\(\int \frac {(c+d x)^{5/2}}{(a+b x)^4} \, dx\) [238]

Optimal result

Integrand size = 17, antiderivative size = 126 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^4} \, dx=-\frac {5 d^2 \sqrt {c+d x}}{8 b^3 (a+b x)}-\frac {5 d (c+d x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(c+d x)^{5/2}}{3 b (a+b x)^3}-\frac {5 d^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{7/2} \sqrt {b c-a d}} \] Output:

-5/8*d^2*(d*x+c)^(1/2)/b^3/(b*x+a)-5/12*d*(d*x+c)^(3/2)/b^2/(b*x+a)^2-1/3* 
(d*x+c)^(5/2)/b/(b*x+a)^3-5/8*d^3*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c) 
^(1/2))/b^(7/2)/(-a*d+b*c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.94 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^4} \, dx=-\frac {\sqrt {c+d x} \left (15 a^2 d^2+10 a b d (c+4 d x)+b^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )\right )}{24 b^3 (a+b x)^3}+\frac {5 d^3 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{8 b^{7/2} \sqrt {-b c+a d}} \] Input:

Integrate[(c + d*x)^(5/2)/(a + b*x)^4,x]
 

Output:

-1/24*(Sqrt[c + d*x]*(15*a^2*d^2 + 10*a*b*d*(c + 4*d*x) + b^2*(8*c^2 + 26* 
c*d*x + 33*d^2*x^2)))/(b^3*(a + b*x)^3) + (5*d^3*ArcTan[(Sqrt[b]*Sqrt[c + 
d*x])/Sqrt[-(b*c) + a*d]])/(8*b^(7/2)*Sqrt[-(b*c) + a*d])
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {51, 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.

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

\(\Big \downarrow \) 51

\(\displaystyle \frac {5 d \int \frac {(c+d x)^{3/2}}{(a+b x)^3}dx}{6 b}-\frac {(c+d x)^{5/2}}{3 b (a+b x)^3}\)

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {5 d \left (\frac {3 d \left (-\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x}}{b (a+b x)}\right )}{4 b}-\frac {(c+d x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(c+d x)^{5/2}}{3 b (a+b x)^3}\)

Input:

Int[(c + d*x)^(5/2)/(a + b*x)^4,x]
 

Output:

-1/3*(c + d*x)^(5/2)/(b*(a + b*x)^3) + (5*d*(-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)))/(6*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 = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.97

method	result	size
pseudoelliptic	\(\frac {\frac {5 d^{3} \left (b x +a \right )^{3} \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8}-\frac {5 \sqrt {\left (a d -b c \right ) b}\, \sqrt {x d +c}\, \left (\left (\frac {11}{5} d^{2} x^{2}+\frac {26}{15} c d x +\frac {8}{15} c^{2}\right ) b^{2}+\frac {2 a d \left (4 x d +c \right ) b}{3}+a^{2} d^{2}\right )}{8}}{b^{3} \left (b x +a \right )^{3} \sqrt {\left (a d -b c \right ) b}}\)	\(122\)
derivativedivides	\(2 d^{3} \left (\frac {-\frac {11 \left (x d +c \right )^{\frac {5}{2}}}{16 b}-\frac {5 \left (a d -b c \right ) \left (x d +c \right )^{\frac {3}{2}}}{6 b^{2}}-\frac {5 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {x d +c}}{16 b^{3}}}{\left (\left (x d +c \right ) b +a d -b c \right )^{3}}+\frac {5 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16 b^{3} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(130\)
default	\(2 d^{3} \left (\frac {-\frac {11 \left (x d +c \right )^{\frac {5}{2}}}{16 b}-\frac {5 \left (a d -b c \right ) \left (x d +c \right )^{\frac {3}{2}}}{6 b^{2}}-\frac {5 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {x d +c}}{16 b^{3}}}{\left (\left (x d +c \right ) b +a d -b c \right )^{3}}+\frac {5 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16 b^{3} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(130\)

Input:

int((d*x+c)^(5/2)/(b*x+a)^4,x,method=_RETURNVERBOSE)
 

Output:

5/8*(d^3*(b*x+a)^3*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))-((a*d-b*c)* 
b)^(1/2)*(d*x+c)^(1/2)*((11/5*d^2*x^2+26/15*c*d*x+8/15*c^2)*b^2+2/3*a*d*(4 
*d*x+c)*b+a^2*d^2))/((a*d-b*c)*b)^(1/2)/b^3/(b*x+a)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (102) = 204\).

Time = 0.09 (sec) , antiderivative size = 563, normalized size of antiderivative = 4.47 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^4} \, dx=\left [\frac {15 \, {\left (b^{3} d^{3} x^{3} + 3 \, a b^{2} d^{3} x^{2} + 3 \, a^{2} b d^{3} x + a^{3} d^{3}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) - 2 \, {\left (8 \, b^{4} c^{3} + 2 \, a b^{3} c^{2} d + 5 \, a^{2} b^{2} c d^{2} - 15 \, a^{3} b d^{3} + 33 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (13 \, b^{4} c^{2} d + 7 \, a b^{3} c d^{2} - 20 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{48 \, {\left (a^{3} b^{5} c - a^{4} b^{4} d + {\left (b^{8} c - a b^{7} d\right )} x^{3} + 3 \, {\left (a b^{7} c - a^{2} b^{6} d\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c - a^{3} b^{5} d\right )} x\right )}}, \frac {15 \, {\left (b^{3} d^{3} x^{3} + 3 \, a b^{2} d^{3} x^{2} + 3 \, a^{2} b d^{3} x + a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - {\left (8 \, b^{4} c^{3} + 2 \, a b^{3} c^{2} d + 5 \, a^{2} b^{2} c d^{2} - 15 \, a^{3} b d^{3} + 33 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (13 \, b^{4} c^{2} d + 7 \, a b^{3} c d^{2} - 20 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{24 \, {\left (a^{3} b^{5} c - a^{4} b^{4} d + {\left (b^{8} c - a b^{7} d\right )} x^{3} + 3 \, {\left (a b^{7} c - a^{2} b^{6} d\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c - a^{3} b^{5} d\right )} x\right )}}\right ] \] Input:

integrate((d*x+c)^(5/2)/(b*x+a)^4,x, algorithm="fricas")
 

Output:

[1/48*(15*(b^3*d^3*x^3 + 3*a*b^2*d^3*x^2 + 3*a^2*b*d^3*x + a^3*d^3)*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*(8*b^4*c^3 + 2*a*b^3*c^2*d + 5*a^2*b^2*c*d^2 - 15*a^3*b 
*d^3 + 33*(b^4*c*d^2 - a*b^3*d^3)*x^2 + 2*(13*b^4*c^2*d + 7*a*b^3*c*d^2 - 
20*a^2*b^2*d^3)*x)*sqrt(d*x + c))/(a^3*b^5*c - a^4*b^4*d + (b^8*c - a*b^7* 
d)*x^3 + 3*(a*b^7*c - a^2*b^6*d)*x^2 + 3*(a^2*b^6*c - a^3*b^5*d)*x), 1/24* 
(15*(b^3*d^3*x^3 + 3*a*b^2*d^3*x^2 + 3*a^2*b*d^3*x + a^3*d^3)*sqrt(-b^2*c 
+ a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (8*b^4 
*c^3 + 2*a*b^3*c^2*d + 5*a^2*b^2*c*d^2 - 15*a^3*b*d^3 + 33*(b^4*c*d^2 - a* 
b^3*d^3)*x^2 + 2*(13*b^4*c^2*d + 7*a*b^3*c*d^2 - 20*a^2*b^2*d^3)*x)*sqrt(d 
*x + c))/(a^3*b^5*c - a^4*b^4*d + (b^8*c - a*b^7*d)*x^3 + 3*(a*b^7*c - a^2 
*b^6*d)*x^2 + 3*(a^2*b^6*c - a^3*b^5*d)*x)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^4} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**(5/2)/(b*x+a)**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)^(5/2)/(b*x+a)^4,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 = 161, normalized size of antiderivative = 1.28 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^4} \, dx=\frac {5 \, d^{3} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{8 \, \sqrt {-b^{2} c + a b d} b^{3}} - \frac {33 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{3} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c d^{3} + 15 \, \sqrt {d x + c} b^{2} c^{2} d^{3} + 40 \, {\left (d x + c\right )}^{\frac {3}{2}} a b d^{4} - 30 \, \sqrt {d x + c} a b c d^{4} + 15 \, \sqrt {d x + c} a^{2} d^{5}}{24 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}^{3} b^{3}} \] Input:

integrate((d*x+c)^(5/2)/(b*x+a)^4,x, algorithm="giac")
 

Output:

5/8*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d) 
*b^3) - 1/24*(33*(d*x + c)^(5/2)*b^2*d^3 - 40*(d*x + c)^(3/2)*b^2*c*d^3 + 
15*sqrt(d*x + c)*b^2*c^2*d^3 + 40*(d*x + c)^(3/2)*a*b*d^4 - 30*sqrt(d*x + 
c)*a*b*c*d^4 + 15*sqrt(d*x + c)*a^2*d^5)/(((d*x + c)*b - b*c + a*d)^3*b^3)
 

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.76 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^4} \, dx=\frac {5\,d^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{8\,b^{7/2}\,\sqrt {a\,d-b\,c}}-\frac {\frac {11\,d^3\,{\left (c+d\,x\right )}^{5/2}}{8\,b}+\frac {5\,d^3\,\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{8\,b^3}+\frac {5\,d^3\,\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,b^2}}{\left (c+d\,x\right )\,\left (3\,a^2\,b\,d^2-6\,a\,b^2\,c\,d+3\,b^3\,c^2\right )+b^3\,{\left (c+d\,x\right )}^3-\left (3\,b^3\,c-3\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^2+a^3\,d^3-b^3\,c^3+3\,a\,b^2\,c^2\,d-3\,a^2\,b\,c\,d^2} \] Input:

int((c + d*x)^(5/2)/(a + b*x)^4,x)
 

Output:

(5*d^3*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(8*b^(7/2)*(a*d 
- b*c)^(1/2)) - ((11*d^3*(c + d*x)^(5/2))/(8*b) + (5*d^3*(c + d*x)^(1/2)*( 
a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(8*b^3) + (5*d^3*(a*d - b*c)*(c + d*x)^(3/ 
2))/(3*b^2))/((c + d*x)*(3*b^3*c^2 + 3*a^2*b*d^2 - 6*a*b^2*c*d) + b^3*(c + 
 d*x)^3 - (3*b^3*c - 3*a*b^2*d)*(c + d*x)^2 + a^3*d^3 - b^3*c^3 + 3*a*b^2* 
c^2*d - 3*a^2*b*c*d^2)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 410, normalized size of antiderivative = 3.25 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^4} \, dx=\frac {15 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{3} d^{3}+45 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} b \,d^{3} x +45 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a \,b^{2} d^{3} x^{2}+15 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{3} d^{3} x^{3}-15 \sqrt {d x +c}\, a^{3} b \,d^{3}+5 \sqrt {d x +c}\, a^{2} b^{2} c \,d^{2}-40 \sqrt {d x +c}\, a^{2} b^{2} d^{3} x +2 \sqrt {d x +c}\, a \,b^{3} c^{2} d +14 \sqrt {d x +c}\, a \,b^{3} c \,d^{2} x -33 \sqrt {d x +c}\, a \,b^{3} d^{3} x^{2}+8 \sqrt {d x +c}\, b^{4} c^{3}+26 \sqrt {d x +c}\, b^{4} c^{2} d x +33 \sqrt {d x +c}\, b^{4} c \,d^{2} x^{2}}{24 b^{4} \left (a \,b^{3} d \,x^{3}-b^{4} c \,x^{3}+3 a^{2} b^{2} d \,x^{2}-3 a \,b^{3} c \,x^{2}+3 a^{3} b d x -3 a^{2} b^{2} c x +a^{4} d -a^{3} b c \right )} \] Input:

int((d*x+c)^(5/2)/(b*x+a)^4,x)
 

Output:

(15*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c 
)))*a**3*d**3 + 45*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b) 
*sqrt(a*d - b*c)))*a**2*b*d**3*x + 45*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c 
 + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a*b**2*d**3*x**2 + 15*sqrt(b)*sqrt(a 
*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*b**3*d**3*x**3 
 - 15*sqrt(c + d*x)*a**3*b*d**3 + 5*sqrt(c + d*x)*a**2*b**2*c*d**2 - 40*sq 
rt(c + d*x)*a**2*b**2*d**3*x + 2*sqrt(c + d*x)*a*b**3*c**2*d + 14*sqrt(c + 
 d*x)*a*b**3*c*d**2*x - 33*sqrt(c + d*x)*a*b**3*d**3*x**2 + 8*sqrt(c + d*x 
)*b**4*c**3 + 26*sqrt(c + d*x)*b**4*c**2*d*x + 33*sqrt(c + d*x)*b**4*c*d** 
2*x**2)/(24*b**4*(a**4*d - a**3*b*c + 3*a**3*b*d*x - 3*a**2*b**2*c*x + 3*a 
**2*b**2*d*x**2 - 3*a*b**3*c*x**2 + a*b**3*d*x**3 - b**4*c*x**3))