\(\int \frac {(c+d x)^{5/2}}{(a+b x)^6} \, dx\) [240]

Optimal result

Integrand size = 17, antiderivative size = 198 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^6} \, dx=-\frac {d^2 \sqrt {c+d x}}{16 b^3 (a+b x)^3}-\frac {d^3 \sqrt {c+d x}}{64 b^3 (b c-a d) (a+b x)^2}+\frac {3 d^4 \sqrt {c+d x}}{128 b^3 (b c-a d)^2 (a+b x)}-\frac {d (c+d x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}-\frac {3 d^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{128 b^{7/2} (b c-a d)^{5/2}} \] Output:

-1/16*d^2*(d*x+c)^(1/2)/b^3/(b*x+a)^3-1/64*d^3*(d*x+c)^(1/2)/b^3/(-a*d+b*c 
)/(b*x+a)^2+3/128*d^4*(d*x+c)^(1/2)/b^3/(-a*d+b*c)^2/(b*x+a)-1/8*d*(d*x+c) 
^(3/2)/b^2/(b*x+a)^4-1/5*(d*x+c)^(5/2)/b/(b*x+a)^5-3/128*d^5*arctanh(b^(1/ 
2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(7/2)/(-a*d+b*c)^(5/2)
 

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.12 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^6} \, dx=-\frac {\sqrt {c+d x} \left (15 a^4 d^4+10 a^3 b d^3 (c+7 d x)+2 a^2 b^2 d^2 \left (4 c^2+23 c d x+64 d^2 x^2\right )-2 a b^3 d \left (88 c^3+256 c^2 d x+233 c d^2 x^2+35 d^3 x^3\right )+b^4 \left (128 c^4+336 c^3 d x+248 c^2 d^2 x^2+10 c d^3 x^3-15 d^4 x^4\right )\right )}{640 b^3 (b c-a d)^2 (a+b x)^5}+\frac {3 d^5 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{128 b^{7/2} (-b c+a d)^{5/2}} \] Input:

Integrate[(c + d*x)^(5/2)/(a + b*x)^6,x]
 

Output:

-1/640*(Sqrt[c + d*x]*(15*a^4*d^4 + 10*a^3*b*d^3*(c + 7*d*x) + 2*a^2*b^2*d 
^2*(4*c^2 + 23*c*d*x + 64*d^2*x^2) - 2*a*b^3*d*(88*c^3 + 256*c^2*d*x + 233 
*c*d^2*x^2 + 35*d^3*x^3) + b^4*(128*c^4 + 336*c^3*d*x + 248*c^2*d^2*x^2 + 
10*c*d^3*x^3 - 15*d^4*x^4)))/(b^3*(b*c - a*d)^2*(a + b*x)^5) + (3*d^5*ArcT 
an[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(128*b^(7/2)*(-(b*c) + a*d 
)^(5/2))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {51, 51, 51, 52, 52, 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.

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 51

\(\displaystyle \frac {d \left (\frac {3 d \int \frac {\sqrt {c+d x}}{(a+b x)^4}dx}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {d \left (\frac {3 d \left (\frac {d \int \frac {1}{(a+b x)^3 \sqrt {c+d x}}dx}{6 b}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {d \left (\frac {3 d \left (\frac {d \left (-\frac {3 d \int \frac {1}{(a+b x)^2 \sqrt {c+d x}}dx}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 b}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {d \left (\frac {3 d \left (\frac {d \left (-\frac {3 d \left (-\frac {d \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{2 (b c-a d)}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 b}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {d \left (\frac {3 d \left (\frac {d \left (-\frac {3 d \left (-\frac {\int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{b c-a d}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 b}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {d \left (\frac {3 d \left (\frac {d \left (-\frac {3 d \left (\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 b}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}\)

Input:

Int[(c + d*x)^(5/2)/(a + b*x)^6,x]
 

Output:

-1/5*(c + d*x)^(5/2)/(b*(a + b*x)^5) + (d*(-1/4*(c + d*x)^(3/2)/(b*(a + b* 
x)^4) + (3*d*(-1/3*Sqrt[c + d*x]/(b*(a + b*x)^3) + (d*(-1/2*Sqrt[c + d*x]/ 
((b*c - a*d)*(a + b*x)^2) - (3*d*(-(Sqrt[c + d*x]/((b*c - a*d)*(a + b*x))) 
 + (d*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(Sqrt[b]*(b*c - a* 
d)^(3/2))))/(4*(b*c - a*d))))/(6*b)))/(8*b)))/(2*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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[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 = 0.59 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(2 d^{5} \left (\frac {\frac {3 b \left (x d +c \right )^{\frac {9}{2}}}{256 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {7 \left (x d +c \right )^{\frac {7}{2}}}{128 \left (a d -b c \right )}-\frac {\left (x d +c \right )^{\frac {5}{2}}}{10 b}-\frac {7 \left (a d -b c \right ) \left (x d +c \right )^{\frac {3}{2}}}{128 b^{2}}-\frac {3 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {x d +c}}{256 b^{3}}}{\left (\left (x d +c \right ) b +a d -b c \right )^{5}}+\frac {3 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{256 b^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {\left (a d -b c \right ) b}}\right )\)	\(205\)
default	\(2 d^{5} \left (\frac {\frac {3 b \left (x d +c \right )^{\frac {9}{2}}}{256 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {7 \left (x d +c \right )^{\frac {7}{2}}}{128 \left (a d -b c \right )}-\frac {\left (x d +c \right )^{\frac {5}{2}}}{10 b}-\frac {7 \left (a d -b c \right ) \left (x d +c \right )^{\frac {3}{2}}}{128 b^{2}}-\frac {3 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {x d +c}}{256 b^{3}}}{\left (\left (x d +c \right ) b +a d -b c \right )^{5}}+\frac {3 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{256 b^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {\left (a d -b c \right ) b}}\right )\)	\(205\)
pseudoelliptic	\(-\frac {3 \left (\left (\left (-b^{4} x^{4}-\frac {14}{3} a \,x^{3} b^{3}+\frac {128}{15} a^{2} b^{2} x^{2}+\frac {14}{3} a^{3} b x +a^{4}\right ) d^{4}+\frac {2 c \left (b^{3} x^{3}-\frac {233}{5} a \,b^{2} x^{2}+\frac {23}{5} a^{2} b x +a^{3}\right ) b \,d^{3}}{3}+\frac {8 b^{2} c^{2} \left (31 b^{2} x^{2}-64 a b x +a^{2}\right ) d^{2}}{15}-\frac {176 c^{3} \left (-\frac {21 b x}{11}+a \right ) b^{3} d}{15}+\frac {128 c^{4} b^{4}}{15}\right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {x d +c}-d^{5} \left (b x +a \right )^{5} \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )\right )}{128 \sqrt {\left (a d -b c \right ) b}\, \left (b x +a \right )^{5} \left (a d -b c \right )^{2} b^{3}}\)	\(219\)

Input:

int((d*x+c)^(5/2)/(b*x+a)^6,x,method=_RETURNVERBOSE)
 

Output:

2*d^5*((3/256*b/(a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(9/2)+7/128/(a*d-b*c)* 
(d*x+c)^(7/2)-1/10*(d*x+c)^(5/2)/b-7/128*(a*d-b*c)/b^2*(d*x+c)^(3/2)-3/256 
/b^3*(a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(1/2))/((d*x+c)*b+a*d-b*c)^5+3/25 
6/b^3/(a^2*d^2-2*a*b*c*d+b^2*c^2)/((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. 662 vs. \(2 (166) = 332\).

Time = 0.15 (sec) , antiderivative size = 1337, normalized size of antiderivative = 6.75 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^6} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^(5/2)/(b*x+a)^6,x, algorithm="fricas")
 

Output:

[1/1280*(15*(b^5*d^5*x^5 + 5*a*b^4*d^5*x^4 + 10*a^2*b^3*d^5*x^3 + 10*a^3*b 
^2*d^5*x^2 + 5*a^4*b*d^5*x + a^5*d^5)*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*(128*b^6*c^ 
5 - 304*a*b^5*c^4*d + 184*a^2*b^4*c^3*d^2 + 2*a^3*b^3*c^2*d^3 + 5*a^4*b^2* 
c*d^4 - 15*a^5*b*d^5 - 15*(b^6*c*d^4 - a*b^5*d^5)*x^4 + 10*(b^6*c^2*d^3 - 
8*a*b^5*c*d^4 + 7*a^2*b^4*d^5)*x^3 + 2*(124*b^6*c^3*d^2 - 357*a*b^5*c^2*d^ 
3 + 297*a^2*b^4*c*d^4 - 64*a^3*b^3*d^5)*x^2 + 2*(168*b^6*c^4*d - 424*a*b^5 
*c^3*d^2 + 279*a^2*b^4*c^2*d^3 + 12*a^3*b^3*c*d^4 - 35*a^4*b^2*d^5)*x)*sqr 
t(d*x + c))/(a^5*b^7*c^3 - 3*a^6*b^6*c^2*d + 3*a^7*b^5*c*d^2 - a^8*b^4*d^3 
 + (b^12*c^3 - 3*a*b^11*c^2*d + 3*a^2*b^10*c*d^2 - a^3*b^9*d^3)*x^5 + 5*(a 
*b^11*c^3 - 3*a^2*b^10*c^2*d + 3*a^3*b^9*c*d^2 - a^4*b^8*d^3)*x^4 + 10*(a^ 
2*b^10*c^3 - 3*a^3*b^9*c^2*d + 3*a^4*b^8*c*d^2 - a^5*b^7*d^3)*x^3 + 10*(a^ 
3*b^9*c^3 - 3*a^4*b^8*c^2*d + 3*a^5*b^7*c*d^2 - a^6*b^6*d^3)*x^2 + 5*(a^4* 
b^8*c^3 - 3*a^5*b^7*c^2*d + 3*a^6*b^6*c*d^2 - a^7*b^5*d^3)*x), 1/640*(15*( 
b^5*d^5*x^5 + 5*a*b^4*d^5*x^4 + 10*a^2*b^3*d^5*x^3 + 10*a^3*b^2*d^5*x^2 + 
5*a^4*b*d^5*x + a^5*d^5)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)* 
sqrt(d*x + c)/(b*d*x + b*c)) - (128*b^6*c^5 - 304*a*b^5*c^4*d + 184*a^2*b^ 
4*c^3*d^2 + 2*a^3*b^3*c^2*d^3 + 5*a^4*b^2*c*d^4 - 15*a^5*b*d^5 - 15*(b^6*c 
*d^4 - a*b^5*d^5)*x^4 + 10*(b^6*c^2*d^3 - 8*a*b^5*c*d^4 + 7*a^2*b^4*d^5)*x 
^3 + 2*(124*b^6*c^3*d^2 - 357*a*b^5*c^2*d^3 + 297*a^2*b^4*c*d^4 - 64*a^...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^6} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**(5/2)/(b*x+a)**6,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^6} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)^(5/2)/(b*x+a)^6,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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (166) = 332\).

Time = 0.14 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.92 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^6} \, dx=\frac {3 \, d^{5} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{128 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {15 \, {\left (d x + c\right )}^{\frac {9}{2}} b^{4} d^{5} - 70 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{4} c d^{5} - 128 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} c^{2} d^{5} + 70 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c^{3} d^{5} - 15 \, \sqrt {d x + c} b^{4} c^{4} d^{5} + 70 \, {\left (d x + c\right )}^{\frac {7}{2}} a b^{3} d^{6} + 256 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{3} c d^{6} - 210 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{3} c^{2} d^{6} + 60 \, \sqrt {d x + c} a b^{3} c^{3} d^{6} - 128 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} b^{2} d^{7} + 210 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b^{2} c d^{7} - 90 \, \sqrt {d x + c} a^{2} b^{2} c^{2} d^{7} - 70 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{3} b d^{8} + 60 \, \sqrt {d x + c} a^{3} b c d^{8} - 15 \, \sqrt {d x + c} a^{4} d^{9}}{640 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{5}} \] Input:

integrate((d*x+c)^(5/2)/(b*x+a)^6,x, algorithm="giac")
 

Output:

3/128*d^5*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^2 - 2*a*b^4 
*c*d + a^2*b^3*d^2)*sqrt(-b^2*c + a*b*d)) + 1/640*(15*(d*x + c)^(9/2)*b^4* 
d^5 - 70*(d*x + c)^(7/2)*b^4*c*d^5 - 128*(d*x + c)^(5/2)*b^4*c^2*d^5 + 70* 
(d*x + c)^(3/2)*b^4*c^3*d^5 - 15*sqrt(d*x + c)*b^4*c^4*d^5 + 70*(d*x + c)^ 
(7/2)*a*b^3*d^6 + 256*(d*x + c)^(5/2)*a*b^3*c*d^6 - 210*(d*x + c)^(3/2)*a* 
b^3*c^2*d^6 + 60*sqrt(d*x + c)*a*b^3*c^3*d^6 - 128*(d*x + c)^(5/2)*a^2*b^2 
*d^7 + 210*(d*x + c)^(3/2)*a^2*b^2*c*d^7 - 90*sqrt(d*x + c)*a^2*b^2*c^2*d^ 
7 - 70*(d*x + c)^(3/2)*a^3*b*d^8 + 60*sqrt(d*x + c)*a^3*b*c*d^8 - 15*sqrt( 
d*x + c)*a^4*d^9)/((b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*((d*x + c)*b - b* 
c + a*d)^5)
 

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 411, normalized size of antiderivative = 2.08 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^6} \, dx=\frac {3\,d^5\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{128\,b^{7/2}\,{\left (a\,d-b\,c\right )}^{5/2}}-\frac {\frac {d^5\,{\left (c+d\,x\right )}^{5/2}}{5\,b}-\frac {7\,d^5\,{\left (c+d\,x\right )}^{7/2}}{64\,\left (a\,d-b\,c\right )}+\frac {3\,d^5\,\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{128\,b^3}+\frac {7\,d^5\,\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}{64\,b^2}-\frac {3\,b\,d^5\,{\left (c+d\,x\right )}^{9/2}}{128\,{\left (a\,d-b\,c\right )}^2}}{b^5\,{\left (c+d\,x\right )}^5-{\left (c+d\,x\right )}^2\,\left (-10\,a^3\,b^2\,d^3+30\,a^2\,b^3\,c\,d^2-30\,a\,b^4\,c^2\,d+10\,b^5\,c^3\right )-\left (5\,b^5\,c-5\,a\,b^4\,d\right )\,{\left (c+d\,x\right )}^4+a^5\,d^5-b^5\,c^5+{\left (c+d\,x\right )}^3\,\left (10\,a^2\,b^3\,d^2-20\,a\,b^4\,c\,d+10\,b^5\,c^2\right )+\left (c+d\,x\right )\,\left (5\,a^4\,b\,d^4-20\,a^3\,b^2\,c\,d^3+30\,a^2\,b^3\,c^2\,d^2-20\,a\,b^4\,c^3\,d+5\,b^5\,c^4\right )-10\,a^2\,b^3\,c^3\,d^2+10\,a^3\,b^2\,c^2\,d^3+5\,a\,b^4\,c^4\,d-5\,a^4\,b\,c\,d^4} \] Input:

int((c + d*x)^(5/2)/(a + b*x)^6,x)
 

Output:

(3*d^5*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(128*b^(7/2)*(a* 
d - b*c)^(5/2)) - ((d^5*(c + d*x)^(5/2))/(5*b) - (7*d^5*(c + d*x)^(7/2))/( 
64*(a*d - b*c)) + (3*d^5*(c + d*x)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/ 
(128*b^3) + (7*d^5*(a*d - b*c)*(c + d*x)^(3/2))/(64*b^2) - (3*b*d^5*(c + d 
*x)^(9/2))/(128*(a*d - b*c)^2))/(b^5*(c + d*x)^5 - (c + d*x)^2*(10*b^5*c^3 
 - 10*a^3*b^2*d^3 + 30*a^2*b^3*c*d^2 - 30*a*b^4*c^2*d) - (5*b^5*c - 5*a*b^ 
4*d)*(c + d*x)^4 + a^5*d^5 - b^5*c^5 + (c + d*x)^3*(10*b^5*c^2 + 10*a^2*b^ 
3*d^2 - 20*a*b^4*c*d) + (c + d*x)*(5*b^5*c^4 + 5*a^4*b*d^4 - 20*a^3*b^2*c* 
d^3 + 30*a^2*b^3*c^2*d^2 - 20*a*b^4*c^3*d) - 10*a^2*b^3*c^3*d^2 + 10*a^3*b 
^2*c^2*d^3 + 5*a*b^4*c^4*d - 5*a^4*b*c*d^4)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 973, normalized size of antiderivative = 4.91 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^6} \, dx =\text {Too large to display} \] Input:

int((d*x+c)^(5/2)/(b*x+a)^6,x)
 

Output:

(15*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c 
)))*a**5*d**5 + 75*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b) 
*sqrt(a*d - b*c)))*a**4*b*d**5*x + 150*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt( 
c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**3*b**2*d**5*x**2 + 150*sqrt(b)*s 
qrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**2*b**3 
*d**5*x**3 + 75*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sq 
rt(a*d - b*c)))*a*b**4*d**5*x**4 + 15*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c 
 + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*b**5*d**5*x**5 - 15*sqrt(c + d*x)*a* 
*5*b*d**5 + 5*sqrt(c + d*x)*a**4*b**2*c*d**4 - 70*sqrt(c + d*x)*a**4*b**2* 
d**5*x + 2*sqrt(c + d*x)*a**3*b**3*c**2*d**3 + 24*sqrt(c + d*x)*a**3*b**3* 
c*d**4*x - 128*sqrt(c + d*x)*a**3*b**3*d**5*x**2 + 184*sqrt(c + d*x)*a**2* 
b**4*c**3*d**2 + 558*sqrt(c + d*x)*a**2*b**4*c**2*d**3*x + 594*sqrt(c + d* 
x)*a**2*b**4*c*d**4*x**2 + 70*sqrt(c + d*x)*a**2*b**4*d**5*x**3 - 304*sqrt 
(c + d*x)*a*b**5*c**4*d - 848*sqrt(c + d*x)*a*b**5*c**3*d**2*x - 714*sqrt( 
c + d*x)*a*b**5*c**2*d**3*x**2 - 80*sqrt(c + d*x)*a*b**5*c*d**4*x**3 + 15* 
sqrt(c + d*x)*a*b**5*d**5*x**4 + 128*sqrt(c + d*x)*b**6*c**5 + 336*sqrt(c 
+ d*x)*b**6*c**4*d*x + 248*sqrt(c + d*x)*b**6*c**3*d**2*x**2 + 10*sqrt(c + 
 d*x)*b**6*c**2*d**3*x**3 - 15*sqrt(c + d*x)*b**6*c*d**4*x**4)/(640*b**4*( 
a**8*d**3 - 3*a**7*b*c*d**2 + 5*a**7*b*d**3*x + 3*a**6*b**2*c**2*d - 15*a* 
*6*b**2*c*d**2*x + 10*a**6*b**2*d**3*x**2 - a**5*b**3*c**3 + 15*a**5*b*...