}}+\frac {x^{3/2}}{3 a \left (a+b x^3\right )}\right )}{4 a b}+\frac {x^{3/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}\)

Input:

Int[(Sqrt[x]*(A + B*x^3))/(a + b*x^3)^3,x]
 

Output:

((A*b - a*B)*x^(3/2))/(6*a*b*(a + b*x^3)^2) + ((3*A*b + a*B)*(x^(3/2)/(3*a 
*(a + b*x^3)) + ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]]/(3*a^(3/2)*Sqrt[b])))/(4 
*a*b)
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 819

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( 
c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 
1) + 1)/(a*n*(p + 1))   Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a 
, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 851

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = 
 Denominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ 
n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && 
FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\frac {\left (3 A b +B a \right ) x^{\frac {9}{2}}}{12 a^{2}}+\frac {\left (5 A b -B a \right ) x^{\frac {3}{2}}}{12 a b}}{\left (b \,x^{3}+a \right )^{2}}+\frac {\left (3 A b +B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 a^{2} b \sqrt {a b}}\)	\(82\)
default	\(\frac {\frac {\left (3 A b +B a \right ) x^{\frac {9}{2}}}{12 a^{2}}+\frac {\left (5 A b -B a \right ) x^{\frac {3}{2}}}{12 a b}}{\left (b \,x^{3}+a \right )^{2}}+\frac {\left (3 A b +B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 a^{2} b \sqrt {a b}}\)	\(82\)

Input:

int(x^(1/2)*(B*x^3+A)/(b*x^3+a)^3,x,method=_RETURNVERBOSE)
 

Output:

2/3*(1/8*(3*A*b+B*a)/a^2*x^(9/2)+1/8*(5*A*b-B*a)/a/b*x^(3/2))/(b*x^3+a)^2+ 
1/12*(3*A*b+B*a)/a^2/b/(a*b)^(1/2)*arctan(b*x^(3/2)/(a*b)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 313, normalized size of antiderivative = 3.01 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\left [-\frac {{\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{6} + B a^{3} + 3 \, A a^{2} b + 2 \, {\left (B a^{2} b + 3 \, A a b^{2}\right )} x^{3}\right )} \sqrt {-a b} \log \left (\frac {b x^{3} - 2 \, \sqrt {-a b} x^{\frac {3}{2}} - a}{b x^{3} + a}\right ) - 2 \, {\left ({\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{4} - {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{3} b^{4} x^{6} + 2 \, a^{4} b^{3} x^{3} + a^{5} b^{2}\right )}}, \frac {{\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{6} + B a^{3} + 3 \, A a^{2} b + 2 \, {\left (B a^{2} b + 3 \, A a b^{2}\right )} x^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right ) + {\left ({\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{4} - {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{12 \, {\left (a^{3} b^{4} x^{6} + 2 \, a^{4} b^{3} x^{3} + a^{5} b^{2}\right )}}\right ] \] Input:

integrate(x^(1/2)*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="fricas")
 

Output:

[-1/24*(((B*a*b^2 + 3*A*b^3)*x^6 + B*a^3 + 3*A*a^2*b + 2*(B*a^2*b + 3*A*a* 
b^2)*x^3)*sqrt(-a*b)*log((b*x^3 - 2*sqrt(-a*b)*x^(3/2) - a)/(b*x^3 + a)) - 
 2*((B*a^2*b^2 + 3*A*a*b^3)*x^4 - (B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(x))/(a^3 
*b^4*x^6 + 2*a^4*b^3*x^3 + a^5*b^2), 1/12*(((B*a*b^2 + 3*A*b^3)*x^6 + B*a^ 
3 + 3*A*a^2*b + 2*(B*a^2*b + 3*A*a*b^2)*x^3)*sqrt(a*b)*arctan(sqrt(a*b)*x^ 
(3/2)/a) + ((B*a^2*b^2 + 3*A*a*b^3)*x^4 - (B*a^3*b - 5*A*a^2*b^2)*x)*sqrt( 
x))/(a^3*b^4*x^6 + 2*a^4*b^3*x^3 + a^5*b^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(1/2)*(B*x**3+A)/(b*x**3+a)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {{\left (B a b + 3 \, A b^{2}\right )} x^{\frac {9}{2}} - {\left (B a^{2} - 5 \, A a b\right )} x^{\frac {3}{2}}}{12 \, {\left (a^{2} b^{3} x^{6} + 2 \, a^{3} b^{2} x^{3} + a^{4} b\right )}} + \frac {{\left (B a + 3 \, A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 \, \sqrt {a b} a^{2} b} \] Input:

integrate(x^(1/2)*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="maxima")
 

Output:

1/12*((B*a*b + 3*A*b^2)*x^(9/2) - (B*a^2 - 5*A*a*b)*x^(3/2))/(a^2*b^3*x^6 
+ 2*a^3*b^2*x^3 + a^4*b) + 1/12*(B*a + 3*A*b)*arctan(b*x^(3/2)/sqrt(a*b))/ 
(sqrt(a*b)*a^2*b)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {{\left (B a + 3 \, A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 \, \sqrt {a b} a^{2} b} + \frac {B a b x^{\frac {9}{2}} + 3 \, A b^{2} x^{\frac {9}{2}} - B a^{2} x^{\frac {3}{2}} + 5 \, A a b x^{\frac {3}{2}}}{12 \, {\left (b x^{3} + a\right )}^{2} a^{2} b} \] Input:

integrate(x^(1/2)*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="giac")
 

Output:

1/12*(B*a + 3*A*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*a^2*b) + 1/12*(B 
*a*b*x^(9/2) + 3*A*b^2*x^(9/2) - B*a^2*x^(3/2) + 5*A*a*b*x^(3/2))/((b*x^3 
+ a)^2*a^2*b)
 

Mupad [B] (verification not implemented)

Time = 0.86 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {\frac {x^{9/2}\,\left (3\,A\,b+B\,a\right )}{12\,a^2}+\frac {x^{3/2}\,\left (5\,A\,b-B\,a\right )}{12\,a\,b}}{a^2+2\,a\,b\,x^3+b^2\,x^6}+\frac {\mathrm {atan}\left (\frac {b^{3/2}\,x^{3/2}\,\left (9\,A^2\,b^3+6\,A\,B\,a\,b^2+B^2\,a^2\,b\right )}{\sqrt {a}\,\left (3\,A\,b+B\,a\right )\,\left (3\,A\,b^3+B\,a\,b^2\right )}\right )\,\left (3\,A\,b+B\,a\right )}{12\,a^{5/2}\,b^{3/2}} \] Input:

int((x^(1/2)*(A + B*x^3))/(a + b*x^3)^3,x)
 

Output:

((x^(9/2)*(3*A*b + B*a))/(12*a^2) + (x^(3/2)*(5*A*b - B*a))/(12*a*b))/(a^2 
 + b^2*x^6 + 2*a*b*x^3) + (atan((b^(3/2)*x^(3/2)*(9*A^2*b^3 + B^2*a^2*b + 
6*A*B*a*b^2))/(a^(1/2)*(3*A*b + B*a)*(3*A*b^3 + B*a*b^2)))*(3*A*b + B*a))/ 
(12*a^(5/2)*b^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.95 \[ \int \frac {\sqrt {x} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {-b^{\frac {1}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}-2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right )-b^{\frac {7}{6}} a^{\frac {1}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}-2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right ) x^{3}+b^{\frac {1}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right )+b^{\frac {7}{6}} a^{\frac {1}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right ) x^{3}-b^{\frac {1}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {\sqrt {x}\, b^{\frac {1}{6}}}{a^{\frac {1}{6}}}\right )-b^{\frac {7}{6}} a^{\frac {1}{6}} \mathit {atan} \left (\frac {\sqrt {x}\, b^{\frac {1}{6}}}{a^{\frac {1}{6}}}\right ) x^{3}+\sqrt {x}\, b^{\frac {2}{3}} a^{\frac {2}{3}} x}{3 b^{\frac {2}{3}} a^{\frac {5}{3}} \left (b \,x^{3}+a \right )} \] Input:

int(x^(1/2)*(B*x^3+A)/(b*x^3+a)^3,x)
 

Output:

( - b**(1/6)*a**(1/6)*atan((b**(1/6)*a**(1/6)*sqrt(3) - 2*sqrt(x)*b**(1/3) 
)/(b**(1/6)*a**(1/6)))*a - b**(1/6)*a**(1/6)*atan((b**(1/6)*a**(1/6)*sqrt( 
3) - 2*sqrt(x)*b**(1/3))/(b**(1/6)*a**(1/6)))*b*x**3 + b**(1/6)*a**(1/6)*a 
tan((b**(1/6)*a**(1/6)*sqrt(3) + 2*sqrt(x)*b**(1/3))/(b**(1/6)*a**(1/6)))* 
a + b**(1/6)*a**(1/6)*atan((b**(1/6)*a**(1/6)*sqrt(3) + 2*sqrt(x)*b**(1/3) 
)/(b**(1/6)*a**(1/6)))*b*x**3 - b**(1/6)*a**(1/6)*atan((sqrt(x)*b**(1/3))/ 
(b**(1/6)*a**(1/6)))*a - b**(1/6)*a**(1/6)*atan((sqrt(x)*b**(1/3))/(b**(1/ 
6)*a**(1/6)))*b*x**3 + sqrt(x)*b**(2/3)*a**(2/3)*x)/(3*b**(2/3)*a**(2/3)*a 
*(a + b*x**3))