\(\int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{x^3} \, dx\) [615]

Optimal result

Integrand size = 32, antiderivative size = 166 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{x^3} \, dx=\frac {2 d^2 \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}-\frac {c^2 \sqrt {b c^2-b d^2 x^2}}{2 x^2 \sqrt {c+d x}}-\frac {7 c d \sqrt {b c^2-b d^2 x^2}}{4 x \sqrt {c+d x}}+\frac {1}{4} \sqrt {b} \sqrt {c} d^2 \text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {b} \sqrt {c} \sqrt {c+d x}}\right ) \] Output:

2*d^2*(-b*d^2*x^2+b*c^2)^(1/2)/(d*x+c)^(1/2)-1/2*c^2*(-b*d^2*x^2+b*c^2)^(1 
/2)/x^2/(d*x+c)^(1/2)-7/4*c*d*(-b*d^2*x^2+b*c^2)^(1/2)/x/(d*x+c)^(1/2)+1/4 
*b^(1/2)*c^(1/2)*d^2*arctanh((-b*d^2*x^2+b*c^2)^(1/2)/b^(1/2)/c^(1/2)/(d*x 
+c)^(1/2))
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.67 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{x^3} \, dx=\frac {1}{4} \sqrt {b \left (c^2-d^2 x^2\right )} \left (\frac {-2 c^2-7 c d x+8 d^2 x^2}{x^2 \sqrt {c+d x}}+\frac {\sqrt {c} d^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {c^2-d^2 x^2}}\right ) \] Input:

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

Output:

(Sqrt[b*(c^2 - d^2*x^2)]*((-2*c^2 - 7*c*d*x + 8*d^2*x^2)/(x^2*Sqrt[c + d*x 
]) + (Sqrt[c]*d^2*ArcTanh[(Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/Sq 
rt[c^2 - d^2*x^2]))/4
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {586, 100, 27, 87, 60, 73, 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.

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

\(\Big \downarrow \) 586

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {\int \frac {b c d (9 c+4 d x) \sqrt {b c-b d x}}{2 x^2}dx}{2 b c}-\frac {c (b c-b d x)^{3/2}}{2 b x^2}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {1}{4} d \int \frac {(9 c+4 d x) \sqrt {b c-b d x}}{x^2}dx-\frac {c (b c-b d x)^{3/2}}{2 b x^2}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {1}{4} d \left (-\frac {1}{2} d \int \frac {\sqrt {b c-b d x}}{x}dx-\frac {9 (b c-b d x)^{3/2}}{b x}\right )-\frac {c (b c-b d x)^{3/2}}{2 b x^2}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {1}{4} d \left (-\frac {1}{2} d \left (b c \int \frac {1}{x \sqrt {b c-b d x}}dx+2 \sqrt {b c-b d x}\right )-\frac {9 (b c-b d x)^{3/2}}{b x}\right )-\frac {c (b c-b d x)^{3/2}}{2 b x^2}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {1}{4} d \left (-\frac {1}{2} d \left (2 \sqrt {b c-b d x}-2 \sqrt {b} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {b c-b d x}}{\sqrt {b} \sqrt {c}}\right )\right )-\frac {9 (b c-b d x)^{3/2}}{b x}\right )-\frac {c (b c-b d x)^{3/2}}{2 b x^2}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

Input:

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

Output:

(Sqrt[b*c^2 - b*d^2*x^2]*(-1/2*(c*(b*c - b*d*x)^(3/2))/(b*x^2) + (d*((-9*( 
b*c - b*d*x)^(3/2))/(b*x) - (d*(2*Sqrt[b*c - b*d*x] - 2*Sqrt[b]*Sqrt[c]*Ar 
cTanh[Sqrt[b*c - b*d*x]/(Sqrt[b]*Sqrt[c])]))/2))/4))/(Sqrt[c + d*x]*Sqrt[b 
*c - b*d*x])
 

Defintions of rubi rules used

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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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]
 

rule 586

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_) 
, x_Symbol] :> Simp[(a + b*x^2)^FracPart[p]/((c + d*x)^FracPart[p]*(a/c + ( 
b*x)/d)^FracPart[p])   Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0]
 

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {\sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}}{\sqrt {b c}}\right ) b c \,d^{2} x^{2}+8 d^{2} x^{2} \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}-7 c d x \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}-2 \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}\, c^{2}\right )}{4 \sqrt {d x +c}\, \sqrt {\left (-d x +c \right ) b}\, x^{2} \sqrt {b c}}\)	\(134\)
risch	\(-\frac {c \left (-d x +c \right ) \left (7 d x +2 c \right ) \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b}{4 x^{2} \sqrt {-b \left (d x -c \right )}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}-\frac {d^{2} \left (-8 \sqrt {-b d x +b c}-\frac {b c \,\operatorname {arctanh}\left (\frac {\sqrt {-b d x +b c}}{\sqrt {b c}}\right )}{\sqrt {b c}}\right ) \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}}{4 \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}\)	\(181\)

Input:

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

Output:

1/4*(b*(-d^2*x^2+c^2))^(1/2)*(arctanh(((-d*x+c)*b)^(1/2)/(b*c)^(1/2))*b*c* 
d^2*x^2+8*d^2*x^2*((-d*x+c)*b)^(1/2)*(b*c)^(1/2)-7*c*d*x*((-d*x+c)*b)^(1/2 
)*(b*c)^(1/2)-2*((-d*x+c)*b)^(1/2)*(b*c)^(1/2)*c^2)/(d*x+c)^(1/2)/((-d*x+c 
)*b)^(1/2)/x^2/(b*c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.68 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{x^3} \, dx=\left [\frac {{\left (d^{3} x^{3} + c d^{2} x^{2}\right )} \sqrt {b c} \log \left (-\frac {b d^{2} x^{2} - b c d x - 2 \, b c^{2} - 2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {b c} \sqrt {d x + c}}{d x^{2} + c x}\right ) + 2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (8 \, d^{2} x^{2} - 7 \, c d x - 2 \, c^{2}\right )} \sqrt {d x + c}}{8 \, {\left (d x^{3} + c x^{2}\right )}}, -\frac {{\left (d^{3} x^{3} + c d^{2} x^{2}\right )} \sqrt {-b c} \arctan \left (\frac {\sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {-b c} \sqrt {d x + c}}{b c d x + b c^{2}}\right ) - \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (8 \, d^{2} x^{2} - 7 \, c d x - 2 \, c^{2}\right )} \sqrt {d x + c}}{4 \, {\left (d x^{3} + c x^{2}\right )}}\right ] \] Input:

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

Output:

[1/8*((d^3*x^3 + c*d^2*x^2)*sqrt(b*c)*log(-(b*d^2*x^2 - b*c*d*x - 2*b*c^2 
- 2*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(b*c)*sqrt(d*x + c))/(d*x^2 + c*x)) + 2*s 
qrt(-b*d^2*x^2 + b*c^2)*(8*d^2*x^2 - 7*c*d*x - 2*c^2)*sqrt(d*x + c))/(d*x^ 
3 + c*x^2), -1/4*((d^3*x^3 + c*d^2*x^2)*sqrt(-b*c)*arctan(sqrt(-b*d^2*x^2 
+ b*c^2)*sqrt(-b*c)*sqrt(d*x + c)/(b*c*d*x + b*c^2)) - sqrt(-b*d^2*x^2 + b 
*c^2)*(8*d^2*x^2 - 7*c*d*x - 2*c^2)*sqrt(d*x + c))/(d*x^3 + c*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(sqrt(-b*d^2*x^2 + b*c^2)*(d*x + c)^(3/2)/x^3, x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.63 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{x^3} \, dx=-\frac {\frac {b c d^{3} \arctan \left (\frac {\sqrt {-b d x + b c}}{\sqrt {-b c}}\right )}{\sqrt {-b c}} - 8 \, \sqrt {-b d x + b c} d^{3} + \frac {9 \, \sqrt {-b d x + b c} b^{2} c^{2} d^{3} - 7 \, {\left (-b d x + b c\right )}^{\frac {3}{2}} b c d^{3}}{b^{2} d^{2} x^{2}}}{4 \, d} \] Input:

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

Output:

-1/4*(b*c*d^3*arctan(sqrt(-b*d*x + b*c)/sqrt(-b*c))/sqrt(-b*c) - 8*sqrt(-b 
*d*x + b*c)*d^3 + (9*sqrt(-b*d*x + b*c)*b^2*c^2*d^3 - 7*(-b*d*x + b*c)^(3/ 
2)*b*c*d^3)/(b^2*d^2*x^2))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.54 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{x^3} \, dx=\frac {\sqrt {b}\, \left (-4 \sqrt {-d x +c}\, c^{2}-14 \sqrt {-d x +c}\, c d x +16 \sqrt {-d x +c}\, d^{2} x^{2}-\sqrt {c}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\right ) d^{2} x^{2}+\sqrt {c}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\right ) d^{2} x^{2}\right )}{8 x^{2}} \] Input:

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

Output:

(sqrt(b)*( - 4*sqrt(c - d*x)*c**2 - 14*sqrt(c - d*x)*c*d*x + 16*sqrt(c - d 
*x)*d**2*x**2 - sqrt(c)*log(sqrt(c - d*x) - sqrt(c))*d**2*x**2 + sqrt(c)*l 
og(sqrt(c - d*x) + sqrt(c))*d**2*x**2))/(8*x**2)