\(\int \frac {x^{7/2} (A+B x)}{b x+c x^2} \, dx\) [81]

Optimal result

Integrand size = 22, antiderivative size = 113 \[ \int \frac {x^{7/2} (A+B x)}{b x+c x^2} \, dx=-\frac {2 b^2 (b B-A c) \sqrt {x}}{c^4}+\frac {2 b (b B-A c) x^{3/2}}{3 c^3}-\frac {2 (b B-A c) x^{5/2}}{5 c^2}+\frac {2 B x^{7/2}}{7 c}+\frac {2 b^{5/2} (b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{9/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int \frac {x^{7/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \sqrt {x} \left (-105 b^3 B+35 b^2 c (3 A+B x)-7 b c^2 x (5 A+3 B x)+3 c^3 x^2 (7 A+5 B x)\right )}{105 c^4}+\frac {2 b^{5/2} (b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{9/2}} \] Input:

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

Output:

(2*Sqrt[x]*(-105*b^3*B + 35*b^2*c*(3*A + B*x) - 7*b*c^2*x*(5*A + 3*B*x) + 
3*c^3*x^2*(7*A + 5*B*x)))/(105*c^4) + (2*b^(5/2)*(b*B - A*c)*ArcTan[(Sqrt[ 
c]*Sqrt[x])/Sqrt[b]])/c^(9/2)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {9, 90, 60, 60, 60, 73, 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.

\(\displaystyle \int \frac {x^{7/2} (A+B x)}{b x+c x^2} \, dx\)

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {x^{5/2} (A+B x)}{b+c x}dx\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {2 B x^{7/2}}{7 c}-\frac {(b B-A c) \int \frac {x^{5/2}}{b+c x}dx}{c}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 B x^{7/2}}{7 c}-\frac {(b B-A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \int \frac {x^{3/2}}{b+c x}dx}{c}\right )}{c}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 B x^{7/2}}{7 c}-\frac {(b B-A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {b \int \frac {\sqrt {x}}{b+c x}dx}{c}\right )}{c}\right )}{c}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 B x^{7/2}}{7 c}-\frac {(b B-A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {b \int \frac {1}{\sqrt {x} (b+c x)}dx}{c}\right )}{c}\right )}{c}\right )}{c}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 B x^{7/2}}{7 c}-\frac {(b B-A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \int \frac {1}{b+c x}d\sqrt {x}}{c}\right )}{c}\right )}{c}\right )}{c}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 B x^{7/2}}{7 c}-\frac {(b B-A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{3/2}}\right )}{c}\right )}{c}\right )}{c}\)

Input:

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

Output:

(2*B*x^(7/2))/(7*c) - ((b*B - A*c)*((2*x^(5/2))/(5*c) - (b*((2*x^(3/2))/(3 
*c) - (b*((2*Sqrt[x])/c - (2*Sqrt[b]*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^ 
(3/2)))/c))/c))/c
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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]
 

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {2 \left (15 B \,c^{3} x^{3}+21 A \,c^{3} x^{2}-21 B b \,c^{2} x^{2}-35 A b \,c^{2} x +35 B \,b^{2} c x +105 A \,b^{2} c -105 B \,b^{3}\right ) \sqrt {x}}{105 c^{4}}-\frac {2 b^{3} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{4} \sqrt {b c}}\)	\(100\)
derivativedivides	\(\frac {\frac {2 B \,c^{3} x^{\frac {7}{2}}}{7}+\frac {2 A \,c^{3} x^{\frac {5}{2}}}{5}-\frac {2 B b \,c^{2} x^{\frac {5}{2}}}{5}-\frac {2 A b \,c^{2} x^{\frac {3}{2}}}{3}+\frac {2 B \,b^{2} c \,x^{\frac {3}{2}}}{3}+2 A \,b^{2} c \sqrt {x}-2 B \,b^{3} \sqrt {x}}{c^{4}}-\frac {2 b^{3} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{4} \sqrt {b c}}\)	\(106\)
default	\(\frac {\frac {2 B \,c^{3} x^{\frac {7}{2}}}{7}+\frac {2 A \,c^{3} x^{\frac {5}{2}}}{5}-\frac {2 B b \,c^{2} x^{\frac {5}{2}}}{5}-\frac {2 A b \,c^{2} x^{\frac {3}{2}}}{3}+\frac {2 B \,b^{2} c \,x^{\frac {3}{2}}}{3}+2 A \,b^{2} c \sqrt {x}-2 B \,b^{3} \sqrt {x}}{c^{4}}-\frac {2 b^{3} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{4} \sqrt {b c}}\)	\(106\)

Input:

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

Output:

2/105*(15*B*c^3*x^3+21*A*c^3*x^2-21*B*b*c^2*x^2-35*A*b*c^2*x+35*B*b^2*c*x+ 
105*A*b^2*c-105*B*b^3)*x^(1/2)/c^4-2*b^3*(A*c-B*b)/c^4/(b*c)^(1/2)*arctan( 
c*x^(1/2)/(b*c)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.03 \[ \int \frac {x^{7/2} (A+B x)}{b x+c x^2} \, dx=\left [-\frac {105 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x - 2 \, c \sqrt {x} \sqrt {-\frac {b}{c}} - b}{c x + b}\right ) - 2 \, {\left (15 \, B c^{3} x^{3} - 105 \, B b^{3} + 105 \, A b^{2} c - 21 \, {\left (B b c^{2} - A c^{3}\right )} x^{2} + 35 \, {\left (B b^{2} c - A b c^{2}\right )} x\right )} \sqrt {x}}{105 \, c^{4}}, \frac {2 \, {\left (105 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) + {\left (15 \, B c^{3} x^{3} - 105 \, B b^{3} + 105 \, A b^{2} c - 21 \, {\left (B b c^{2} - A c^{3}\right )} x^{2} + 35 \, {\left (B b^{2} c - A b c^{2}\right )} x\right )} \sqrt {x}\right )}}{105 \, c^{4}}\right ] \] Input:

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

Output:

[-1/105*(105*(B*b^3 - A*b^2*c)*sqrt(-b/c)*log((c*x - 2*c*sqrt(x)*sqrt(-b/c 
) - b)/(c*x + b)) - 2*(15*B*c^3*x^3 - 105*B*b^3 + 105*A*b^2*c - 21*(B*b*c^ 
2 - A*c^3)*x^2 + 35*(B*b^2*c - A*b*c^2)*x)*sqrt(x))/c^4, 2/105*(105*(B*b^3 
 - A*b^2*c)*sqrt(b/c)*arctan(c*sqrt(x)*sqrt(b/c)/b) + (15*B*c^3*x^3 - 105* 
B*b^3 + 105*A*b^2*c - 21*(B*b*c^2 - A*c^3)*x^2 + 35*(B*b^2*c - A*b*c^2)*x) 
*sqrt(x))/c^4]
                                                                                    
                                                                                    
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (109) = 218\).

Time = 10.79 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.60 \[ \int \frac {x^{7/2} (A+B x)}{b x+c x^2} \, dx=\begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {7}{2}}}{7}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {\frac {2 A x^{\frac {7}{2}}}{7} + \frac {2 B x^{\frac {9}{2}}}{9}}{b} & \text {for}\: c = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {7}{2}}}{7}}{c} & \text {for}\: b = 0 \\- \frac {A b^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c^{4} \sqrt {- \frac {b}{c}}} + \frac {A b^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c^{4} \sqrt {- \frac {b}{c}}} + \frac {2 A b^{2} \sqrt {x}}{c^{3}} - \frac {2 A b x^{\frac {3}{2}}}{3 c^{2}} + \frac {2 A x^{\frac {5}{2}}}{5 c} + \frac {B b^{4} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c^{5} \sqrt {- \frac {b}{c}}} - \frac {B b^{4} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c^{5} \sqrt {- \frac {b}{c}}} - \frac {2 B b^{3} \sqrt {x}}{c^{4}} + \frac {2 B b^{2} x^{\frac {3}{2}}}{3 c^{3}} - \frac {2 B b x^{\frac {5}{2}}}{5 c^{2}} + \frac {2 B x^{\frac {7}{2}}}{7 c} & \text {otherwise} \end {cases} \] Input:

integrate(x**(7/2)*(B*x+A)/(c*x**2+b*x),x)
 

Output:

Piecewise((zoo*(2*A*x**(5/2)/5 + 2*B*x**(7/2)/7), Eq(b, 0) & Eq(c, 0)), (( 
2*A*x**(7/2)/7 + 2*B*x**(9/2)/9)/b, Eq(c, 0)), ((2*A*x**(5/2)/5 + 2*B*x**( 
7/2)/7)/c, Eq(b, 0)), (-A*b**3*log(sqrt(x) - sqrt(-b/c))/(c**4*sqrt(-b/c)) 
 + A*b**3*log(sqrt(x) + sqrt(-b/c))/(c**4*sqrt(-b/c)) + 2*A*b**2*sqrt(x)/c 
**3 - 2*A*b*x**(3/2)/(3*c**2) + 2*A*x**(5/2)/(5*c) + B*b**4*log(sqrt(x) - 
sqrt(-b/c))/(c**5*sqrt(-b/c)) - B*b**4*log(sqrt(x) + sqrt(-b/c))/(c**5*sqr 
t(-b/c)) - 2*B*b**3*sqrt(x)/c**4 + 2*B*b**2*x**(3/2)/(3*c**3) - 2*B*b*x**( 
5/2)/(5*c**2) + 2*B*x**(7/2)/(7*c), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.93 \[ \int \frac {x^{7/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \, {\left (B b^{4} - A b^{3} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{4}} + \frac {2 \, {\left (15 \, B c^{3} x^{\frac {7}{2}} - 21 \, {\left (B b c^{2} - A c^{3}\right )} x^{\frac {5}{2}} + 35 \, {\left (B b^{2} c - A b c^{2}\right )} x^{\frac {3}{2}} - 105 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {x}\right )}}{105 \, c^{4}} \] Input:

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

Output:

2*(B*b^4 - A*b^3*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^4) + 2/105*(1 
5*B*c^3*x^(7/2) - 21*(B*b*c^2 - A*c^3)*x^(5/2) + 35*(B*b^2*c - A*b*c^2)*x^ 
(3/2) - 105*(B*b^3 - A*b^2*c)*sqrt(x))/c^4
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.02 \[ \int \frac {x^{7/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \, {\left (B b^{4} - A b^{3} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{4}} + \frac {2 \, {\left (15 \, B c^{6} x^{\frac {7}{2}} - 21 \, B b c^{5} x^{\frac {5}{2}} + 21 \, A c^{6} x^{\frac {5}{2}} + 35 \, B b^{2} c^{4} x^{\frac {3}{2}} - 35 \, A b c^{5} x^{\frac {3}{2}} - 105 \, B b^{3} c^{3} \sqrt {x} + 105 \, A b^{2} c^{4} \sqrt {x}\right )}}{105 \, c^{7}} \] Input:

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

Output:

2*(B*b^4 - A*b^3*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^4) + 2/105*(1 
5*B*c^6*x^(7/2) - 21*B*b*c^5*x^(5/2) + 21*A*c^6*x^(5/2) + 35*B*b^2*c^4*x^( 
3/2) - 35*A*b*c^5*x^(3/2) - 105*B*b^3*c^3*sqrt(x) + 105*A*b^2*c^4*sqrt(x)) 
/c^7
 

Mupad [B] (verification not implemented)

Time = 5.14 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.11 \[ \int \frac {x^{7/2} (A+B x)}{b x+c x^2} \, dx=x^{5/2}\,\left (\frac {2\,A}{5\,c}-\frac {2\,B\,b}{5\,c^2}\right )+\frac {2\,B\,x^{7/2}}{7\,c}+\frac {b^2\,\sqrt {x}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )}{c^2}+\frac {2\,b^{5/2}\,\mathrm {atan}\left (\frac {b^{5/2}\,\sqrt {c}\,\sqrt {x}\,\left (A\,c-B\,b\right )}{B\,b^4-A\,b^3\,c}\right )\,\left (A\,c-B\,b\right )}{c^{9/2}}-\frac {b\,x^{3/2}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )}{3\,c} \] Input:

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

Output:

x^(5/2)*((2*A)/(5*c) - (2*B*b)/(5*c^2)) + (2*B*x^(7/2))/(7*c) + (b^2*x^(1/ 
2)*((2*A)/c - (2*B*b)/c^2))/c^2 + (2*b^(5/2)*atan((b^(5/2)*c^(1/2)*x^(1/2) 
*(A*c - B*b))/(B*b^4 - A*b^3*c))*(A*c - B*b))/c^(9/2) - (b*x^(3/2)*((2*A)/ 
c - (2*B*b)/c^2))/(3*c)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.12 \[ \int \frac {x^{7/2} (A+B x)}{b x+c x^2} \, dx=\frac {-2 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c +2 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{4}+2 \sqrt {x}\, a \,b^{2} c^{2}-\frac {2 \sqrt {x}\, a b \,c^{3} x}{3}+\frac {2 \sqrt {x}\, a \,c^{4} x^{2}}{5}-2 \sqrt {x}\, b^{4} c +\frac {2 \sqrt {x}\, b^{3} c^{2} x}{3}-\frac {2 \sqrt {x}\, b^{2} c^{3} x^{2}}{5}+\frac {2 \sqrt {x}\, b \,c^{4} x^{3}}{7}}{c^{5}} \] Input:

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

Output:

(2*( - 105*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*b**2*c + 
105*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*b**4 + 105*sqrt(x) 
*a*b**2*c**2 - 35*sqrt(x)*a*b*c**3*x + 21*sqrt(x)*a*c**4*x**2 - 105*sqrt(x 
)*b**4*c + 35*sqrt(x)*b**3*c**2*x - 21*sqrt(x)*b**2*c**3*x**2 + 15*sqrt(x) 
*b*c**4*x**3))/(105*c**5)