\(\int \frac {x^2}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx\) [730]

Optimal result

Integrand size = 32, antiderivative size = 104 \[ \int \frac {x^2}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\frac {2 \left (b c^2-b d^2 x^2\right )^{3/2}}{3 b^2 d^3 (c+d x)^{3/2}}-\frac {\sqrt {2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}{\sqrt {b c^2-b d^2 x^2}}\right )}{\sqrt {b} d^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.19 \[ \int \frac {x^2}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\frac {2 (c-d x)^2 (c+d x)-3 \sqrt {2} c^{3/2} \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{3 d^3 \sqrt {c+d x} \sqrt {b \left (c^2-d^2 x^2\right )}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.43, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {581, 27, 600, 458, 471, 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.

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

\(\Big \downarrow \) 581

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 600

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

\(\Big \downarrow \) 458

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

\(\Big \downarrow \) 471

\(\displaystyle \frac {c \left (6 c d \int \frac {1}{\frac {d^2 \left (b c^2-b d^2 x^2\right )}{c+d x}-2 b c d^2}d\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}+\frac {4 \sqrt {b c^2-b d^2 x^2}}{b d \sqrt {c+d x}}\right )}{3 d^2}-\frac {2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}{3 b d^3}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 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 458

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c 
, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
 

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

rule 581

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 
2*p + 1))), x] + Simp[1/(d^m*(m + n + 2*p + 1))   Int[(c + d*x)^n*(a + b*x^ 
2)^p*ExpandToSum[d^m*(m + n + 2*p + 1)*x^m - (m + n + 2*p + 1)*(c + d*x)^m 
+ c*(c + d*x)^(m - 2)*(c*(m + n - 1) + c*(m + n + 2*p + 1) + 2*d*(m + n + p 
)*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] & 
& IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[2*p] || ILtQ[m + n, 0] 
)
 

rule 600

Int[((A_.) + (B_.)*(x_))/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2] 
), x_Symbol] :> Simp[B/d   Int[Sqrt[c + d*x]/Sqrt[a + b*x^2], x], x] - Simp 
[(B*c - A*d)/d   Int[1/(Sqrt[c + d*x]*Sqrt[a + b*x^2]), x], x] /; FreeQ[{a, 
 b, c, d, A, B}, x] && NegQ[b/a]
 

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11

method	result	size
default	\(-\frac {\sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (3 b \,c^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right )+2 d x \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}-2 c \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}\right )}{3 b \sqrt {d x +c}\, \sqrt {\left (-d x +c \right ) b}\, d^{3} \sqrt {b c}}\)	\(115\)
risch	\(\frac {2 \left (-d x +c \right )^{2} \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}}{3 d^{3} \sqrt {-b \left (d x -c \right )}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}-\frac {c^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-b d x +b c}\, \sqrt {2}}{2 \sqrt {b c}}\right ) \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}}{d^{3} \sqrt {b c}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}\)	\(165\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.62 \[ \int \frac {x^2}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (b c d x + b c^{2}\right )} \sqrt {\frac {c}{b}} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x - 3 \, c^{2} + 2 \, \sqrt {2} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {\frac {c}{b}}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 4 \, \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} {\left (d x - c\right )}}{6 \, {\left (b d^{4} x + b c d^{3}\right )}}, \frac {3 \, \sqrt {2} {\left (b c d x + b c^{2}\right )} \sqrt {-\frac {c}{b}} \arctan \left (\frac {\sqrt {2} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {-\frac {c}{b}}}{2 \, {\left (c d x + c^{2}\right )}}\right ) - 2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} {\left (d x - c\right )}}{3 \, {\left (b d^{4} x + b c d^{3}\right )}}\right ] \] Input:

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

Output:

[1/6*(3*sqrt(2)*(b*c*d*x + b*c^2)*sqrt(c/b)*log(-(d^2*x^2 - 2*c*d*x - 3*c^ 
2 + 2*sqrt(2)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*sqrt(c/b))/(d^2*x^2 + 
 2*c*d*x + c^2)) - 4*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*(d*x - c))/(b* 
d^4*x + b*c*d^3), 1/3*(3*sqrt(2)*(b*c*d*x + b*c^2)*sqrt(-c/b)*arctan(1/2*s 
qrt(2)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*sqrt(-c/b)/(c*d*x + c^2)) - 
2*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*(d*x - c))/(b*d^4*x + b*c*d^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

\[ \int \frac {x^2}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\int { \frac {x^{2}}{\sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c}} \,d x } \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.62 \[ \int \frac {x^2}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\frac {\frac {3 \, \sqrt {2} c^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-b d x + b c}}{2 \, \sqrt {-b c}}\right )}{\sqrt {-b c} d^{2}} + \frac {2 \, {\left (-b d x + b c\right )}^{\frac {3}{2}}}{b^{2} d^{2}}}{3 \, d} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.72 \[ \int \frac {x^2}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}} \, dx=\frac {\sqrt {b}\, \left (4 \sqrt {-d x +c}\, c -4 \sqrt {-d x +c}\, d x +3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c -3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c \right )}{6 b \,d^{3}} \] Input:

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

Output:

(sqrt(b)*(4*sqrt(c - d*x)*c - 4*sqrt(c - d*x)*d*x + 3*sqrt(c)*sqrt(2)*log( 
sqrt(c - d*x) - sqrt(c)*sqrt(2))*c - 3*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) + 
 sqrt(c)*sqrt(2))*c))/(6*b*d**3)