\(\int \frac {(A+B x) (d+e x)^{7/2}}{a+b x} \, dx\) [148]

Optimal result

Integrand size = 22, antiderivative size = 198 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{a+b x} \, dx=\frac {2 (A b-a B) (b d-a e)^3 \sqrt {d+e x}}{b^5}+\frac {2 (A b-a B) (b d-a e)^2 (d+e x)^{3/2}}{3 b^4}+\frac {2 (A b-a B) (b d-a e) (d+e x)^{5/2}}{5 b^3}+\frac {2 (A b-a B) (d+e x)^{7/2}}{7 b^2}+\frac {2 B (d+e x)^{9/2}}{9 b e}-\frac {2 (A b-a B) (b d-a e)^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{11/2}} \] Output:

2*(A*b-B*a)*(-a*e+b*d)^3*(e*x+d)^(1/2)/b^5+2/3*(A*b-B*a)*(-a*e+b*d)^2*(e*x 
+d)^(3/2)/b^4+2/5*(A*b-B*a)*(-a*e+b*d)*(e*x+d)^(5/2)/b^3+2/7*(A*b-B*a)*(e* 
x+d)^(7/2)/b^2+2/9*B*(e*x+d)^(9/2)/b/e-2*(A*b-B*a)*(-a*e+b*d)^(7/2)*arctan 
h(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(11/2)
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.33 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{a+b x} \, dx=\frac {2 \sqrt {d+e x} \left (315 a^4 B e^4-105 a^3 b e^3 (10 B d+3 A e+B e x)+21 a^2 b^2 e^2 \left (5 A e (10 d+e x)+B \left (58 d^2+16 d e x+3 e^2 x^2\right )\right )-3 a b^3 e \left (7 A e \left (58 d^2+16 d e x+3 e^2 x^2\right )+B \left (176 d^3+122 d^2 e x+66 d e^2 x^2+15 e^3 x^3\right )\right )+b^4 \left (35 B (d+e x)^4+3 A e \left (176 d^3+122 d^2 e x+66 d e^2 x^2+15 e^3 x^3\right )\right )\right )}{315 b^5 e}+\frac {2 (A b-a B) (-b d+a e)^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{b^{11/2}} \] Input:

Integrate[((A + B*x)*(d + e*x)^(7/2))/(a + b*x),x]
 

Output:

(2*Sqrt[d + e*x]*(315*a^4*B*e^4 - 105*a^3*b*e^3*(10*B*d + 3*A*e + B*e*x) + 
 21*a^2*b^2*e^2*(5*A*e*(10*d + e*x) + B*(58*d^2 + 16*d*e*x + 3*e^2*x^2)) - 
 3*a*b^3*e*(7*A*e*(58*d^2 + 16*d*e*x + 3*e^2*x^2) + B*(176*d^3 + 122*d^2*e 
*x + 66*d*e^2*x^2 + 15*e^3*x^3)) + b^4*(35*B*(d + e*x)^4 + 3*A*e*(176*d^3 
+ 122*d^2*e*x + 66*d*e^2*x^2 + 15*e^3*x^3))))/(315*b^5*e) + (2*(A*b - a*B) 
*(-(b*d) + a*e)^(7/2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/ 
b^(11/2)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {90, 60, 60, 60, 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.

\(\displaystyle \int \frac {(A+B x) (d+e x)^{7/2}}{a+b x} \, dx\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {(A b-a B) \int \frac {(d+e x)^{7/2}}{a+b x}dx}{b}+\frac {2 B (d+e x)^{9/2}}{9 b e}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \int \frac {(d+e x)^{5/2}}{a+b x}dx}{b}+\frac {2 (d+e x)^{7/2}}{7 b}\right )}{b}+\frac {2 B (d+e x)^{9/2}}{9 b e}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {(d+e x)^{3/2}}{a+b x}dx}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{b}+\frac {2 (d+e x)^{7/2}}{7 b}\right )}{b}+\frac {2 B (d+e x)^{9/2}}{9 b e}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{b}+\frac {2 (d+e x)^{7/2}}{7 b}\right )}{b}+\frac {2 B (d+e x)^{9/2}}{9 b e}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{b}+\frac {2 (d+e x)^{7/2}}{7 b}\right )}{b}+\frac {2 B (d+e x)^{9/2}}{9 b e}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{b}+\frac {2 (d+e x)^{7/2}}{7 b}\right )}{b}+\frac {2 B (d+e x)^{9/2}}{9 b e}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{b}+\frac {2 (d+e x)^{7/2}}{7 b}\right )}{b}+\frac {2 B (d+e x)^{9/2}}{9 b e}\)

Input:

Int[((A + B*x)*(d + e*x)^(7/2))/(a + b*x),x]
 

Output:

(2*B*(d + e*x)^(9/2))/(9*b*e) + ((A*b - a*B)*((2*(d + e*x)^(7/2))/(7*b) + 
((b*d - a*e)*((2*(d + e*x)^(5/2))/(5*b) + ((b*d - a*e)*((2*(d + e*x)^(3/2) 
)/(3*b) + ((b*d - a*e)*((2*Sqrt[d + e*x])/b - (2*Sqrt[b*d - a*e]*ArcTanh[( 
Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/b))/b))/b))/b
 

Defintions of rubi rules used

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 90

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

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 = 0.74 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.36

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (\frac {\left (-\frac {\left (e x +d \right )^{4} B}{3}-\frac {176 A \left (\frac {15}{176} e^{3} x^{3}+\frac {3}{8} e^{2} d \,x^{2}+\frac {61}{88} d^{2} x e +d^{3}\right ) e}{35}\right ) b^{4}}{3}+\frac {58 a e \left (\frac {\left (\frac {15}{2} e^{3} x^{3}+33 e^{2} d \,x^{2}+61 d^{2} x e +88 d^{3}\right ) B}{203}+A e \left (\frac {3}{58} e^{2} x^{2}+\frac {8}{29} d e x +d^{2}\right )\right ) b^{3}}{15}-\frac {10 a^{2} e^{2} \left (\frac {\left (\frac {3}{2} e^{2} x^{2}+8 d e x +29 d^{2}\right ) B}{25}+A e \left (\frac {e x}{10}+d \right )\right ) b^{2}}{3}+a^{3} \left (\frac {\left (e x +10 d \right ) B}{3}+A e \right ) e^{3} b -B \,a^{4} e^{4}\right ) \sqrt {\left (a e -d b \right ) b}\, \sqrt {e x +d}-e \left (a e -d b \right )^{4} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )\right )}{\sqrt {\left (a e -d b \right ) b}\, e \,b^{5}}\)	\(269\)
risch	\(-\frac {2 \left (-35 B \,e^{4} b^{4} x^{4}-45 A \,b^{4} e^{4} x^{3}+45 B a \,b^{3} e^{4} x^{3}-140 B \,b^{4} d \,e^{3} x^{3}+63 A a \,b^{3} e^{4} x^{2}-198 A \,b^{4} d \,e^{3} x^{2}-63 B \,a^{2} b^{2} e^{4} x^{2}+198 B a \,b^{3} d \,e^{3} x^{2}-210 B \,b^{4} d^{2} e^{2} x^{2}-105 A \,a^{2} b^{2} e^{4} x +336 A a \,b^{3} d \,e^{3} x -366 A \,b^{4} d^{2} e^{2} x +105 B \,a^{3} b \,e^{4} x -336 B \,a^{2} b^{2} d \,e^{3} x +366 B a \,b^{3} d^{2} e^{2} x -140 B \,b^{4} d^{3} e x +315 A \,a^{3} b \,e^{4}-1050 A \,a^{2} b^{2} d \,e^{3}+1218 A a \,b^{3} d^{2} e^{2}-528 A \,b^{4} d^{3} e -315 B \,a^{4} e^{4}+1050 B \,a^{3} b d \,e^{3}-1218 B \,a^{2} b^{2} d^{2} e^{2}+528 B a \,b^{3} d^{3} e -35 B \,b^{4} d^{4}\right ) \sqrt {e x +d}}{315 e \,b^{5}}+\frac {2 \left (A \,a^{4} b \,e^{4}-4 A \,a^{3} b^{2} d \,e^{3}+6 A \,a^{2} b^{3} d^{2} e^{2}-4 A a \,b^{4} d^{3} e +A \,d^{4} b^{5}-B \,a^{5} e^{4}+4 B \,a^{4} b d \,e^{3}-6 B \,a^{3} b^{2} d^{2} e^{2}+4 B \,a^{2} b^{3} d^{3} e -B a \,b^{4} d^{4}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{b^{5} \sqrt {\left (a e -d b \right ) b}}\)	\(481\)
derivativedivides	\(\frac {-\frac {2 \left (-\frac {B \left (e x +d \right )^{\frac {9}{2}} b^{4}}{9}-\frac {A \,b^{4} e \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {B a \,b^{3} e \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {A a \,b^{3} e^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {A \,b^{4} d e \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {B \,a^{2} b^{2} e^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {B a \,b^{3} d e \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {A \,a^{2} b^{2} e^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 A a \,b^{3} d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {A \,b^{4} d^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {B \,a^{3} b \,e^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {2 B \,a^{2} b^{2} d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {B a \,b^{3} d^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+A \,a^{3} b \,e^{4} \sqrt {e x +d}-3 A \,a^{2} b^{2} d \,e^{3} \sqrt {e x +d}+3 A a \,b^{3} d^{2} e^{2} \sqrt {e x +d}-A \,b^{4} d^{3} e \sqrt {e x +d}-B \,a^{4} e^{4} \sqrt {e x +d}+3 B \,a^{3} b d \,e^{3} \sqrt {e x +d}-3 B \,a^{2} b^{2} d^{2} e^{2} \sqrt {e x +d}+B a \,b^{3} d^{3} e \sqrt {e x +d}\right )}{b^{5}}+\frac {2 e \left (A \,a^{4} b \,e^{4}-4 A \,a^{3} b^{2} d \,e^{3}+6 A \,a^{2} b^{3} d^{2} e^{2}-4 A a \,b^{4} d^{3} e +A \,d^{4} b^{5}-B \,a^{5} e^{4}+4 B \,a^{4} b d \,e^{3}-6 B \,a^{3} b^{2} d^{2} e^{2}+4 B \,a^{2} b^{3} d^{3} e -B a \,b^{4} d^{4}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{b^{5} \sqrt {\left (a e -d b \right ) b}}}{e}\)	\(531\)
default	\(\frac {-\frac {2 \left (-\frac {B \left (e x +d \right )^{\frac {9}{2}} b^{4}}{9}-\frac {A \,b^{4} e \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {B a \,b^{3} e \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {A a \,b^{3} e^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {A \,b^{4} d e \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {B \,a^{2} b^{2} e^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {B a \,b^{3} d e \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {A \,a^{2} b^{2} e^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 A a \,b^{3} d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {A \,b^{4} d^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {B \,a^{3} b \,e^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {2 B \,a^{2} b^{2} d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {B a \,b^{3} d^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+A \,a^{3} b \,e^{4} \sqrt {e x +d}-3 A \,a^{2} b^{2} d \,e^{3} \sqrt {e x +d}+3 A a \,b^{3} d^{2} e^{2} \sqrt {e x +d}-A \,b^{4} d^{3} e \sqrt {e x +d}-B \,a^{4} e^{4} \sqrt {e x +d}+3 B \,a^{3} b d \,e^{3} \sqrt {e x +d}-3 B \,a^{2} b^{2} d^{2} e^{2} \sqrt {e x +d}+B a \,b^{3} d^{3} e \sqrt {e x +d}\right )}{b^{5}}+\frac {2 e \left (A \,a^{4} b \,e^{4}-4 A \,a^{3} b^{2} d \,e^{3}+6 A \,a^{2} b^{3} d^{2} e^{2}-4 A a \,b^{4} d^{3} e +A \,d^{4} b^{5}-B \,a^{5} e^{4}+4 B \,a^{4} b d \,e^{3}-6 B \,a^{3} b^{2} d^{2} e^{2}+4 B \,a^{2} b^{3} d^{3} e -B a \,b^{4} d^{4}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{b^{5} \sqrt {\left (a e -d b \right ) b}}}{e}\)	\(531\)

Input:

int((B*x+A)*(e*x+d)^(7/2)/(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

-2*((1/3*(-1/3*(e*x+d)^4*B-176/35*A*(15/176*e^3*x^3+3/8*e^2*d*x^2+61/88*d^ 
2*x*e+d^3)*e)*b^4+58/15*a*e*(1/203*(15/2*e^3*x^3+33*e^2*d*x^2+61*d^2*x*e+8 
8*d^3)*B+A*e*(3/58*e^2*x^2+8/29*d*e*x+d^2))*b^3-10/3*a^2*e^2*(1/25*(3/2*e^ 
2*x^2+8*d*e*x+29*d^2)*B+A*e*(1/10*e*x+d))*b^2+a^3*(1/3*(e*x+10*d)*B+A*e)*e 
^3*b-B*a^4*e^4)*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)-e*(a*e-b*d)^4*(A*b-B*a)* 
arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)))/((a*e-b*d)*b)^(1/2)/e/b^5
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (170) = 340\).

Time = 0.10 (sec) , antiderivative size = 865, normalized size of antiderivative = 4.37 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{a+b x} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(e*x+d)^(7/2)/(b*x+a),x, algorithm="fricas")
 

Output:

[1/315*(315*((B*a*b^3 - A*b^4)*d^3*e - 3*(B*a^2*b^2 - A*a*b^3)*d^2*e^2 + 3 
*(B*a^3*b - A*a^2*b^2)*d*e^3 - (B*a^4 - A*a^3*b)*e^4)*sqrt((b*d - a*e)/b)* 
log((b*e*x + 2*b*d - a*e + 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a 
)) + 2*(35*B*b^4*e^4*x^4 + 35*B*b^4*d^4 - 528*(B*a*b^3 - A*b^4)*d^3*e + 12 
18*(B*a^2*b^2 - A*a*b^3)*d^2*e^2 - 1050*(B*a^3*b - A*a^2*b^2)*d*e^3 + 315* 
(B*a^4 - A*a^3*b)*e^4 + 5*(28*B*b^4*d*e^3 - 9*(B*a*b^3 - A*b^4)*e^4)*x^3 + 
 3*(70*B*b^4*d^2*e^2 - 66*(B*a*b^3 - A*b^4)*d*e^3 + 21*(B*a^2*b^2 - A*a*b^ 
3)*e^4)*x^2 + (140*B*b^4*d^3*e - 366*(B*a*b^3 - A*b^4)*d^2*e^2 + 336*(B*a^ 
2*b^2 - A*a*b^3)*d*e^3 - 105*(B*a^3*b - A*a^2*b^2)*e^4)*x)*sqrt(e*x + d))/ 
(b^5*e), 2/315*(315*((B*a*b^3 - A*b^4)*d^3*e - 3*(B*a^2*b^2 - A*a*b^3)*d^2 
*e^2 + 3*(B*a^3*b - A*a^2*b^2)*d*e^3 - (B*a^4 - A*a^3*b)*e^4)*sqrt(-(b*d - 
 a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) + (35*B 
*b^4*e^4*x^4 + 35*B*b^4*d^4 - 528*(B*a*b^3 - A*b^4)*d^3*e + 1218*(B*a^2*b^ 
2 - A*a*b^3)*d^2*e^2 - 1050*(B*a^3*b - A*a^2*b^2)*d*e^3 + 315*(B*a^4 - A*a 
^3*b)*e^4 + 5*(28*B*b^4*d*e^3 - 9*(B*a*b^3 - A*b^4)*e^4)*x^3 + 3*(70*B*b^4 
*d^2*e^2 - 66*(B*a*b^3 - A*b^4)*d*e^3 + 21*(B*a^2*b^2 - A*a*b^3)*e^4)*x^2 
+ (140*B*b^4*d^3*e - 366*(B*a*b^3 - A*b^4)*d^2*e^2 + 336*(B*a^2*b^2 - A*a* 
b^3)*d*e^3 - 105*(B*a^3*b - A*a^2*b^2)*e^4)*x)*sqrt(e*x + d))/(b^5*e)]
 

Sympy [A] (verification not implemented)

Time = 4.24 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.90 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{a+b x} \, dx=\begin {cases} \frac {2 \left (\frac {B \left (d + e x\right )^{\frac {9}{2}}}{9 b} + \frac {\left (d + e x\right )^{\frac {7}{2}} \left (A b e - B a e\right )}{7 b^{2}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (- A a b e^{2} + A b^{2} d e + B a^{2} e^{2} - B a b d e\right )}{5 b^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (A a^{2} b e^{3} - 2 A a b^{2} d e^{2} + A b^{3} d^{2} e - B a^{3} e^{3} + 2 B a^{2} b d e^{2} - B a b^{2} d^{2} e\right )}{3 b^{4}} + \frac {\sqrt {d + e x} \left (- A a^{3} b e^{4} + 3 A a^{2} b^{2} d e^{3} - 3 A a b^{3} d^{2} e^{2} + A b^{4} d^{3} e + B a^{4} e^{4} - 3 B a^{3} b d e^{3} + 3 B a^{2} b^{2} d^{2} e^{2} - B a b^{3} d^{3} e\right )}{b^{5}} - \frac {e \left (- A b + B a\right ) \left (a e - b d\right )^{4} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{b^{6} \sqrt {\frac {a e - b d}{b}}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {7}{2}} \left (\frac {B x}{b} - \frac {\left (- A b + B a\right ) \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((B*x+A)*(e*x+d)**(7/2)/(b*x+a),x)
 

Output:

Piecewise((2*(B*(d + e*x)**(9/2)/(9*b) + (d + e*x)**(7/2)*(A*b*e - B*a*e)/ 
(7*b**2) + (d + e*x)**(5/2)*(-A*a*b*e**2 + A*b**2*d*e + B*a**2*e**2 - B*a* 
b*d*e)/(5*b**3) + (d + e*x)**(3/2)*(A*a**2*b*e**3 - 2*A*a*b**2*d*e**2 + A* 
b**3*d**2*e - B*a**3*e**3 + 2*B*a**2*b*d*e**2 - B*a*b**2*d**2*e)/(3*b**4) 
+ sqrt(d + e*x)*(-A*a**3*b*e**4 + 3*A*a**2*b**2*d*e**3 - 3*A*a*b**3*d**2*e 
**2 + A*b**4*d**3*e + B*a**4*e**4 - 3*B*a**3*b*d*e**3 + 3*B*a**2*b**2*d**2 
*e**2 - B*a*b**3*d**3*e)/b**5 - e*(-A*b + B*a)*(a*e - b*d)**4*atan(sqrt(d 
+ e*x)/sqrt((a*e - b*d)/b))/(b**6*sqrt((a*e - b*d)/b)))/e, Ne(e, 0)), (d** 
(7/2)*(B*x/b - (-A*b + B*a)*Piecewise((x/a, Eq(b, 0)), (log(a + b*x)/b, Tr 
ue))/b), True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B x) (d+e x)^{7/2}}{a+b x} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((B*x+A)*(e*x+d)^(7/2)/(b*x+a),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*e-b*d>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. 560 vs. \(2 (170) = 340\).

Time = 0.13 (sec) , antiderivative size = 560, normalized size of antiderivative = 2.83 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{a+b x} \, dx=-\frac {2 \, {\left (B a b^{4} d^{4} - A b^{5} d^{4} - 4 \, B a^{2} b^{3} d^{3} e + 4 \, A a b^{4} d^{3} e + 6 \, B a^{3} b^{2} d^{2} e^{2} - 6 \, A a^{2} b^{3} d^{2} e^{2} - 4 \, B a^{4} b d e^{3} + 4 \, A a^{3} b^{2} d e^{3} + B a^{5} e^{4} - A a^{4} b e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{5}} + \frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} B b^{8} e^{8} - 45 \, {\left (e x + d\right )}^{\frac {7}{2}} B a b^{7} e^{9} + 45 \, {\left (e x + d\right )}^{\frac {7}{2}} A b^{8} e^{9} - 63 \, {\left (e x + d\right )}^{\frac {5}{2}} B a b^{7} d e^{9} + 63 \, {\left (e x + d\right )}^{\frac {5}{2}} A b^{8} d e^{9} - 105 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{7} d^{2} e^{9} + 105 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{8} d^{2} e^{9} - 315 \, \sqrt {e x + d} B a b^{7} d^{3} e^{9} + 315 \, \sqrt {e x + d} A b^{8} d^{3} e^{9} + 63 \, {\left (e x + d\right )}^{\frac {5}{2}} B a^{2} b^{6} e^{10} - 63 \, {\left (e x + d\right )}^{\frac {5}{2}} A a b^{7} e^{10} + 210 \, {\left (e x + d\right )}^{\frac {3}{2}} B a^{2} b^{6} d e^{10} - 210 \, {\left (e x + d\right )}^{\frac {3}{2}} A a b^{7} d e^{10} + 945 \, \sqrt {e x + d} B a^{2} b^{6} d^{2} e^{10} - 945 \, \sqrt {e x + d} A a b^{7} d^{2} e^{10} - 105 \, {\left (e x + d\right )}^{\frac {3}{2}} B a^{3} b^{5} e^{11} + 105 \, {\left (e x + d\right )}^{\frac {3}{2}} A a^{2} b^{6} e^{11} - 945 \, \sqrt {e x + d} B a^{3} b^{5} d e^{11} + 945 \, \sqrt {e x + d} A a^{2} b^{6} d e^{11} + 315 \, \sqrt {e x + d} B a^{4} b^{4} e^{12} - 315 \, \sqrt {e x + d} A a^{3} b^{5} e^{12}\right )}}{315 \, b^{9} e^{9}} \] Input:

integrate((B*x+A)*(e*x+d)^(7/2)/(b*x+a),x, algorithm="giac")
 

Output:

-2*(B*a*b^4*d^4 - A*b^5*d^4 - 4*B*a^2*b^3*d^3*e + 4*A*a*b^4*d^3*e + 6*B*a^ 
3*b^2*d^2*e^2 - 6*A*a^2*b^3*d^2*e^2 - 4*B*a^4*b*d*e^3 + 4*A*a^3*b^2*d*e^3 
+ B*a^5*e^4 - A*a^4*b*e^4)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(s 
qrt(-b^2*d + a*b*e)*b^5) + 2/315*(35*(e*x + d)^(9/2)*B*b^8*e^8 - 45*(e*x + 
 d)^(7/2)*B*a*b^7*e^9 + 45*(e*x + d)^(7/2)*A*b^8*e^9 - 63*(e*x + d)^(5/2)* 
B*a*b^7*d*e^9 + 63*(e*x + d)^(5/2)*A*b^8*d*e^9 - 105*(e*x + d)^(3/2)*B*a*b 
^7*d^2*e^9 + 105*(e*x + d)^(3/2)*A*b^8*d^2*e^9 - 315*sqrt(e*x + d)*B*a*b^7 
*d^3*e^9 + 315*sqrt(e*x + d)*A*b^8*d^3*e^9 + 63*(e*x + d)^(5/2)*B*a^2*b^6* 
e^10 - 63*(e*x + d)^(5/2)*A*a*b^7*e^10 + 210*(e*x + d)^(3/2)*B*a^2*b^6*d*e 
^10 - 210*(e*x + d)^(3/2)*A*a*b^7*d*e^10 + 945*sqrt(e*x + d)*B*a^2*b^6*d^2 
*e^10 - 945*sqrt(e*x + d)*A*a*b^7*d^2*e^10 - 105*(e*x + d)^(3/2)*B*a^3*b^5 
*e^11 + 105*(e*x + d)^(3/2)*A*a^2*b^6*e^11 - 945*sqrt(e*x + d)*B*a^3*b^5*d 
*e^11 + 945*sqrt(e*x + d)*A*a^2*b^6*d*e^11 + 315*sqrt(e*x + d)*B*a^4*b^4*e 
^12 - 315*sqrt(e*x + d)*A*a^3*b^5*e^12)/(b^9*e^9)
 

Mupad [B] (verification not implemented)

Time = 1.03 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.15 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{a+b x} \, dx=\left (\frac {2\,A\,e-2\,B\,d}{7\,b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{7\,b^2\,e^2}\right )\,{\left (d+e\,x\right )}^{7/2}+\frac {2\,B\,{\left (d+e\,x\right )}^{9/2}}{9\,b\,e}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{7/2}\,\sqrt {d+e\,x}}{-B\,a^5\,e^4+4\,B\,a^4\,b\,d\,e^3+A\,a^4\,b\,e^4-6\,B\,a^3\,b^2\,d^2\,e^2-4\,A\,a^3\,b^2\,d\,e^3+4\,B\,a^2\,b^3\,d^3\,e+6\,A\,a^2\,b^3\,d^2\,e^2-B\,a\,b^4\,d^4-4\,A\,a\,b^4\,d^3\,e+A\,b^5\,d^4}\right )\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{7/2}}{b^{11/2}}+\frac {\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,{\left (a\,e^2-b\,d\,e\right )}^2\,{\left (d+e\,x\right )}^{3/2}}{3\,b^2\,e^2}-\frac {\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,{\left (a\,e^2-b\,d\,e\right )}^3\,\sqrt {d+e\,x}}{b^3\,e^3}-\frac {\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,\left (a\,e^2-b\,d\,e\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,b\,e} \] Input:

int(((A + B*x)*(d + e*x)^(7/2))/(a + b*x),x)
 

Output:

((2*A*e - 2*B*d)/(7*b*e) - (2*B*(a*e^2 - b*d*e))/(7*b^2*e^2))*(d + e*x)^(7 
/2) + (2*B*(d + e*x)^(9/2))/(9*b*e) + (2*atan((b^(1/2)*(A*b - B*a)*(a*e - 
b*d)^(7/2)*(d + e*x)^(1/2))/(A*b^5*d^4 - B*a^5*e^4 + A*a^4*b*e^4 - B*a*b^4 
*d^4 - 4*A*a^3*b^2*d*e^3 + 4*B*a^2*b^3*d^3*e + 6*A*a^2*b^3*d^2*e^2 - 6*B*a 
^3*b^2*d^2*e^2 - 4*A*a*b^4*d^3*e + 4*B*a^4*b*d*e^3))*(A*b - B*a)*(a*e - b* 
d)^(7/2))/b^(11/2) + (((2*A*e - 2*B*d)/(b*e) - (2*B*(a*e^2 - b*d*e))/(b^2* 
e^2))*(a*e^2 - b*d*e)^2*(d + e*x)^(3/2))/(3*b^2*e^2) - (((2*A*e - 2*B*d)/( 
b*e) - (2*B*(a*e^2 - b*d*e))/(b^2*e^2))*(a*e^2 - b*d*e)^3*(d + e*x)^(1/2)) 
/(b^3*e^3) - (((2*A*e - 2*B*d)/(b*e) - (2*B*(a*e^2 - b*d*e))/(b^2*e^2))*(a 
*e^2 - b*d*e)*(d + e*x)^(5/2))/(5*b*e)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.25 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{a+b x} \, dx=\frac {2 \sqrt {e x +d}\, \left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right )}{9 e} \] Input:

int((B*x+A)*(e*x+d)^(7/2)/(b*x+a),x)
 

Output:

(2*sqrt(d + e*x)*(d**4 + 4*d**3*e*x + 6*d**2*e**2*x**2 + 4*d*e**3*x**3 + e 
**4*x**4))/(9*e)