\(\int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx\) [226]

Optimal result

Integrand size = 17, antiderivative size = 100 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=-\frac {3 d \sqrt {c+d x}}{4 b^2 (a+b x)}-\frac {(c+d x)^{3/2}}{2 b (a+b x)^2}-\frac {3 d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{5/2} \sqrt {b c-a d}} \] Output:

-3/4*d*(d*x+c)^(1/2)/b^2/(b*x+a)-1/2*(d*x+c)^(3/2)/b/(b*x+a)^2-3/4*d^2*arc 
tanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(5/2)/(-a*d+b*c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=-\frac {\sqrt {c+d x} (2 b c+3 a d+5 b d x)}{4 b^2 (a+b x)^2}+\frac {3 d^2 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{4 b^{5/2} \sqrt {-b c+a d}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {51, 51, 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.

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-1/2*(c + d*x)^(3/2)/(b*(a + b*x)^2) + (3*d*(-(Sqrt[c + d*x]/(b*(a + b*x)) 
) - (d*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b^(3/2)*Sqrt[b*c 
 - a*d])))/(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 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 = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {d^{2} \left (-\frac {\sqrt {x d +c}\, \left (5 b d x +3 a d +2 b c \right )}{d^{2} \left (b x +a \right )^{2}}+\frac {3 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{4 b^{2}}\)	\(79\)
derivativedivides	\(2 d^{2} \left (\frac {-\frac {5 \left (x d +c \right )^{\frac {3}{2}}}{8 b}-\frac {3 \left (a d -b c \right ) \sqrt {x d +c}}{8 b^{2}}}{\left (\left (x d +c \right ) b +a d -b c \right )^{2}}+\frac {3 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 b^{2} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(97\)
default	\(2 d^{2} \left (\frac {-\frac {5 \left (x d +c \right )^{\frac {3}{2}}}{8 b}-\frac {3 \left (a d -b c \right ) \sqrt {x d +c}}{8 b^{2}}}{\left (\left (x d +c \right ) b +a d -b c \right )^{2}}+\frac {3 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 b^{2} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(97\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (80) = 160\).

Time = 0.10 (sec) , antiderivative size = 383, normalized size of antiderivative = 3.83 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=\left [\frac {3 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) - 2 \, {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} + 5 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{8 \, {\left (a^{2} b^{4} c - a^{3} b^{3} d + {\left (b^{6} c - a b^{5} d\right )} x^{2} + 2 \, {\left (a b^{5} c - a^{2} b^{4} d\right )} x\right )}}, \frac {3 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} + 5 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{4 \, {\left (a^{2} b^{4} c - a^{3} b^{3} d + {\left (b^{6} c - a b^{5} d\right )} x^{2} + 2 \, {\left (a b^{5} c - a^{2} b^{4} d\right )} x\right )}}\right ] \] Input:

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

Output:

[1/8*(3*(b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt(b^2*c - a*b*d)*log((b*d 
*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) - 2*(2* 
b^3*c^2 + a*b^2*c*d - 3*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^2)*x)*sqrt(d*x + 
c))/(a^2*b^4*c - a^3*b^3*d + (b^6*c - a*b^5*d)*x^2 + 2*(a*b^5*c - a^2*b^4* 
d)*x), 1/4*(3*(b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt(-b^2*c + a*b*d)*a 
rctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (2*b^3*c^2 + a*b 
^2*c*d - 3*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^2)*x)*sqrt(d*x + c))/(a^2*b^4* 
c - a^3*b^3*d + (b^6*c - a*b^5*d)*x^2 + 2*(a*b^5*c - a^2*b^4*d)*x)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((d*x+c)^(3/2)/(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 = 0.12 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=\frac {3 \, d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, \sqrt {-b^{2} c + a b d} b^{2}} - \frac {5 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} - 3 \, \sqrt {d x + c} b c d^{2} + 3 \, \sqrt {d x + c} a d^{3}}{4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2} b^{2}} \] Input:

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

Output:

3/4*d^2*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d) 
*b^2) - 1/4*(5*(d*x + c)^(3/2)*b*d^2 - 3*sqrt(d*x + c)*b*c*d^2 + 3*sqrt(d* 
x + c)*a*d^3)/(((d*x + c)*b - b*c + a*d)^2*b^2)
 

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.35 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=\frac {3\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{4\,b^{5/2}\,\sqrt {a\,d-b\,c}}-\frac {\frac {5\,d^2\,{\left (c+d\,x\right )}^{3/2}}{4\,b}+\frac {3\,d^2\,\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}{4\,b^2}}{b^2\,{\left (c+d\,x\right )}^2-\left (2\,b^2\,c-2\,a\,b\,d\right )\,\left (c+d\,x\right )+a^2\,d^2+b^2\,c^2-2\,a\,b\,c\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.59 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx=\frac {3 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} d^{2}+6 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a b \,d^{2} x +3 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{2} d^{2} x^{2}-3 \sqrt {d x +c}\, a^{2} b \,d^{2}+\sqrt {d x +c}\, a \,b^{2} c d -5 \sqrt {d x +c}\, a \,b^{2} d^{2} x +2 \sqrt {d x +c}\, b^{3} c^{2}+5 \sqrt {d x +c}\, b^{3} c d x}{4 b^{3} \left (a \,b^{2} d \,x^{2}-b^{3} c \,x^{2}+2 a^{2} b d x -2 a \,b^{2} c x +a^{3} d -a^{2} b c \right )} \] Input:

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

Output:

(3*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c) 
))*a**2*d**2 + 6*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*s 
qrt(a*d - b*c)))*a*b*d**2*x + 3*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 - 3*sqrt(c + d*x)*a**2*b*d* 
*2 + sqrt(c + d*x)*a*b**2*c*d - 5*sqrt(c + d*x)*a*b**2*d**2*x + 2*sqrt(c + 
 d*x)*b**3*c**2 + 5*sqrt(c + d*x)*b**3*c*d*x)/(4*b**3*(a**3*d - a**2*b*c + 
 2*a**2*b*d*x - 2*a*b**2*c*x + a*b**2*d*x**2 - b**3*c*x**2))