\(\int \frac {(c+d x) (a x+b x^2)^{3/2}}{x^5} \, dx\) [63]

Optimal result

Integrand size = 22, antiderivative size = 98 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x^5} \, dx=-\frac {2 a d \sqrt {a x+b x^2}}{3 x^2}-\frac {8 b d \sqrt {a x+b x^2}}{3 x}-\frac {2 c \left (a x+b x^2\right )^{5/2}}{5 a x^5}+2 b^{3/2} d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) \] Output:

-2/3*a*d*(b*x^2+a*x)^(1/2)/x^2-8/3*b*d*(b*x^2+a*x)^(1/2)/x-2/5*c*(b*x^2+a* 
x)^(5/2)/a/x^5+2*b^(3/2)*d*arctanh(b^(1/2)*x/(b*x^2+a*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.15 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x^5} \, dx=-\frac {2 \sqrt {x (a+b x)} \left (\sqrt {a+b x} \left (3 b^2 c x^2+a^2 (3 c+5 d x)+2 a b x (3 c+10 d x)\right )+15 a b^{3/2} d x^{5/2} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )\right )}{15 a x^3 \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*(a*x + b*x^2)^(3/2))/x^5,x]
 

Output:

(-2*Sqrt[x*(a + b*x)]*(Sqrt[a + b*x]*(3*b^2*c*x^2 + a^2*(3*c + 5*d*x) + 2* 
a*b*x*(3*c + 10*d*x)) + 15*a*b^(3/2)*d*x^(5/2)*Log[-(Sqrt[b]*Sqrt[x]) + Sq 
rt[a + b*x]]))/(15*a*x^3*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1220, 1130, 1125, 25, 27, 1091, 219}

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.

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1130

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

\(\Big \downarrow \) 1125

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

Int[((c + d*x)*(a*x + b*x^2)^(3/2))/x^5,x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1125

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ 
Symbol] :> Simp[-2*e^(2*m + 3)*(Sqrt[a + b*x + c*x^2]/((-2*c*d + b*e)^(m + 
2)*(d + e*x))), x] - Simp[e^(2*m + 2)   Int[(1/Sqrt[a + b*x + c*x^2])*Expan 
dToSum[((-2*c*d + b*e)^(-m - 1) - ((-c)*d + b*e + c*e*x)^(-m - 1))/(d + e*x 
), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && ILtQ[m, 0] && EqQ[m + p, -3/2]
 

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 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 = 0.48 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(\frac {2 d a \,b^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) x^{3}-\frac {2 \left (\left (\frac {5 d x}{3}+c \right ) a^{2}+2 \left (\frac {10 d x}{3}+c \right ) x b a +b^{2} c \,x^{2}\right ) \sqrt {x \left (b x +a \right )}}{5}}{a \,x^{3}}\)	\(78\)
risch	\(-\frac {2 \left (b x +a \right ) \left (20 a b d \,x^{2}+3 b^{2} c \,x^{2}+5 a^{2} d x +6 a b c x +3 a^{2} c \right )}{15 x^{2} \sqrt {x \left (b x +a \right )}\, a}+b^{\frac {3}{2}} d \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )\)	\(90\)
default	\(-\frac {2 c \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5 a \,x^{5}}+d \left (-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{3 a \,x^{4}}+\frac {2 b \left (-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{a \,x^{3}}+\frac {4 b \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{a \,x^{2}}-\frac {6 b \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{3}+\frac {a \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{2}\right )}{a}\right )}{a}\right )}{3 a}\right )\)	\(174\)

Input:

int((d*x+c)*(b*x^2+a*x)^(3/2)/x^5,x,method=_RETURNVERBOSE)
 

Output:

2/5*(5*d*a*b^(3/2)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))*x^3-((5/3*d*x+c)*a 
^2+2*(10/3*d*x+c)*x*b*a+b^2*c*x^2)*(x*(b*x+a))^(1/2))/a/x^3
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.93 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x^5} \, dx=\left [\frac {15 \, a b^{\frac {3}{2}} d x^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (3 \, a^{2} c + {\left (3 \, b^{2} c + 20 \, a b d\right )} x^{2} + {\left (6 \, a b c + 5 \, a^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{15 \, a x^{3}}, -\frac {2 \, {\left (15 \, a \sqrt {-b} b d x^{3} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) + {\left (3 \, a^{2} c + {\left (3 \, b^{2} c + 20 \, a b d\right )} x^{2} + {\left (6 \, a b c + 5 \, a^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}\right )}}{15 \, a x^{3}}\right ] \] Input:

integrate((d*x+c)*(b*x^2+a*x)^(3/2)/x^5,x, algorithm="fricas")
 

Output:

[1/15*(15*a*b^(3/2)*d*x^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 2 
*(3*a^2*c + (3*b^2*c + 20*a*b*d)*x^2 + (6*a*b*c + 5*a^2*d)*x)*sqrt(b*x^2 + 
 a*x))/(a*x^3), -2/15*(15*a*sqrt(-b)*b*d*x^3*arctan(sqrt(b*x^2 + a*x)*sqrt 
(-b)/(b*x + a)) + (3*a^2*c + (3*b^2*c + 20*a*b*d)*x^2 + (6*a*b*c + 5*a^2*d 
)*x)*sqrt(b*x^2 + a*x))/(a*x^3)]
 

Sympy [F]

\[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x^5} \, dx=\int \frac {\left (x \left (a + b x\right )\right )^{\frac {3}{2}} \left (c + d x\right )}{x^{5}}\, dx \] Input:

integrate((d*x+c)*(b*x**2+a*x)**(3/2)/x**5,x)
 

Output:

Integral((x*(a + b*x))**(3/2)*(c + d*x)/x**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.61 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x^5} \, dx=b^{\frac {3}{2}} d \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {2 \, \sqrt {b x^{2} + a x} b^{2} c}{5 \, a x} - \frac {7 \, \sqrt {b x^{2} + a x} b d}{3 \, x} + \frac {\sqrt {b x^{2} + a x} b c}{5 \, x^{2}} - \frac {\sqrt {b x^{2} + a x} a d}{3 \, x^{2}} + \frac {3 \, \sqrt {b x^{2} + a x} a c}{5 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} d}{3 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} c}{x^{4}} \] Input:

integrate((d*x+c)*(b*x^2+a*x)^(3/2)/x^5,x, algorithm="maxima")
 

Output:

b^(3/2)*d*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 2/5*sqrt(b*x^2 + 
a*x)*b^2*c/(a*x) - 7/3*sqrt(b*x^2 + a*x)*b*d/x + 1/5*sqrt(b*x^2 + a*x)*b*c 
/x^2 - 1/3*sqrt(b*x^2 + a*x)*a*d/x^2 + 3/5*sqrt(b*x^2 + a*x)*a*c/x^3 - 1/3 
*(b*x^2 + a*x)^(3/2)*d/x^3 - (b*x^2 + a*x)^(3/2)*c/x^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (80) = 160\).

Time = 0.16 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.64 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x^5} \, dx=-b^{\frac {3}{2}} d \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right ) + \frac {2 \, {\left (15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} b^{2} c + 30 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} a b d + 30 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} a b^{\frac {3}{2}} c + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} a^{2} \sqrt {b} d + 30 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{2} b c + 5 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{3} d + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{3} \sqrt {b} c + 3 \, a^{4} c\right )}}{15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{5}} \] Input:

integrate((d*x+c)*(b*x^2+a*x)^(3/2)/x^5,x, algorithm="giac")
 

Output:

-b^(3/2)*d*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a)) + 2/15* 
(15*(sqrt(b)*x - sqrt(b*x^2 + a*x))^4*b^2*c + 30*(sqrt(b)*x - sqrt(b*x^2 + 
 a*x))^4*a*b*d + 30*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*a*b^(3/2)*c + 15*(sq 
rt(b)*x - sqrt(b*x^2 + a*x))^3*a^2*sqrt(b)*d + 30*(sqrt(b)*x - sqrt(b*x^2 
+ a*x))^2*a^2*b*c + 5*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a^3*d + 15*(sqrt(b 
)*x - sqrt(b*x^2 + a*x))*a^3*sqrt(b)*c + 3*a^4*c)/(sqrt(b)*x - sqrt(b*x^2 
+ a*x))^5
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x^5} \, dx=\int \frac {{\left (b\,x^2+a\,x\right )}^{3/2}\,\left (c+d\,x\right )}{x^5} \,d x \] Input:

int(((a*x + b*x^2)^(3/2)*(c + d*x))/x^5,x)
                                                                                    
                                                                                    
 

Output:

int(((a*x + b*x^2)^(3/2)*(c + d*x))/x^5, x)
 

Reduce [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.37 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x^5} \, dx=\frac {-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{2} c}{5}-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{2} d x}{3}-\frac {4 \sqrt {x}\, \sqrt {b x +a}\, a b c x}{5}-\frac {8 \sqrt {x}\, \sqrt {b x +a}\, a b d \,x^{2}}{3}-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, b^{2} c \,x^{2}}{5}+2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a b d \,x^{3}+\frac {16 \sqrt {b}\, a b d \,x^{3}}{15}-\frac {2 \sqrt {b}\, b^{2} c \,x^{3}}{5}}{a \,x^{3}} \] Input:

int((d*x+c)*(b*x^2+a*x)^(3/2)/x^5,x)
 

Output:

(2*( - 3*sqrt(x)*sqrt(a + b*x)*a**2*c - 5*sqrt(x)*sqrt(a + b*x)*a**2*d*x - 
 6*sqrt(x)*sqrt(a + b*x)*a*b*c*x - 20*sqrt(x)*sqrt(a + b*x)*a*b*d*x**2 - 3 
*sqrt(x)*sqrt(a + b*x)*b**2*c*x**2 + 15*sqrt(b)*log((sqrt(a + b*x) + sqrt( 
x)*sqrt(b))/sqrt(a))*a*b*d*x**3 + 8*sqrt(b)*a*b*d*x**3 - 3*sqrt(b)*b**2*c* 
x**3))/(15*a*x**3)