\(\int \frac {x^{13/2} (A+B x^3)}{(a+b x^3)^3} \, dx\) [149]

Optimal result

Integrand size = 22, antiderivative size = 112 \[ \int \frac {x^{13/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {2 B x^{3/2}}{3 b^3}-\frac {(A b-a B) x^{9/2}}{6 b^2 \left (a+b x^3\right )^2}-\frac {(3 A b-7 a B) x^{3/2}}{12 b^3 \left (a+b x^3\right )}+\frac {(A b-5 a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2}} \] Output:

2/3*B*x^(3/2)/b^3-1/6*(A*b-B*a)*x^(9/2)/b^2/(b*x^3+a)^2-1/12*(3*A*b-7*B*a) 
*x^(3/2)/b^3/(b*x^3+a)+1/4*(A*b-5*B*a)*arctan(b^(1/2)*x^(3/2)/a^(1/2))/a^( 
1/2)/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {x^{13/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {x^{3/2} \left (-3 a A b+15 a^2 B-5 A b^2 x^3+25 a b B x^3+8 b^2 B x^6\right )}{12 b^3 \left (a+b x^3\right )^2}+\frac {(A b-5 a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2}} \] Input:

Integrate[(x^(13/2)*(A + B*x^3))/(a + b*x^3)^3,x]
 

Output:

(x^(3/2)*(-3*a*A*b + 15*a^2*B - 5*A*b^2*x^3 + 25*a*b*B*x^3 + 8*b^2*B*x^6)) 
/(12*b^3*(a + b*x^3)^2) + ((A*b - 5*a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]] 
)/(4*Sqrt[a]*b^(7/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {957, 817, 843, 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.

\(\displaystyle \int \frac {x^{13/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {x^{15/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-5 a B) \int \frac {x^{13/2}}{\left (b x^3+a\right )^2}dx}{4 a b}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {x^{15/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-5 a B) \left (\frac {3 \int \frac {x^{7/2}}{b x^3+a}dx}{2 b}-\frac {x^{9/2}}{3 b \left (a+b x^3\right )}\right )}{4 a b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^{15/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-5 a B) \left (\frac {3 \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{b x^3+a}dx}{b}\right )}{2 b}-\frac {x^{9/2}}{3 b \left (a+b x^3\right )}\right )}{4 a b}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {x^{15/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-5 a B) \left (\frac {3 \left (\frac {2 x^{3/2}}{3 b}-\frac {2 a \int \frac {x}{b x^3+a}d\sqrt {x}}{b}\right )}{2 b}-\frac {x^{9/2}}{3 b \left (a+b x^3\right )}\right )}{4 a b}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {x^{15/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-5 a B) \left (\frac {3 \left (\frac {2 x^{3/2}}{3 b}-\frac {2 a \int \frac {1}{a+b x}dx^{3/2}}{3 b}\right )}{2 b}-\frac {x^{9/2}}{3 b \left (a+b x^3\right )}\right )}{4 a b}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {x^{15/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-5 a B) \left (\frac {3 \left (\frac {2 x^{3/2}}{3 b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 b^{3/2}}\right )}{2 b}-\frac {x^{9/2}}{3 b \left (a+b x^3\right )}\right )}{4 a b}\)

Input:

Int[(x^(13/2)*(A + B*x^3))/(a + b*x^3)^3,x]
 

Output:

((A*b - a*B)*x^(15/2))/(6*a*b*(a + b*x^3)^2) - ((A*b - 5*a*B)*(-1/3*x^(9/2 
)/(b*(a + b*x^3)) + (3*((2*x^(3/2))/(3*b) - (2*Sqrt[a]*ArcTan[(Sqrt[b]*x^( 
3/2))/Sqrt[a]])/(3*b^(3/2))))/(2*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 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 843

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*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 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 = 0.99 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {2 B \,x^{\frac {3}{2}}}{3 b^{3}}+\frac {\frac {2 \left (\left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{\frac {9}{2}}-\frac {a \left (3 A b -7 B a \right ) x^{\frac {3}{2}}}{8}\right )}{3 \left (b \,x^{3}+a \right )^{2}}+\frac {\left (A b -5 B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}}{b^{3}}\)	\(85\)
default	\(\frac {2 B \,x^{\frac {3}{2}}}{3 b^{3}}+\frac {\frac {2 \left (\left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{\frac {9}{2}}-\frac {a \left (3 A b -7 B a \right ) x^{\frac {3}{2}}}{8}\right )}{3 \left (b \,x^{3}+a \right )^{2}}+\frac {\left (A b -5 B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}}{b^{3}}\)	\(85\)
risch	\(\frac {2 B \,x^{\frac {3}{2}}}{3 b^{3}}+\frac {\frac {\frac {2 \left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{\frac {9}{2}}}{3}-\frac {a \left (3 A b -7 B a \right ) x^{\frac {3}{2}}}{12}}{\left (b \,x^{3}+a \right )^{2}}+\frac {\left (A b -5 B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}}{b^{3}}\)	\(85\)

Input:

int(x^(13/2)*(B*x^3+A)/(b*x^3+a)^3,x,method=_RETURNVERBOSE)
 

Output:

2/3*B*x^(3/2)/b^3+2/3/b^3*(((-5/8*b^2*A+9/8*a*b*B)*x^(9/2)-1/8*a*(3*A*b-7* 
B*a)*x^(3/2))/(b*x^3+a)^2+3/8*(A*b-5*B*a)/(a*b)^(1/2)*arctan(b*x^(3/2)/(a* 
b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 344, normalized size of antiderivative = 3.07 \[ \int \frac {x^{13/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\left [\frac {3 \, {\left ({\left (5 \, B a b^{2} - A b^{3}\right )} x^{6} + 5 \, B a^{3} - A a^{2} b + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x^{3}\right )} \sqrt {-a b} \log \left (\frac {b x^{3} - 2 \, \sqrt {-a b} x^{\frac {3}{2}} - a}{b x^{3} + a}\right ) + 2 \, {\left (8 \, B a b^{3} x^{7} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{4} + 3 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (a b^{6} x^{6} + 2 \, a^{2} b^{5} x^{3} + a^{3} b^{4}\right )}}, -\frac {3 \, {\left ({\left (5 \, B a b^{2} - A b^{3}\right )} x^{6} + 5 \, B a^{3} - A a^{2} b + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right ) - {\left (8 \, B a b^{3} x^{7} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{4} + 3 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{12 \, {\left (a b^{6} x^{6} + 2 \, a^{2} b^{5} x^{3} + a^{3} b^{4}\right )}}\right ] \] Input:

integrate(x^(13/2)*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="fricas")
 

Output:

[1/24*(3*((5*B*a*b^2 - A*b^3)*x^6 + 5*B*a^3 - A*a^2*b + 2*(5*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*(8*B*a*b^3*x^7 + 5*(5*B*a^2*b^2 - A*a*b^3)*x^4 + 3*(5*B*a^3*b - A*a^2* 
b^2)*x)*sqrt(x))/(a*b^6*x^6 + 2*a^2*b^5*x^3 + a^3*b^4), -1/12*(3*((5*B*a*b 
^2 - A*b^3)*x^6 + 5*B*a^3 - A*a^2*b + 2*(5*B*a^2*b - A*a*b^2)*x^3)*sqrt(a* 
b)*arctan(sqrt(a*b)*x^(3/2)/a) - (8*B*a*b^3*x^7 + 5*(5*B*a^2*b^2 - A*a*b^3 
)*x^4 + 3*(5*B*a^3*b - A*a^2*b^2)*x)*sqrt(x))/(a*b^6*x^6 + 2*a^2*b^5*x^3 + 
 a^3*b^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{13/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(13/2)*(B*x**3+A)/(b*x**3+a)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.90 \[ \int \frac {x^{13/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {{\left (9 \, B a b - 5 \, A b^{2}\right )} x^{\frac {9}{2}} + {\left (7 \, B a^{2} - 3 \, A a b\right )} x^{\frac {3}{2}}}{12 \, {\left (b^{5} x^{6} + 2 \, a b^{4} x^{3} + a^{2} b^{3}\right )}} + \frac {2 \, B x^{\frac {3}{2}}}{3 \, b^{3}} - \frac {{\left (5 \, B a - A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} \] Input:

integrate(x^(13/2)*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="maxima")
 

Output:

1/12*((9*B*a*b - 5*A*b^2)*x^(9/2) + (7*B*a^2 - 3*A*a*b)*x^(3/2))/(b^5*x^6 
+ 2*a*b^4*x^3 + a^2*b^3) + 2/3*B*x^(3/2)/b^3 - 1/4*(5*B*a - A*b)*arctan(b* 
x^(3/2)/sqrt(a*b))/(sqrt(a*b)*b^3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79 \[ \int \frac {x^{13/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {2 \, B x^{\frac {3}{2}}}{3 \, b^{3}} - \frac {{\left (5 \, B a - A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {9 \, B a b x^{\frac {9}{2}} - 5 \, A b^{2} x^{\frac {9}{2}} + 7 \, B a^{2} x^{\frac {3}{2}} - 3 \, A a b x^{\frac {3}{2}}}{12 \, {\left (b x^{3} + a\right )}^{2} b^{3}} \] Input:

integrate(x^(13/2)*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="giac")
 

Output:

2/3*B*x^(3/2)/b^3 - 1/4*(5*B*a - A*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a* 
b)*b^3) + 1/12*(9*B*a*b*x^(9/2) - 5*A*b^2*x^(9/2) + 7*B*a^2*x^(3/2) - 3*A* 
a*b*x^(3/2))/((b*x^3 + a)^2*b^3)
 

Mupad [B] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.29 \[ \int \frac {x^{13/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {x^{3/2}\,\left (\frac {7\,B\,a^2}{12}-\frac {A\,a\,b}{4}\right )-x^{9/2}\,\left (\frac {5\,A\,b^2}{12}-\frac {3\,B\,a\,b}{4}\right )}{a^2\,b^3+2\,a\,b^4\,x^3+b^5\,x^6}+\frac {2\,B\,x^{3/2}}{3\,b^3}-\frac {\mathrm {atan}\left (\frac {27\,\sqrt {a}\,\sqrt {b}\,x^{3/2}\,\left (A^2\,b^2-10\,A\,B\,a\,b+25\,B^2\,a^2\right )}{\left (135\,B\,a^2-27\,A\,a\,b\right )\,\left (A\,b-5\,B\,a\right )}\right )\,\left (A\,b-5\,B\,a\right )}{4\,\sqrt {a}\,b^{7/2}} \] Input:

int((x^(13/2)*(A + B*x^3))/(a + b*x^3)^3,x)
 

Output:

(x^(3/2)*((7*B*a^2)/12 - (A*a*b)/4) - x^(9/2)*((5*A*b^2)/12 - (3*B*a*b)/4) 
)/(a^2*b^3 + b^5*x^6 + 2*a*b^4*x^3) + (2*B*x^(3/2))/(3*b^3) - (atan((27*a^ 
(1/2)*b^(1/2)*x^(3/2)*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/((135*B*a^2 - 2 
7*A*a*b)*(A*b - 5*B*a)))*(A*b - 5*B*a))/(4*a^(1/2)*b^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.96 \[ \int \frac {x^{13/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {3 b^{\frac {1}{6}} a^{\frac {13}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}-2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right )+3 b^{\frac {7}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}-2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right ) x^{3}-3 b^{\frac {1}{6}} a^{\frac {13}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right )-3 b^{\frac {7}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right ) x^{3}+3 b^{\frac {1}{6}} a^{\frac {13}{6}} \mathit {atan} \left (\frac {\sqrt {x}\, b^{\frac {1}{6}}}{a^{\frac {1}{6}}}\right )+3 b^{\frac {7}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {\sqrt {x}\, b^{\frac {1}{6}}}{a^{\frac {1}{6}}}\right ) x^{3}+3 \sqrt {x}\, b^{\frac {2}{3}} a^{\frac {5}{3}} x +2 \sqrt {x}\, b^{\frac {5}{3}} a^{\frac {2}{3}} x^{4}}{3 b^{\frac {8}{3}} a^{\frac {2}{3}} \left (b \,x^{3}+a \right )} \] Input:

int(x^(13/2)*(B*x^3+A)/(b*x^3+a)^3,x)
 

Output:

(3*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**2 + 3*b**(1/6)*a**(1/6)*atan((b**(1/6)*a**(1/6)*s 
qrt(3) - 2*sqrt(x)*b**(1/3))/(b**(1/6)*a**(1/6)))*a*b*x**3 - 3*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**2 - 3*b**(1/6)*a**(1/6)*atan((b**(1/6)*a**(1/6)*sqrt(3) + 2*sqr 
t(x)*b**(1/3))/(b**(1/6)*a**(1/6)))*a*b*x**3 + 3*b**(1/6)*a**(1/6)*atan((s 
qrt(x)*b**(1/3))/(b**(1/6)*a**(1/6)))*a**2 + 3*b**(1/6)*a**(1/6)*atan((sqr 
t(x)*b**(1/3))/(b**(1/6)*a**(1/6)))*a*b*x**3 + 3*sqrt(x)*b**(2/3)*a**(2/3) 
*a*x + 2*sqrt(x)*b**(2/3)*a**(2/3)*b*x**4)/(3*b**(2/3)*a**(2/3)*b**2*(a + 
b*x**3))