\(\int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx\) [237]

Optimal result

Integrand size = 17, antiderivative size = 119 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\frac {15 d^2 \sqrt {c+d x}}{4 b^3}-\frac {5 d (c+d x)^{3/2}}{4 b^2 (a+b x)}-\frac {(c+d x)^{5/2}}{2 b (a+b x)^2}-\frac {15 d^2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{7/2}} \] Output:

15/4*d^2*(d*x+c)^(1/2)/b^3-5/4*d*(d*x+c)^(3/2)/b^2/(b*x+a)-1/2*(d*x+c)^(5/ 
2)/b/(b*x+a)^2-15/4*d^2*(-a*d+b*c)^(1/2)*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a 
*d+b*c)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\frac {\sqrt {c+d x} \left (15 a^2 d^2-5 a b d (c-5 d x)+b^2 \left (-2 c^2-9 c d x+8 d^2 x^2\right )\right )}{4 b^3 (a+b x)^2}-\frac {15 d^2 \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{4 b^{7/2}} \] Input:

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

Output:

(Sqrt[c + d*x]*(15*a^2*d^2 - 5*a*b*d*(c - 5*d*x) + b^2*(-2*c^2 - 9*c*d*x + 
 8*d^2*x^2)))/(4*b^3*(a + b*x)^2) - (15*d^2*Sqrt[-(b*c) + a*d]*ArcTan[(Sqr 
t[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(4*b^(7/2))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {51, 51, 60, 73, 221}

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.

\(\displaystyle \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {5 d \int \frac {(c+d x)^{3/2}}{(a+b x)^2}dx}{4 b}-\frac {(c+d x)^{5/2}}{2 b (a+b x)^2}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {5 d \left (\frac {3 d \int \frac {\sqrt {c+d x}}{a+b x}dx}{2 b}-\frac {(c+d x)^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {(c+d x)^{5/2}}{2 b (a+b x)^2}\)

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {5 d \left (\frac {3 d \left (\frac {2 \sqrt {c+d x}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )}{2 b}-\frac {(c+d x)^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {(c+d x)^{5/2}}{2 b (a+b x)^2}\)

Input:

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

Output:

-1/2*(c + d*x)^(5/2)/(b*(a + b*x)^2) + (5*d*(-((c + d*x)^(3/2)/(b*(a + b*x 
))) + (3*d*((2*Sqrt[c + d*x])/b - (2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt 
[c + d*x])/Sqrt[b*c - a*d]])/b^(3/2)))/(2*b)))/(4*b)
 

Defintions of rubi rules used

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

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 221

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {2 d^{2} \sqrt {x d +c}}{b^{3}}-\frac {\left (2 a d -2 b c \right ) d^{2} \left (\frac {-\frac {9 b \left (x d +c \right )^{\frac {3}{2}}}{8}+\left (-\frac {7 a d}{8}+\frac {7 b c}{8}\right ) \sqrt {x d +c}}{\left (\left (x d +c \right ) b +a d -b c \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{3}}\)	\(117\)
pseudoelliptic	\(-\frac {15 \left (d^{2} \left (b x +a \right )^{2} \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )-\left (\left (\frac {8}{15} d^{2} x^{2}-\frac {3}{5} c d x -\frac {2}{15} c^{2}\right ) b^{2}-\frac {a d \left (-5 x d +c \right ) b}{3}+a^{2} d^{2}\right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {x d +c}\right )}{4 \sqrt {\left (a d -b c \right ) b}\, b^{3} \left (b x +a \right )^{2}}\)	\(130\)
derivativedivides	\(2 d^{2} \left (\frac {\sqrt {x d +c}}{b^{3}}-\frac {\frac {\left (-\frac {9}{8} a b d +\frac {9}{8} b^{2} c \right ) \left (x d +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} d^{2}+\frac {7}{4} a b c d -\frac {7}{8} b^{2} c^{2}\right ) \sqrt {x d +c}}{\left (\left (x d +c \right ) b +a d -b c \right )^{2}}+\frac {15 \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}}{b^{3}}\right )\)	\(138\)
default	\(2 d^{2} \left (\frac {\sqrt {x d +c}}{b^{3}}-\frac {\frac {\left (-\frac {9}{8} a b d +\frac {9}{8} b^{2} c \right ) \left (x d +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} d^{2}+\frac {7}{4} a b c d -\frac {7}{8} b^{2} c^{2}\right ) \sqrt {x d +c}}{\left (\left (x d +c \right ) b +a d -b c \right )^{2}}+\frac {15 \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}}{b^{3}}\right )\)	\(138\)

Input:

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

Output:

2*d^2*(d*x+c)^(1/2)/b^3-1/b^3*(2*a*d-2*b*c)*d^2*((-9/8*b*(d*x+c)^(3/2)+(-7 
/8*a*d+7/8*b*c)*(d*x+c)^(1/2))/((d*x+c)*b+a*d-b*c)^2+15/8/((a*d-b*c)*b)^(1 
/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.89 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\left [\frac {15 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (8 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} - 5 \, a b c d + 15 \, a^{2} d^{2} - {\left (9 \, b^{2} c d - 25 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (8 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} - 5 \, a b c d + 15 \, a^{2} d^{2} - {\left (9 \, b^{2} c d - 25 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \] Input:

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

Output:

[1/8*(15*(b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt((b*c - a*d)/b)*log((b* 
d*x + 2*b*c - a*d - 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2* 
(8*b^2*d^2*x^2 - 2*b^2*c^2 - 5*a*b*c*d + 15*a^2*d^2 - (9*b^2*c*d - 25*a*b* 
d^2)*x)*sqrt(d*x + c))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*d^2* 
x^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b* 
sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (8*b^2*d^2*x^2 - 2*b^2*c^2 - 5*a*b*c*d 
 + 15*a^2*d^2 - (9*b^2*c*d - 25*a*b*d^2)*x)*sqrt(d*x + c))/(b^5*x^2 + 2*a* 
b^4*x + a^2*b^3)]
 

Sympy [F(-1)]

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

integrate((d*x+c)**(5/2)/(b*x+a)**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.44 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\frac {2 \, \sqrt {d x + c} d^{2}}{b^{3}} + \frac {15 \, {\left (b c d^{2} - a d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, \sqrt {-b^{2} c + a b d} b^{3}} - \frac {9 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c d^{2} - 7 \, \sqrt {d x + c} b^{2} c^{2} d^{2} - 9 \, {\left (d x + c\right )}^{\frac {3}{2}} a b d^{3} + 14 \, \sqrt {d x + c} a b c d^{3} - 7 \, \sqrt {d x + c} a^{2} d^{4}}{4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2} b^{3}} \] Input:

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

Output:

2*sqrt(d*x + c)*d^2/b^3 + 15/4*(b*c*d^2 - a*d^3)*arctan(sqrt(d*x + c)*b/sq 
rt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^3) - 1/4*(9*(d*x + c)^(3/2)*b^ 
2*c*d^2 - 7*sqrt(d*x + c)*b^2*c^2*d^2 - 9*(d*x + c)^(3/2)*a*b*d^3 + 14*sqr 
t(d*x + c)*a*b*c*d^3 - 7*sqrt(d*x + c)*a^2*d^4)/(((d*x + c)*b - b*c + a*d) 
^2*b^3)
 

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.67 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\frac {2\,d^2\,\sqrt {c+d\,x}}{b^3}-\frac {\left (\frac {9\,b^2\,c\,d^2}{4}-\frac {9\,a\,b\,d^3}{4}\right )\,{\left (c+d\,x\right )}^{3/2}-\sqrt {c+d\,x}\,\left (\frac {7\,a^2\,d^4}{4}-\frac {7\,a\,b\,c\,d^3}{2}+\frac {7\,b^2\,c^2\,d^2}{4}\right )}{b^5\,{\left (c+d\,x\right )}^2-\left (2\,b^5\,c-2\,a\,b^4\,d\right )\,\left (c+d\,x\right )+b^5\,c^2+a^2\,b^3\,d^2-2\,a\,b^4\,c\,d}-\frac {15\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,d^2\,\sqrt {a\,d-b\,c}\,\sqrt {c+d\,x}}{a\,d^3-b\,c\,d^2}\right )\,\sqrt {a\,d-b\,c}}{4\,b^{7/2}} \] Input:

int((c + d*x)^(5/2)/(a + b*x)^3,x)
 

Output:

(2*d^2*(c + d*x)^(1/2))/b^3 - (((9*b^2*c*d^2)/4 - (9*a*b*d^3)/4)*(c + d*x) 
^(3/2) - (c + d*x)^(1/2)*((7*a^2*d^4)/4 + (7*b^2*c^2*d^2)/4 - (7*a*b*c*d^3 
)/2))/(b^5*(c + d*x)^2 - (2*b^5*c - 2*a*b^4*d)*(c + d*x) + b^5*c^2 + a^2*b 
^3*d^2 - 2*a*b^4*c*d) - (15*d^2*atan((b^(1/2)*d^2*(a*d - b*c)^(1/2)*(c + d 
*x)^(1/2))/(a*d^3 - b*c*d^2))*(a*d - b*c)^(1/2))/(4*b^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.07 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} d^{2}-30 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a b \,d^{2} x -15 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{2} d^{2} x^{2}+15 \sqrt {d x +c}\, a^{2} b \,d^{2}-5 \sqrt {d x +c}\, a \,b^{2} c d +25 \sqrt {d x +c}\, a \,b^{2} d^{2} x -2 \sqrt {d x +c}\, b^{3} c^{2}-9 \sqrt {d x +c}\, b^{3} c d x +8 \sqrt {d x +c}\, b^{3} d^{2} x^{2}}{4 b^{4} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:

int((d*x+c)^(5/2)/(b*x+a)^3,x)
 

Output:

( - 15*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - 
b*c)))*a**2*d**2 - 30*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt 
(b)*sqrt(a*d - b*c)))*a*b*d**2*x - 15*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c 
 + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*b**2*d**2*x**2 + 15*sqrt(c + d*x)*a* 
*2*b*d**2 - 5*sqrt(c + d*x)*a*b**2*c*d + 25*sqrt(c + d*x)*a*b**2*d**2*x - 
2*sqrt(c + d*x)*b**3*c**2 - 9*sqrt(c + d*x)*b**3*c*d*x + 8*sqrt(c + d*x)*b 
**3*d**2*x**2)/(4*b**4*(a**2 + 2*a*b*x + b**2*x**2))