\(\int \frac {\sqrt {c-d x} (c+d x)^{3/2}}{x^3} \, dx\) [254]

Optimal result

Integrand size = 23, antiderivative size = 110 \[ \int \frac {\sqrt {c-d x} (c+d x)^{3/2}}{x^3} \, dx=-\frac {c \sqrt {c-d x} \sqrt {c+d x}}{2 x^2}-\frac {d \sqrt {c-d x} \sqrt {c+d x}}{x}-2 d^2 \arctan \left (\frac {\sqrt {c+d x}}{\sqrt {c-d x}}\right )+\frac {1}{2} d^2 \text {arctanh}\left (\frac {\sqrt {c-d x} \sqrt {c+d x}}{c}\right ) \] Output:

-1/2*c*(-d*x+c)^(1/2)*(d*x+c)^(1/2)/x^2-d*(-d*x+c)^(1/2)*(d*x+c)^(1/2)/x-2 
*d^2*arctan((d*x+c)^(1/2)/(-d*x+c)^(1/2))+1/2*d^2*arctanh((-d*x+c)^(1/2)*( 
d*x+c)^(1/2)/c)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {c-d x} (c+d x)^{3/2}}{x^3} \, dx=-\frac {\sqrt {c-d x} \sqrt {c+d x} (c+2 d x)}{2 x^2}+2 d^2 \arctan \left (\frac {\sqrt {c-d x}}{\sqrt {c+d x}}\right )+d^2 \text {arctanh}\left (\frac {\sqrt {c-d x}}{\sqrt {c+d x}}\right ) \] Input:

Integrate[(Sqrt[c - d*x]*(c + d*x)^(3/2))/x^3,x]
 

Output:

-1/2*(Sqrt[c - d*x]*Sqrt[c + d*x]*(c + 2*d*x))/x^2 + 2*d^2*ArcTan[Sqrt[c - 
 d*x]/Sqrt[c + d*x]] + d^2*ArcTanh[Sqrt[c - d*x]/Sqrt[c + d*x]]
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {108, 27, 166, 25, 27, 175, 45, 103, 218, 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.

\(\displaystyle \int \frac {\sqrt {c-d x} (c+d x)^{3/2}}{x^3} \, dx\)

\(\Big \downarrow \) 108

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 166

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

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

\(\Big \downarrow \) 45

\(\displaystyle \frac {1}{2} d \left (-d \left (c \int \frac {1}{x \sqrt {c-d x} \sqrt {c+d x}}dx+4 d \int \frac {1}{-\frac {(c-d x) d}{c+d x}-d}d\frac {\sqrt {c-d x}}{\sqrt {c+d x}}\right )-\frac {\sqrt {c-d x} \sqrt {c+d x}}{x}\right )-\frac {\sqrt {c-d x} (c+d x)^{3/2}}{2 x^2}\)

\(\Big \downarrow \) 103

\(\displaystyle \frac {1}{2} d \left (-d \left (4 d \int \frac {1}{-\frac {(c-d x) d}{c+d x}-d}d\frac {\sqrt {c-d x}}{\sqrt {c+d x}}-c d \int \frac {1}{c^2 d-d (c-d x) (c+d x)}d\left (\sqrt {c-d x} \sqrt {c+d x}\right )\right )-\frac {\sqrt {c-d x} \sqrt {c+d x}}{x}\right )-\frac {\sqrt {c-d x} (c+d x)^{3/2}}{2 x^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{2} d \left (-d \left (-c d \int \frac {1}{c^2 d-d (c-d x) (c+d x)}d\left (\sqrt {c-d x} \sqrt {c+d x}\right )-4 \arctan \left (\frac {\sqrt {c-d x}}{\sqrt {c+d x}}\right )\right )-\frac {\sqrt {c-d x} \sqrt {c+d x}}{x}\right )-\frac {\sqrt {c-d x} (c+d x)^{3/2}}{2 x^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} d \left (-d \left (-4 \arctan \left (\frac {\sqrt {c-d x}}{\sqrt {c+d x}}\right )-\text {arctanh}\left (\frac {\sqrt {c-d x} \sqrt {c+d x}}{c}\right )\right )-\frac {\sqrt {c-d x} \sqrt {c+d x}}{x}\right )-\frac {\sqrt {c-d x} (c+d x)^{3/2}}{2 x^2}\)

Input:

Int[(Sqrt[c - d*x]*(c + d*x)^(3/2))/x^3,x]
 

Output:

-1/2*(Sqrt[c - d*x]*(c + d*x)^(3/2))/x^2 + (d*(-((Sqrt[c - d*x]*Sqrt[c + d 
*x])/x) - d*(-4*ArcTan[Sqrt[c - d*x]/Sqrt[c + d*x]] - ArcTanh[(Sqrt[c - d* 
x]*Sqrt[c + d*x])/c])))/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 45

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] &&  !GtQ[c, 0]
 

rule 103

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

rule 108

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

rule 166

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - 
a*f)*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* 
c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h 
)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, 
e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
 

rule 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

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 = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.22

method	result	size
risch	\(-\frac {\sqrt {-x d +c}\, \sqrt {x d +c}\, \left (2 x d +c \right )}{2 x^{2}}+\frac {\left (\frac {d^{2} c \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{2 \sqrt {c^{2}}}-\frac {d^{3} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}\right ) \sqrt {\left (-x d +c \right ) \left (x d +c \right )}}{\sqrt {-x d +c}\, \sqrt {x d +c}}\)	\(134\)
default	\(-\frac {\sqrt {-x d +c}\, \sqrt {x d +c}\, \left (2 \,\operatorname {csgn}\left (c \right ) \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right ) d^{2} x^{2}-\operatorname {csgn}\left (d \right ) \ln \left (\frac {2 c \left (\sqrt {-d^{2} x^{2}+c^{2}}\, \operatorname {csgn}\left (c \right )+c \right )}{x}\right ) d^{2} x^{2}+2 \sqrt {-d^{2} x^{2}+c^{2}}\, \operatorname {csgn}\left (c \right ) \operatorname {csgn}\left (d \right ) d x +\sqrt {-d^{2} x^{2}+c^{2}}\, \operatorname {csgn}\left (c \right ) \operatorname {csgn}\left (d \right ) c \right ) \operatorname {csgn}\left (c \right ) \operatorname {csgn}\left (d \right )}{2 \sqrt {-d^{2} x^{2}+c^{2}}\, x^{2}}\)	\(148\)

Input:

int((-d*x+c)^(1/2)*(d*x+c)^(3/2)/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2*(-d*x+c)^(1/2)*(d*x+c)^(1/2)*(2*d*x+c)/x^2+(1/2*d^2*c/(c^2)^(1/2)*ln( 
(2*c^2+2*(c^2)^(1/2)*(-d^2*x^2+c^2)^(1/2))/x)-d^3/(d^2)^(1/2)*arctan((d^2) 
^(1/2)*x/(-d^2*x^2+c^2)^(1/2)))*((-d*x+c)*(d*x+c))^(1/2)/(-d*x+c)^(1/2)/(d 
*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {c-d x} (c+d x)^{3/2}}{x^3} \, dx=\frac {4 \, d^{2} x^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-d x + c} - c}{d x}\right ) - d^{2} x^{2} \log \left (\frac {\sqrt {d x + c} \sqrt {-d x + c} - c}{x}\right ) - {\left (2 \, d x + c\right )} \sqrt {d x + c} \sqrt {-d x + c}}{2 \, x^{2}} \] Input:

integrate((-d*x+c)^(1/2)*(d*x+c)^(3/2)/x^3,x, algorithm="fricas")
 

Output:

1/2*(4*d^2*x^2*arctan((sqrt(d*x + c)*sqrt(-d*x + c) - c)/(d*x)) - d^2*x^2* 
log((sqrt(d*x + c)*sqrt(-d*x + c) - c)/x) - (2*d*x + c)*sqrt(d*x + c)*sqrt 
(-d*x + c))/x^2
 

Sympy [F]

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

integrate((-d*x+c)**(1/2)*(d*x+c)**(3/2)/x**3,x)
 

Output:

Integral(sqrt(c - d*x)*(c + d*x)**(3/2)/x**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {c-d x} (c+d x)^{3/2}}{x^3} \, dx=-d^{2} \arcsin \left (\frac {d x}{c}\right ) + \frac {1}{2} \, d^{2} \log \left (\frac {2 \, c^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} c}{{\left | x \right |}}\right ) - \frac {\sqrt {-d^{2} x^{2} + c^{2}} d^{2}}{2 \, c} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} d}{x} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}}}{2 \, c x^{2}} \] Input:

integrate((-d*x+c)^(1/2)*(d*x+c)^(3/2)/x^3,x, algorithm="maxima")
 

Output:

-d^2*arcsin(d*x/c) + 1/2*d^2*log(2*c^2/abs(x) + 2*sqrt(-d^2*x^2 + c^2)*c/a 
bs(x)) - 1/2*sqrt(-d^2*x^2 + c^2)*d^2/c - sqrt(-d^2*x^2 + c^2)*d/x - 1/2*( 
-d^2*x^2 + c^2)^(3/2)/(c*x^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 390 vs. \(2 (90) = 180\).

Time = 0.39 (sec) , antiderivative size = 390, normalized size of antiderivative = 3.55 \[ \int \frac {\sqrt {c-d x} (c+d x)^{3/2}}{x^3} \, dx=-\frac {2 \, {\left (\pi + 2 \, \arctan \left (\frac {\sqrt {d x + c} {\left (\frac {{\left (\sqrt {2} \sqrt {c} - \sqrt {-d x + c}\right )}^{2}}{d x + c} - 1\right )}}{2 \, {\left (\sqrt {2} \sqrt {c} - \sqrt {-d x + c}\right )}}\right )\right )} d^{3} - d^{3} \log \left ({\left | -\frac {\sqrt {2} \sqrt {c} - \sqrt {-d x + c}}{\sqrt {d x + c}} + \frac {\sqrt {d x + c}}{\sqrt {2} \sqrt {c} - \sqrt {-d x + c}} + 2 \right |}\right ) + d^{3} \log \left ({\left | -\frac {\sqrt {2} \sqrt {c} - \sqrt {-d x + c}}{\sqrt {d x + c}} + \frac {\sqrt {d x + c}}{\sqrt {2} \sqrt {c} - \sqrt {-d x + c}} - 2 \right |}\right ) + \frac {4 \, {\left (d^{3} {\left (\frac {\sqrt {2} \sqrt {c} - \sqrt {-d x + c}}{\sqrt {d x + c}} - \frac {\sqrt {d x + c}}{\sqrt {2} \sqrt {c} - \sqrt {-d x + c}}\right )}^{3} - 12 \, d^{3} {\left (\frac {\sqrt {2} \sqrt {c} - \sqrt {-d x + c}}{\sqrt {d x + c}} - \frac {\sqrt {d x + c}}{\sqrt {2} \sqrt {c} - \sqrt {-d x + c}}\right )}\right )}}{{\left ({\left (\frac {\sqrt {2} \sqrt {c} - \sqrt {-d x + c}}{\sqrt {d x + c}} - \frac {\sqrt {d x + c}}{\sqrt {2} \sqrt {c} - \sqrt {-d x + c}}\right )}^{2} - 4\right )}^{2}}}{2 \, d} \] Input:

integrate((-d*x+c)^(1/2)*(d*x+c)^(3/2)/x^3,x, algorithm="giac")
 

Output:

-1/2*(2*(pi + 2*arctan(1/2*sqrt(d*x + c)*((sqrt(2)*sqrt(c) - sqrt(-d*x + c 
))^2/(d*x + c) - 1)/(sqrt(2)*sqrt(c) - sqrt(-d*x + c))))*d^3 - d^3*log(abs 
(-(sqrt(2)*sqrt(c) - sqrt(-d*x + c))/sqrt(d*x + c) + sqrt(d*x + c)/(sqrt(2 
)*sqrt(c) - sqrt(-d*x + c)) + 2)) + d^3*log(abs(-(sqrt(2)*sqrt(c) - sqrt(- 
d*x + c))/sqrt(d*x + c) + sqrt(d*x + c)/(sqrt(2)*sqrt(c) - sqrt(-d*x + c)) 
 - 2)) + 4*(d^3*((sqrt(2)*sqrt(c) - sqrt(-d*x + c))/sqrt(d*x + c) - sqrt(d 
*x + c)/(sqrt(2)*sqrt(c) - sqrt(-d*x + c)))^3 - 12*d^3*((sqrt(2)*sqrt(c) - 
 sqrt(-d*x + c))/sqrt(d*x + c) - sqrt(d*x + c)/(sqrt(2)*sqrt(c) - sqrt(-d* 
x + c))))/(((sqrt(2)*sqrt(c) - sqrt(-d*x + c))/sqrt(d*x + c) - sqrt(d*x + 
c)/(sqrt(2)*sqrt(c) - sqrt(-d*x + c)))^2 - 4)^2)/d
 

Mupad [F(-1)]

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

int(((c + d*x)^(3/2)*(c - d*x)^(1/2))/x^3,x)
 

Output:

int(((c + d*x)^(3/2)*(c - d*x)^(1/2))/x^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.80 \[ \int \frac {\sqrt {c-d x} (c+d x)^{3/2}}{x^3} \, dx=\frac {4 \mathit {asin} \left (\frac {\sqrt {-d x +c}}{\sqrt {c}\, \sqrt {2}}\right ) d^{2} x^{2}-\sqrt {d x +c}\, \sqrt {-d x +c}\, c -2 \sqrt {d x +c}\, \sqrt {-d x +c}\, d x +\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )-1\right ) d^{2} x^{2}-\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )+1\right ) d^{2} x^{2}+\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )-1\right ) d^{2} x^{2}-\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )+1\right ) d^{2} x^{2}}{2 x^{2}} \] Input:

int((-d*x+c)^(1/2)*(d*x+c)^(3/2)/x^3,x)
 

Output:

(4*asin(sqrt(c - d*x)/(sqrt(c)*sqrt(2)))*d**2*x**2 - sqrt(c + d*x)*sqrt(c 
- d*x)*c - 2*sqrt(c + d*x)*sqrt(c - d*x)*d*x + log( - sqrt(2) + tan(asin(s 
qrt(c - d*x)/(sqrt(c)*sqrt(2)))/2) - 1)*d**2*x**2 - log( - sqrt(2) + tan(a 
sin(sqrt(c - d*x)/(sqrt(c)*sqrt(2)))/2) + 1)*d**2*x**2 + log(sqrt(2) + tan 
(asin(sqrt(c - d*x)/(sqrt(c)*sqrt(2)))/2) - 1)*d**2*x**2 - log(sqrt(2) + t 
an(asin(sqrt(c - d*x)/(sqrt(c)*sqrt(2)))/2) + 1)*d**2*x**2)/(2*x**2)