\(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{13/2}} \, dx\) [201]

Optimal result

Integrand size = 24, antiderivative size = 156 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{13/2}} \, dx=-\frac {c (14 b B+5 A c) \sqrt {b x+c x^2}}{8 x^{3/2}}+\frac {2 B c^2 \sqrt {b x+c x^2}}{\sqrt {x}}-\frac {(6 b B+5 A c) \left (b x+c x^2\right )^{3/2}}{12 x^{7/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{3 x^{11/2}}-\frac {5 c^2 (6 b B+A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 \sqrt {b}} \] Output:

-1/8*c*(5*A*c+14*B*b)*(c*x^2+b*x)^(1/2)/x^(3/2)+2*B*c^2*(c*x^2+b*x)^(1/2)/ 
x^(1/2)-1/12*(5*A*c+6*B*b)*(c*x^2+b*x)^(3/2)/x^(7/2)-1/3*A*(c*x^2+b*x)^(5/ 
2)/x^(11/2)-5/8*c^2*(A*c+6*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2)) 
/b^(1/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{13/2}} \, dx=-\frac {\sqrt {x (b+c x)} \left (\sqrt {b} \sqrt {b+c x} \left (6 B x \left (2 b^2+9 b c x-8 c^2 x^2\right )+A \left (8 b^2+26 b c x+33 c^2 x^2\right )\right )+15 c^2 (6 b B+A c) x^3 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{24 \sqrt {b} x^{7/2} \sqrt {b+c x}} \] Input:

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(13/2),x]
 

Output:

-1/24*(Sqrt[x*(b + c*x)]*(Sqrt[b]*Sqrt[b + c*x]*(6*B*x*(2*b^2 + 9*b*c*x - 
8*c^2*x^2) + A*(8*b^2 + 26*b*c*x + 33*c^2*x^2)) + 15*c^2*(6*b*B + A*c)*x^3 
*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(Sqrt[b]*x^(7/2)*Sqrt[b + c*x])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96, 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, 1131, 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.

\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{13/2}} \, dx\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(A c+6 b B) \int \frac {\left (c x^2+b x\right )^{5/2}}{x^{11/2}}dx}{6 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(A c+6 b B) \left (\frac {5}{4} c \int \frac {\left (c x^2+b x\right )^{3/2}}{x^{7/2}}dx-\frac {\left (b x+c x^2\right )^{5/2}}{2 x^{9/2}}\right )}{6 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(A c+6 b B) \left (\frac {5}{4} c \left (\frac {3}{2} c \int \frac {\sqrt {c x^2+b x}}{x^{3/2}}dx-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{2 x^{9/2}}\right )}{6 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}\)

\(\Big \downarrow \) 1131

\(\displaystyle \frac {(A c+6 b B) \left (\frac {5}{4} c \left (\frac {3}{2} c \left (b \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx+\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{2 x^{9/2}}\right )}{6 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {(A c+6 b B) \left (\frac {5}{4} c \left (\frac {3}{2} c \left (2 b \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}+\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{2 x^{9/2}}\right )}{6 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {(A c+6 b B) \left (\frac {5}{4} c \left (\frac {3}{2} c \left (\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}-2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )\right )-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{2 x^{9/2}}\right )}{6 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}\)

Input:

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(13/2),x]
 

Output:

-1/3*(A*(b*x + c*x^2)^(7/2))/(b*x^(13/2)) + ((6*b*B + A*c)*(-1/2*(b*x + c* 
x^2)^(5/2)/x^(9/2) + (5*c*(-((b*x + c*x^2)^(3/2)/x^(5/2)) + (3*c*((2*Sqrt[ 
b*x + c*x^2])/Sqrt[x] - 2*Sqrt[b]*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[ 
x])]))/2))/4))/(6*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 1131

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 + 2*p + 1))), x 
] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1)))   Int[(d + e*x)^(m + 1)*(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne 
Q[m + 2*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 = 0.98 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (33 A \,c^{2} x^{2}+54 x^{2} B b c +26 A b c x +12 x B \,b^{2}+8 b^{2} A \right )}{24 x^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}+\frac {c^{2} \left (32 B \sqrt {c x +b}-\frac {2 \left (5 A c +30 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{16 \sqrt {x \left (c x +b \right )}}\)	\(119\)
default	\(-\frac {\sqrt {x \left (c x +b \right )}\, \left (15 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{3} x^{3}+90 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{2} x^{3}-48 B \,c^{2} x^{3} \sqrt {c x +b}\, \sqrt {b}+33 A \,c^{2} x^{2} \sqrt {c x +b}\, \sqrt {b}+54 B \,b^{\frac {3}{2}} c \,x^{2} \sqrt {c x +b}+26 A \,b^{\frac {3}{2}} c x \sqrt {c x +b}+12 B \,b^{\frac {5}{2}} x \sqrt {c x +b}+8 A \,b^{\frac {5}{2}} \sqrt {c x +b}\right )}{24 x^{\frac {7}{2}} \sqrt {c x +b}\, \sqrt {b}}\)	\(166\)

Input:

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(13/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(c*x+b)*(33*A*c^2*x^2+54*B*b*c*x^2+26*A*b*c*x+12*B*b^2*x+8*A*b^2)/x^ 
(5/2)/(x*(c*x+b))^(1/2)+1/16*c^2*(32*B*(c*x+b)^(1/2)-2*(5*A*c+30*B*b)/b^(1 
/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 = 0.10 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.67 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{13/2}} \, dx=\left [\frac {15 \, {\left (6 \, B b c^{2} + A c^{3}\right )} \sqrt {b} x^{4} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (48 \, B b c^{2} x^{3} - 8 \, A b^{3} - 3 \, {\left (18 \, B b^{2} c + 11 \, A b c^{2}\right )} x^{2} - 2 \, {\left (6 \, B b^{3} + 13 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, b x^{4}}, \frac {15 \, {\left (6 \, B b c^{2} + A c^{3}\right )} \sqrt {-b} x^{4} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (48 \, B b c^{2} x^{3} - 8 \, A b^{3} - 3 \, {\left (18 \, B b^{2} c + 11 \, A b c^{2}\right )} x^{2} - 2 \, {\left (6 \, B b^{3} + 13 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, b x^{4}}\right ] \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(13/2),x, algorithm="fricas")
 

Output:

[1/48*(15*(6*B*b*c^2 + A*c^3)*sqrt(b)*x^4*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x 
^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(48*B*b*c^2*x^3 - 8*A*b^3 - 3*(18*B*b^ 
2*c + 11*A*b*c^2)*x^2 - 2*(6*B*b^3 + 13*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt 
(x))/(b*x^4), 1/24*(15*(6*B*b*c^2 + A*c^3)*sqrt(-b)*x^4*arctan(sqrt(c*x^2 
+ b*x)*sqrt(-b)/(b*sqrt(x))) + (48*B*b*c^2*x^3 - 8*A*b^3 - 3*(18*B*b^2*c + 
 11*A*b*c^2)*x^2 - 2*(6*B*b^3 + 13*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/ 
(b*x^4)]
 

Sympy [F]

\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{13/2}} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{x^{\frac {13}{2}}}\, dx \] Input:

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x**(13/2),x)
 

Output:

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**(13/2), x)
 

Maxima [F]

\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{13/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{x^{\frac {13}{2}}} \,d x } \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(13/2),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/x^(13/2), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{13/2}} \, dx=\frac {1}{24} \, c^{3} {\left (\frac {48 \, \sqrt {c x + b} B}{c} + \frac {15 \, {\left (6 \, B b + A c\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} c} - \frac {54 \, {\left (c x + b\right )}^{\frac {5}{2}} B b - 96 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{2} + 42 \, \sqrt {c x + b} B b^{3} + 33 \, {\left (c x + b\right )}^{\frac {5}{2}} A c - 40 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c + 15 \, \sqrt {c x + b} A b^{2} c}{c^{4} x^{3}}\right )} \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(13/2),x, algorithm="giac")
 

Output:

1/24*c^3*(48*sqrt(c*x + b)*B/c + 15*(6*B*b + A*c)*arctan(sqrt(c*x + b)/sqr 
t(-b))/(sqrt(-b)*c) - (54*(c*x + b)^(5/2)*B*b - 96*(c*x + b)^(3/2)*B*b^2 + 
 42*sqrt(c*x + b)*B*b^3 + 33*(c*x + b)^(5/2)*A*c - 40*(c*x + b)^(3/2)*A*b* 
c + 15*sqrt(c*x + b)*A*b^2*c)/(c^4*x^3))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{13/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^{13/2}} \,d x \] Input:

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(13/2),x)
 

Output:

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(13/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.20 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{13/2}} \, dx=\frac {-16 \sqrt {c x +b}\, a \,b^{3}-52 \sqrt {c x +b}\, a \,b^{2} c x -66 \sqrt {c x +b}\, a b \,c^{2} x^{2}-24 \sqrt {c x +b}\, b^{4} x -108 \sqrt {c x +b}\, b^{3} c \,x^{2}+96 \sqrt {c x +b}\, b^{2} c^{2} x^{3}+15 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a \,c^{3} x^{3}+90 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{2} c^{2} x^{3}-15 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a \,c^{3} x^{3}-90 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{2} c^{2} x^{3}}{48 b \,x^{3}} \] Input:

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(13/2),x)
 

Output:

( - 16*sqrt(b + c*x)*a*b**3 - 52*sqrt(b + c*x)*a*b**2*c*x - 66*sqrt(b + c* 
x)*a*b*c**2*x**2 - 24*sqrt(b + c*x)*b**4*x - 108*sqrt(b + c*x)*b**3*c*x**2 
 + 96*sqrt(b + c*x)*b**2*c**2*x**3 + 15*sqrt(b)*log(sqrt(b + c*x) - sqrt(b 
))*a*c**3*x**3 + 90*sqrt(b)*log(sqrt(b + c*x) - sqrt(b))*b**2*c**2*x**3 - 
15*sqrt(b)*log(sqrt(b + c*x) + sqrt(b))*a*c**3*x**3 - 90*sqrt(b)*log(sqrt( 
b + c*x) + sqrt(b))*b**2*c**2*x**3)/(48*b*x**3)