3.1.18 \(\int \frac {(b x+c x^2)^{3/2}}{x^4} \, dx\) [18]

3.1.18.1 Optimal result

Integrand size = 17, antiderivative size = 68 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=-\frac {2 c \sqrt {b x+c x^2}}{x}-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}+2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \]

output

-2/3*(c*x^2+b*x)^(3/2)/x^3+2*c^(3/2)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))- 
2*c*(c*x^2+b*x)^(1/2)/x
 

3.1.18.2 Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.16 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=-\frac {2 \sqrt {x (b+c x)} \left (\sqrt {b+c x} (b+4 c x)+3 c^{3/2} x^{3/2} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )\right )}{3 x^2 \sqrt {b+c x}} \]

input

Integrate[(b*x + c*x^2)^(3/2)/x^4,x]
 

output

(-2*Sqrt[x*(b + c*x)]*(Sqrt[b + c*x]*(b + 4*c*x) + 3*c^(3/2)*x^(3/2)*Log[- 
(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]]))/(3*x^2*Sqrt[b + c*x])
 

3.1.18.3 Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {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.

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

\(\Big \downarrow \) 1130

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

\(\Big \downarrow \) 1125

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

input

Int[(b*x + c*x^2)^(3/2)/x^4,x]
 

output

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

3.1.18.3.1 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]
 

3.1.18.4 Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(\frac {6 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) x^{2}-2 \sqrt {x \left (c x +b \right )}\, \left (4 c x +b \right )}{3 x^{2}}\)	\(49\)
risch	\(-\frac {2 \left (c x +b \right ) \left (4 c x +b \right )}{3 x \sqrt {x \left (c x +b \right )}}+c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )\)	\(55\)
default	\(-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{3 b \,x^{4}}+\frac {2 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{3}}+\frac {4 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{2}}-\frac {6 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )}{b}\right )}{b}\right )}{3 b}\)	\(151\)

input

int((c*x^2+b*x)^(3/2)/x^4,x,method=_RETURNVERBOSE)
 

output

1/3*(6*c^(3/2)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))*x^2-2*(x*(c*x+b))^(1/2 
)*(4*c*x+b))/x^2
 

3.1.18.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.71 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, c^{\frac {3}{2}} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, \sqrt {c x^{2} + b x} {\left (4 \, c x + b\right )}}{3 \, x^{2}}, -\frac {2 \, {\left (3 \, \sqrt {-c} c x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} {\left (4 \, c x + b\right )}\right )}}{3 \, x^{2}}\right ] \]

input

integrate((c*x^2+b*x)^(3/2)/x^4,x, algorithm="fricas")
 

output

[1/3*(3*c^(3/2)*x^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*sqrt( 
c*x^2 + b*x)*(4*c*x + b))/x^2, -2/3*(3*sqrt(-c)*c*x^2*arctan(sqrt(c*x^2 + 
b*x)*sqrt(-c)/(c*x)) + sqrt(c*x^2 + b*x)*(4*c*x + b))/x^2]
 

3.1.18.6 Sympy [F]

\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{4}}\, dx \]

input

integrate((c*x**2+b*x)**(3/2)/x**4,x)
 

output

Integral((x*(b + c*x))**(3/2)/x**4, x)
 

3.1.18.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.15 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=c^{\frac {3}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{2} + b x} c}{3 \, x} - \frac {\sqrt {c x^{2} + b x} b}{3 \, x^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{3 \, x^{3}} \]

input

integrate((c*x^2+b*x)^(3/2)/x^4,x, algorithm="maxima")
 

output

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

3.1.18.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (56) = 112\).

Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.69 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=-c^{\frac {3}{2}} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right ) + \frac {2 \, {\left (6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b c + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{2} \sqrt {c} + b^{3}\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3}} \]

input

integrate((c*x^2+b*x)^(3/2)/x^4,x, algorithm="giac")
 

output

-c^(3/2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b)) + 2/3*(6 
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b*c + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x)) 
*b^2*sqrt(c) + b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^3
 

3.1.18.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^4} \,d x \]

input

int((b*x + c*x^2)^(3/2)/x^4,x)
 

output

int((b*x + c*x^2)^(3/2)/x^4, x)