\(\int \frac {x^{7/2} (A+B x^3)}{(a+b x^3)^3} \, dx\) [150]

Optimal result

Integrand size = 22, antiderivative size = 101 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=-\frac {(A b-a B) x^{3/2}}{6 b^2 \left (a+b x^3\right )^2}+\frac {(A b-5 a B) x^{3/2}}{12 a b^2 \left (a+b x^3\right )}+\frac {(A b+3 a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{3/2} b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.91 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=-\frac {x^{3/2} \left (a A b+3 a^2 B-A b^2 x^3+5 a b B x^3\right )}{12 a b^2 \left (a+b x^3\right )^2}+\frac {(A b+3 a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{3/2} b^{5/2}} \] Input:

Integrate[(x^(7/2)*(A + B*x^3))/(a + b*x^3)^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {957, 817, 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.

\(\displaystyle \int \frac {x^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {(3 a B+A b) \int \frac {x^{7/2}}{\left (b x^3+a\right )^2}dx}{4 a b}+\frac {x^{9/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {(3 a B+A b) \left (\frac {\int \frac {\sqrt {x}}{b x^3+a}dx}{2 b}-\frac {x^{3/2}}{3 b \left (a+b x^3\right )}\right )}{4 a b}+\frac {x^{9/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {(3 a B+A b) \left (\frac {\int \frac {x}{b x^3+a}d\sqrt {x}}{b}-\frac {x^{3/2}}{3 b \left (a+b x^3\right )}\right )}{4 a b}+\frac {x^{9/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(3 a B+A b) \left (\frac {\int \frac {1}{a+b x}dx^{3/2}}{3 b}-\frac {x^{3/2}}{3 b \left (a+b x^3\right )}\right )}{4 a b}+\frac {x^{9/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(3 a B+A b) \left (\frac {\arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 \sqrt {a} b^{3/2}}-\frac {x^{3/2}}{3 b \left (a+b x^3\right )}\right )}{4 a b}+\frac {x^{9/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}\)

Input:

Int[(x^(7/2)*(A + B*x^3))/(a + b*x^3)^3,x]
 

Output:

((A*b - a*B)*x^(9/2))/(6*a*b*(a + b*x^3)^2) + ((A*b + 3*a*B)*(-1/3*x^(3/2) 
/(b*(a + b*x^3)) + ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]]/(3*Sqrt[a]*b^(3/2)))) 
/(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 817

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( 
n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 0] && 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 = 0.94 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {\frac {\left (A b -5 B a \right ) x^{\frac {9}{2}}}{12 a b}-\frac {\left (A b +3 B a \right ) x^{\frac {3}{2}}}{12 b^{2}}}{\left (b \,x^{3}+a \right )^{2}}+\frac {\left (A b +3 B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 b^{2} a \sqrt {a b}}\)	\(81\)
default	\(\frac {\frac {\left (A b -5 B a \right ) x^{\frac {9}{2}}}{12 a b}-\frac {\left (A b +3 B a \right ) x^{\frac {3}{2}}}{12 b^{2}}}{\left (b \,x^{3}+a \right )^{2}}+\frac {\left (A b +3 B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 b^{2} a \sqrt {a b}}\)	\(81\)

Input:

int(x^(7/2)*(B*x^3+A)/(b*x^3+a)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 314, normalized size of antiderivative = 3.11 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\left [-\frac {{\left ({\left (3 \, B a b^{2} + A b^{3}\right )} x^{6} + 3 \, B a^{3} + A a^{2} b + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x^{3}\right )} \sqrt {-a b} \log \left (\frac {b x^{3} - 2 \, \sqrt {-a b} x^{\frac {3}{2}} - a}{b x^{3} + a}\right ) + 2 \, {\left ({\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{4} + {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{2} b^{5} x^{6} + 2 \, a^{3} b^{4} x^{3} + a^{4} b^{3}\right )}}, \frac {{\left ({\left (3 \, B a b^{2} + A b^{3}\right )} x^{6} + 3 \, B a^{3} + A a^{2} b + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right ) - {\left ({\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{4} + {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{12 \, {\left (a^{2} b^{5} x^{6} + 2 \, a^{3} b^{4} x^{3} + a^{4} b^{3}\right )}}\right ] \] Input:

integrate(x^(7/2)*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(7/2)*(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 = 0.95 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=-\frac {{\left (5 \, B a b - A b^{2}\right )} x^{\frac {9}{2}} + {\left (3 \, B a^{2} + A a b\right )} x^{\frac {3}{2}}}{12 \, {\left (a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}} + \frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 \, \sqrt {a b} a b^{2}} \] Input:

integrate(x^(7/2)*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.83 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 \, \sqrt {a b} a b^{2}} - \frac {5 \, B a b x^{\frac {9}{2}} - A b^{2} x^{\frac {9}{2}} + 3 \, B a^{2} x^{\frac {3}{2}} + A a b x^{\frac {3}{2}}}{12 \, {\left (b x^{3} + a\right )}^{2} a b^{2}} \] Input:

integrate(x^(7/2)*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.86 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {9\,b^{3/2}\,x^{3/2}\,\left (A^2\,b^2+6\,A\,B\,a\,b+9\,B^2\,a^2\right )}{\sqrt {a}\,\left (9\,A\,b^2+27\,B\,a\,b\right )\,\left (A\,b+3\,B\,a\right )}\right )\,\left (A\,b+3\,B\,a\right )}{12\,a^{3/2}\,b^{5/2}}-\frac {\frac {x^{3/2}\,\left (A\,b+3\,B\,a\right )}{12\,b^2}-\frac {x^{9/2}\,\left (A\,b-5\,B\,a\right )}{12\,a\,b}}{a^2+2\,a\,b\,x^3+b^2\,x^6} \] Input:

int((x^(7/2)*(A + B*x^3))/(a + b*x^3)^3,x)
 

Output:

(atan((9*b^(3/2)*x^(3/2)*(A^2*b^2 + 9*B^2*a^2 + 6*A*B*a*b))/(a^(1/2)*(9*A* 
b^2 + 27*B*a*b)*(A*b + 3*B*a)))*(A*b + 3*B*a))/(12*a^(3/2)*b^(5/2)) - ((x^ 
(3/2)*(A*b + 3*B*a))/(12*b^2) - (x^(9/2)*(A*b - 5*B*a))/(12*a*b))/(a^2 + b 
^2*x^6 + 2*a*b*x^3)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.02 \[ \int \frac {x^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {-b^{\frac {1}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}-2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right )-b^{\frac {7}{6}} a^{\frac {1}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}-2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right ) x^{3}+b^{\frac {1}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right )+b^{\frac {7}{6}} a^{\frac {1}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right ) x^{3}-b^{\frac {1}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {\sqrt {x}\, b^{\frac {1}{6}}}{a^{\frac {1}{6}}}\right )-b^{\frac {7}{6}} a^{\frac {1}{6}} \mathit {atan} \left (\frac {\sqrt {x}\, b^{\frac {1}{6}}}{a^{\frac {1}{6}}}\right ) x^{3}-\sqrt {x}\, b^{\frac {2}{3}} a^{\frac {2}{3}} x}{3 b^{\frac {5}{3}} a^{\frac {2}{3}} \left (b \,x^{3}+a \right )} \] Input:

int(x^(7/2)*(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)*b 
*(a + b*x**3))