\(\int \frac {A+B x^3}{x^{5/2} (a+b x^3)^3} \, dx\) [152]

Optimal result

Integrand size = 22, antiderivative size = 113 \[ \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^3} \, dx=-\frac {2 A}{3 a^3 x^{3/2}}-\frac {(A b-a B) x^{3/2}}{6 a^2 \left (a+b x^3\right )^2}-\frac {(7 A b-3 a B) x^{3/2}}{12 a^3 \left (a+b x^3\right )}-\frac {(5 A b-a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b}} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^3} \, dx=\frac {-8 a^2 A-25 a A b x^3+5 a^2 B x^3-15 A b^2 x^6+3 a b B x^6}{12 a^3 x^{3/2} \left (a+b x^3\right )^2}+\frac {(-5 A b+a B) \arctan \left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b}} \] Input:

Integrate[(A + B*x^3)/(x^(5/2)*(a + b*x^3)^3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 125, 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, 819, 847, 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.

\(\displaystyle \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^3} \, dx\)

\(\Big \downarrow \) 957

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

\(\Big \downarrow \) 819

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 851

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 218

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

Input:

Int[(A + B*x^3)/(x^(5/2)*(a + b*x^3)^3),x]
 

Output:

(A*b - a*B)/(6*a*b*x^(3/2)*(a + b*x^3)^2) + ((5*A*b - a*B)*(1/(3*a*x^(3/2) 
*(a + b*x^3)) + (3*(-2/(3*a*x^(3/2)) - (2*Sqrt[b]*ArcTan[(Sqrt[b]*x^(3/2)) 
/Sqrt[a]])/(3*a^(3/2))))/(2*a)))/(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 847

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -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 = 0.78 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(-\frac {2 A}{3 a^{3} x^{\frac {3}{2}}}-\frac {2 \left (\frac {\left (\frac {7}{8} b^{2} A -\frac {3}{8} a b B \right ) x^{\frac {9}{2}}+\frac {a \left (9 A b -5 B a \right ) x^{\frac {3}{2}}}{8}}{\left (b \,x^{3}+a \right )^{2}}+\frac {3 \left (5 A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{3 a^{3}}\)	\(86\)
default	\(-\frac {2 A}{3 a^{3} x^{\frac {3}{2}}}-\frac {2 \left (\frac {\left (\frac {7}{8} b^{2} A -\frac {3}{8} a b B \right ) x^{\frac {9}{2}}+\frac {a \left (9 A b -5 B a \right ) x^{\frac {3}{2}}}{8}}{\left (b \,x^{3}+a \right )^{2}}+\frac {3 \left (5 A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{3 a^{3}}\)	\(86\)
risch	\(-\frac {2 A}{3 a^{3} x^{\frac {3}{2}}}-\frac {\frac {\frac {2 \left (\frac {7}{8} b^{2} A -\frac {3}{8} a b B \right ) x^{\frac {9}{2}}}{3}+\frac {a \left (9 A b -5 B a \right ) x^{\frac {3}{2}}}{12}}{\left (b \,x^{3}+a \right )^{2}}+\frac {\left (5 A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}}{a^{3}}\)	\(87\)

Input:

int((B*x^3+A)/x^(5/2)/(b*x^3+a)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.07 \[ \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^3} \, dx=\left [\frac {3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{8} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{5} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{3} + 2 \, \sqrt {-a b} x^{\frac {3}{2}} - a}{b x^{3} + a}\right ) + 2 \, {\left (3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{6} - 8 \, A a^{3} b + 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {x}}{24 \, {\left (a^{4} b^{3} x^{8} + 2 \, a^{5} b^{2} x^{5} + a^{6} b x^{2}\right )}}, \frac {3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{8} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{5} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right ) + {\left (3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{6} - 8 \, A a^{3} b + 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {x}}{12 \, {\left (a^{4} b^{3} x^{8} + 2 \, a^{5} b^{2} x^{5} + a^{6} b x^{2}\right )}}\right ] \] Input:

integrate((B*x^3+A)/x^(5/2)/(b*x^3+a)^3,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^3} \, dx=\text {Timed out} \] Input:

integrate((B*x**3+A)/x**(5/2)/(b*x**3+a)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^3} \, dx=\frac {3 \, {\left (B a b - 5 \, A b^{2}\right )} x^{6} + 5 \, {\left (B a^{2} - 5 \, A a b\right )} x^{3} - 8 \, A a^{2}}{12 \, {\left (a^{3} b^{2} x^{\frac {15}{2}} + 2 \, a^{4} b x^{\frac {9}{2}} + a^{5} x^{\frac {3}{2}}\right )}} + \frac {{\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} \] Input:

integrate((B*x^3+A)/x^(5/2)/(b*x^3+a)^3,x, algorithm="maxima")
 

Output:

1/12*(3*(B*a*b - 5*A*b^2)*x^6 + 5*(B*a^2 - 5*A*a*b)*x^3 - 8*A*a^2)/(a^3*b^ 
2*x^(15/2) + 2*a^4*b*x^(9/2) + a^5*x^(3/2)) + 1/4*(B*a - 5*A*b)*arctan(b*x 
^(3/2)/sqrt(a*b))/(sqrt(a*b)*a^3)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^3} \, dx=\frac {{\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} - \frac {2 \, A}{3 \, a^{3} x^{\frac {3}{2}}} + \frac {3 \, B a b x^{\frac {9}{2}} - 7 \, A b^{2} x^{\frac {9}{2}} + 5 \, B a^{2} x^{\frac {3}{2}} - 9 \, A a b x^{\frac {3}{2}}}{12 \, {\left (b x^{3} + a\right )}^{2} a^{3}} \] Input:

integrate((B*x^3+A)/x^(5/2)/(b*x^3+a)^3,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.89 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.44 \[ \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^3} \, dx=-\frac {\frac {2\,A}{3\,a}+\frac {5\,x^3\,\left (5\,A\,b-B\,a\right )}{12\,a^2}+\frac {b\,x^6\,\left (5\,A\,b-B\,a\right )}{4\,a^3}}{a^2\,x^{3/2}+b^2\,x^{15/2}+2\,a\,b\,x^{9/2}}-\frac {\mathrm {atan}\left (\frac {8\,a^{7/2}\,\sqrt {b}\,x^{3/2}\,\left (86400\,A^2\,a^9\,b^5-34560\,A\,B\,a^{10}\,b^4+3456\,B^2\,a^{11}\,b^3\right )}{\left (5\,A\,b-B\,a\right )\,\left (138240\,A\,a^{13}\,b^4-27648\,B\,a^{14}\,b^3\right )}\right )\,\left (5\,A\,b-B\,a\right )}{4\,a^{7/2}\,\sqrt {b}} \] Input:

int((A + B*x^3)/(x^(5/2)*(a + b*x^3)^3),x)
 

Output:

- ((2*A)/(3*a) + (5*x^3*(5*A*b - B*a))/(12*a^2) + (b*x^6*(5*A*b - B*a))/(4 
*a^3))/(a^2*x^(3/2) + b^2*x^(15/2) + 2*a*b*x^(9/2)) - (atan((8*a^(7/2)*b^( 
1/2)*x^(3/2)*(86400*A^2*a^9*b^5 + 3456*B^2*a^11*b^3 - 34560*A*B*a^10*b^4)) 
/((5*A*b - B*a)*(138240*A*a^13*b^4 - 27648*B*a^14*b^3)))*(5*A*b - B*a))/(4 
*a^(7/2)*b^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.09 \[ \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^3} \, dx=\frac {3 \sqrt {x}\, b^{\frac {7}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}-2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right ) x +3 \sqrt {x}\, b^{\frac {13}{6}} a^{\frac {1}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}-2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right ) x^{4}-3 \sqrt {x}\, b^{\frac {7}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right ) x -3 \sqrt {x}\, b^{\frac {13}{6}} a^{\frac {1}{6}} \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+2 \sqrt {x}\, b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right ) x^{4}+3 \sqrt {x}\, b^{\frac {7}{6}} a^{\frac {7}{6}} \mathit {atan} \left (\frac {\sqrt {x}\, b^{\frac {1}{6}}}{a^{\frac {1}{6}}}\right ) x +3 \sqrt {x}\, b^{\frac {13}{6}} a^{\frac {1}{6}} \mathit {atan} \left (\frac {\sqrt {x}\, b^{\frac {1}{6}}}{a^{\frac {1}{6}}}\right ) x^{4}-2 b^{\frac {2}{3}} a^{\frac {5}{3}}-3 b^{\frac {5}{3}} a^{\frac {2}{3}} x^{3}}{3 \sqrt {x}\, b^{\frac {2}{3}} a^{\frac {8}{3}} x \left (b \,x^{3}+a \right )} \] Input:

int((B*x^3+A)/x^(5/2)/(b*x^3+a)^3,x)
 

Output:

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