\(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{15/2}} \, dx\) [202]

Optimal result

Integrand size = 24, antiderivative size = 170 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{15/2}} \, dx=-\frac {c (24 b B+5 A c) \sqrt {b x+c x^2}}{32 x^{5/2}}-\frac {c^2 (88 b B+5 A c) \sqrt {b x+c x^2}}{64 b x^{3/2}}-\frac {(8 b B+5 A c) \left (b x+c x^2\right )^{3/2}}{24 x^{9/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}-\frac {5 c^3 (8 b B-A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{3/2}} \] Output:

-1/32*c*(5*A*c+24*B*b)*(c*x^2+b*x)^(1/2)/x^(5/2)-1/64*c^2*(5*A*c+88*B*b)*( 
c*x^2+b*x)^(1/2)/b/x^(3/2)-1/24*(5*A*c+8*B*b)*(c*x^2+b*x)^(3/2)/x^(9/2)-1/ 
4*A*(c*x^2+b*x)^(5/2)/x^(13/2)-5/64*c^3*(-A*c+8*B*b)*arctanh((c*x^2+b*x)^( 
1/2)/b^(1/2)/x^(1/2))/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.82 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{15/2}} \, dx=\frac {\sqrt {x (b+c x)} \left (-\sqrt {b} \sqrt {b+c x} \left (8 b B x \left (8 b^2+26 b c x+33 c^2 x^2\right )+A \left (48 b^3+136 b^2 c x+118 b c^2 x^2+15 c^3 x^3\right )\right )+15 c^3 (-8 b B+A c) x^4 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{192 b^{3/2} x^{9/2} \sqrt {b+c x}} \] Input:

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(15/2),x]
 

Output:

(Sqrt[x*(b + c*x)]*(-(Sqrt[b]*Sqrt[b + c*x]*(8*b*B*x*(8*b^2 + 26*b*c*x + 3 
3*c^2*x^2) + A*(48*b^3 + 136*b^2*c*x + 118*b*c^2*x^2 + 15*c^3*x^3))) + 15* 
c^3*(-8*b*B + A*c)*x^4*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(192*b^(3/2)*x^(9/ 
2)*Sqrt[b + c*x])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1220, 1130, 1130, 1130, 1136, 220}

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.

\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{15/2}} \, dx\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(8 b B-A c) \int \frac {\left (c x^2+b x\right )^{5/2}}{x^{13/2}}dx}{8 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(8 b B-A c) \left (\frac {5}{6} c \int \frac {\left (c x^2+b x\right )^{3/2}}{x^{9/2}}dx-\frac {\left (b x+c x^2\right )^{5/2}}{3 x^{11/2}}\right )}{8 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(8 b B-A c) \left (\frac {5}{6} c \left (\frac {3}{4} c \int \frac {\sqrt {c x^2+b x}}{x^{5/2}}dx-\frac {\left (b x+c x^2\right )^{3/2}}{2 x^{7/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{3 x^{11/2}}\right )}{8 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(8 b B-A c) \left (\frac {5}{6} c \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx-\frac {\sqrt {b x+c x^2}}{x^{3/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{2 x^{7/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{3 x^{11/2}}\right )}{8 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {(8 b B-A c) \left (\frac {5}{6} c \left (\frac {3}{4} c \left (c \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}-\frac {\sqrt {b x+c x^2}}{x^{3/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{2 x^{7/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{3 x^{11/2}}\right )}{8 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {(8 b B-A c) \left (\frac {5}{6} c \left (\frac {3}{4} c \left (-\frac {c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}}-\frac {\sqrt {b x+c x^2}}{x^{3/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{2 x^{7/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{3 x^{11/2}}\right )}{8 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}\)

Input:

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(15/2),x]
 

Output:

-1/4*(A*(b*x + c*x^2)^(7/2))/(b*x^(15/2)) + ((8*b*B - A*c)*(-1/3*(b*x + c* 
x^2)^(5/2)/x^(11/2) + (5*c*(-1/2*(b*x + c*x^2)^(3/2)/x^(7/2) + (3*c*(-(Sqr 
t[b*x + c*x^2]/x^(3/2)) - (c*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])]) 
/Sqrt[b]))/4))/6))/(8*b)
 

Defintions of rubi rules used

rule 220

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

rule 1130

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + p + 1))), x] 
- Simp[c*(p/(e^2*(m + p + 1)))   Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p 
 - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & 
& IntegerQ[2*p]
 

rule 1136

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x 
_Symbol] :> Simp[2*e   Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + 
 b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 
- b*d*e + a*e^2, 0]
 

rule 1220

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x 
^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e 
*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1))   Int[(d + e*x 
)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0 
]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 
]
 

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (15 A \,c^{3} x^{3}+264 x^{3} B b \,c^{2}+118 A b \,c^{2} x^{2}+208 x^{2} B \,b^{2} c +136 A \,b^{2} c x +64 x B \,b^{3}+48 A \,b^{3}\right )}{192 x^{\frac {7}{2}} b \sqrt {x \left (c x +b \right )}}+\frac {5 c^{3} \left (A c -8 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{64 b^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}\)	\(132\)
default	\(\frac {\sqrt {x \left (c x +b \right )}\, \left (15 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{4} x^{4}-120 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{3} x^{4}-15 A \,c^{3} x^{3} \sqrt {c x +b}\, \sqrt {b}-264 B \,b^{\frac {3}{2}} c^{2} x^{3} \sqrt {c x +b}-118 A \,b^{\frac {3}{2}} c^{2} x^{2} \sqrt {c x +b}-208 B \,b^{\frac {5}{2}} c \,x^{2} \sqrt {c x +b}-136 A \,b^{\frac {5}{2}} c x \sqrt {c x +b}-64 B \,b^{\frac {7}{2}} x \sqrt {c x +b}-48 A \,b^{\frac {7}{2}} \sqrt {c x +b}\right )}{192 b^{\frac {3}{2}} x^{\frac {9}{2}} \sqrt {c x +b}}\)	\(185\)

Input:

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(15/2),x,method=_RETURNVERBOSE)
 

Output:

-1/192*(c*x+b)*(15*A*c^3*x^3+264*B*b*c^2*x^3+118*A*b*c^2*x^2+208*B*b^2*c*x 
^2+136*A*b^2*c*x+64*B*b^3*x+48*A*b^3)/x^(7/2)/b/(x*(c*x+b))^(1/2)+5/64*c^3 
*(A*c-8*B*b)/b^(3/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^(1/2)/ 
(x*(c*x+b))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.72 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{15/2}} \, dx=\left [-\frac {15 \, {\left (8 \, B b c^{3} - A c^{4}\right )} \sqrt {b} x^{5} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (48 \, A b^{4} + 3 \, {\left (88 \, B b^{2} c^{2} + 5 \, A b c^{3}\right )} x^{3} + 2 \, {\left (104 \, B b^{3} c + 59 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (8 \, B b^{4} + 17 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{384 \, b^{2} x^{5}}, \frac {15 \, {\left (8 \, B b c^{3} - A c^{4}\right )} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (48 \, A b^{4} + 3 \, {\left (88 \, B b^{2} c^{2} + 5 \, A b c^{3}\right )} x^{3} + 2 \, {\left (104 \, B b^{3} c + 59 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (8 \, B b^{4} + 17 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{192 \, b^{2} x^{5}}\right ] \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(15/2),x, algorithm="fricas")
 

Output:

[-1/384*(15*(8*B*b*c^3 - A*c^4)*sqrt(b)*x^5*log(-(c*x^2 + 2*b*x + 2*sqrt(c 
*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(48*A*b^4 + 3*(88*B*b^2*c^2 + 5*A*b* 
c^3)*x^3 + 2*(104*B*b^3*c + 59*A*b^2*c^2)*x^2 + 8*(8*B*b^4 + 17*A*b^3*c)*x 
)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^5), 1/192*(15*(8*B*b*c^3 - A*c^4)*sqrt 
(-b)*x^5*arctan(sqrt(c*x^2 + b*x)*sqrt(-b)/(b*sqrt(x))) - (48*A*b^4 + 3*(8 
8*B*b^2*c^2 + 5*A*b*c^3)*x^3 + 2*(104*B*b^3*c + 59*A*b^2*c^2)*x^2 + 8*(8*B 
*b^4 + 17*A*b^3*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^5)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{15/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x**(15/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{15/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{x^{\frac {15}{2}}} \,d x } \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(15/2),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/x^(15/2), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{15/2}} \, dx=\frac {\frac {15 \, {\left (8 \, B b c^{4} - A c^{5}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {264 \, {\left (c x + b\right )}^{\frac {7}{2}} B b c^{4} - 584 \, {\left (c x + b\right )}^{\frac {5}{2}} B b^{2} c^{4} + 440 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{3} c^{4} - 120 \, \sqrt {c x + b} B b^{4} c^{4} + 15 \, {\left (c x + b\right )}^{\frac {7}{2}} A c^{5} + 73 \, {\left (c x + b\right )}^{\frac {5}{2}} A b c^{5} - 55 \, {\left (c x + b\right )}^{\frac {3}{2}} A b^{2} c^{5} + 15 \, \sqrt {c x + b} A b^{3} c^{5}}{b c^{4} x^{4}}}{192 \, c} \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(15/2),x, algorithm="giac")
 

Output:

1/192*(15*(8*B*b*c^4 - A*c^5)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b) 
- (264*(c*x + b)^(7/2)*B*b*c^4 - 584*(c*x + b)^(5/2)*B*b^2*c^4 + 440*(c*x 
+ b)^(3/2)*B*b^3*c^4 - 120*sqrt(c*x + b)*B*b^4*c^4 + 15*(c*x + b)^(7/2)*A* 
c^5 + 73*(c*x + b)^(5/2)*A*b*c^5 - 55*(c*x + b)^(3/2)*A*b^2*c^5 + 15*sqrt( 
c*x + b)*A*b^3*c^5)/(b*c^4*x^4))/c
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{15/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^{15/2}} \,d x \] Input:

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(15/2),x)
 

Output:

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(15/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{15/2}} \, dx=\frac {-96 \sqrt {c x +b}\, a \,b^{4}-272 \sqrt {c x +b}\, a \,b^{3} c x -236 \sqrt {c x +b}\, a \,b^{2} c^{2} x^{2}-30 \sqrt {c x +b}\, a b \,c^{3} x^{3}-128 \sqrt {c x +b}\, b^{5} x -416 \sqrt {c x +b}\, b^{4} c \,x^{2}-528 \sqrt {c x +b}\, b^{3} c^{2} x^{3}-15 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a \,c^{4} x^{4}+120 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{2} c^{3} x^{4}+15 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a \,c^{4} x^{4}-120 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{2} c^{3} x^{4}}{384 b^{2} x^{4}} \] Input:

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(15/2),x)
 

Output:

( - 96*sqrt(b + c*x)*a*b**4 - 272*sqrt(b + c*x)*a*b**3*c*x - 236*sqrt(b + 
c*x)*a*b**2*c**2*x**2 - 30*sqrt(b + c*x)*a*b*c**3*x**3 - 128*sqrt(b + c*x) 
*b**5*x - 416*sqrt(b + c*x)*b**4*c*x**2 - 528*sqrt(b + c*x)*b**3*c**2*x**3 
 - 15*sqrt(b)*log(sqrt(b + c*x) - sqrt(b))*a*c**4*x**4 + 120*sqrt(b)*log(s 
qrt(b + c*x) - sqrt(b))*b**2*c**3*x**4 + 15*sqrt(b)*log(sqrt(b + c*x) + sq 
rt(b))*a*c**4*x**4 - 120*sqrt(b)*log(sqrt(b + c*x) + sqrt(b))*b**2*c**3*x* 
*4)/(384*b**2*x**4)