\(\int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx\) [488]

Optimal result

Integrand size = 22, antiderivative size = 185 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx=-\frac {2 (b e-a f)}{7 f (d e-c f) (e+f x)^{7/2}}-\frac {2 (b c-a d)}{5 (d e-c f)^2 (e+f x)^{5/2}}-\frac {2 d (b c-a d)}{3 (d e-c f)^3 (e+f x)^{3/2}}-\frac {2 d^2 (b c-a d)}{(d e-c f)^4 \sqrt {e+f x}}+\frac {2 d^{5/2} (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{9/2}} \] Output:

1/7*(2*a*f-2*b*e)/f/(-c*f+d*e)/(f*x+e)^(7/2)-2/5*(-a*d+b*c)/(-c*f+d*e)^2/( 
f*x+e)^(5/2)-2/3*d*(-a*d+b*c)/(-c*f+d*e)^3/(f*x+e)^(3/2)-2*d^2*(-a*d+b*c)/ 
(-c*f+d*e)^4/(f*x+e)^(1/2)+2*d^(5/2)*(-a*d+b*c)*arctanh(d^(1/2)*(f*x+e)^(1 
/2)/(-c*f+d*e)^(1/2))/(-c*f+d*e)^(9/2)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.43 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx=\frac {-2 b \left (15 d^3 e^4+3 c^3 f^3 (2 e+7 f x)-c^2 d f^2 \left (32 e^2+112 e f x+35 f^2 x^2\right )+c d^2 f \left (116 e^3+406 e^2 f x+350 e f^2 x^2+105 f^3 x^3\right )\right )+2 a f \left (-15 c^3 f^3+3 c^2 d f^2 (22 e+7 f x)-c d^2 f \left (122 e^2+112 e f x+35 f^2 x^2\right )+d^3 \left (176 e^3+406 e^2 f x+350 e f^2 x^2+105 f^3 x^3\right )\right )}{105 f (d e-c f)^4 (e+f x)^{7/2}}+\frac {2 d^{5/2} (-b c+a d) \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{(-d e+c f)^{9/2}} \] Input:

Integrate[(a + b*x)/((c + d*x)*(e + f*x)^(9/2)),x]
 

Output:

(-2*b*(15*d^3*e^4 + 3*c^3*f^3*(2*e + 7*f*x) - c^2*d*f^2*(32*e^2 + 112*e*f* 
x + 35*f^2*x^2) + c*d^2*f*(116*e^3 + 406*e^2*f*x + 350*e*f^2*x^2 + 105*f^3 
*x^3)) + 2*a*f*(-15*c^3*f^3 + 3*c^2*d*f^2*(22*e + 7*f*x) - c*d^2*f*(122*e^ 
2 + 112*e*f*x + 35*f^2*x^2) + d^3*(176*e^3 + 406*e^2*f*x + 350*e*f^2*x^2 + 
 105*f^3*x^3)))/(105*f*(d*e - c*f)^4*(e + f*x)^(7/2)) + (2*d^(5/2)*(-(b*c) 
 + a*d)*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/(-(d*e) + c*f) 
^(9/2)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {87, 61, 61, 61, 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.

\(\displaystyle \int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx\)

\(\Big \downarrow \) 87

\(\displaystyle -\frac {(b c-a d) \int \frac {1}{(c+d x) (e+f x)^{7/2}}dx}{d e-c f}-\frac {2 (b e-a f)}{7 f (e+f x)^{7/2} (d e-c f)}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {(b c-a d) \left (\frac {d \int \frac {1}{(c+d x) (e+f x)^{5/2}}dx}{d e-c f}+\frac {2}{5 (e+f x)^{5/2} (d e-c f)}\right )}{d e-c f}-\frac {2 (b e-a f)}{7 f (e+f x)^{7/2} (d e-c f)}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {(b c-a d) \left (\frac {d \left (\frac {d \int \frac {1}{(c+d x) (e+f x)^{3/2}}dx}{d e-c f}+\frac {2}{3 (e+f x)^{3/2} (d e-c f)}\right )}{d e-c f}+\frac {2}{5 (e+f x)^{5/2} (d e-c f)}\right )}{d e-c f}-\frac {2 (b e-a f)}{7 f (e+f x)^{7/2} (d e-c f)}\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(b c-a d) \left (\frac {d \left (\frac {d \left (\frac {2}{\sqrt {e+f x} (d e-c f)}-\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{3/2}}\right )}{d e-c f}+\frac {2}{3 (e+f x)^{3/2} (d e-c f)}\right )}{d e-c f}+\frac {2}{5 (e+f x)^{5/2} (d e-c f)}\right )}{d e-c f}-\frac {2 (b e-a f)}{7 f (e+f x)^{7/2} (d e-c f)}\)

Input:

Int[(a + b*x)/((c + d*x)*(e + f*x)^(9/2)),x]
 

Output:

(-2*(b*e - a*f))/(7*f*(d*e - c*f)*(e + f*x)^(7/2)) - ((b*c - a*d)*(2/(5*(d 
*e - c*f)*(e + f*x)^(5/2)) + (d*(2/(3*(d*e - c*f)*(e + f*x)^(3/2)) + (d*(2 
/((d*e - c*f)*Sqrt[e + f*x]) - (2*Sqrt[d]*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/ 
Sqrt[d*e - c*f]])/(d*e - c*f)^(3/2)))/(d*e - c*f)))/(d*e - c*f)))/(d*e - c 
*f)
 

Defintions of rubi rules used

rule 61

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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 87

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

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 = 0.49 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {2 d^{3} f \left (a d -b c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{4} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a f -b e \right )}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}}}-\frac {2 f \left (a d -b c \right ) d}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}+\frac {2 f \left (a d -b c \right )}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}}}+\frac {2 f \left (a d -b c \right ) d^{2}}{\left (c f -d e \right )^{4} \sqrt {f x +e}}}{f}\)	\(178\)
default	\(\frac {\frac {2 d^{3} f \left (a d -b c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{4} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a f -b e \right )}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}}}-\frac {2 f \left (a d -b c \right ) d}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}+\frac {2 f \left (a d -b c \right )}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}}}+\frac {2 f \left (a d -b c \right ) d^{2}}{\left (c f -d e \right )^{4} \sqrt {f x +e}}}{f}\)	\(178\)
pseudoelliptic	\(-\frac {2 \left (-7 d^{3} f \left (f x +e \right )^{\frac {7}{2}} \left (a d -b c \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )+\left (\left (-7 a \,d^{3} x^{3}+\frac {7 c \,x^{2} \left (3 b x +a \right ) d^{2}}{3}-\frac {7 c^{2} x \left (\frac {5 b x}{3}+a \right ) d}{5}+c^{3} \left (\frac {7 b x}{5}+a \right )\right ) f^{4}-\frac {22 \left (\frac {175 a \,d^{3} x^{2}}{33}-\frac {56 c \left (\frac {25 b x}{8}+a \right ) x \,d^{2}}{33}+c^{2} \left (\frac {56 b x}{33}+a \right ) d -\frac {c^{3} b}{11}\right ) e \,f^{3}}{5}+\frac {122 \left (-\frac {203 x a \,d^{2}}{61}+c \left (\frac {203 b x}{61}+a \right ) d -\frac {16 b \,c^{2}}{61}\right ) d \,e^{2} f^{2}}{15}-\frac {176 d^{2} \left (a d -\frac {29 b c}{44}\right ) e^{3} f}{15}+b \,d^{3} e^{4}\right ) \sqrt {\left (c f -d e \right ) d}\right )}{7 \left (f x +e \right )^{\frac {7}{2}} \sqrt {\left (c f -d e \right ) d}\, f \left (c f -d e \right )^{4}}\)	\(248\)

Input:

int((b*x+a)/(d*x+c)/(f*x+e)^(9/2),x,method=_RETURNVERBOSE)
 

Output:

2/f*(-1/7*(a*f-b*e)/(c*f-d*e)/(f*x+e)^(7/2)-1/3*f*(a*d-b*c)/(c*f-d*e)^3*d/ 
(f*x+e)^(3/2)+1/5*f*(a*d-b*c)/(c*f-d*e)^2/(f*x+e)^(5/2)+f*(a*d-b*c)/(c*f-d 
*e)^4*d^2/(f*x+e)^(1/2)+d^3*f*(a*d-b*c)/(c*f-d*e)^4/((c*f-d*e)*d)^(1/2)*ar 
ctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 673 vs. \(2 (161) = 322\).

Time = 0.19 (sec) , antiderivative size = 1376, normalized size of antiderivative = 7.44 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)/(d*x+c)/(f*x+e)^(9/2),x, algorithm="fricas")
 

Output:

[-1/105*(105*((b*c*d^2 - a*d^3)*f^5*x^4 + 4*(b*c*d^2 - a*d^3)*e*f^4*x^3 + 
6*(b*c*d^2 - a*d^3)*e^2*f^3*x^2 + 4*(b*c*d^2 - a*d^3)*e^3*f^2*x + (b*c*d^2 
 - a*d^3)*e^4*f)*sqrt(d/(d*e - c*f))*log((d*f*x + 2*d*e - c*f - 2*(d*e - c 
*f)*sqrt(f*x + e)*sqrt(d/(d*e - c*f)))/(d*x + c)) + 2*(15*b*d^3*e^4 + 15*a 
*c^3*f^4 + 105*(b*c*d^2 - a*d^3)*f^4*x^3 + 4*(29*b*c*d^2 - 44*a*d^3)*e^3*f 
 - 2*(16*b*c^2*d - 61*a*c*d^2)*e^2*f^2 + 6*(b*c^3 - 11*a*c^2*d)*e*f^3 + 35 
*(10*(b*c*d^2 - a*d^3)*e*f^3 - (b*c^2*d - a*c*d^2)*f^4)*x^2 + 7*(58*(b*c*d 
^2 - a*d^3)*e^2*f^2 - 16*(b*c^2*d - a*c*d^2)*e*f^3 + 3*(b*c^3 - a*c^2*d)*f 
^4)*x)*sqrt(f*x + e))/(d^4*e^8*f - 4*c*d^3*e^7*f^2 + 6*c^2*d^2*e^6*f^3 - 4 
*c^3*d*e^5*f^4 + c^4*e^4*f^5 + (d^4*e^4*f^5 - 4*c*d^3*e^3*f^6 + 6*c^2*d^2* 
e^2*f^7 - 4*c^3*d*e*f^8 + c^4*f^9)*x^4 + 4*(d^4*e^5*f^4 - 4*c*d^3*e^4*f^5 
+ 6*c^2*d^2*e^3*f^6 - 4*c^3*d*e^2*f^7 + c^4*e*f^8)*x^3 + 6*(d^4*e^6*f^3 - 
4*c*d^3*e^5*f^4 + 6*c^2*d^2*e^4*f^5 - 4*c^3*d*e^3*f^6 + c^4*e^2*f^7)*x^2 + 
 4*(d^4*e^7*f^2 - 4*c*d^3*e^6*f^3 + 6*c^2*d^2*e^5*f^4 - 4*c^3*d*e^4*f^5 + 
c^4*e^3*f^6)*x), -2/105*(105*((b*c*d^2 - a*d^3)*f^5*x^4 + 4*(b*c*d^2 - a*d 
^3)*e*f^4*x^3 + 6*(b*c*d^2 - a*d^3)*e^2*f^3*x^2 + 4*(b*c*d^2 - a*d^3)*e^3* 
f^2*x + (b*c*d^2 - a*d^3)*e^4*f)*sqrt(-d/(d*e - c*f))*arctan(sqrt(f*x + e) 
*sqrt(-d/(d*e - c*f))) + (15*b*d^3*e^4 + 15*a*c^3*f^4 + 105*(b*c*d^2 - a*d 
^3)*f^4*x^3 + 4*(29*b*c*d^2 - 44*a*d^3)*e^3*f - 2*(16*b*c^2*d - 61*a*c*d^2 
)*e^2*f^2 + 6*(b*c^3 - 11*a*c^2*d)*e*f^3 + 35*(10*(b*c*d^2 - a*d^3)*e*f...
 

Sympy [A] (verification not implemented)

Time = 6.12 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.10 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx=\begin {cases} \frac {2 \left (\frac {d^{2} f \left (a d - b c\right )}{\sqrt {e + f x} \left (c f - d e\right )^{4}} + \frac {d^{2} f \left (a d - b c\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{\sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )^{4}} - \frac {d f \left (a d - b c\right )}{3 \left (e + f x\right )^{\frac {3}{2}} \left (c f - d e\right )^{3}} + \frac {f \left (a d - b c\right )}{5 \left (e + f x\right )^{\frac {5}{2}} \left (c f - d e\right )^{2}} - \frac {a f - b e}{7 \left (e + f x\right )^{\frac {7}{2}} \left (c f - d e\right )}\right )}{f} & \text {for}\: f \neq 0 \\\frac {\frac {b x}{d} + \frac {\left (a d - b c\right ) \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d}}{e^{\frac {9}{2}}} & \text {otherwise} \end {cases} \] Input:

integrate((b*x+a)/(d*x+c)/(f*x+e)**(9/2),x)
 

Output:

Piecewise((2*(d**2*f*(a*d - b*c)/(sqrt(e + f*x)*(c*f - d*e)**4) + d**2*f*( 
a*d - b*c)*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))/(sqrt((c*f - d*e)/d)*(c 
*f - d*e)**4) - d*f*(a*d - b*c)/(3*(e + f*x)**(3/2)*(c*f - d*e)**3) + f*(a 
*d - b*c)/(5*(e + f*x)**(5/2)*(c*f - d*e)**2) - (a*f - b*e)/(7*(e + f*x)** 
(7/2)*(c*f - d*e)))/f, Ne(f, 0)), ((b*x/d + (a*d - b*c)*Piecewise((x/c, Eq 
(d, 0)), (log(c + d*x)/d, True))/d)/e**(9/2), True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)/(d*x+c)/(f*x+e)^(9/2),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(c*f-d*e>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. 439 vs. \(2 (161) = 322\).

Time = 0.13 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.37 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx=-\frac {2 \, {\left (b c d^{3} - a d^{4}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{{\left (d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 6 \, c^{2} d^{2} e^{2} f^{2} - 4 \, c^{3} d e f^{3} + c^{4} f^{4}\right )} \sqrt {-d^{2} e + c d f}} - \frac {2 \, {\left (15 \, b d^{3} e^{4} + 105 \, {\left (f x + e\right )}^{3} b c d^{2} f - 105 \, {\left (f x + e\right )}^{3} a d^{3} f + 35 \, {\left (f x + e\right )}^{2} b c d^{2} e f - 35 \, {\left (f x + e\right )}^{2} a d^{3} e f + 21 \, {\left (f x + e\right )} b c d^{2} e^{2} f - 21 \, {\left (f x + e\right )} a d^{3} e^{2} f - 45 \, b c d^{2} e^{3} f - 15 \, a d^{3} e^{3} f - 35 \, {\left (f x + e\right )}^{2} b c^{2} d f^{2} + 35 \, {\left (f x + e\right )}^{2} a c d^{2} f^{2} - 42 \, {\left (f x + e\right )} b c^{2} d e f^{2} + 42 \, {\left (f x + e\right )} a c d^{2} e f^{2} + 45 \, b c^{2} d e^{2} f^{2} + 45 \, a c d^{2} e^{2} f^{2} + 21 \, {\left (f x + e\right )} b c^{3} f^{3} - 21 \, {\left (f x + e\right )} a c^{2} d f^{3} - 15 \, b c^{3} e f^{3} - 45 \, a c^{2} d e f^{3} + 15 \, a c^{3} f^{4}\right )}}{105 \, {\left (d^{4} e^{4} f - 4 \, c d^{3} e^{3} f^{2} + 6 \, c^{2} d^{2} e^{2} f^{3} - 4 \, c^{3} d e f^{4} + c^{4} f^{5}\right )} {\left (f x + e\right )}^{\frac {7}{2}}} \] Input:

integrate((b*x+a)/(d*x+c)/(f*x+e)^(9/2),x, algorithm="giac")
 

Output:

-2*(b*c*d^3 - a*d^4)*arctan(sqrt(f*x + e)*d/sqrt(-d^2*e + c*d*f))/((d^4*e^ 
4 - 4*c*d^3*e^3*f + 6*c^2*d^2*e^2*f^2 - 4*c^3*d*e*f^3 + c^4*f^4)*sqrt(-d^2 
*e + c*d*f)) - 2/105*(15*b*d^3*e^4 + 105*(f*x + e)^3*b*c*d^2*f - 105*(f*x 
+ e)^3*a*d^3*f + 35*(f*x + e)^2*b*c*d^2*e*f - 35*(f*x + e)^2*a*d^3*e*f + 2 
1*(f*x + e)*b*c*d^2*e^2*f - 21*(f*x + e)*a*d^3*e^2*f - 45*b*c*d^2*e^3*f - 
15*a*d^3*e^3*f - 35*(f*x + e)^2*b*c^2*d*f^2 + 35*(f*x + e)^2*a*c*d^2*f^2 - 
 42*(f*x + e)*b*c^2*d*e*f^2 + 42*(f*x + e)*a*c*d^2*e*f^2 + 45*b*c^2*d*e^2* 
f^2 + 45*a*c*d^2*e^2*f^2 + 21*(f*x + e)*b*c^3*f^3 - 21*(f*x + e)*a*c^2*d*f 
^3 - 15*b*c^3*e*f^3 - 45*a*c^2*d*e*f^3 + 15*a*c^3*f^4)/((d^4*e^4*f - 4*c*d 
^3*e^3*f^2 + 6*c^2*d^2*e^2*f^3 - 4*c^3*d*e*f^4 + c^4*f^5)*(f*x + e)^(7/2))
 

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.18 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx=\frac {2\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,\left (c^4\,f^4-4\,c^3\,d\,e\,f^3+6\,c^2\,d^2\,e^2\,f^2-4\,c\,d^3\,e^3\,f+d^4\,e^4\right )}{{\left (c\,f-d\,e\right )}^{9/2}}\right )\,\left (a\,d-b\,c\right )}{{\left (c\,f-d\,e\right )}^{9/2}}-\frac {\frac {2\,\left (a\,f-b\,e\right )}{7\,\left (c\,f-d\,e\right )}-\frac {2\,\left (e+f\,x\right )\,\left (a\,d\,f-b\,c\,f\right )}{5\,{\left (c\,f-d\,e\right )}^2}-\frac {2\,d^2\,{\left (e+f\,x\right )}^3\,\left (a\,d\,f-b\,c\,f\right )}{{\left (c\,f-d\,e\right )}^4}+\frac {2\,d\,{\left (e+f\,x\right )}^2\,\left (a\,d\,f-b\,c\,f\right )}{3\,{\left (c\,f-d\,e\right )}^3}}{f\,{\left (e+f\,x\right )}^{7/2}} \] Input:

int((a + b*x)/((e + f*x)^(9/2)*(c + d*x)),x)
 

Output:

(2*d^(5/2)*atan((d^(1/2)*(e + f*x)^(1/2)*(c^4*f^4 + d^4*e^4 + 6*c^2*d^2*e^ 
2*f^2 - 4*c*d^3*e^3*f - 4*c^3*d*e*f^3))/(c*f - d*e)^(9/2))*(a*d - b*c))/(c 
*f - d*e)^(9/2) - ((2*(a*f - b*e))/(7*(c*f - d*e)) - (2*(e + f*x)*(a*d*f - 
 b*c*f))/(5*(c*f - d*e)^2) - (2*d^2*(e + f*x)^3*(a*d*f - b*c*f))/(c*f - d* 
e)^4 + (2*d*(e + f*x)^2*(a*d*f - b*c*f))/(3*(c*f - d*e)^3))/(f*(e + f*x)^( 
7/2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 1131, normalized size of antiderivative = 6.11 \[ \int \frac {a+b x}{(c+d x) (e+f x)^{9/2}} \, dx =\text {Too large to display} \] Input:

int((b*x+a)/(d*x+c)/(f*x+e)^(9/2),x)
 

Output:

(2*(105*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt 
(d)*sqrt(c*f - d*e)))*a*d**3*e**3*f + 315*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - 
 d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*d**3*e**2*f**2*x 
 + 315*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt( 
d)*sqrt(c*f - d*e)))*a*d**3*e*f**3*x**2 + 105*sqrt(d)*sqrt(e + f*x)*sqrt(c 
*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*a*d**3*f**4*x* 
*3 - 105*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqr 
t(d)*sqrt(c*f - d*e)))*b*c*d**2*e**3*f - 315*sqrt(d)*sqrt(e + f*x)*sqrt(c* 
f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*c*d**2*e**2*f 
**2*x - 315*sqrt(d)*sqrt(e + f*x)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/( 
sqrt(d)*sqrt(c*f - d*e)))*b*c*d**2*e*f**3*x**2 - 105*sqrt(d)*sqrt(e + f*x) 
*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b*c*d** 
2*f**4*x**3 - 15*a*c**4*f**5 + 81*a*c**3*d*e*f**4 + 21*a*c**3*d*f**5*x - 1 
88*a*c**2*d**2*e**2*f**3 - 133*a*c**2*d**2*e*f**4*x - 35*a*c**2*d**2*f**5* 
x**2 + 298*a*c*d**3*e**3*f**2 + 518*a*c*d**3*e**2*f**3*x + 385*a*c*d**3*e* 
f**4*x**2 + 105*a*c*d**3*f**5*x**3 - 176*a*d**4*e**4*f - 406*a*d**4*e**3*f 
**2*x - 350*a*d**4*e**2*f**3*x**2 - 105*a*d**4*e*f**4*x**3 - 6*b*c**4*e*f* 
*4 - 21*b*c**4*f**5*x + 38*b*c**3*d*e**2*f**3 + 133*b*c**3*d*e*f**4*x + 35 
*b*c**3*d*f**5*x**2 - 148*b*c**2*d**2*e**3*f**2 - 518*b*c**2*d**2*e**2*f** 
3*x - 385*b*c**2*d**2*e*f**4*x**2 - 105*b*c**2*d**2*f**5*x**3 + 101*b*c...